# Math Help - Need help with 3 Calculus Extra Credit Problems

1. ## Need help with 3 Calculus Extra Credit Problems

I'm in intro to calculus and I need help setting this equation up:

Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium. A cold drink is removed from a refrigerator on a hot summer day and placed in a room where the temperature is 80°F. Express the temperature of the drink as a function of time (minutes) if the temperature of the drink was 40°F when it left the refrigerator and was 50°F after 20 minutes in the room.

Thanks!

2. Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium.

$\frac{dT}{dt} = k(T-A)$

A = room temperature (a constant)
T = temperature of the cold drink at any time t in minutes
k = proportionality constant

separate variables and integrate.

3. Ok, with your help and help from another problem, I have this. Is it the next step?

dT/dt= -k(T-TM)

Where T(t) is the temperature of the drink, and TM is the temperature of the surrounding solution.

So:

dT/dt= -(T-80)

This doesn't seem right, I don't know how to incorporate the other numbers and variables. Grrr calculus

4. $\frac{dT}{dt} = k(T-80)$

separate variables ...

$\frac{dT}{T-80} = k \, dt$

integrate ...

$\ln|T-80| = kt + C_1$

$|T-80| = e^{kt + C_1}$

$|T-80| = e^{C_1} \cdot e^{kt}$

$T - 80 = C_2 \cdot e^{kt}$

$T = 80 + C_2 \cdot e^{kt}$

The calculus is done, so I'm stopping at this point. The rest is algebra ... you were given two temperatures at two different times. With that info, you can determine the constants $C_2$ and $k$ and finalize the temperature as a function of time.

5. I greatly appreciate your help. I have:

Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40
How do I find k?

Second value (50 degrees at time 20)
T= 80 + Ce^kt
50= 80 + Ce^k*20
-30= Ce^20k
Again finding k has stumped me.

What does k represent and how do i find it?

6. Originally Posted by Sm10389
I greatly appreciate your help. I have:

Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40
good.

How do I find k?
Second value (50 degrees at time 20)
how about substituting in -40 for C ? then find k with the second value.
T= 80 + Ce^kt
do it.

7. you my friend are a genius.

so....

Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40

Second value (50 degrees at time 20)
T= 80 + Ce^kt
50= 80 + Ce^k*20
-30= -40e^20k
3/4= e^20k
ln3/4=20k
k= ln(3/4)/80

so would this be my final answer?

T= 80 + -40e^ln((3/4)/20)t

8. ## skeeter

how does that look

9. Originally Posted by Sm10389
how does that look
check it out yourself ... graph the result in your graphing utility and see if the given info matches up.

10. i just did, and it is not correct. where did i goof up?

11. so would this be my final answer?

T= 80 + -40e^ln((3/4)/20)t
looks like you have ...

$k = \ln\left(\frac{\frac{3}{4}}{20}\right)$

... which it ain't.

should be ...

$k = \frac{\ln\left(\frac{3}{4}\right)}{20}$

12. Thank you, I had that too, I just did not use the parentheses correctly in my calculator.

The next step of the problem is to calculate it if the drink were warmer then room temperature (>80).

I know how to set it up like the last one, but we would only be given one point (0, 85) for example.

How would I calculate k here?

13. $k$ will remain the same because the rate of heat transfer will remain the same

... however, you'll have to recalculate $C_2$.

14. ## QUESTION #2

More from the Introduction to Differential Equations-

Investment plan- an investor makes regular deposits totaling D dollars each year into an account that earns interest at the annual rate r compounded continuously.

A: Explain why the account grows at the rate ( dV/dt = rV + D ) where V(t) is the value of the account 2 years after the initial deposit. Solve this differential equation to express V(t) in terms of r and D.

I came up with this:
V(t)= (C/r)*e^rt - (D/r)

I am sure it is correct. This is the next part:
Amanda wants to retire in 20 years. To build up a retirement fund, she makes regular annual deposits of \$8,000. If the prevailing interest rate stays constant at 4% compounded continuously, how much will she have in her account at the end of the 20 year period?

I know how to do everything but:
Find C
Figure out how compounding continuously would affect the equation.


# The Proof of Infinitude of Pythagorean Triples $(x,x+1,z)$

Proof that there exists infinity positive integers triple $x^2+y^2=z^2$ that $x,y$ are consecutive integers, then exhibit five of them.

This is a question in my number theory textbook, the given hint is that
"If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$"
I wondered how someone come up with this idea.

My solution is letting $x=2st, y=s^2-t^2, z=s^2+t^2$ by $s>t, \gcd(s,t)=1$.then consider two cases: $y=x+1$ and $y=x-1$
Case 1: $y=x+1$
Gives me $(s-t)^2-2t^2=1$ then I found this is the form of Pell's equation, I then found \begin{align}s&=5,29,169,985,5741\\t&=2,12,20,408,2378\end{align}then yields five triples $$(20,21,29),(696,697,985),(23660,23661,33461),(803760,803761,1136689),(27304196,27304197,38613965)$$ Case 2:$y=x-1$
Using the same method, I come up with Pell's equation $(s+t)^2-2s^2=1$, after solve that I also get five triples: $$(4,3,5),(120,119,169),(4060,4059,5741),(137904,137903,195025),(4684660,4684659,6625109)$$

I have wondered why the gaps between my solution are quite big, with my curiosity, I start using question's hint and exhibit ten of the triples:$$(3,4,5),(20,21,29),(119,120,169),(696,697,985),(4059,4060,5741),(23660,23661,33461),(137903,137904,195025),(803760,803761,1136689),(4684659,4684660,6625109),(27304196,27304197,38613965)$$ These are actually the same as using solutions alternatively from both cases. But I don't know is this true after these ten triples

Basically the problem was solved, but I would glad to see if someone provide me a procedure to come up with the statement
"If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$", and prove that there are no missing triplet between it.

--After edit--

Thanks to @Dr Peter McGowan !, by the matrix
$$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2\\ 2 & 2 & 3 \end{bmatrix} \begin{bmatrix} x\\x+1\\z \end{bmatrix} = \begin{bmatrix} 3x+2z+2\\3x+2z+1\\4x+3z+2 \end{bmatrix}$$ gives me the hinted statement.

• Hint : If $(a/b)$ is a solution of the Pell-equation $a^2-2b^2=-1$ , then the next solution is $(3a+4b/2a+3b)$ – Peter Apr 21 '18 at 7:43
• Wow, how to know that? – kelvin hong 方 Apr 21 '18 at 7:54
• artofproblemsolving.com/community/c3046h1049346__2 – individ Apr 21 '18 at 8:50
• @individ thanks, but a relevant proof is better. – kelvin hong 方 Apr 21 '18 at 10:33
• Go here. It will take you to a question of mine where I prove the infinitude of Pythagorean Triples... but not using Pell equations, however. Nonetheless, this post might serve more use if $z=x+1$ as opposed to $y$, since I show that $$(2v^2+2v)^2+(2v+1)^2=(2v^2+2v+1)^2\;\forall v.$$ Still, it might increase your understanding on Pythagorean Triples :) – Mr Pie Jun 20 '18 at 12:20

## 3 Answers

You are on the right track. The simplest solution is to recall that all irreducible Pythagorean triples for a rooted ternary tree beginning with $(3, 4, 5)$ triangle. B Berggren discovered that all others can be derived from this most primitive triple. F J M Barning set these out as three matrices that when pre-multiplied by a "vector" of a Pythagorean triple produces another. For the case of consecutive legs we have, starting with $(x_1, y_1, z_1)$, we may calculate the next triple as follows:

\begin {align} x_2&=x_1+2y_1+2z_1 \\ y_2&=2x_1+y_1+2z_1 \\ z_2&=2x_1+2y_1+3z_1 \end {align}

The hint you were given is a variation on the above more general formula specific for consecutive leg lengths. It is an easy proof by induction to show that the formulas are correct. The first few are:

$(3, 4, 5); (20, 21, 29); (119, 120, 169); (696, 697, 985); (4059, 4060, 5741); (23660, 23661, 33461)$; etc. Obviously, this can be continued indefinitely.

The sequence rises geometrically. A simple explicit formula is available for these solutions that are (as you have already guessed) alternating solutions to Pell's equation.

