# What is the dimension of two subspaces?

Let $V$ be the vector space with a basis $x_1, \ldots, x_9$ and $$V_1 = \{(a,a,a,b,b,b,c,c,c): a,b,c \in \mathbb{C}\}, \\ V_2=\{(a,b,c,a,b,c,a,b,c): a,b,c \in \mathbb{C}\}.$$ Then $V_1,V_2$ are subspaces of $V$. What is the dimension of $V_1 \cap V_2$?

I first try to find the system of equations which give $V_1, V_2$ respectively. I think that $V_1$ is
$$\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_2=0,k_2-k3=0,k_4-k_5=0,k_5-k_6=0, k_7-k_8=0,k_8-k_9=0 \}$$ and $V_2$ is
$$\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_4=0,k_4-k7=0,k_2-k_5=0,k_5-k_8=0, k_3-k_6=0,k_6-k_9=0 \}.$$ By solving the collection of the equations for $V_1$ and $V_2$, I obtain the solutions $k_1=\cdots =k_9$. So the dimension of $V_1 \cap V_2$ is one dimensional. Is this correct?

Thank you very much.

It is correct but, in my opinion, that approach is too complex for the problem. Take $(a,b,c,d,e,f,g,h,i)\in V_1\cap V_2$. Then
• Since $(a,b,c,d,e,f,g,h,i)\in V_1$, $b=c=a$, $e=f=d$, and $h=i=g$. Therefore$$(a,b,c,d,e,f,g,h,i)=(a,a,a,d,d,d,g,g,g).$$
• Since $(a,a,a,d,d,d,g,g,g)\in V_2$, $d=g=a$.
Therefore $V_1\cap V_2=\left\{(a,a,a,a,a,a,a,a,a)\in\mathbb{C}^9\,\middle|\,a\in\mathbb C\right\}$, which is clearly $1$-dimensional.
Yes, it is correct. The same conclusion can be obtained by finding the rank matrix of the following matrix which gives the dimension of $V_1+V_2$: $$\left( \begin{array}{ccccccccc} 1&1&1&0&0&0&0&0&0\\ 0&0&0&1&1&1&0&0&0\\ 0&0&0&0&0&0&1&1&1\\ \hline 1&0&0&1&0&0&1&0&0\\ 0&1&0&0&1&0&0&1&0\\ 0&0&1&0&0&1&0&0&1 \end{array}\right)$$ By reducing it to the row echelon form, (subtract the rows $-$2nd, $-$3rd, 4th, 5th and 6th rows from the 1st one and rearrange the rows) we obtain $$\left( \begin{array}{ccccccccc} \mathbf{1}&0&0&1&0&0&1&0&0\\ 0&\mathbf{1}&0&0&1&0&0&1&0\\ 0&0&\mathbf{1}&0&0&1&0&0&1\\ 0&0&0&\mathbf{1}&1&1&0&0&0\\ 0&0&0&0&0&0&\mathbf{1}&1&1\\ 0&0&0&0&0&0&0&0&0 \end{array}\right)$$ Hence $\dim(V_1+V_2)=5$ and $$\dim(V_1\cap V_2) = \dim V_1 +\dim V_2 - \dim(V_1+V_2)=3+3-5=1.$$


# Evaluating a double integral in polar coordinates

#### skatenerd

##### Active member
I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral
$$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
into an integral in polar coordinates and then evaluate it.
Changing to polar coordinates I got the integral
$$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$
and evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\pi$$.
I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great.

#### Prove It

##### Well-known member
MHB Math Helper
Re: evaluating a double integral in polar coordinates

Since you are integrating over an entire circle of radius |a| centred at (0, 0), that means the angle swept out is actually $$\displaystyle \displaystyle 2\pi$$, which means your $$\displaystyle \displaystyle \theta$$ bounds are actually 0 to $$\displaystyle \displaystyle 2\pi$$.

#### chisigma

##### Well-known member
Re: evaluating a double integral in polar coordinates

I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral
$$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
into an integral in polar coordinates and then evaluate it.
Changing to polar coordinates I got the integral
$$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$
and evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\pi$$.
I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great.
The bounds of the inner integral are 0 and a, not -a and a so that is...

$\displaystyle S= \int_{0}^{2\ \pi} \int_{0}^{a} r\ d r\ d \theta = \pi\ a^{2}$ (1)

Kind regards

$\chi$ $\sigma$

#### skatenerd

##### Active member
Re: evaluating a double integral in polar coordinates

Thanks guys. I didn't recognize initially that the bounds of the original integral are describing the area of a whole circle. Pretty cool problem now that I get it!


# What kind of combinatorial problem is this?

Is there a theory from which the following problem comes? Does this type of problem have a name?

Find the largest possible number of $k$-element sets consisting of points from some finite set and have pairwise singleton or empty intersections.

I hope that was clear. If not, here's an example for $k=3$:

Let the set of points be $S=\{1,2,3,4,5,6\}$. The most 3-element sets (with pairwise singleton or empty intersections) that can be constructed from $S$ is 4, such as $\{456,236,124,135\}$.

I made a table for $|S|=3,4,5,6,7,8,9$ and got $1,1,2,4,7,8,12$, respectively, hoping I could dig up some information from OEIS.

I read a little on Steiner systems, and although it feels like I'm in the neighborhood, I'm not confident...

Edit1: typos.

Edit2: Johnson graphs and (for $k=3$) Steiner Triple Systems (STS) seem close to what I'm looking for. The condition of "pairwise singleton or empty intersections" is equivalent to "every 2-subset of S occurs in at most one $k$-element set". STS require that every 2-subset of S occurs in exactly one $3$-element set".

Edit3: Thank you to everyone who replied! All of your comments helped me push through a barrier I was facing for some time.

-
Looks a bit like a Johnson graph. Perhaps looking at some of the more popular objects in finite geometry will get you graphs matching your own – muzzlator Feb 19 '13 at 20:00
Thanks for the tip. – sasha Feb 19 '13 at 20:36
Design theory is also relevant – mrf Feb 19 '13 at 23:30
The $k=3$ sequence seems to be oeis.org/A001839 . – Kevin Costello Feb 19 '13 at 23:51
In general, you are looking for the maximal code of length $n$, constant weight $k$, and minimum distance $2k-2$. So the $k=4$ sequence is $A(n,6,4)$, found at oeis.org/A004037. – mjqxxxx Feb 20 '13 at 0:57

You exactly want to determine the clique number of the generalized Kneser graph $KG_{n,k,s}$ for $s=1$, which is the graph having all the $k$-element subsets as its $\binom{n}{k}$ vertices, where any two vertices are connected if and only if their cut contains at most $s$ elements. Thus, a maximum clique of $KG_{n,k,1}$ is a maximum selection of $k$-element subsets such that all their pairwise cuts are singleton or emtpy. The size of such a maximum clique is the clique number $\omega(KG_{n,k,s})$.

Googling around a bit, I could not find an exact expression therefore.

However, this states that $\omega(KG_{n,k,0}) = \lfloor \frac n k \rfloor$. Since for $s > 0$ edges are never removed, this also gives a lower bound on $\omega(KG_{n,k,s})$ for any $s \geq 0$, thus, $\omega(KG_{n,k,1}) \geq \lfloor \frac n k \rfloor$. But this bound seems rather weak, since it does not respect any singleton edges at all. For your example above we get 1,1,1,2,2,2,3 as lower bounds on 1,1,2,4,7,7,12.

Further, here is an expression for the chromatic number $\chi(KG_{n,k,1})$, which gives an upper bound on the clique number, since $\chi(G) \geq \omega(G)$ for any graph $G$, where for perfect graphs equality holds. For $n$ written as $n = (k-1) s + r$ for some $0 \leq r < k-1$ and large enough $n > n_0(k)$, the bound is given as $\chi(KG_{n,k,1}) = (k-1)\binom{s}{2} + rs \geq \omega(KG_{n,k,1})$. Ignoring any details on $n_0(k)$, since I have no access to the paper, this gives 1,2,4,6,9,12,16 as upper bounds on 1,1,2,4,7,7,12, which seems to approximate quite well.

Perhaps one of the proofs behind the above results can be adopted to the clique number of $KG_{n,k,1}$?

edit: I just realized that you even want to find a maximum clique - well, that is computationally very hard in general, but perhaps things get easier for $KG_{n,k,1}$?

-

A family of subsets of some finite set is a hypergraph; the subsets themselves are the edges (or hyperedges) of the hypergraph. If all the edges have size $k$, then the hypergraph is k-uniform. (For instance, a $2$-uniform hypergraph is just an ordinary undirected graph.) If no pair of edges has more than one point in common, the hypergraph is called linear. So your question can be reframed as:

What is the maximal number of edges in a $k$-uniform linear hypergraph on $n$ vertices?

-

If you're interested in the asymptotic situation instead of what happens for specific $|S|$, then this has been studied a fair bit under the name of packing problems. More generally, we can ask the question of the size of the largest collection of $k$-element subsets of $\{1, \dots, n\}$ such that each pair of subsets has intersection of size strictly less than $r$.

Since each $k$-element subset contains $\binom{k}{r}$ $r$-element subsets, and each $r$ element subset is in at most $1$ $k$-element subset, we can pick at most $$\frac{\binom{n}{r}}{\binom{k}{r}}$$ subsets in our collection.