• Wow, although I have seen that matrix before but don't realize it can be so useful! I have found the related matrix and actually come out with the desired result.Thanks a lot!!!! – kelvin hong 方 Apr 21 '18 at 10:43

$x,x+1,z$ is a Pythagorean triple iff $(2x+1)^2+1=2z^2$.

Let $u=2x+1$. Then $u^2-2z^2=-1$, a negative Pell equation whose solution lies in considering the units of $\mathbb Z[\sqrt 2]$ of norm $-1$.

It is clear that $\omega=1+\sqrt 2$ is a fundamental unit with norm $-1$. Therefore, all the other solutions of $u^2-2z^2=-1$ come from odd powers of $\omega$.

Thus, if $(u_k,z_k)$ is a solution of $u^2-2z^2=-1$, then the next one is given by

\begin{align} u_{k+1}+z_{k+1}\sqrt 2&=(u_k+z_k\sqrt 2)\omega^2 \\ &=(u_k+z_k\sqrt 2)(3+2\sqrt 2) \\ &=(3u_k+4z_k)+(2u_k+3z_k)\sqrt 2 \end{align} So, $u_{k+1}= 3u_k+4z_k$ and $z_{k+1}=2u_k+3z_k$.

Now let $u_{k+1}=2x_{k+1}+1$.

Then $$x_{k+1}=\frac{u_{k+1}-1}{2}=\frac{(3u_k+4z_k)-1}{2}=\frac{3(2x_k+1)+4z_k-1}{2}=3x_k+2z_k+1$$ and $$z_{k+1}=4x_k+3z_k+2$$ as claimed.

• Wow, thanks for your solution! – kelvin hong 方 Apr 28 '18 at 2:54

Pythagorean triples where $$|A-B|=1$$ are scarce and get more rare with altitude but they can continue to be found, indefinitely, given arbitrary precision. (There are only $$22$$ of them with $$A,B,C<10.2\text{ quadrillion.})$$ Proof that there are in infinite number of them is shown by the following functions which accept and yield natural numbers without end.

Another way of putting it is that we are eliminating (subtracting) a countably infinite set of triples from another countably infinite set of triples and even when you subtract a supposedly larger $$\aleph_0$$ set from a smaller $$\aleph_0$$ set (odd numbers minus natural numbers) the results are still infinite because they can be mapped. Here we have all triples minus extraneous triples.

To find them in a timely manner, we can use the following function to find the right combination of $$(m,n)$$ to use in Euclid's formula. The $$\pm$$ is used because the triples alternate: $$A>B$$ and $$A. $$\text{With infinite integers input }(n=\sqrt{2*m^2\pm1}-m)\text{ will always output an integer for some }m.$$ For all $$M\in\mathbb{N}$$, whenever this function yields an integer $$>0$$, we have the $$(m,n)$$ we need for a triple. The following $$19$$ were generated (in $$4$$ seconds) in a loop of $$m=1\text{ to }16,000,000$$ and $$f(m,n)$$ shows the variable values needed to generate each triple using Euclid's formula. The $$g(n,k)$$ shows the values (generated in $$1$$ second) for an alternate set of functions in a loop of $$m=1\text{ to }5,000,000$$. $$A=m^2-n^2\qquad B=2mn\qquad C=M^2+n^2$$ $$f(2,1)=g(1,1)=(3,4,5)$$ $$f(5,2)=g(2,2)=(21,20,29)$$ $$f(12,5)=g(4,5)=(119,120,169)$$ $$f(29,12)=g(9,12)=(697,696,985)$$ $$f(70,29)=g(21,29)=(4059,4060,5741)$$ $$f(169,70)=g(50,70)=(23661,23660,33461)$$ $$f(408,169)=g(120,169)=(137903,137904,195025)$$ $$f(985,408)=g(289,408)=(803761,803760,1136689)$$ $$f(2378,985)=g(697,985)=(4684659,4684660,6625109)$$ $$f(5741,2378)=g(1682,2378)=(27304197,27304196,38613965)$$ $$f(13860,5741)=g(4060,5741)=(159140519,159140520,225058681)$$ $$f(33461,13860)=g(9801,13860)=(927538921,927538920,1311738121)$$ $$f(80782,33461)=g(23661,33461)=(5406093003,5406093004,7645370045)$$ $$f(195025,80782)=g(57122,80782)=(31509019101,31509019100,44560482149)$$ $$f(470832,195025)=g(137904,195025)=(183648021599,183648021600,259717522849)$$ $$f(1136689,470832)=g(332929,470832)=(1070379110497,1070379110496,1513744654945)$$ $$f(2744210,1136689)=g(803761,1136689)=(6238626641379,6238626641380,8822750406821)$$ $$f(6625109,2744210)=g(1940450,2744210)=(36361380737781,36361380737780,51422757785981)$$ $$f(15994428,6625109)=g(4684660,6625109)=(211929657785303,211929657785304,299713796309065)$$

The alternate formula mentioned above is slightly faster since it deals only with the subset of triples where $$GCD(A,B,C)=(2m-1)^2,m\in\mathbb{N}$$. We have the needed $$(n,k)$$ values whenever the following yields a positive integer.

$$k=\sqrt{\frac{(2n-1)^2\pm1}{2}}$$ Having found a needed $$(n,k)$$ the following will generate a needed triple. $$A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$