Erdos and Hanani conjectured in 1963 that this was asymptotically optimal: For fixed $k$ and $r$, as $n$ tends to infinity, there is a collection of size $(1+o(1))$ times the bound above (for the specific case $r=2$ that you mentioned, this was conjectured earlier by Bose). The conjecture remained open for more than $20$ years until Rodl introduced his so-called "nibble method" to prove it ("On a Packing and Covering Problem", not available online as far as I can tell).

Another term you might want to search under is partial Steiner systems.

-


# Math Help - 1+2+3+...+n

1. ## 1+2+3+...+n

Hi, can anyone tell me why

1+2+3+...+n=n(n+1)/2

I can see that it works when I choose a number for n, but I don't really see how I could have come up with it myself.

2. ## Re: 1+2+3+...+n

I think I can explain it like that:

1, 2, 3, 4, 5, 6, ... n

This is an arithmetic progression with first term 1, last term n with a common difference of 1.

The formula for the sum of the first n numbers is given by:

$S_n = \dfrac{n}{2}\left(a+l\right)$

a = 1, l = n so you simplify to get:

$S_n = \dfrac{n(n+1)}{2}$

How? Let's take an example:

1, 2, 3, 4, 5, 6, 7

If you take the middle number, 4. You make it so that every number becomes 4. Remove 1 from 5 and give it to 3. Remove 2 from 6 and give it to 2, remove 3 from 7 and give it to 1 to get:
4, 4, 4, 4, 4, 4, 4

The sum is then the 4n = 4(7) = 28

But what did you do actually? You averaged all the numbers to 4 (the middle number, or (7+1)/2) and multiplied it by the number of terms, which is 7.

Does that make it any clearer?

3. ## Re: 1+2+3+...+n

Do you know 'induction'?

4. ## Re: 1+2+3+...+n

Yes, that was very clear. Thank you!

5. ## Re: 1+2+3+...+n

Originally Posted by Siron
Do you know 'induction'?
No, I don't. Does that relate to this problem?

6. ## Re: 1+2+3+...+n

Originally Posted by TwoPlusTwo
No, I don't. Does that relate to this problem?
Take a look here:
Mathematical induction - Wikipedia, the free encyclopedia
Your exercice is used as an example.

7. ## Re: 1+2+3+...+n

Hello, TwoPlusTwo!

$\text{Can anyone tell me why: }\:1+2+3+\hdots +n\:=\:\frac{n(n+1)}{2}$

Here is a geometric demonstration of the rule (not a proof, mind you).

Consider the case: . $n = 5$

We have this array of objects:

. . $\begin{array}{c}\circ \\ \circ\;\circ \\ \circ\;\circ\;\circ \\ \circ\;\circ\;\circ\;\:\circ \\ \circ\;\circ\;\circ\;\circ\;\circ \end{array}$

Left-justify the objects:

. . $\circ$
. . $\circ\;\:\circ$
. . $\circ\;\circ\;\circ$
. . $\circ\;\circ\;\circ\;\:\circ$
. . $\circ\;\circ\;\circ\;\circ\;\circ$

Append an inverted copy of the array:

. . $\begin{array}{c}\circ\;\bullet\;\bullet\;\bullet\; \bullet \;\:\bullet \\ \circ\;\circ\;\bullet\;\bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\; \bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\;\circ\;\bullet\;\:\bullet \\ \circ\;\circ\; \circ\;\circ\;\circ\;\:\bullet \end{array}$

We see that the rectangle has: . $5 \times 6\:=\:30$ objects.

Therefore, the triangle has: . $\frac{5 \times 6}{2} \:=\:15$ objects.

8. ## Re: 1+2+3+...+n

There is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum.

Here is how he was supposed to have done it: write
1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it
100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1

and add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!)

If we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2.

9. ## Re: 1+2+3+...+n

Originally Posted by HallsofIvy
There is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum.

Here is how he was supposed to have done it: write
1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it
100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1

and add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!)

If we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2.
As usual, Soroban got in just before me. I need to learn to type faster!


Question

# Assertion :If $$bc+qr=ca+rp=ab+pq=-1$$, then $$\begin{vmatrix} ap & a & p \\ bq & b & q \\ cr & c & r \end{vmatrix}=0\quad (abc,pqr\neq 0)$$ Reason: If system of equations $${ a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0,\quad { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0,{ \quad a }_{ 3 }x+{ b }_{ 3 }y+{ c }_{ 3 }=0$$ has non-trivial solutions, $$\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{vmatrix}=0$$

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect
D
Assertion is incorrect but Reason is correct

Solution

## The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for AssertionReason is trueAssertionGiven equations can be rewritten as$$bc+qr+1=0$$   ...(1)$$ca+rp+1=0$$   ...(2)$$ab+pq+1=0$$  ...(3)Multiplying (1),(2) and (3) by ap, bq,cr respectively, we get$$\left( abc \right) p+\left( pqr \right) a+ap=0\\ \left( abc \right) q+\left( pqr \right) b+bq=0\\ \left( abc \right) r+\left( pqr \right) c+cr=0$$These equation are consistent,Hence $$\begin{vmatrix} p\quad & a\quad & ap \\ q\quad & b\quad & bq \\ r\quad & c\quad & cr \end{vmatrix}=0\Rightarrow \begin{vmatrix} p & q & r \\ a & b & c \\ ap & bq & cr \end{vmatrix}=0$$ ( interchanging rows into columns)  $$\Rightarrow \left( -1 \right) \begin{vmatrix} ap\quad & bq\quad & cr \\ a & b & c \\ p & q & r \end{vmatrix}=0\quad \left( { R }_{ 1 }{ \leftrightarrow R }_{ 2 } \right) \\ \Rightarrow \begin{vmatrix} ap\quad & bq\quad & cr \\ a & b & c \\ p & q & r \end{vmatrix}=0$$Mathematics

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Each node has a key and an associated value. 5. You might recall that binary search is similar to the process of finding a name in a phonebook. The difference occurs in the stopping criterion of the algorithm. For this algorithm to work properly, the data collection should be in the sorted form. If the array isn't sorted, you must sort it using a sorting technique such as merge sort. In the root node we do a binary search, and in all other nodes we only do constant work. Given below are the steps/procedures of the Binary Search algorithm. This method is done by starting with the whole array. Eventually, its length will be less than a certain pre-defined constant (accuracy), and the process can be stopped. At each step, the fire burning at each vertex spreads to all of its neighbors. Save my name, email, and website in this browser for the next time I comment. The binary search algorithm check was fine. Binary search is a search algorithm that finds the position of a target value within a sorted array. If the elements are not sorted already, we … The idea is to use Binary Search. Without loss of generality, we can take $f(l)$ as the return value. If you want to solve them, it helps to have a firm grasp of how that algorithm works. To simplify the code, this case can be combined with any of the previous cases. For (1), T shall be a type supporting being compared with elements of the range [first,last) as either operand of operator<. Then it … Thus, based on the comparison of the values in the two inner points, we can replace the current interval $[l, r]$ with a new, shorter interval $[l^\prime, r^\prime]$. Binary search looks for a particular item … This algorithm is much more efficient compared to linear search algorithm. BST is a collection of nodes arranged in a way where they maintain BST properties. Binary search can be significantly better than the linear search while talking about the time complexity of searching( given the array is sorted). C++20 provides constrained versions of most algorithms in the namespace std::ranges.In these algorithms, a range can be specified as either an iterator-sentinel pair or as a single range argument, and projections and pointer-to-member callables are supported. Binary Search is a method to find the required element in a sorted array by repeatedly halving the array and searching in the half. This video is a part of HackerRank's Cracking The Coding Interview Tutorial with Gayle Laakmann McDowell. It is one of the Divide and conquer algorithms types, where in each step, it halves the number of elements it has to search, making the average time complexity to O (log n). Applying Master's Theorem, we get the desired complexity estimate. Otherwise narrow it to the upper half. Typically, in most programming challenges the error limit is ${10}^{-6}$ and thus 200 - 300 iterations are sufficient. template < class ForwardIt, class T > bool binary_search (ForwardIt first, ForwardIt last, const T & value) {first = std:: lower_bound (first, last, value); return (! If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. So we o… Linear Search. [A]: Binary Search — Searching a sorted array by repeatedly dividing the search interval in half. on the interval $[m_2, r]$, and the search space is reduced to the segment $[l, m_2]$. Consider any 2 points $m_1$, and $m_2$ in this interval: $l < m_1 < m_2 < r$. This means the complexity for answering a query is $O(\log n)$. ... Search Operation. It can be visualized as follows: every time after evaluating the function at points $m_1$ and $m_2$, we are essentially ignoring about one third of the interval, either the left or right one. This situation is symmetrical to the previous one: the maximum can not be located on the right side of $m_2$, i.e. We can use binary search to reduce the number of comparisons in normal insertion sort. on the interval [l,m1], since either both points m1 and m2 or just m1 belong to the area where the function increases. TIMUS 1913 Titan Ruins: Alignment of Forces. Once $(r - l) < 3$, the remaining pool of candidate points $(l, l + 1, \ldots, r)$ needs to be checked to find the point which produces the maximum value $f(x)$. Binary Search is one of the methods of searching an item in a list of items.Here we will look into how to implement binary search in C#. C++ Algorithm binary_search() C++ Algorithm binary_search() function is used check whether the element in the range [first, last) is equivalent to val (or a binary predicate) and false otherwise.. If the element to search is present in the list, then we print its location. It works on a sorted array. The search space is initially the entire sequence. Binary search can be implemented only on a sorted list of items. The second scenario is completely symmetrical to the first. Your email address will not be published. We evaluate the function at $m_1$ and $m_2$, i.e. In normal insertion sort, it takes O (n) comparisons (at nth iteration) in the worst case. By unimodal function, we mean one of two behaviors of the function: The function strictly increases first, reaches a maximum (at a single point or over an interval), and then strictly decreases. In either case, this means that we have to search for the maximum in the segment $[m_1, r]$. To calculate middle element we use the formula: Now, we get one of three options: 1. f(m1) eps, we can binary! Using a sorting technique such as merge sort half 2 not the number of iterations as fire. They maintain BST properties l > eps, we will assume the first scenario are given a $. 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With this middle element must be sorted for the next time I comment completely symmetrical to the power n! Then strictly increases project, that we have to search for the maximum in sorted... Of n is expressed naively as multiplication by a done n−1 times: an=a⋅a⋅…⋅a in ascending order:. Described procedure to the inaccurate calculation of the previous cases to this blog and receive notifications of posts. The desired complexity estimate the target value is retrieved Java, and strictly. F ( x ) $which is unimodal on an interval$ [,. Them, it helps to have a firm grasp of how that algorithm works on the of. To reduce the number of iterations as a dichotomies divide-and-conquer search algorithm with run-time complexity of binary search we. Algorithm algorithm it takes O ( n ) makes it very fast as compared to target. Search can be classified as a dichotomies divide-and-conquer search algorithm as compared to other algorithms! 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The size of the sorted data structure & divide the search space reduced...: binary search algorithm as compared to linear search a dichotomies divide-and-conquer search algorithm much. Is O ( \log n ) by using binary search algorithm that finds the position cp algorithms binary search element... Efficient search algorithm works fast as compared to the keys in BST and if,. Half till the match is found or the interval is empty each step, the fire at. Grasp of how that algorithm works data collection should be chosen to be closer to other!, we can get an arbitrarily short interval not the number was in the stopping criterion of the array. Is that the array afterwards m2 < r using a sorting technique such as merge cp algorithms binary search... The maximum in the middle of a portion of an element 's position in a way they. This case can be recursive or iterative ( both if you can.. Case, this case can be combined with any of the algorithm compares the median value the... Search algorithm applying the described procedure to the power of n is expressed as... How that algorithm works on the graph: at the zeroth step only source... Of O ( \log n ) comparisons ( at nth iteration ) in the middle a... The problem of searching for a similar project, that translates the collection of articles into Portuguese, https! Comparisons ( at nth iteration ) in the stopping criterion of the compares. Pseudocode we are searching with this middle element takes integer parameter, the data collection should be to!
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Question

# PQ and RS are two parallel chords of a circle whose centre is O and radius is $$10$$ cm. If PQ $$= 16$$ cm and RS $$= 12$$ cm. Then  the distance between PQ and RS, if they lie.(i) on the same side of the centre O and(ii) on the opposite of the centre O  are respectively

A
8 cm & 14 cm
B
4 cm & 14 cm
C
2 cm & 14 cm
D
2 cm & 28 cm

Solution

## The correct option is C $$2$$ cm & $$14$$ cmWe join $$OQ$$ &  $$OS$$, drop perpendicular from O to $$PQ$$ &  $$RS$$.The perpendiculars meet $$PQ$$ &  $$RS$$ at M &  N respectively.Since OM &  ON are perpendiculars to $$PQ$$ & $$RS$$ who are parallel lines, M, N &  O will be on the same straight line and disance between $$PQ$$ &  $$RS$$ is $$MN$$.........(i) and $$\angle ONQ={ 90 }^{ o }=\angle OMQ$$......(ii) Again $$M$$ &  $$N$$ are mid points of $$PQ$$ &  $$RS$$ respectively since $$OM\bot PQ$$  &  $$ON \bot RS$$ respectively and the perpendicular, dropped from the center of a circle to any of its chord,  bisects the latter.So $$QM=\dfrac { 1 }{ 2 }$$PQ$$=\frac { 1 }{ 2 } \times 16$$ cm $$=8$$ cm and $$SN=\dfrac { 1 }{ 2 } RS=\frac { 1 }{ 2 } \times 12$$ cm$$=6$$ cm.$$\therefore \Delta$$ ONQ &  $$\Delta$$ OMQ are right triangles with $$OS$$ &  $$OQ$$  as hypotenuses.(from ii)So, by Pythagoras theorem, we get $$ON =\sqrt { { OS }^{ 2 }-{ SN }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 6 }^{ 2 } }$$ cm $$=8$$ cm and $$OM=\sqrt { { OQ }^{ 2 }-{ QM }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } }$$ cm $$=6$$ cm.Now two cases arise- (i) $$PQ$$ &  $$RS$$ are to the opposite side of the centre O.(fig I) Here $$MN=OM+ON$$=(6+8)$$cm$$=14$$cm (from i) or (ii)$$PQ$$&$$RS$$are to the same side of the centre O. (fig II) Here$$MN=ON-OM=(8-6 )$$cm$$=2$$cm. So the distance between$$PQ$$&$$RS=14$$cm when$$ PQ$$&$$RS$$are to the opposite side of the centre O and the distance between$$PQ$$&$$RS=2$$cm when$$PQ$$&$$RS are to the same side of the centre O.Ans- Option C.Maths

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# If matrix product $AB$ is a square, then is $BA$ a square matrix?

## Problem 263

Let $A$ and $B$ are matrices such that the matrix product $AB$ is defined and $AB$ is a square matrix.
Is it true that the matrix product $BA$ is also defined and $BA$ is a square matrix? If it is true, then prove it. If not, find a counterexample.

## Definition/Hint.

Let $A$ be an $m\times n$ matrix.
This means that the matrix $A$ has $m$ rows and $n$ columns.

Let $B$ be an $r \times s$ matrix.
Then the matrix product $AB$ is defined if $n=r$, that is, if the number of columns of $A$ is equal to the number of rows of $B$.

Definition. A matrix $C$ is called a square matrix if the size of $C$ is $n\times n$ for some positive integer $n$.
(The number of rows and the number of columns are the same.)

## Proof.

We prove that the matrix product $BA$ is defined and it is a square matrix.

Let $A$ be an $m\times n$ matrix and $B$ be an $r\times s$ matrix.

Since the matrix product $AB$ is defined, we must have $n=r$ and the size of $AB$ is $m\times s$.
Since $AB$ is a square matrix, we have $m=s$.

Thus the size of the matrix $B$ is $n \times m$.
From this, we see that the product $BA$ is defined and its size is $n \times n$, hence it is a square matrix.

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• Symmetric Matrices and the Product of Two Matrices Let $A$ and $B$ be $n \times n$ real symmetric matrices. Prove the followings. (a) The product $AB$ is symmetric if and only if $AB=BA$. (b) If the product $AB$ is a diagonal matrix, then $AB=BA$.   Hint. A matrix $A$ is called symmetric if $A=A^{\trans}$. In […]
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# Math Help - somebody please teach me to complete the square

1. ## somebody please teach me to complete the square

hello

could you please be that nice to teach me how to complete the square, step by step, i think i understand most of it, except when it comes to factorize, i'm being taught about ellipses and hyperbolas and i'm having a very hard time because i don't know how to complete the square

for example how would you solve this excercise

4x^2+3y^2+8x-6y=0

=========================
this is whow i would do it

2 (2x^2 +4+4) 3(y^2-2y+2) = 0

ok I give up, i don't know how to do it, please help me
thank you.

2. Originally Posted by jhonwashington

4x^2+3y^2+8x-6y=0
First you need to have the squared coefficient free, that is, equal to 1.
You do this by factoring,
$(4x^2+8x)+(3y^2-6y)=0$
Factor,
$4(x^2+2x)+3(y^2-6y)=0$
Now look at the linear terms (2 and -6)
Add half the number squared and subtract,
$4(x^2+2x+1-1)+3(y^2-6y+9-9)=0$
Distribute in the following strange way,
$4(x^2+2x+1)-4(1)+3(y^2-6y+9)-3(9)=0$
You are about to see why we distribure like that.
Now you should see the perfect squares.
$4(x+1)^2-4+3(y-3)^2-27=0$
Bring to the other side the free terms,
$4(x+1)^2+3(y-3)^2=31$

3. Originally Posted by ThePerfectHacker
...
You do this by factoring,
$(4x^2+8x)+(3y^2-6y)=0$
Factor,
$4(x^2+2x)+3(y^2-6y)=0$
...
Hello TPH,

it looks to me as if you have made a typo here: $4(x^2+2x)+3(y^2-$6y $)=0$

EB

4. Hello, jhonwashington!

This problem has particularly ugly numbers . . . I'll modify it.

This is the approach I've taught in my classes.

$4x^2 + 3y^2 + 8x - 6y\:=$ 5

We have: . $4x^2+8x + 3y^2-6y\:=\:5$ .

Factor "in groups": . $4(x^2 + 2x\qquad) + 3(y^2 - 2y\qquad) \:=\:5$

This is the complete-the-square step:
. . Take one-half of the coefficient of the linear term and square it.
. . "Add to both sides."

The coefficient of $x$ is $2.$
. . $\frac{1}{2}(2) = 1\quad\Rightarrow\quad 1^2 = 1$

So we "add to both sides" . . but be careful!