# Solving Cubic Equations Formula

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SOLVING THE CUBIC EQUATION The cubic algebraic equation ax 3+bx 2+cx+d=0 was first solved by Tartaglia but made public by Cardano in his book Ars Magna(1545) after being sworn to secrecy concerning the solution method by the. how to solve cubic equation in faster way http://youtu. Use this calculator to solve polynomial equations with an order of 3 such as ax 3 + bx 2 + cx + d = 0 for x including complex solutions. It also gives the imaginary roots. An online cube equation calculation. Mathews Mathematics Department California State University Fullerton There are several formulas for solving the cubic equation. Cubic equations and the nature of their roots A cubic equation has the form ax3 +bx2 +cx+d = 0. Then are real. Let the roots of x2 px+ q= 0 be and , so that p= + and q=. Manages the issue of inherent in the power basis representation of the polynomial in floating point. In the chapter "Classiﬁcation of Conics", we saw that any quadratic equa-tion in two variables can be modiﬁed to one of a few easy equations to un-derstand. Gerolamo Cardano published a method to solve a cubic equation in 1545. Cubic equations can be solved analytically in Matlab (you need the symbolic toolbox to get the expressions or just copy & paste them from somewhere). It is defined as third degree polynomial equation. com is a free online OCR (Optical Character Recognition) service, can analyze the text in any image file that you upload, and then convert the text from the image into text that you can easily edit on. Various forms of van der Waals equation of state Cubic Equations of State Note that the van der Waals (vdw) equation of state is cubic in volume. The former editor of a monthly computing and technology magazine, his work has appeared in The Guardian, GQ and Time Out. Try finding a different solution. What is the formula to solve a cubic equation? [duplicate] There is no general algebraic formula to solve a polynomial equation that has degree 5 or higher in. Solving the Cubic Equation for Dummies. Often, our goal is to solve an ODE, i. In this case it expresses the solution through cubic roots of complex numbers. Equivalently, the cubic formula tells us the solutions of equations of the form ax3 +bx2 +cx+d =0. Differentiated Learning Objectives. Equation (12) may also be explicitly factored by attempting to pull out a term of the form from the cubic equation, leaving behind a quadratic equation which can then be factored using the Quadratic Formula. Fior challenge: Tartaglia pro-poses some questions, and Fior responds by giving Tartaglia 30 depressed cubics to solve. 00000001, initial_guess=0. There is no obvious way that “completing the cube” makes the solution into a matter of just taking cube roots in the same way that “completing the square” solves the quadratic in terms of square roots. Cubic equation calculator Fill in the coefficients a, b, c, and d in the equation a x 3 + b x 2 + c x + d = 0 and click the Solve button. Cubic Equation Calculator. 0 is equal to ax squared plus bx plus c. The model equation is A = slope * C + intercept. However, I have tried plotting the equation for these values, and can clearly see there should be 3 real roots. Using the same trick as above we can transform this into a cubic equation in which the coeﬃcient of x2 vanishes: put x = y − 1. roots but exclusive to cubic polynomials. After reading this chapter, you should be able to: 1. We have developed an energy balance equation for the universe. Once the equation has been entered, the calculator uses the Newton-Raphson numerical method to solve the equation. image source. Solving linear equations in two variables is straightforward (back substitute). To solve this equation means to write down a formula for its roots, where the formula should be an expression built out of the coefficients a, b and c and fixed real numbers (that is, numbers that do not depend on a, b and c) using only addition, subtraction. Bibtex entry for this abstract Preferred format for this abstract (see Preferences ). Just out of interest, is it possible to solve cubic equations with complex solutions, based purely on iterative methodologies? @Strange_Man I think the only way you can do that, is if you write your own complex number class. net is going to be the perfect site to pay a visit to!. While it might not be as straightforward as solving a quadratic equation, there are a couple of methods you can use to find the solution to a cubic equation without resorting to pages and pages of detailed algebra. But your function does not have a squared term; it already is a DEPRESSED cubic so that substitution is irrelevant. I understand that there are two cubic formulas that solve the cubic equation that, which one to use depends on the characterstic value of n. We’ll discuss discriminants some other time. Third Degree Polynomial Formula. Solution: Let us use the division method for solving a cubic equation. Free equations calculator - solve linear, quadratic, polynomial, radical, exponential and logarithmic equations with all the steps. The known Cardano’s formulas for solution of this kind equations are very difficult and almost aren’t used in practice. In fact, the last part is missing and without this part, one cannot implement it into an algorithm. Therefore, u and v satisfy the binomial equations. I shall try to give some examples. This Online Equation Solver solves every equations with set of given variables. Therefore, we can solve for x. Cubic equations of state are called such because they can be rewritten as a cubic function of molar volume. This is the graph of the equation 2x 3 +0x 2 +0x+0. The canonical form of cubic equation is. Let us take an example to understand the process easily. Now divide the equation with x-1 which will give you the quadratic equation. Often, our goal is to solve an ODE, i. The Cubic Equation. Improve your math knowledge with free questions in "Solve a quadratic equation using the quadratic formula" and thousands of other math skills. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. A Level > Arithmetic sequences A Level > Binomial expansion A Level > Differentiation A Level > Factor and remainder theorem A Level > Fibonacci sequences A Level > Geometric sequences A Level > Integration A Level > Logs A Level > Mechanics A Level > Mid-ordinate rule A Level > Partial fractions A Level > Point of inflection A Level. Some numerical methods for solving a univariate polynomial equation p(x) = 0 work by reducing this problem to computing the eigenvalues of the companion ma- trix of p(x), which is deﬁned as follows. This process is equivalent to making Vieta's substitution, but does a slightly better job of motivating Vieta's magic'' substitution. Finally the solutions of the pressed cubic equation is the combination of the cubic roots of the resolvent. use e- as the symbol for an electron. The conventional strategy followed for solving a cubic equation involved its reduction to a quadratic equation and then. Let's use the following equation. SOLVING THE CUBIC EQUATION The cubic algebraic equation ax 3+bx 2+cx+d=0 was first solved by Tartaglia but made public by Cardano in his book Ars Magna(1545) after being sworn to secrecy concerning the solution method by the. This algorithm uses polynomial fitting for a decomposition of the given cubic into a product of a quadratic and a linear factor. Britannica does not currently have an article on this topic. Torres’ Approach) on the Drexel University Website. Solving a Cubic Equation With Excel 2016 Although complex numbers appear to be surreal and seem to involve something like Zeno's paradoxes they are quite useful. an equation in which the highest power of the unknown quantity is a cube. A cubic equation can have 3 real roots or 1 real root and a complex conjugate pair. The roots of this equation can be solved using the below cubic equation formula. formulas for solving cubic equations of the form x3 +3px =2q, x3 +2q =3px, and x3 =3px +2q. math worksheet solving quadratic and cubic. In fact, the last part is missing and without this part, one cannot implement it into an algorithm. The cell you want to change to reach that value is the variable you're solving for, x. Given that is a cubic function with zeros at -3, 2, and 7, find an equation for given that f(3)=5. How to Find the Exact Solution of a General Cubic Equation In this chapter, we are going to find the exact solution of a general cubic equation. edu/~toh/spectrum/CurveFitting. How to Solve a Cubic Equation Part 1 - The Shape of the Discriminant James F. Cubic equations of state are called such because they can be rewritten as a cubic function of molar volume. Students will: solve simple square root equations, listing both solutions where appropriate. How can we solve equations such as the cubic equation shown here? x 3 − x 2 - 4x + 4 = 0. find the exact solution of a general cubic equation. For cylinders and prisms, the formula is the area of the base multiplied by the height. Learn more about cubic equation Symbolic Math Toolbox. Four years later (1545), Cardan published a book called Ars Magna, which contained the solution to the cubic equation and Ferrari’s solution to the quartic. First divide by the leading term, making the polynomial monic. find the exact solution of a general cubic equation. Inequalities-- "The. Cubic equations and the nature of their roots A cubic equation has the form ax3 +bx2 +cx+d = 0. the cubic formula, which thereby solves the cubic equation, nding both real and imaginary roots of the equation. Therefore the sextic. His formula applies to depressed cubics, but, as shown in § Depressed cubic, it allows solving all cubic equations. Factor theorem solving cubic equations 1. Posted in Based on a Context Tagged A Level > Factor and remainder theorem, Algebra > Equations > Finding roots, Algebra > Equations > Iteration, Algebra > Functions > Composite functions Post navigation. 1 Miscellaneous Algebraic Approaches to the Cubic and Quartic For about 100 years after Cardano, \everybody" wanted to say something. It's easy to calculate y for any. Then are real. The app is able to find both real and complex roots. For the solution of the cubic equation we take a trigonometric Viete method, C++ code takes about two dozen lines. h header, poly34. Because the y value is 0,. However the coe cients of (2) are not explicitly available. He applies the cubic formula for this form of the equation and arrives at this "mess":! x= 3 (2+ "121)+ 3 (2""121) If you set your TI to complex mode, you can confirm that this complex formula is, in fact, equal to 4. Multiply the three lengths For a square, all three sides are the same. Let Vdenote the quotient of the polynomial ring modulo the ideal hp(x)igenerated by the polynomial p(x). You would usually use iteration when you cannot solve the equation any other way. Vieta's formulae are used to solve equations as thus, first step is to divide all coefficients by "a". The solution was first published by Girolamo Cardano (1501-1576) in his Algebra book Ars Magna. The solution of cubic and quartic equations In the 16th century in Italy, there occurred the ﬁrst progress on polynomial equations beyond the quadratic case. Set ˆ= : Then ˆ2 = 2 2 + 2 = ( + )2 4 = p2 4qand ˆ= p p2 4q: So = 1 2 + + ! = 1 2 (p+ ˆ) Solving the cubic equation Let’s suppose we. The depressed cubic equation. The cubic formula tells us the roots of polynomials of the form ax3 +bx2 + cx + d. They are not many students who are actually remembering the formulas taught in school. Exercise 10. ax 3 + bx 2 + cx + d = 0. Cubic Equation Formula, cubic equation, Depressing the Cubic Equation, cubic equation solver, how to solve cubic equations, solving cubic equations. So I thought I could. Learn more about cubic equation Symbolic Math Toolbox. Cubic Equation Formula. Linear Equation Solver-- Solve three equations in three unknowns. To understand this example, you should have the knowledge of following Python programming topics:. Wholesale Lingerie cheap Lingerie. For example, we have the formula y = 3x 2 - 12x + 9. In the module, Polynomials, a factoring method will be developed to solve cubic equations that have rational roots. A cubic function is one of the most challenging types of polynomial equation you may have to solve by hand. how to solve cubic equation on excel You are not going to be able to solve cubic equations with a formula. 14k Yellow Gold CZ Cubic Zirconia Cross Hinged Hoop Earrings Hinged Earrings Hinged Zirconia Zirconia Cross 14k Hoop CZ Cubic Gold Yellow NewOCR. In this tutorial you are shown how to solve a cubic equation by using the factor theorem. However, most of the theory is also valid if they belong to a field of characteristic other than 2 or 3. Our objective is to find two roots of the quartic equation The other two roots (real or complex) can then be found by polynomial division and the quadratic formula. I have used the Newton-Raphson method to solve Cubic equations of the form by first iteratively finding one solution, and then reducing the polynomial to a quadratic and solving for it using the quadratic formula. -2-Solving cubics is not as simple as solving quadratics. The Polynomial equations don’t contain a negative power of its variables. Title: Hyperbolic identity for solving the cubic equation: Authors: Rochon, Paul: Publication: American Journal of Physics, Volume 54, Issue 2, pp. In addition, Ferrari was also able to discover the solution to the quartic equation, but it also required the use of the depressed cubic. Use this x-coordinate and plug it into either of the original equations for the lines and solve for y. The remaining unknown can then be calculated. Solving a Polynomial Equation Solve 2x5 + 24x = 14x3. When we deal with the cubic equation one surprising result is that often we have to express the roots of the equation in terms of complex numbers although the roots are real. cubic root of unity. In the By Changing Cell box, type A3. how to solve cubic equation on excel You are not going to be able to solve cubic equations with a formula. roots([1 -3 2]) and Matlab will give you the roots of the polynomial equation. After reading this chapter, you should be able to: 1. Let ax³ + bx² + cx + d = 0 be any cubic equation and α,β,γ are roots. Solution Techniques for Cubic Expressions and Root Finding Print Hopefully, we have convinced you that the use of cubic equations of state can represent a very meaningful and advantageous way of modeling the PVT behavior of petroleum fluids. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. Algorithm for solving cubic equations. asks tricky questions. The cubic formula gives the roots of any cubic equation Solving the depressed cubic equation with Vieta's substitution. 3 2 ax bx cx d + + + = 0 (1). Use this calculator to solve polynomial equations with an order of 3 such as ax 3 + bx 2 + cx + d = 0 for x including complex solutions. 5, the y value is 0. That means, reducing the equation to the one where the maximum power of the equation is 2. First, suppose. Computing such square roots again leads to cubic equations. …treatment of the solution of cubic equations. A quadratic equation can be solved by using the quadratic formula. Cubic equations and the nature of their roots. By the fundamental theorem of algebra, cubic equation always has 3 3 3 roots, some of which might be equal. Finding the center of mass doesn't require solving the equation—it is merely −1/3 the coefficient of the quadratic term. The van der Waals (from his thesis of 1873) equation is a cubic in the molar volume. And we generally deal with x's, in this. There are several ways to solve cubic equation. Possible Outcomes When Solving a Cubic Equation If you consider all the cases, there are three possible outcomes when solving a cubic equation: 1. Naturally, the first solutions were geometric, for ancient Egyptians and Greeks knew nothing of algebra. Furthermore, there are examples of cubic equations with real coefficients and three distinct real solutions for which the cubic formula requires one to calculate a root of a complex number. The Polynomial equations don’t contain a negative power of its variables. The polynomial x4+ax3+bx2+ cx+dhas roots. solving a cubic equation. The amount of profit (in millions) made by Scandal Math, a company that writes math problems based on tabloid articles, can be found by the equation P(n) = −n2 + 10n, where n is the number of textbooks sold (also in millions). at first, has a lower rate of growth than the linear equation f(x) =50x; at first, has a slower rate of growth than a cubic function like f(x) = x 3, but eventually the growth rate of an exponential function f(x) = 2 x, increases more and more -- until the exponential growth function has the greatest value and rate of growth!. Use a ruler to measure each side in inches. Fitting a cubic to data Let us now try to fit a cubic polynomial to some data. Given a quadratic of the form ax2+bx+c, one can ﬁnd the two roots in terms of radicals as-b p b2-4ac 2a. Write an equation for the cubic polynomial function whose graph has zeroes at 2, 3, and 5.


1. Oct 23, 2014

### bamajon1974

I want to de-nest the following radical:
(1) $$\sqrt{3+2\sqrt{2}}$$
Into the general simplified form:
(2) $$a+b\sqrt{2}$$
Equating (1) with (2),
(3) $$\sqrt{3+2\sqrt{2}} = a+b\sqrt{2}$$
and squaring both sides:
(4) $$3+2\sqrt{2} = a^2 + 2b^2 + 2ab\sqrt{2}$$
generates a system of two equations with two unknowns, a and b, after equating the rational and irrational parts:
(5) $$3 = a^2 + 2b^2$$
(6) $$2\sqrt{2} = 2ab\sqrt{2}$$
Simplifying (6) and solving for b:
(7) $$b=\frac{1}{a}$$
Substituting (7) into (5) yields:
(8) $$3 = a^2 + 2\frac{1}{a^2}$$
Clearing the denominator and moving non-zero terms to one side generates a quartic equation:
(9) $$0 = a^4 - 3a^2 +2$$
(10) $$x=a^2, x^2 = (a^2)^2 = a^4, x = \pm \sqrt{a}$$
(11) $$0 = x^2 - 3x + 2$$
The square root of the discriminant is an integer, 1, which presumably makes simplification of the nested radical possible. Finding the roots, x, of (11) gives
(12)$$x=1, x =\sqrt{2}$$
Substituting the roots in (12) into (10) gives a:
(13) $$a = \pm \sqrt{2} , a = \pm 1$$
Then b is found from (7):
(14) $$b = \pm \frac{1}{\sqrt{2}} , b = \pm 1$$
Using the positive values of a and b, the de-nested radical is:
(15) $$\sqrt{3+2\sqrt{2}} = 1+\sqrt{2}$$

My questions are:
(1) Is this approach for simplifying nested radical correct?

(2) The positive values of a and b produce the correct simplified form in (15) while the negative values of a and b do not. Is there a way to figure out which roots from (9) are correct and which to reject other than calculating the numerical value of (15) with both positive and negative a's and b's to see which is equal to the nested form?

(3) I have also seen the general simplified de-nested form as:
(16) $$\sqrt{a} + \sqrt{b}$$
Going through an analogous process as above, it generates the correct simplified form in (15) as well. Is one form (2) or (15) better than the other? Or is it just a personal preference which one to use?

Thanks!

2. Oct 23, 2014

### symbolipoint

The square root radical symbol means, the expression under the radical raised to the one-half power.

$$\sqrt{3+2\sqrt{2}}$$
$$(3+2\sqrt{2})^(1/2)$$
$$(3+2(2)^(1/2))^(1/2)$$

( I KNOW what I'm trying to do but still cannot make it appear correctly)

3. Oct 23, 2014

### SteamKing

Staff Emeritus
You mean like this? Just enclose your expression for the exponent in a pair of curly braces {}.

4. Oct 24, 2014

### Mentallic

I'll give you a chance to look at these again. You've ended up with the correct answer however because you made two errors that both cancelled each other out, so that was a lucky one on your part ;)

Yes.

You'll often have situations arise whereby you need to find, say, the length of an object, but the equations you're dealing with are quadratics and will thus give you two solutions, a positive and a negative one. It suffices to toss away the negative value because you're solving a problem that only makes sense with a positive answer.

It's similar in this case. You're trying to simplify a square root. You know that square roots don't give negative results so you're allowed to simply toss out the result $-1-\sqrt{2}$ because that is clearly negative and can't be the answer. You don't need to look at the numerical values of each result with a calculator if you simply use a bit of intuition.

I prefer yours merely because the $\sqrt{2}$ result cancels and you end up with a nice result in (7). Neither are incorrect, but also both are assumptions. You assumed that the answer is of the form $a+b\sqrt{2}$ and the "general" simplified de-nested form $\sqrt{a}+\sqrt{b}$ assumes that the result is the sum of two square roots. What if it were $\sqrt{a}+\sqrt{b}+\sqrt{c}$ instead? Or something even more complicated?
So yes, I prefer your assumption over the other because yours is easier to work with.

5. Oct 24, 2014

### bamajon1974

I see my error:
12)
$$x=1, x=\sqrt{2}$$
Should be
$$x=1, x=2$$
My bad. Although in my defense that was an error I didn't catch after typing so much unfamiliar latex code into these lines.

I had neglected to mention that in the if you try to de-nest
$$\sqrt{3+2\sqrt{2}}$$
by assuming it equals the other simplified form (16)
$$\sqrt{a} + \sqrt{b}$$
Equating the nested radical with the assumed simplified form and squaring both sides:
$$3 + 2\sqrt{2} = a + b + 2\sqrt{ab}$$
Equating rational and irrational parts generates 2 equations with 2 unknowns:
$$3 = a + b$$
$$2\sqrt{2} = 2\sqrt{ab}$$
Solving for b in the second equation, substituting into the first and moving all non-zero terms to one side of the equation generates a quadratic:
$$0=a^2+3a+2$$
of which the roots, a, are
$$a= 1, a=2$$
So a and b are
$$a=1, b=2$$
or
$$a=2, b=1$$
So the simplified form is
$$1+\sqrt{2}$$
identical to above, but I didn't have any negative roots to discard. So in this particular case, the second form was a little easier.