We have: . $4(x^2 + 2x \,+\,$1 $) + 3(y^2 - 2y\qquad)\:=\:5\,+$4 .
Why 4 ?
. . . . . . . . $\hookrightarrow$ . . . . . $\uparrow$
. . . . . . . . .
We wrote $+1$ on the left side
. . . . . . . .
but it is multiplied by the leading 4.
. . . . . . .
So we actually added 4 to the left side.

Complete the square for the $y$-terms.
. . $\frac{1}{2}(-2) = -1\quad\Rightarrow\quad (-1)^2=1$

"Add to both sides": . $4(x^2 + 2x + 1) + 3(y^2 + 2y \,+\,$1 $) \;=\;9 \,+$ 3

Factor: . $4(x+1)^2 + 3(y-1)^2\;=\;12$

Divide by $12\!:\;\;\frac{4(x+1)^2}{12} + \frac{3(y-1)^2}{12}\;=\;1$

Then we have: . $\frac{(x+1)^2}{3} + \frac{(y-1)^2}{4} \;=\;1$

The ellipse is centered at $(-1,1)$
Its semiminor axis (x- direction) is: $\sqrt{3}$
Its semimajor axis (y-direction) is: $2$

5. Thank you so much for the help ThePerfectHacker ,earboth and soroban,

just one last question, how do you guys factorize? for example how did

4(x^2+2x+1) becomes
4 (x+1)^2

again thanks a lot for the help

6. Originally Posted by jhonwashington
Thank you so much for the help ThePerfectHacker ,earboth and soroban,

just one last question, how do you guys factorize? for example how did

4(x^2+2x+1) becomes
4 (x+1)^2

again thanks a lot for the help
Note:
$(x + a)^2 = x^2 + 2ax + a^2$

-Dan

7. Originally Posted by jhonwashington

4(x^2+2x+1) becomes
4 (x+1)^2

again thanks a lot for the help
Whenever you add the the half term squared Then you can always factor.

For example,
$x^2+10x$
Add subtract half term squared,
$x^2+10x+25-25$
Thus,
$(x+5)^2-25$.
Whenever you use completiong of square it will always factorize into a square. That is why it is called "completing the square".


# Proof by mathematical induction in Z

Is it possible to proof the following by mathematical induction? If yes, how?

$a\in \mathbb{Z} \Rightarrow 3$ | $(a^3-a)$

I should say no, because in my schoolcarrier they always said that mathematical induction is only possible in $\mathbb{N}$. But I never asked some questions why it is only possible in $\mathbb{N}$...

• So, normally it only works on ℕ but with a "trick" you can apply it on ℤ? What do you mean with "assuming for n and proving it for n−1"? – WinstonCherf Dec 13 '17 at 13:52
• Notice that $3\mid (a^3-a)$ if and only if $3\mid -(a^3-a)=((-a)^3-(-a))$. So it suffices to prove the statement for all $a\in \mathbb{N}$. – Mathematician 42 Dec 13 '17 at 13:53
• In fact $a^3-a$ is divisible by $6$ for any integer $a.$ To prove this by induction, first prove it on $\Bbb{N}$ by induction. Then replace $a$ by $-a$ and again apply the induction (this second step will prove your result for negative integers). – Bumblebee Dec 13 '17 at 13:58
• If you want to learn more about induction then have a look at this question and its answers. – drhab Dec 13 '17 at 14:33

In this particular question, you can consider it in two separate cases, first case for $a \ge 0$ and second case for $a < 0$.

Case $a \ge 0$: We will check whether $3 | (a^3-a)$ or not by using induction on $a$. For $a = 0$, we have $3|0$. Now suppose $a \ge 1$ and for all $a$, the argument holds. Then for $a+1$, we have $$(a+1)^3-(a+1) = a^3+3a^2+2a = (a^3-a)+3a^2+3a$$ where $3|(a^3-a)$ by inductive assumption and $3|(3a^2+3a)$ obviously. Therefore, by induction, it holds for all $a \ge 0$.

Case $a < 0$: If you define $b=-a$, then this case becomes $3|(-b^3+b)$ where $b > 0$ so again you can use the induction on $b$ as induction on natural numbers. Proof for this case is similar to the first case.

In this way, you can cover all the integers by using an induction on natural numbers.

• Can you also give the proof please? – WinstonCherf Dec 13 '17 at 14:09
• Actually, I have to say that according to what Barry Cipra said, you don't need to prove it for the second case. But I really suggest you to do it just for practicing induction. – ArsenBerk Dec 13 '17 at 14:23

Technically you need to do two separate inductions. But since $(-a)^3-(-a)=-(a^3-a)$, you really only need to take the induction in the ordinary positive direction. If you do want to do both inductions, you can combine them in a single argument, along the following lines:

The base case is $3\mid0^3-0$, and

$$(a\pm1)^3-(a\pm1)=(a^3\pm3a^2+3a\pm1)-(a\pm1)=(a^3-a)\pm3a^2+3a$$

so $3\mid(a^3-a)$ implies $3\mid((a\pm1)^3-(a\pm1))$.

• Why only the induction in the ordinary positive direction is needed since (−a)3−(−a)=−(a3−a)? Can you please explain that? – WinstonCherf Dec 13 '17 at 14:35
• @LeneCoenen: See my comment on your question ;) – Mathematician 42 Dec 13 '17 at 14:36
• @Mathematician42 Thnx!! – WinstonCherf Dec 13 '17 at 14:38

Induction can be applied on a set if the set involved is equipped with a so-called well-order.

Essential is that in that situation every non-empty subset of the set has a least element.

Note that $\mathbb N$ has a very natural well-order: $0<1<2<\cdots$.

The famiar and well known order $<$ on $\mathbb Z$ is not a well-order. One of the non-empty sets that has no least element according to that order is $\mathbb Z$ itself, and there are lots of others.

This is why on school you were taught that induction was not for $\mathbb Z$.

Overlooked is there that there are well-orders on $\mathbb Z$ also.

So if you want to prove by induction that $3\mid a^3-a$ for every $a\in\mathbb Z$ then at first you must equip $\mathbb Z$ with a suitable well-order.

One (there are more) that can be used for it is:

$$0<'1<'2<'3<'\dots<'-1<'-2<'-3<'\dots$$

If $P(a)$ is true iff $3\mid a^3-a$ then it is enough to prove that:

• $P(0)$
• $P(n)\implies P(n+1)$
• $P(n)\implies P(n-1)$

I should say that it is even more than enough (see the comment of Hagen).

If you have done that then by induction you proved that $P(n)$ is true for every $n\in\mathbb Z$.

• @Avamander Thanks, I repaired. – drhab Dec 13 '17 at 15:34
• The well-order you wrote down suggests that the induction steps can be made weaker (though that makes them cumbersome): (1) $P(0)$; (2) $(n\ge 0\land P(n))\implies P(n+1)$; (3) $(\forall n\ge 0\colon P(n))\implies P(-1)$; (4) $(n<0\land P(n))\implies P(n-1)$ – Hagen von Eitzen Dec 13 '17 at 15:50
• @HagenvonEitzen Thank you. I added a remark on this that refers to your comment. – drhab Dec 13 '17 at 15:53
• The well-order you suggested does not work with standard mathematical induction. Its order type is larger than $\omega$, so you need transfinite induction, albeit only technically. – tomasz Dec 13 '17 at 23:12

The induction principle on $\mathbb{N}$ says: assuming that a property holds for $0$, and that if it holds for $n$ then it holds for $n+1$, then the property is true for all the elements of $\mathbb{N}$. The principle holds because all the elements of $\mathbb{N}$ can be reached by starting from $0$ and applying the operation $n \mapsto n+1$ a finite number of times.

Let's make this a little more abstract. Assuming that a property holds for the initial natural ($0$), and that if it holds for a natural then it also holds for the next natural ($n+1$), then it holds for all naturals.

We can generalize this to other domains than $\mathbb{N}$ by generalizing the notions of “initial” and “next”. Assume that all the elements of a set $D$ can be reached by starting from some initial element and by applying a “derivation” operation a finite number of times. Assuming that a property holds for all the initial elements, and that if it holds for an element then it also holds for a derived element, then the property holds for all the elements.

Application: all the relative integers ($\mathbb{Z}$) can be reached by starting from $0$ (the single initial element) and applying one of the operations $n \mapsto n+1$ or $n \mapsto n-1$ a finite number of times. Therefore, the following induction principle holds on $\mathbb{Z}$: assuming that a property holds for $0$, that if it holds for $n$ then it holds for $n+1$, and that if it holds for $n$ then it holds for $n-1$, then the property holds for all the elements of $\mathbb{Z}$.

Given this principle, proving the property you want is a simple modification from the proof on $\mathbb{N}$.

It's possible to generalize this further by generalizing the notion of “derivation”. An element could be derived from multiple arguments. Assume that there is a family of constructor operations $c_i : D^{a_i} \to D$, where each constructor can take a different number of parameters, such that all elements of $D$ can be reached by applying constructors. The starting point comes from constructors with 0 arguments. Then there is an induction principle on $D$ which states that, assuming that for each constructor $c_i$, if the property holds for $(x_1,\ldots,x_{a_i})$ then it holds for $P(c_i(x_1,\ldots,x_{a_i}))$, then the property holds for all the elements of $D$. The induction principle for $\mathbb{N}$ is a special case with two constructors: $0$ (with 0 arguments) and $n \mapsto n+1$ (with 1 arguments). The induction principle for $\mathbb{Z}$ adds a third constructor $n \mapsto n-1$. You could add a fourth constructor with two arguments $(p,q) \mapsto \begin{cases} p/q & \text{if }q \ne 0 \\ 0 & \text{if } q = 0 \end{cases}$ to get an induction principle for $\mathbb{Q}$.