I see why rejecting the negative roots is justified as well.


# Proof critique: Induction

#### sweatingbear

##### Member
We wish to show that $3^n > n^3 \, , \ \forall n \geqslant 4$. Base case $n = 4$ yields

$3^4 = 81 > 4^3 = 64$

Assume the inequality holds for $n = p$ i.e. $3^p > p^3$ for $p \geqslant 4$. Then

$3^{p+1} > 3p^3$

$p \geqslant 4$ implies $3p^3 \geq 192$, as well as $(p+1)^3 \geqslant 125$. Thus $3p^3 > (p+1)^3$ for $p \geqslant 4$ and we have

$3^{p+1} > 3p^3 > (p+1)^3$

which concludes the proof.

Feedback, forum?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$p \geqslant 4$ implies $3p^3 \geq 192$, as well as $(p+1)^3 \geqslant 125$. Thus $3p^3 > (p+1)^3$ for $p \geqslant 4$ and we have
One question :

if x>5 and y>2 then x>y ?

#### sweatingbear

##### Member
One question :

if x>5 and y>2 then x>y ?
I was actually a bit uncertain about that. How else would one go about?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
We need to prove that

$$\displaystyle 3^{p+1}> (p+1)^3$$ assuming that $$\displaystyle 3^p>p^3\,\,\,\, \forall p\geq 4$$

$$\displaystyle \tag{1} 3^{p+1}>3p^3\geq p^3+p^3$$

Lemma :

$$\displaystyle p^3-3p^2-3p-1 \geq 0$$

Take the derivative

$$\displaystyle 3p^2-6p-3 =3(p^2-2p-1)=3(p^2-2p+1-2)=3(p-1)^2-6$$

The function is positive for $p=4$ and increases for $$\displaystyle p\geq 4$$ so the lemma is satisfied .

Hence we have

$$\displaystyle p^3 \geq 3p^2+3p+1 \,\,\,\,\forall p\geq 4$$

Using this in (1) we get

$$\displaystyle 3^{p+1}>p^3+3p^2+3p+1=(p+1)^3 \,\,\,\square$$

#### Deveno

##### Well-known member
MHB Math Scholar
Here is how I would do this proof:

(inductive step only).

Suppose that $$\displaystyle 3^k > k^3, k > 3$$.

Then:

$$\displaystyle 3^{k+1} = 3(3^k) > 3k^3$$.

If we can show that:

$$\displaystyle 3k^3 > (k+1)^3$$, we are done.

Equivalently, we must show that:

$$\displaystyle 3k^3 > k^3 + 3k^2 + 3k + 1$$, so that:

$$\displaystyle 2k^3 - 3k^2 - 3k - 1 > 0$$.

Note that $$\displaystyle 2k^3 - 3k^2 - 3k - 1 > 2k^3 - 3k^2 - 5k + 3$$

if $$\displaystyle 2k - 4 > 0$$, that is if $$\displaystyle k > 2$$, which is true.

But $$\displaystyle 2k^2 - 3k^2 - 5k + 3 = (2k - 1)(k^2 - k - 3)$$.

Now $$\displaystyle 2k - 1 > 0$$ for any $$\displaystyle k > 0$$, so we are down to showing

$$\displaystyle k^2 - k - 3 > 0$$ whenever $$\displaystyle k > 3$$.

Since $$\displaystyle k^2 - k > 3$$ is the same as $$\displaystyle k(k-1) > 3$$, we have:

$$\displaystyle k(k-1) > 3(2) = 6 > 3$$.

Thus we conclude that $$\displaystyle k^2 - k - 3 > 0$$ and so:

$$\displaystyle 3^{k+1} = 3(3^k) > 3k^3 > (k+1)^3$$.

With all due respect to Zaid, I wanted to give a purely algebraic proof.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Equivalently, we must show that:

$$\displaystyle 3k^3 > k^3 + 3k^2 + 3k + 1$$
Starting from this point, we could continue as follows. We need to show that $3k^2+3k+1<2k^3$.

$3k^2+3k+1<3k^2+3k^2+k^2=7k^2$ since $k>1$. Now, $7k^2<2k^3\iff 7<2k$, and the last inequality is true since $k\ge4$.

#### Deveno

##### Well-known member
MHB Math Scholar
Indeed, we just need to find something that is less than $$\displaystyle 2k^3$$ and larger than $$\displaystyle 3k^2 + 3k + 1$$ that "factors nice" (so we can apply what we know specifically about $$\displaystyle k$$). Very nice solution!


# Partitions of n that generate all numbers smaller than n

Consider a partition $$n=n_1+n_2+\cdots+n_k$$ such that each number $1,\cdots, n$ can be obtained by adding some of the numbers $n_1,n_2,\cdots,n_k$. For example, $$9=4+3+1+1,$$

and every number $1,2,\cdots,9$ be ca written as a sum of some of the numbers $4,3,1,1$. This other partition $$9=6+1+1+1$$ fails the desired property, as $4$ (and $5$) cannot be given by any sum of $6,1,1,1$.

Question: Can we charaterize which partitions of an arbitrary $n$ have this property? We clearly need at least one $1$ among $n_1,n_2,\cdots,n_k$, and intuitively we need many small numbers $n_i$. But I can't see much beyond this. Any idea or reference will be appreciated.

• If at least half of the number are 1's, you can do it. But the condition is certainly not necessary. – MathChat Jan 26 '17 at 6:03
• I once studied this problem and found a constructive partition method. Here is the brief. We are given a positive integer $n$. STEP ONE: if $n$ is an even number, partition it into $A=\frac{n}{2}$ and $B=\frac{n}{2}$; otherwise, partition it into $A=\frac{n+1}{2}$ and $B=\frac{n-1}{2}$. STEP TWO: re-partition $B$ into $A_1$ and $B_1$. STEP THREE: re-partition $B_1$......Until we get $1$. I didn't prove this method always works but I believe it is valid. – apprenant Jan 26 '17 at 6:04
• EXAMPLE: $13 \rightarrow (7,6) \rightarrow (7,3,3) \rightarrow (7,3,2,1)$. May it be helpful. – apprenant Jan 26 '17 at 6:06
• @apprenant. This method looks interesting. BTW, is this problem well known? – ALEXIS Jan 26 '17 at 6:14
• You can work up the other way, too, by powers of two, e.g $13 \to 1,2,4,6$ – Joffan Jan 26 '17 at 6:17

Let $\lambda$ be a partition of $n$. The required condition is that $\lambda$ contain partitions $\lambda_i$ of each $1 \le i < n$. Clearly if $\lambda$ contains a partition of $j$ then it also contains a partition of $(n - j)$, being the multiset difference $\lambda - \lambda_j$.

Therefore the first thing to notice is that $\lambda$ cannot contain any element $a > \lceil \frac{n}{2} \rceil$, for if it did then $\{a\}$ cannot be part of a partition of $\lceil \frac{n}{2} \rceil$ and $\lambda - \{a\}$ is a partition of $(n - a) < (n - \lceil \frac{n}{2} \rceil) < \lceil \frac{n}{2} \rceil$ cannot contain a partition of $\lceil \frac{n}{2} \rceil$.

Now, suppose that the largest element of $\lambda$ is $m$. It is certainly sufficient that $\lambda - \{m\}$ should satisfy the corresponding condition of providing partitions for each $1 \le i < (n - m)$. Proof: $\lambda - \{m\}$ is a partition of $(n - m)$ and provides partitions for each smaller natural number, so it remains to construct partitions $\lambda_i$ for $(n - m) < i < n$. We can do this by taking partitions from $\lambda - \{m\}$ for $(n - 2m) < j < n - m$ and then adding $\{m\}$ to each one. This can only fail if $j < 0$, which can only happen if $(n - 2m) < -1$. Since $m \le \lceil \frac{n}{2} \rceil$ we have $n - 2m \ge n - 2\lceil \frac{n}{2} \rceil$, which is $0$ if $n$ is even and $-1$ if $n$ is odd, so all cases are covered.

The interesting question is whether it's necessary that $\lambda - \{m\}$ should satisfy the same condition. Clearly it must contain partitions of $1 \le i < m$, since $\{m\}$ doesn't participate in them. And by the simple principle of taking complements in $\lambda$ it's clear that for each $m \le i < n$ the remnant $\lambda - \{m\}$ must either contain partitions of $(i - m)$ and $n - i$; or $i$ and $n - m - i$. Is that sufficient?

My intuition is that it's necessary, and testing on small examples (up to $n = 30$) supports that, but I haven't proved it.

In the Online Encyclopedia of Integer Sequences it's A126796 and a comment claims the characterisation

A partition is complete iff each part is no more than 1 more than the sum of all smaller parts. (This includes the smallest part, which thus must be 1.) - Franklin T. Adams-Watters, Mar 22 2007

This is just a quick and dirty list of the first examples, for $n$ up to $10$. Feel free to edit, extend, or amend.

$$1\\1+1\\ 1+1+1\quad1+2\\ 1+1+1+1\quad1+1+2\\ 1^5\quad1+1+1+2\quad1+2+2\quad1+1+3\\ 1^6\quad1^4+2\quad1^2+2+2\quad1^3+3\quad1+2+3\\ 1^7\quad1^5+2\quad1^3+2+2\quad1+2^3\quad1^4+3\quad1^2+2+3\quad1^3+4\quad1+2+4\\ 1^8\quad1^6+2\quad1^4+2+2\quad1^2+2+2+2\quad1^5+3\quad1^3+2+3\quad1^2+3+3\quad1+2+2+3\quad1^4+4\quad1+1+2+4\\ 1^9\quad1^7+2\quad1^5+2+2\quad1^3+2^3\quad1+2^4\quad1^6+3\quad1^4+2+3\quad1^2+2+2+3\quad1^3+3+3\quad1+2+3+3\quad1^5+4\quad1^3+2+4\quad1+2+2+4\quad1+1+3+4\quad1^4+5\quad1^2+2+5\\ 1^{10}\quad1^8+2\quad1^6+2^2\quad1^4+2^3\quad1^2+2^4\quad1^7+3\quad1^5+2+3\quad1^3+2^2+3\quad1+2^3+3\quad1^4+3^2\quad1^2+2+3^2\quad1^6+4\quad1^4+2+4\quad1^2+2^2+4\quad1^3+3+4\quad1+2+3+4\quad1^5+5\quad1^3+2+5\quad1+2^2+5\quad1+1+3+5$$

I hope the meaning of the superscript notation is clear, and I hope someone will check that I didn't make any mistakes or overlook anything. The list so far gives the sequence

$$1,1,2,2,4,5,8,10,16,20,\ldots$$

which (after correcting a pair of mistakes in the original posting here) Peter Taylor found in the OEIS.

• I get 1, 1, 1, 2, 2, 4, 5, 8, 10, 16, 20, 31, 39, 55, 71, 100, 125, 173, 218, 291 starting at index 0. A126796 – Peter Taylor Jan 26 '17 at 17:07
• @PeterTaylor, thank you! I see now what I missed: $7=1^3+4$ and $10=1+1+3+5$. Darn! I swear, I checked and double checked all my counts. I guess I needed to triple check.... – Barry Cipra Jan 26 '17 at 19:12


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# Are there primes of every possible number of digits?

That is, is it the case that for every natural number $n$, there is a prime number of $n$ digits? Or, is there some $n$ such that no primes of $n$-digits exist?

I am wondering this because of this Project Euler problem: https://projecteuler.net/problem=37. I find it very surprising that there are only a finite number of truncatable primes (and even more surprising that there are only 11)! However, I was thinking that result would make total sense if there is an $n$ such that there are no $n$-digit primes, since any $k$-digit truncatable prime implies the existence of at least one $n$-digit prime for every $n\leq k$.

If not, does anyone have insight into an intuitive reason why there are finitely many trunctable primes (and such a small number at that)? Thanks!

• Anyway, yes: for all $n$ there are a lot of primes having $n$ digits. Sep 10, 2016 at 22:34
• Bertrand's postulate (an ill-chosen name) says there is always a prime strictly between $n$ and $2n$ for $n\gt 1$. Sep 10, 2016 at 23:03
• Think about the reverse. If you have an $n$-digit prime, how many 'chances' do you have to extend it to an $(n+1)$-digit prime? The odds being able to do so quickly turn against you.
– user14972
Sep 11, 2016 at 0:18
• Just a side-comment - ..and even more surprising that there are only 11 ...Maybe I am wrong regrading your meaning , but I believe that there are 15 ( and not 11 ) both-sides truncatable primes .. Wiki entry on truncatable prime. Sep 11, 2016 at 3:05
• Via the Wikipedia article I found M. El Bachraoui's 2006 ["Primes in the Interval [2n, 3n]"](m-hikari.com/ijcms-password/ijcms-password13-16-2006/…) which relies on elementary methods and three results from an undergraduate textbook. Sep 12, 2016 at 1:52

## 2 Answers

Yes, there is always such a prime. Bertrand's postulate states that for any $k>3$, there is a prime between $k$ and $2k-2$. This specifically means that there is a prime between $10^n$ and $10\cdot 10^n$.

To commemorate $50$ upvotes, here are some additional details: Bertrand's postulate has been proven, so what I've written here is not just conjecture. Also, the result can be strengthened in the following sense (by the prime number theorem): For any $\epsilon > 0$, there is a $K$ such that for any $k > K$, there is a prime between $k$ and $(1+\epsilon)k$. For instance, for $\epsilon = 1/5$, we have $K = 24$ and for $\epsilon = \frac{1}{16597}$ the value of $K$ is $2010759$ (numbers gotten from Wikipedia).

• With the side note that Bertrand's postulate is a (proved) theorem. Sep 11, 2016 at 22:29
• Just another note: those interested in this sort of thing should look for papers by Pierre Dusart - he has proven many of the best approximations of this form. Sep 14, 2016 at 3:09

While the answer using Bertrand's postulate is correct, it may be misleading. Since it only guarantees one prime between $N$ and $2N$, you might expect only three or four primes with a particular number of digits. This is very far from the truth.

The primes do become scarcer among larger numbers, but only very gradually. An important result dignified with the name of the Prime Number Theorem'' says (roughly) that the probability of a random number of around the size of $N$ being prime is approximately $1/\ln(N)$.

To take a concrete example, for $N = 10^{22}$, $1/\ln(N)$ is about $0.02$, so one would expect only about $2\%$ of $22$-digit numbers to be prime. In some sense, $2\%$ is small, but since there are $9\cdot 10^{21}$ numbers with $22$ digits, that means about $1.8\cdot 10^{20}$ of them are prime; not just three or four! (In fact, there are exactly $180,340,017,203,297,174,362$ primes with $22$ digits.)

In short, the number of $n$-digit numbers increases with $n$ much faster than the density of primes decreases, so the number of $n$-digit primes increases rapidly as $n$ increases.

• The prime number theorem will give you a bound on the number of primes between $10^n$ and $10^{n+1}$. But is the bound tight enough to prove that the number of such primes is a strictly growing function of $n$? Sep 11, 2016 at 8:33
• @kasperd There are some known (explicit) estimates on the error term in the prime number theorem, I can imagine they are strong enough to show this, albeit possibly only for large $n$. Sep 11, 2016 at 8:50
• The prime number theorem on its own would allow for very large gaps between primes, but not so large that there are no primes between $10^n$ and $10^{n+1}$ when n is large enough. On the other hand, it is a limit, so it says nothing about small primes. Sep 11, 2016 at 18:49
• The bounds from Wikipedia $\frac{x}{\log x + 2} < \pi(x) < \frac{x}{\log x - 4}$ for $x> 55$ can be used to show that there is always a prime with $n$ digits for $n\ge 3$. Oct 13, 2020 at 13:22


# How many poker hands have two pairs?

I'm trying to calculate how many poker hands called Two Pair, there are. Such a hand consists of one pair of one rank, another pair of another rank and one card of a third rank. A poker hand consists of 5 cards.

I have two methods that I thought would work equally well. Turns out only one of them yields the correct answer. I was wondering if anyone here knows why the second solution gives the wrong answer.