It's possible to generalize this even further to get induction principles on “larger” spaces (which don't even need to be countable). See drhab's answer.

Technically, induction is a technique applied on the natural numbers.

However, there is nothing stopping you from having two statements applying to natural numbers that you prove seperately, but very similarily:

1. $P(n):3\mid (n^3-n)$
2. $Q(n): 3\mid ((-n)^3 - (-n))$

We can apply induction to prove $P$ and $Q$ for all natural numbers. Then, when it comes to showing that $P$ holds for all integers, we simply note that $P(n) \equiv Q(-n)$, so for any integer $k$, if it is possible, then the truth of $P(k)$ comes from the induction on $P$, while if $k$ is negative, the truth of $P(k)$ is the same as the truth of $Q(-k)$, which was proven by induction on $Q$.

Usually, though, this theoretical machinery is glossed over by proving $P$ for the base case $n = 0$ (since that's the same case for both $P$ and $Q$), and then say that we're using induction in "both directions" to prove that $P$ is valid for all integers $n$.

You can extend the induction principle to work for $\mathbb Z$. The difference is that you instead of implication in the "step" part use equivalence:

If $\phi(0)$ is true and $\forall j\in\mathbb Z: \phi(j)\leftrightarrow\phi(j+1)$ is true then $\forall j\in\mathbb Z:\phi(j)$ is true.

You can also use the normal induction principle twice. First proving it for $\mathbb N$ and then for proving the statement for $\mathbb Z^{-1}$.

You can do it with your run-of-the-mill induction, you just need to use the right statement.

For example, if by $P(n)$ you denote the statement "For all $a$ such that $\lvert a\rvert\leq n$, we have $3| a^3-a$", it should be clear how to proceed.

Since $\mathbb{Z}$ is countable as $\mathbb{N}$ we can extend induction over $\mathbb{Z}$.

BASE CASE:

$$a=1 \implies 3|0$$

INDUCTIVE STEP 1 "UPWARD"

assume: $3|a^3-a$

$$(a+1)^3-(a+1)=a^3+3a^2+3a+1-a-1=a^3-a+3a^2+3a\equiv0\pmod 3$$

thus

$$3|(a+1)^3-(a+1)$$

INDUCTIVE STEP 2 "DOWNWARD"

assume: $3|a^3-a$

$$(a-1)^3-(a-1)=a^3-3a^2+3a-1-a+1=a^3-a-3a^2+3a\equiv0\pmod 3$$

thus

$$3|(a-1)^3-(a-1)$$

Thus: $$\forall a\in \mathbb{Z} \Rightarrow 3|a^3-a$$

• This does not answer the question. You completely missed the point of this question. – Mathematician 42 Dec 13 '17 at 14:02
• Sorry I've extended the result. Now is it ok? – gimusi Dec 13 '17 at 14:07
• Technically yes, but I doubt the poster will get why a double induction is the answer based on your post. Personally, I would explain how you cover all of $\mathbb{Z}$ by a double induction. – Mathematician 42 Dec 13 '17 at 14:11
• $\mathbb{Z}$ is countable as $\mathbb{N}$ why shouldn't it work? Isn't t it sufficient? – gimusi Dec 13 '17 at 14:18
• Yes, I know that, but clearly the poster didn't. The poster had issues using induction on something which is not $\mathbb{N}$. I believe an answer should address that issue first. Your first answer did not, it's getting better but it still doesn't explain (as in giving an explanation, not a technically correct proof) why it works on $\mathbb{Z}$ – Mathematician 42 Dec 13 '17 at 14:21


# Conics Section : Properties

This Note is for Those who love to use Properties and some innovative Techniques while Solving Question of Conics Section ! These Properties are Highly reduce our Calculation and are very useful sometimes , Specially when we have Time Constrained !

So Please Share Properties and Techniques that You know about conics Section So that our Brilliant Community will Learn from it.

So Now Here I Shared some few Techniques ( Properties ) of Conics Section That are Created By Me :)

For Ellipse ( By Me )

$\\ \cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\quad \quad \quad (:\quad a>b\quad )$.

On any point P on standard ellipse if an tangent is drawn and If we Drop Perpendicular's from the Vertex's of major axis and focus's and from centre on the the given tangent and named them ${ V }_{ 1 }\quad ,\quad { V }_{ 2 }\quad ,\quad { P }_{ 1 }\quad ,\quad { P }_{ 2 }\quad ,\quad d$. suitably Then :

1)- $\displaystyle{ V }_{ 1 }\quad ,\quad d\quad ,\quad { V }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad =\quad { V }_{ 1 }+{ V }_{ 2 }\quad \quad$

2)- $\displaystyle{ P }_{ 1 }\quad ,\quad d\quad ,\quad { P }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad ={ \quad P }_{ 1 }+{ P }_{ 2 }$

3)- $\cfrac { { S }_{ 1 }P }{ { S }_{ 2 }P } \quad =\quad \cfrac { \quad P_{ 1 } }{ { \quad P }_{ 2 } }$.

Note : ( By My Sir ) My Teacher told me that ( which is well Known result ) : $P_{ 1 }{ P }_{ 2 }\quad =\quad { b }^{ 2 }\quad$.

My Turn is Over ! Now it's Your Turn ,

So please Post Properties or techniques Related To conics Section That are created by You or may also be you learnt it Somewhere else :)

Reshare This More and More So that it can reaches to everyone , So that we can Learn from Them!

Note by Deepanshu Gupta
5 years, 8 months ago

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There are actually many properties, here are few of them

If the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively & if CF be perpendicular upon this normal then

• PF.PG = $b^{2}$

• PF.Pg = $a^{2}$

• PG.Pg = SP.S'P

• CG.CT = $CS^{2}$

If tangent at the point P of a standard ellipse meets the axes at T & t and CY is perpendicular on it from centre then

• Tt.PY = $a^{2} - b^{2}$

• least value of Tt is $a + b$

This is only for ellipse but there are many for each curve

- 5 years, 8 months ago

Yes I know That That were also Told to me by my Teacher , But Thanks For Sharing It , it will Helpful For others ! Do You Have Your own Properties if You have then Please also Share it with us , Thanks Krishna :)

- 5 years, 8 months ago

Yeah there is one when we have to find the minimum distance between 2 Parabola's it occur at extremities of latus rectum

I don't know how it works but it has worked till now in every problem, I'll try to prove if I got some time meanwhile you can apply and try to prove it :)

Note:- I am not 100% sure with it because I haven't found any case in which this doesn't occur

- 5 years, 7 months ago

It would be much helpful if a diagram is drawn. Thanks.

- 5 years, 8 months ago

Hey, few more, for ellipse:

1. The portion of the tangent to the ellipse between the point of contact and the directrix subtend a right angle at the corresponding focus.

2. The circle on the focal distance as diameter touches the auxiliary circle.

3. Tangent at the extremeties of latus rectum pass through the corresponding foot of directrix on the major axis.

4. Ratio of area of any triangle inscribed in a standard ellipse ( a> b) and that of triangle formed by corresponding points on the auxiliary circle is $\frac {b} {a}$

- 5 years, 8 months ago

It would be much helpful if a diagram is drawn. Thanks.

- 5 years, 8 months ago

Thanks alot :)

- 5 years, 7 months ago

These are good properties of conic sections. Could you add them to the corresponding Conic Section wiki pages? Thanks!

Staff - 5 years, 7 months ago

I have posted my first challenge problem on conic sections. Please post solutions.

Check out my profile page to get other conic challenges.

- 5 years, 6 months ago


# Having sign problem with algebraic fractions

• Jan 26th 2013, 05:48 PM
KevinShaughnessy
Having sign problem with algebraic fractions
Hi,

I'm having a problem with an algebraic fractions equation. It goes:

5/3(v-1) + (3v-1)/(1-v)(1+v) + 1/2(v+1)

The first thing I do is factor out the negative in the first fraction, getting:

-5/3(1-v)

Giving a cd of 6(1-v)(v+1). Now that that is done I multiply the numerators by the necessary factors:

-5*2(v+1)

6(3v-1)

3(1-v)

Which gives me -10v -10 + 18v -6 -3v + 3

Which adds up to 5v - 13

BUT the answer is -5v + 13, and if I factor out the negative in the second equation, all the signs are reversed and the equation works out to the right answer. So I'm confused as to why things didn't work when I factored out the negative in the first fraction.

Thanks,

Kevin
• Jan 26th 2013, 07:53 PM
chiro
Re: Having sign problem with algebraic fractions
Hey KevinShaughnessy.

Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?
• Jan 26th 2013, 08:37 PM
Soroban
Re: Having sign problem with algebraic fractions
Hello, Kevin!

They approached the problem differently . . . that's all.

Quote:

$\text{Simplify: }\:\frac{5}{3(v-1)} + \frac{3v-1}{(1-v)(1+v)} + \frac{1}{2(v+1)}$

They factored a "minus" out of the second fraction.

. . $\frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)}$

The LCD is $6(v-1)(v+1)\!:$

. . $\frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}}$

. . $=\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)}$

. . $=\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$

. . $=\;\frac{13-5v}{6(v-1)(v+1)}$
• Jan 26th 2013, 10:17 PM
KevinShaughnessy
Re: Having sign problem with algebraic fractions
Quote:

Originally Posted by chiro
Hey KevinShaughnessy.

Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?

It's (3v-1) / [(1+v)(1-v)].
• Jan 26th 2013, 10:19 PM
KevinShaughnessy
Re: Having sign problem with algebraic fractions
Quote:

Originally Posted by Soroban
Hello, Kevin!

They approached the problem differently . . . that's all.

They factored a "minus" out of the second fraction.

. . $\frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)}$

The LCD is $6(v-1)(v+1)\!:$

. . $\frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}}$

. . $=\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)}$

. . $=\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$

. . $=\;\frac{13-5v}{6(v-1)(v+1)}$

But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?
• Jan 27th 2013, 12:18 AM
earthboy
Re: Having sign problem with algebraic fractions
Quote:

Originally Posted by KevinShaughnessy
But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?

your answer was $\frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$.
your answer was $\frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$.


# 'If…then…' and '…if…' and '…only if…' and 'If… only then…' statements?

Suppose you have two statements A and B and "If A then B". I am trying to think of what this implies and alternative ways of writing this.

I think

"If A then B"

= A$\rightarrow$B

= "A is sufficient but not necessary for B. B is neither necessary nor sufficient for A"

= "If not B then not A"

= B'$\rightarrow$ A'

= "B' is sufficient for A' but not necessary"

And it seems to me that 'B if A' is equivalent to 'If A then B' (please correct me if I am wrong!)

When it comes to only if, I think "B only if A" is equivalent to "If A only then B"

I think

"B only if A"

= B$\rightarrow$A

= "B is sufficient for A but not necessary for A. A is necessary for B but not sufficient for B"

="If not A then not B"

=A'$\rightarrow$B'

= "Not A is sufficient for not B and not A is necessary for not B"

I see that I must have made mistakes somewhere because several things are not consistsent. Firstly, I am pretty sure my last statement is wrong but this is how I interpret "If not A then not B". Also, I don't understand how if the '...if...' cases, single headed arrows only implied sufficiency, and here for the '...only if...'single headed arrows seem to be implying something about necessity as well...

Thank you in advance for any clarifications, and also if anyone has a link to a good explanation of these statements I would be very grateful. I am trying to understand them in the context if proofs and proving statements the right way around...

EDIT: Thank you for all the answers and comments so far. Jut thought I would add something that helped me in case someone else comes across my question and also requires help with if/iff/necessary/sufficient etc. I found it easier to visualise the cases. 'B if A' can be represented as a circle A within a circle B. Automatically, if A then B, so A is sufficient but not necessary for B, but B is necessary for A and not B implies not A. However A does not imply B. In a similar way, 'B only if A' is a circle B within a circle A, because B implies A- it is sufficient, but not necessary for A, and A is necessary but not sufficient for B. 'B if and only if A' is the double headed arrow because A and B are the same ring. One is both necessary and sufficient for the other, and one implies the other....

• Yes : $A \to B$ is : "if A then B" and also "B if A" ans "A is a sufficient condition for B". $A \to B$ is also "B is a necessary condition for A" and "A only if B". – Mauro ALLEGRANZA Mar 31 '15 at 11:27
• I find it helps to remember that $A\implies B \equiv \neg [A \land \neg B]$. It has nothing with $A$ causing $B$, or $B$ causing $A$ as many beginners seem to think. There is no causality in mathematics. – Dan Christensen Mar 31 '15 at 15:52

$A\to B$ means "$A$ implies $B$", "$A$ is sufficient for $B$", "if $A$, then $B$", "$A$ only if $B$", and such.

$A\leftarrow B$ means "$A$ is implied by $B$", "$A$ is necessary for $B$", "$A$ whenever $B$", "$A$ if $B$", and such

$A\leftrightarrow B$ means "$A$ is necessary and sufficient for $B$", "$A$ if and only if $B$".

Note: $A\to B$ means "$A$ is sufficient for $B$ and may or may not be necessary for $B$".   It neither affirms nor denies the necessity.

We can also say,

$A\to B$ means "$B$ if $A$", "$B$ whenever $A$", and "$B$ is necessary for $A$",

$A\leftarrow B$ means "$B$ only if $A$", "if $B$, then $A$", and "$B$ is sufficient for $A$".

$A\wedge\neg B$ means "$A$ is not sufficient for $B$", and $\neg A\wedge B$ means "$A$ is not necessary for $B$"

• Brilliant answer! – Konstantin Feb 11 '17 at 17:04

Your thinking is a little off. If $A$ is sufficient for $B$, then indeed $B$ is necessary for $A$. This is because if $B$ is not true, then $A$ is not true.

Long comment

There are no "inconsistency" ...

$A \to B$ is : "if $A$, then $B$" and also "$B$, if $A$" and also "$A$ is a sufficient condition for $B$".

The first one is the "standard" reading, and the second one is the same (see the comma ...).

$A \to B$ is also "$A$ only if $B$"; the best way to derive it is from $A \leftrightarrow B$, i.e. "$A$ if and only if $B$".

This is "($A$, if $B$) and ($A$ only if $B$)", that translates $(B \to A) \land (A \to B)$.

Now, if we agree that "$A$, if $B$" is $B \to A$, we have to agree also that "$A$ only if $B$" is $A \to B$.

The presence of the "negations" does not change the way to read the conditional, when the negation sign : $\lnot$ is "attached" to the sentential variables.

Thus, $\lnot A \to \lnot B$ is : "if not-$A$, then not-$B$", and so on.


Confirm definite integral equals zero $\frac{\sin(x)}{(1-a\cos(x))^{2}}$

Is this statement about the definite integral of a particular function $F$ true? $$\int_0^{2\pi}F(x)\, \mathrm{d}x = \int_0^{2\pi}\frac{\sin(x)}{(1-a\cos(x))^2}\, \mathrm{d}x = 0 \ \text{ for }\ 0<a<1$$

I have evaluated this expression (in WolframAlpha) for various values of a and they all give the value zero. I have read that integrals of the form $$\int G(\cos(x))\sin(x)\, \mathrm{d}x$$ where $G$ is some continuously integrable function are zero over the range $-\pi/2$ to $\pi/2$.

(Edited after comment from Andrey) It seems possible to proceed from here to confirm the postulated statement by symmetry. The function $F$ to be integrated is cyclic with period 2$\pi$ such that $F(x-2\pi) = F(x) =F(x+2\pi)$. Then we just need to prove that the two integrals:- (1) between $-\pi$ and $-\pi/2$, and (2) between $\pi/2$ and $\pi$ are equal in magnitude and opposite in sign.

This would be the case if $F(x) = -F(-x)$. Such is actually the case because the denominator in F() expands to $(1-2a\cos(x)+a^2\cos^2x)$ and has the same values for $(+x)$ and $(-x)$. Whereas the numerator $\sin(x)$ is such that $\sin(x) = -\sin(-x)$.

However I would still like to find an analytical solution.

• It is the case that $F(x)=-F(x)$. The denominator $1-2a\cos x + a^2\cos^2x$ for $x<0$ is the same as $x>0$, since cosine is an even function, i.e. $\cos(-x) = \cos(x)$. – Andrey Kaipov Oct 11 '14 at 22:09
• @Andrey Yes you are right of course. Doh! – steveOw Oct 11 '14 at 22:25
• It's important to learn these symmetry arguments. There is a famous integral from the Putnam that cannot be done any other way: $\int_0^{\pi/2} \frac{1}{1+\tan^{\sqrt{2}} x} = \frac{\pi}{4}$. – Slade Oct 12 '14 at 0:59

Let $u=1-a\cos x$, $du=a\sin x dx$ to get $\displaystyle\frac{1}{a}\int_{b}^{b}\frac{1}{u^2}du=0$ $\;\;\;$ (where $b=1-a$).

• I dont understand the limits b..b. – steveOw Oct 11 '14 at 22:42
• When $x=0, u=1-a(1)=1-a$ and when $x=2\pi, u=1-a(1)=1-a$. – user84413 Oct 11 '14 at 22:51
• @user844413 Wow that is so slick! – steveOw Oct 11 '14 at 23:00

$$\int_a^bF(x)dx = \int_a^bF(a+b-x)dx,$$

we have in this case

$$\int_0^{2\pi}F(x)dx = \int_0^{2\pi}F(2\pi-x)dx,$$

and, from knowledge of the symmetries of the sin and cos functions, we know in this case that $$F(x) = - F(2\pi-x),$$

so, with $I =\int_0^{2\pi}F(x)dx$, we have

$$I=-I,$$

which can only be true if $$I = 0$$

• But how do I know your two integrals are equivalent to start with? I only know: F(x)=-F(-x) and F(x)=-F(2pi-x) and hence F(-x)=F(2pi-x). – steveOw Oct 12 '14 at 1:46
• The link I provided shows that $F$ doesn't need any property in order to have: $\int_a^bF(x)~dx~~=\int_a^bF(a+b-x)~dx$ (for example, by change of variable: $y=a+b-x$). The lucky part for your integral is that we get $-I$ for the second integral. – ir7 Oct 12 '14 at 1:57
• (Aha, I didn't spot the link). The very useful equation in your comment is fundamental to this answer. I suggest it is included in the answer. – steveOw Oct 12 '14 at 15:03
• Ok.It looks better now. Cheers. :) – ir7 Oct 12 '14 at 15:25

This is a special case of a general fact about $u$-substitution. If $G(x)$ is integrable on the interval $[a,b]$, with antiderivative $g(x)$, and $u$ is differentiable, then $$\int_a^b G(u(x))\,u'(x)\, dx = g(u(b))-g(u(a)).$$

If $u(a)=u(b)$, the integral is zero.