Solution 1 (Correct):

We choose 2 ranks out of 13, which can be done in $\binom{13}{2}$ ways.

For the first rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.

For the second rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.

The last card can be chosen in $44$ different ways.

So the total number of hands is $\binom{13}{2}\cdot \binom{4}{2}\cdot \binom{4}{2}\cdot 44=123,552$

Solution 2 (Incorrect):

We choose 3 ranks out of 13, which can be done in $\binom{13}{3}$ ways.

For the first rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.

For the second rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.

For the third rank we choose 1 suit out of 4, which can be done in $4$ ways.

So the total number of hands is $\binom{13}{3}\cdot \binom{4}{2}\cdot \binom{4}{2}\cdot 4=41,184$

This is just a third of the correct number of hands. Why the second solution is wrong unfortunately seems to elude me......

• In attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3. – Angina Seng Jul 26 '18 at 18:23
• @LordSharktheUnknown: that looks like an answer to me. It is exactly what was asked. – Ross Millikan Jul 26 '18 at 18:35
• Combinatorics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too. – amWhy Jul 26 '18 at 18:56
• Thanks, but I still don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three for the singleton? – Stargazer Jul 26 '18 at 19:02
• Never mind, I understand it now that I thought some more about it. – Stargazer Jul 26 '18 at 19:12

After you've chosen which three ranks are in the hand, you need to choose either (a) which two of the three ranks to make the pairs, or (b) which one of the ranks to make the singleton. The number of ways to do these are ${3 \choose 2}$ and ${3 \choose 1}$, respectively, and of course each equals $3$.

Assume your choose three ranks R1, R2, R3.

In the first solution: You choose (R1, R2) first $\binom{13}{2}$ then assign the suits. Lastly, you choose R3 from the rest (remaining 44). The reason why $\binom{13}{2}$ is chosen to avoid double count since order does not matter for this two ranks: as (R1,R1,R2,R2,R3) is the same as (R2,R2,R1,R1,R3). (writing R1 R1 meaning two suits rank 1, pardon my laziness). This is correct.

In the second solution:

You choose (R1,R2,R3) first then assign the suits. But using $\binom{13}{3}$ means the order of the three ranks do not matter. In other words, you are treating the three hands (R1,R1,R2,R2,R3), (R3,R3,R2,R2,R1), (R1,R1,R3,R3,R2) as one hand only. Thus you undercount 3 times.

Your first method is to count ways to choose two from thirteen ranks for the pairs, two from four suits for each of those, and one from fourty-four cards to be the loner (or one from eleven ranks and one from four suits).   That is okay. $$\binom{13}2\binom 42^2\binom{44}1 \\\binom{13}2\binom 42^2\binom {11}1\binom 41$$

Your second method is to count ways to choose three from thirteen ranks, two from four suits for the pairs, one from four suits for the singleton, and—wait—which two from those three selected ranks are to be the pairs?   Ah, that is better.$$\binom {13}3\binom 42^2\binom 41\binom 32$$

...and of course $\binom{13}{3}\binom 32=\frac{13!}{3!10!}\frac{3!}{2!1!}=\frac{13!}{2!11!}\frac{11!}{10!1!}=\binom {13}2\binom{11}1$


# Integration Problem

## Homework Statement

$$\int {\frac{sin^{2}x}{1+sin^{2}x}dx}$$

## Homework Equations

Let t = tan x/2, then dx = 2/(1+t^2) and sin x = 2t / (1+t^2)

## The Attempt at a Solution

I got up to the point where $$\int {\frac{8t^{2}}{(1+6t^{2}+t^{4})(1+t^{2})} dt}$$. Not sure if I'm on the right track and if I am, do I use partial fractions after this?

The final answer is attached. Can't really make out the handwriting :/

#### Attachments

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Last edited:

Yes, it looks like partial fractions is the way to go after your substitution.

Hmm after I do partial fractions, I get
$$\int {\frac{2}{1+t^{2}} + {\frac{-2t^{2}-2}{(1+6t^{2}+t^{4})} dt}$$

After this, I do not know what's the next step. Kindly advise. Thanks.

you are summing 2 functions of t , one of these two look very much like a derivative of a certain function..

If you mean 2tan^-1 t, I can get this part. But what about the 2nd function?


# set of odd integers proof

• Feb 24th 2010, 02:21 PM
james121515
set of odd integers proof
I am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far?

$\mbox{If }A=\{x \in \mathbb{Z}~|~x = 2k+1\mbox{ for some }k \in \mathbb{Z}\}$ and $B=\{y \in \mathbb{Z}~|~y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $A=B$

$\mbox{\textbf{Proof.}}$ Let $x\in A$. then $\exists~k \in \mathbb{Z}\mbox{ such that }x=2k+1$. Equivalently,
$\Longrightarrow x=2k+1+1-1$
$\Longrightarrow x=2k+2-1$
$\Longrightarrow x=2(k+1)-1$

Since $k \in\mathbb{Z} \Longrightarrow k+1 \in \mathbb{Z}$ $x = 2(k+1)-1 \Longrightarrow x \in B$. Therefore, $A\subseteq B$
• Feb 24th 2010, 02:36 PM
Plato
Quote:

Originally Posted by james121515
I am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far?

$\mbox{If }A=\{x \in \mathbb{Z}~|~x = 2k+1\mbox{ for some }k \in \mathbb{Z}\}$ and $B=\{y \in \mathbb{Z}~|~y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $A=B$

$\mbox{\textbf{Proof.}}$ Let $x\in A$. then $\exists~k \in \mathbb{Z}\mbox{ such that }x=2k+1$. Equivalently,
$\Longrightarrow x=2k+1+1-1$
$\Longrightarrow x=2k+2-1$
$\Longrightarrow x=2(k+1)-1$

Since $k \in\mathbb{Z} \Longrightarrow k+1 \in \mathbb{Z}$ $x = 2(k+1)-1 \Longrightarrow x \in B$. Therefore, $A\subseteq B$

That 'way' is correct.
By symmetry you are done.

$x=2k+1=2(k+1)-1$
• Feb 24th 2010, 09:36 PM
james121515

So you are saying that due to symmetry, there is no need to show the other "right to left" containment due to symmetry?

-James
• Feb 25th 2010, 08:21 AM
Plato
Quote:

Originally Posted by james121515
$x=2k+1=2(k+1)-1$


# What is the general solution to the equation $\sin x + \sqrt{3}\cos x = \sqrt2$

I need to find the general solution to the equation

$$\sin(x) + \sqrt3\cos(x)=\sqrt2$$

So I went ahead and divided by $$2$$, thus getting the form

$$\cos(x-\frac{\pi}{6})=\cos(\frac{\pi}{4})$$

Thus the general solution to this would be $$x = 2n\pi \pm\frac{\pi}{4}+\frac{\pi}{6}$$

Which simplifies out to be,

$$x = 2n\pi +\frac{5\pi}{12}$$ $$x = 2n\pi -\frac{\pi}{12}$$

But the answer doesn't have the 2nd solution as a solution to the given equation. Did I go wrong somewhere?

• Your answer seems to be the correct one. For example $x=-\frac {\pi} {12}$ does satisfy the given equation. Mar 5 '20 at 6:39
• You solution is correct. May be they skip the second one. Mar 5 '20 at 6:43

As Kavi Rama Murthy's comment indicates, you haven't done anything wrong that I can see. You can quite easily very that $$x = 2n\pi - \frac{\pi}{12}$$ is a solution (coming from using $$\cos\left(-\frac{\pi}{4}\right)$$ on the right), as well as the first one you specify of $$x = 2n\pi + \frac{5\pi}{12}$$ (coming from using $$\cos\left(\frac{\pi}{4}\right)$$ on the right). Thus, it seems the answer has an oversight.
• @Techie5879 Unless there's some stated restriction on what $x$ could be, they are both valid options, so it seems the multiple-choice test has a mistake in it. Mar 5 '20 at 6:42


# Why is $1/i$ equal to $-i$?

When I entered the value $$\frac{1}{i}$$ in my calculator, I received the answer as $-i$ whereas I was expecting the answer as $i^{-1}$. Even google calculator shows the same answer (Click here to check it out).

Is there a fault in my calculator or $\frac{1}{i}$ really equals $-i$? If it does then how?

• Hint $i^2 = -1$
– Mann
May 11, 2015 at 12:14
• Multiply by $i/i$. May 11, 2015 at 12:14
• Hint $$z=\frac{1}{i}\iff zi=1\implies \dots$$ May 11, 2015 at 12:56
• Three down votes for someone exhibiting natural mathematical curiosity and having the wherewithal to ask about it is shameful. May 11, 2015 at 14:50
• Excellent question I wondered that myself when I read it. I could say $+1$ but given the context of the question I should say $+i$! May 13, 2015 at 1:04

$$\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$$

Note that $i(-i)=1$. By definition, this means that $(1/i)=-i$.

The notation "$i$ raised to the power $-1$" denotes the element that multiplied by $i$ gives the multiplicative identity: $1$.

In fact, $-i$ satisfies that since

$$(-i)\cdot i= -(i\cdot i)= -(-1) =1$$

That notation holds in general. For example, $2^{-1}=\frac{1}{2}$ since $\frac{1}{2}$ is the number that gives $1$ when multiplied by $2$.

• I appreciate that this answer gives context to the calculation. +1 ! May 11, 2015 at 15:04

There are multiple ways of writing out a given complex number, or a number in general. Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example.

The complex numbers are a field. This means that every non-$0$ element has a multiplicative inverse, and that inverse is unique.

While $1/i = i^{-1}$ is true (pretty much by definition), if we have a value $c$ such that $c * i = 1$ then $c = i^{-1}$.

This is because we know that inverses in the complex numbers are unique.

As it happens, $(-i) * i = -(i*i) = -(-1) = 1$. So $-i = i^{-1}$.

As fractions (or powers) are usually considered "less simple" than simple negation, when the calculator displays $i^{-1}$ it simplifies it to $-i$.

$-i$ is the multiplicative inverse of $i$ in the field of complex numbers, i.e. $-i * i = 1$, or $i^{-1} = -i$.

$$\frac{1}{i}=\frac{i^4}{i}=i^3=i^2\cdot i = -i$$

I always like to point out that this fits well into a pattern you see when "rationalising the denominator", if the denominator is a root: $$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{1}{2}\sqrt{2}$$ $$\frac{1}{\sqrt{17}} = \frac{1}{\sqrt{17}}\cdot \frac{\sqrt{17}}{\sqrt{17}} = \frac{1}{17}\sqrt{17}$$ $$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}}\cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{1}{a}\sqrt{a}$$ $$\frac{1}{i} = \frac{1}{\sqrt{-1}} = \frac{1}{\sqrt{-1}}\cdot \frac{\sqrt{-1}}{\sqrt{-1}} = \frac{1}{-1}\sqrt{-1} = - i.$$ In this vein, it is almost more suggestive to write $$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\frac{1}{\sqrt{17}} = \frac{\sqrt{17}}{17}$$ $$\frac{1}{i} = \frac{i}{-1}.$$

By the definition of the inverse $$\frac1i\cdot i=1.$$

This agrees with

$$(-i)\cdot i=1.$$

Any complex number is fully described by its magnitude and phase (argument) via the complex exponential.

$$X = |X|e^{i\arg{X}}$$

It is useful to write complex numbers in this form when multiplying and dividing as we can make use of exponent rules. Division in this instance simplifies to dividing the magnitudes and subtracting the phases.

Before we compute this division, lets calculate the magnitude and phase of $$1$$ and $$i$$. It is quite obvious that the magnitudes of both numbers are $$1$$ (i.e. $$|1|=|i|=1$$). And by definition the phases are:

$$\arg{1} = 0$$ $$\arg{i} = \frac{\pi}{2}$$

Our two complex exponentials are therefore:

$$1 = e^{i0}$$ $$i = e^{i\frac{\pi}{2}}$$

Now we perform the division making use of the exponent rules:

$$\frac{1}{i}=\frac{e^{i0}}{e^{i\frac{\pi}{2}}}=e^{-i\frac{\pi}{2}}$$

If you consult the unit circle (since the magnitude is 1), you will find that a phase of $$-\frac{\pi}{2}$$ corresponds to $$−i$$. Alternatively you can apply Euler's formula:

$$e^{-i\frac{\pi}{2}} = \cos\left(-\frac{\pi}{2}\right) +i\sin\left(-\frac{\pi}{2}\right) =-i$$

I want to add the method that I like.

$$\frac{1}{i}$$ $$=\frac{1}{cis(\frac{\pi}{2})}$$ $$= cis(- \frac{\pi}{2})$$ $$=-i$$

Where $$cis(x)= \cos(x)+i \sin(x)$$


# Sum of Harmonic Series

It is well known that the sum of a harmonic series does not have a closed form. Here is a formula which gives us a good approximation.

We need to find the sum of the following series

$\dfrac{1}{a}+\dfrac{1}{a+d}+\dfrac{1}{a+2d}+\ldots+\dfrac{1}{a+(n-1)d}$

Consider the function $$f(x)=\frac{1}{x}$$, we intend to take middle riemann sums with rectangles of width $$d$$ starting from $$x=a$$ to $$x=a+(n-1)d$$.

Each rectangle in the figure has a width $$d$$. The height of the $$i\text{th}$$ rectangle is $$\frac{1}{a+(i-1)d}$$. The sum of the area of the rectangles is approximately equal to the area under the curve.

Area under f(x) from $$x=a-\frac{d}{2}$$ to $$x=a+\left(n-\frac{1}{2}\right)d \approx\displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}$$

$\large\Rightarrow \int_{a-\frac{d}{2}}^{a+\left(n-\frac{1}{2}\right)d} \dfrac{\mathrm{d}x}{x}\approx \displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}$

Let $$S_n =\displaystyle\sum_{n=1}^{n} \frac{1}{a+(n-1)d}$$

$\large\ln\dfrac{2a+(2n-1)d}{2a-d}\approx d\times S_n$

$\large\boxed{\Rightarrow s_n\approx\dfrac{\ln\dfrac{2a+(2n-1)d}{2a-d}}{d}}$

Note

• Apologies for the shabby graph.

• $$d\neq 0$$

Note by Aneesh Kundu
2 years, 6 months ago

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@Aneesh Kundu I have just added your formula to Harmonic Progression wiki. I have also added important points from your discussion with Atul. You can also contribute to the wiki.

- 2 years, 1 month ago

How can area under that curve=d×Sn ??here d is denoted as width

- 2 years, 2 months ago

Area under the curve $A=\frac{1}{a} d +\frac{1}{a+d} d+ \frac{1}{a+2d} d \ldots$ $\frac{A}{d}= ( \frac{1}{a}+ \frac{1}{a+d} \ldots )$ $A=d\dot S_{n}$

- 2 years, 2 months ago

Ohoo now i understand clearly. Thanks bro...

- 2 years, 2 months ago

your above expression will be incorrect when $\boxed{2a=d}$

- 2 years, 5 months ago

In this case calculate the sum from $$a_2$$ to $$a_n$$, using the given formula and then add $$a_1$$ to both sides.

- 2 years, 5 months ago

Hey how you have assigned limit of $$x$$ can you please clarify

- 2 years, 6 months ago

$$x$$ varies from $$a-\frac{d}{2}$$ to $$a+\left(n-\frac{1}{2}\right)d$$.

- 2 years, 5 months ago

yaa ,I got this but also you can't use this formula for finding sum of similar terms

i.e. $$S_n= \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+.... \frac {1}{2}(n^{th} term)$$ as common difference is $$0$$ so it will be in indeterminate form

- 2 years, 5 months ago

Thanks for the suggestion, I added this point in the note.

- 2 years, 5 months ago

I mean to is it original(your own)???

- 2 years, 5 months ago

Nope, its not purely original. I was reading about the convergence tests and I happened came across the Integral test, which inspired this note.

- 2 years, 5 months ago

it's really fantastic

- 2 years, 5 months ago

Thanks. :)

- 2 years, 5 months ago

by the way is it real???

- 2 years, 5 months ago

This formula gives really good approximations when $$d\rightarrow 0$$ or for large values of $$n$$ with a not so big $$d$$ .

- 2 years, 5 months ago

Just followed you :-)

- 2 years, 5 months ago

Yes Absolutely

- 2 years, 5 months ago

Thanks! This is a very good and useful note.

- 2 years, 6 months ago

Thanks. :)

- 2 years, 5 months ago