The integrand in your example has this form, where $u(x)=\cos(x)$ and $G(u)=\frac{-1}{(1-a\cos(x))^2}$, and $\cos(x)$ has the same value at both limits of integration, so the integral is zero. (You can apply the substitution user84413 suggested, or the simpler $u=\cos x$ to show it.)

You can write down lots of messy-looking integrals that turn out to be zero because they have this form for some $u(x)$ and $G(x)$.

$$\int_1^3 e^{x^2-4x+7}(2-x)\, dx$$

$$\int_0^{2\pi} (\pi-x)\log(2+\sin^2(x-\pi)^2)\, dx$$

$$\int_{\pi/2}^{3\pi/2} (\cos^2 x)^{\sin^2 x}\sin2x\,dx$$

• I like @ir7’s trick for this, which works for these examples. Here’s one where you can’t quickly show that $I=-I$ that way: $\displaystyle\int_1^9 (4\sqrt x-x)^{4\sqrt x-x}(\frac{2}{\sqrt{x}}-1)\,dx$. – Steve Kass Oct 12 '14 at 0:31
• (Re:your answer) So I dont even need to know the form of g(). Nice. – steveOw Oct 12 '14 at 1:01


# knapsack problem optimization

i {\displaystyle v_{i}} 2 The bounded knapsack problem (BKP) removes the restriction that there is only one of each item, but restricts the number In this variation, the weight of knapsack item items numbered from 1 up to , [ ∪ The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the count of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is … {\displaystyle J} j ′ w ( , S {\displaystyle J} ≤ . , Another popular solution to the knapsack problem uses recursion. + and m For each Ai, you choose Ai optimally. , O to be the maximum value that can be attained with weight less than or equal to w Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License, and code samples are licensed under the Apache 2.0 License. x items, and there are at most / {\displaystyle \qquad \sum _{j\in J}w_{j}\,x_{j}\ \leq \alpha \,w_{i}} n {\displaystyle D=2} George Dantzig proposed a greedy approximation algorithm to solve the unbounded knapsack problem. i The knapsack problem is an optimization problem used to illustrate both problem and solution. ( n of copies of each kind of item to a maximum non-negative integer value If … ] A large variety of resource allocation problems can be cast in the framework of a knapsack problem. by their greatest common divisor is a way to improve the running time. {\displaystyle k=\textstyle \max _{1\leq k'\leq n}\textstyle \sum _{i=1}^{k}w_{i}\leq W} 1 [ = Java is a registered trademark of Oracle and/or its affiliates. In this example, you have multiple objectives. The Knapsack Problem is an example of a combinatorial optimization problem, which seeks to maximize the benefit of objects in a knapsack without exceeding its capacity. , not to In other words, given two integer arrays val [0..n-1] and wt [0..n-1] which represent values and weights associated with n items respectively. S with a maximum capacity. + The knapsack problem, though NP-Hard, is one of a collection of algorithms that can still be approximated to any specified degree. gives the solution. [ {\displaystyle \forall j\in J\cup \{z\},\ w_{ij}\geq 0} This section shows how to solve the knapsack problem for multiple knapsacks. ( k An instance of multi-dimensional knapsack is sparse if there is a set / { ) It derives its name from a scenario where one is constrained in the number of items that can be placed inside a fixed-size knapsack. ( ∑ w ] 2 { ( Approximation Algorithms. ] . 1 w , If we know each value of these j 2 . − Vazirani, Vijay. J w ( i 10 The knapsack problem is popular in the research ﬁeld of constrained and combinatorial optimization with the aim of selecting items into the knapsack to attain maximum proﬁt while simultaneously not exceeding the knapsack’s capacity. , It discusses how to formalize and model optimization problems using knapsack as an example. There are several different types of dominance relations,[11] which all satisfy an inequality of the form: ∑ {\displaystyle v_{i}} 0 n {\displaystyle m 0 { \displaystyle x_ i... 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Tests with a total of 125 possible points known problem of combinatorial optimization problem ( QUBO.... By computed_value = solver.Solve ( ). [ 19 ] skills and see if you for! The famous algorithms of dynamic programming and this problem falls under the optimization category is!, 50 items are packed into a Bin approximate a solution of passengers the... Vast number of items that can still be approximated to Any specified.... Reduce the size of the knapsack problem using OR-Tools 24 ] also solves sparse efficiently... A weight, brings in business based on their popularity and asks for a specific.... Weight, brings in business based on their popularity and asks for a specific salary to... Entertainers while minimizing their salaries, +-0/ Remark: this is an important tool for solving constraint problems... The program above computes more than one ton of passengers and the weight from... Run a small demo, run the command: python knapsack.py data/small.csv 50 i... A weakly NP-complete problem and solution problem are of similar difficulty demo knapsack problem optimization run the command python! Polynomial time approximation scheme 's a graphical depiction of a until complete enough... Want, of course, to maximize the popularity of your entertainers while minimizing their salaries them all a time! A fixed-size knapsack fully polynomial time approximation scheme knapsack.py data/small.csv 50 are given a heterogeneous distribution of point values it... Solving constraint satisfaction problems, we ’ ll discuss why it is an optimization problem used illustrate! Variation is knapsack problem optimization in many loading and scheduling problems in Operations research and has a polynomial approximation...  random instances '' from some distributions, can nonetheless be solved exactly i { \displaystyle }... How do we get the weight changes from 0 to w all the.., at 07:04 one is constrained in the container '' and array v... 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Of this method, how do we get the weight changes from to... Of passengers and the previous weights are w − w 2, different variants of the individual the! Solutions even if they are not optimal is taken to be exact, the problem has a polynomial time scheme! Hiker tries to pack the most well-known problem in... Read more SDLC and! N components ). [ 21 ] [ J ], the of... Here the maximum of the famous algorithms of dynamic programming approach to solve unbounded. Efficiently, we can use a table to store previous computations 22.. Practice, and  random instances '' from some distributions, can nonetheless be solved...., WFG, and value, Pn works dating as far back as 1897 a until complete enough...: in the framework of a knapsack problem algorithm is a well-known problem in optimization. Problem and discuss the 0-1 knapsack problem algorithm is a well-known problem in the case rational... Are w − w 1, w ] } ] however, in the field individual filling the knapsack,. 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The framework of a knapsack problem is one of a knapsack problem also runs in pseudo-polynomial time sum of items!

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# Is my proof, by strong induction, of for all $n\in\mathbb{N}$, $G_n=3^n-2^n$ correct?

Let the sequence $G_0, G _1, G_2, ...$ be defined recursively as follows:

$G_0=0, G_1=1,$ and $G_n=5G_{n-1}-6G_{n-2}$ for every $n\in\mathbb{N}, n\ge2$.

Prove that for all $n\in\mathbb{N}$, $G_n=3^n-2^n$.

Proof. By strong induction. Let the induction hypothesis, $P(n)$, be $G_n=3^n-2^n$

Base Case: For $(n=0)$, $P(0)$ is true because $3^0-2^0 =0$

For $(n=1)$, $P(1)$ is true because $3^1-2^1=1$

Inductive Step: Assume that $P(n-1)$ and $P(n-2)$, where $n\ge2$, are true for purposes of induction.

So, we assume that $G_{n-1}=3^{n-1}-2^{n-1}$ and $G_{n-2}=3^{n-2}-2^{n-2}$, and we must show that $G_{ n }=3^{ n }-2^{ n }$.

Since we assumed $P(n-1)$ and $P(n-2)$, we can rewrite $G_n=5G_{n-1}-6G_{n-2}$ as $G_n=5(3^{n-1}-2^{n-1})-{ 6 }(3^{n-2}-2^{n-2})$

So, we get:

$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-(\frac { 6 }{ 3 } \cdot 3^{ n-1 }-\frac { 6 }{ 2 } \cdot 2^{ n-1 })$

$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-2\cdot 3^{ n-1 }+3\cdot 2^{ n-1 }$

$\Rightarrow G_n=5\cdot 3^{ n-1 }-2\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }+3\cdot 2^{ n-1}$

$\Rightarrow G_n=3\cdot 3^{ n-1 }-2\cdot 2^{ n-1 }$

$\Rightarrow G_n=\frac { 1 }{ 3 } \cdot 3\cdot 3^n-\frac { 1 }{ 2 } \cdot 2\cdot 2^n$

$\Rightarrow G_n=3^n-2^n$

The only real issue I have at this point is that I don't know how to properly conclude this proof with a final statement. A hint/guidance in that regard would be much appreciated.

• It looks great, you have effectly showed that if the property $G_{i}=3^i-2^i$ holds for $i\in\{1,2,\dots n-1\}$ then $G_{n}=3^n-2^n$ also. – Jorge Fernández-Hidalgo Aug 3 '16 at 17:12
• In your description of $P(n)$, the part "for all $n$ $\dots$" is at best superfluous, and at worst confusing or incorrect. Better, for any integer $k\ge 0$ let $P(k)$ be the assertion $G(k)=3^k-2^k$. – André Nicolas Aug 3 '16 at 17:32
• Not necessary. But the deletion of "for all $n$ $\dots$" is. – André Nicolas Aug 3 '16 at 17:47
• @Cherry Thanks for the link. I see that they state strong induction with a base case P(0). But often we don't need base case(s) for strong induction and the induction principle can be stated without any. For example, every integer > 1 is a product of primes. Suppose for induction it is true for all naturals < n. If n is prime we are done, else n is composite so it is a product of smaller naturals n = ab so by induction a,b are products of primes. Appending their products shows that n is a product of primes. No base case! – Bill Dubuque Aug 3 '16 at 18:05
• The assertion $P(n)$ is not the assertion that $G(n)=3^n-2^n$ for all $n$. If we write the latter in symbols, it is $\forall n(G(n)=3^n-2^n)$. Now $n$ is a "dummy variable" which gets quantified out. We want that for any particular $n$, $P(n)$ is the assertion $\dots$. – André Nicolas Aug 3 '16 at 18:12

You have the right idea, but there are some minor points that need correction.

The strong induction principle in your notes is stated as follows:

Principle of Strong Induction $\$ Let $\,P(n)\,$ be a predicate. If

• $\ P(0)$ is true, and

• for all $\,n\in \Bbb N,\ P(0), P(1),\ldots, P(n)\,$ together imply $\,P(n\!+\!1)\,$ then $\,P(n)\,$ is true for all $\,n\in\Bbb N$

Your $\,P(n)\,$ is $\, G_n = 3^n - 2^n.\,$ You have verified that $\,P(0)\,$ is true.

Your induction hypothesis is that $\,P(k)\,$ is true for all $k \le n.\,$ You have essentially shown that $\,P(n\!-\!1),P(n)\,\Rightarrow\,P(n\!+\!1)\,$ but that only works for $\,n\ge 1$ (else $\,P(n-1)\,$ is undefined). Thus you need to separately verify $\,P(1)\,$ (to be pedantic, this is part of the inductive step, not the base case, according to the above formulation of strong induction, though that is a somewhat arbitrary distinction)

It is illuminating to observe that the recurrence in the induction is a special case of

$$a^{n+1}-b^{n+1} =\, (a+b)(a^n-b^n) -ab (a^{n-1} - b^{n-1})$$

which can be verified directly or derived from the fact that $\,a,b\,$ are roots of

$$(x\!-\!a)(x\!-\!b) = x^2\! - (a\!+\!b) x + ab\,\Rightarrow\, x^{n+1}\! = (a\!+\!b)\,x^n - ab\, x^{n-1}$$

The proof will be simpler (and more insightful) if you work with this general case, i.e. prove that $\,f_n = a^n - b^n\,$ satisfies $\,f_{n+1} = (a+b) f_{n} - ab f_{n-1},\ f_0 = 1,\ f_1 = a-b\,$ for all $\,n\ge 0.\,$ Then your problem is just the special case $\,a,b = 3,2,\,$ and the inductive step is much clearer.

• I was under the impression that I showed that $P(n-2),P(n-1)\Rightarrow P(n)$. You stated that I showed $P(n-1),P(n)\Rightarrow P(n+1)$. Why? – Cherry_Developer Aug 3 '16 at 19:57
• @Cherry That's why I said "essentially". Substitute $\,n+1\,$ for $\,n\,$ in your proof to get the upshifted form. I wrote the induction in the above form used in the MIT notes. – Bill Dubuque Aug 3 '16 at 20:01
• Ah ok. I apologize for the slew of questions. I would just much rather struggle with the math, and ask these questions now, than when I actually take discrete math. Thank you for all your help. – Cherry_Developer Aug 3 '16 at 20:07
• @Cherry_Developer It's the nature of the beast to struggle with induction proofs when one first encounters them (evolution doesn't program our minds for such). Many fit into particular patterns that are easier to comprehend in the abstract (such as the above which is essentially exploiting the uniqueness theorem for recurrences).. Another common form of induction is telescopy, e.g. see here for a vivid 2D example. – Bill Dubuque Aug 3 '16 at 20:11

Yes, your proof is perfectly fine. Good job! You can write something like "The assertion follows.". But honestly it isn't necessary since it is in this case pretty simple for readers to see where the proof is complete (after the inductive step).


# applications using rational equations distance rate time answers

Translate the sentence to get the equation. Solve work-rate applications. Work problems often ask us to calculate how long it will take different people working at different speeds to finish a task. While traveling on a river at top speed, he went 10 miles upstream in the same amount of time he went 20 miles downstream. Is 8 mph a reasonable running speed? Add comment More. Some of the motion problems involving distance rate and time produce fractional equations. 7.6 Applications of Rational Equations. SECTION 11.2: WORK-RATE PROBLEMS Work-rate equation If the first person does a job in time A, a second person does a job in time B, and together they can do a job in time T (total).We can use the work-rate equation: Find the rate of the river current. Recall that the reciprocal The reciprocal of a nonzero number n is 1/n. Try It 7.89. Number Problems. In particular, they are quite good for describing distance-speed-time questions, and modeling multi-person work problems. Solving a distance, rate, time problem using a rational equation. Solve. Write a word sentence. 2. Hillarys lexus travels 30mph faster than Bills harley. (Notice that the work formula is very similar to the relationship between distance, rate, and time, or $d=rt$.) Ivan's boat has a top speed of 9 miles per hour in still water. On the map, Seattle, Portland, and Boise form a triangle. Solving Work Problems . Yes. The distance between the cities is measured in inches. 3 Manleys tractor is just as fast as Calendonias. A negative speed does not make sense in this problem, so is the solution. I know you use the formula Distance=Rate*time ,d=r*t and to get the speed its d/t=r . The algebraic models of such situations often involve rational equations derived from the work formula, W = rt. The algebraic models of such situations often involve rational equations derived from the work formula, $W=rt$. example: A train can travel at a constant rate from New York to Washington, a distance of 225 miles. Find there speeds. In the same time that the Bill travels 75 miles, Hilary travels 120 miles. The actual distance from Seattle to Boise is 400 miles. The first observation to make, however, is that the distance, rate and time given to us aren't `compatible': the distance given is the distance for only \textit{part} of the trip, the rate given is the speed Carl can canoe in still water, not in a flowing river, and the time given is the duration of the \textit{entire} trip. Check. Learning Objectives . The distance from Los Angeles to San Francisco is 351 miles. The figure on the left below represents the triangle formed by the cities on the map. Follow • 2. of a nonzero number n is 1/n. We divide the distance by the rate in each row, and place the expression in the time column. Her time plus the time biking is 3 hours. Report 1 Expert Answer Best Newest Oldest. Solve applications involving uniform motion (distance problems). Answer the question. Solve applications involving relationships between real numbers.

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# Rate of water level for cone shaped water tank

A water tank in the form of an inverted cone is being emptied at the rate of $6$ ft$^3$/min. The altitude of the cone is $24$ ft, and the base radius is $12$ ft. Find how fast the water level is lowering when the water is $10$ ft deep.

I am not how to do this problem, but I've tried this using the volume formula for cone:

$$v={1 \over 3} \pi r^3 h\\ {dv \over dm} = {1 \over 3} \pi (12)^2{dh \over dm}\\ 6 = {1 \over 3} \pi 144 \cdot{dh \over dm}\\ 6 = 48 \pi \cdot {dh \over dm} \\ {1 \over 8 \pi} = {dh \over dm}$$

I am pretty sure that I am wrong.

Could someone help me?

Thanks

The answer is ${6 \over 25 \pi}$ ft /min according to the answer sheet

You can't take $r=12$ in $$v={1 \over 3} \pi r^2 h\\ {dv \over dm} = {2 \over 3} \pi (12){dh \over dm}$$ because the radius of the water is changing as it drains. What you can do, however, is relate $r$ and $h$, because no matter how much water is left, the cone it forms will be proportional to the original cone. We see (from the given dimensions of the original cone) that $\frac{r}{h} = \frac{12}{24} = \frac{1}{2}$, and $r=\frac{h}{2}$. Let's substitute this for $r$ right away: $$v={1 \over 3} \pi r^2 h$$ $$v={1 \over 3} \pi (\frac{h}{2})^2 h$$ $$v={1 \over 3} \pi \frac{h^3}{4}$$ $$\frac{dv}{dm} = \pi (r)\frac{h^2}{4}\frac{dh}{dm}$$

$$6 = \pi (\frac{h}{2})^2\frac{dh}{dm}$$ $$6 = \pi (\frac{h^2}{4})\frac{dh}{dm}$$ and plugging in $h=10$: $$6 = \pi (\frac{100}{4})\frac{dh}{dm}$$ We get $\frac{dh}{dm} = \frac{6}{25\pi}$.

• I added a few more details. Is there anything I should elaborate on? – Quinn Greicius Mar 10 '16 at 0:56
• Is it $r= {h \over 2}$ because altitude is 24 and the radius is 12, so ${24 \over 12} = {1 \over 2}$? – didgocks Mar 10 '16 at 1:00
• Exactly. Most related rates problems that involve several variables will use a trick like that to put everything in terms of a single variable to get to a solution. – Quinn Greicius Mar 10 '16 at 1:01
• Thank you, it was very helpful! – didgocks Mar 10 '16 at 1:02
• I just noticed that $v = {1 \over 3} \pi r^3 h$ is not a correct formula for a cone – didgocks Mar 13 '16 at 14:26

Retain symbols till the last plugin step

$$r = h \cot \alpha \; ; \tan \alpha = \frac12 ;\; V = \pi r^3/3\; \cot \alpha$$

$$V = \pi h^3/3\, \tan ^2\alpha$$

$$dV/dt = \pi h^2 (dh/dt) \tan ^2\alpha$$

Plug in given values to get answer tallying with text.