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 April 3rd, 2014, 12:31 AM #1 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,182 Thanks: 181 Revival-Math Q&A The Fiboncci sequence begins with the numbers 1,1 and then each subsequent number is the sum of the two previous numbers. 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , ... Prove that every 5th Fibonacci number is divisible by 5 Q&A RULES 1) Anyone can post an answer. 2) The one who posted the problem decides the winner. 3) The winner must post the next question or relinquish control. Thanks from MarkFL, mathbalarka, eddybob123 and 2 others
 April 3rd, 2014, 01:07 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs I will use induction. Our base case $P_1$ is obviously true since: $\displaystyle F_{5}=5$ Thus, our induction hypothesis $P_n$ is: $\displaystyle F_{5n}=5m$ where $\displaystyle m,n\in\mathbb{N}$ Now, we if use the recursive definition of the sequence and observe that: $\displaystyle F_{5(n+1)}-F_{5n}=F_{5n+4}+F_{5n+3}-F_{5n}$ $\displaystyle F_{5(n+1)}-F_{5n}=F_{5n+3}+2F_{5n+2}+F_{5n+1}-F_{5n}$ $\displaystyle F_{5(n+1)}-F_{5n}=3F_{5n+2}+2F_{5n+1}-F_{5n}$ $\displaystyle F_{5(n+1)}-F_{5n}=5F_{5n+1}+2F_{5n}$ Adding this to the hypothesis, we obtain: $\displaystyle F_{5(n+1)}=5F_{5n+1}+3\cdot5m=5\left(F_{5n+1}+3m \right)$ We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction. Thanks from agentredlum and mathbalarka Last edited by MarkFL; April 3rd, 2014 at 10:23 AM.
April 3rd, 2014, 01:58 AM   #3
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Quote:
 Originally Posted by MarkFL $\displaystyle F_{5(n+1)}-F_{5n}=3_{5n+2}+2F_{5n+1}-F_{5n}$ $\displaystyle F_{5(n+1)}-F_{5n}=5_{5n+1}+F_{5n}$
Sorry to nitpick but it’s $\displaystyle F_{5(n+1)}-F_{5n}=5_{5n+1}+\color{red}2\color{black}F_{5n}$ and we should have
$$F_{5(n+1)}\ =\ 5F_{5n+1}+3F_{5n}\ =\ 5\left(F_{5n+1}+3m \right)$$
For example,
$$F_{10}\ =\ 55\ =\ 40+15\ =\ 5F_6+3F_5$$
$$F_{15}\ =\ 610\ =\ 445+165\ =\ 5F_{11}+3F_{10}$$

April 3rd, 2014, 08:24 AM   #4
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Quote:
 Originally Posted by Olinguito Sorry to nitpick but it’s $\displaystyle F_{5(n+1)}-F_{5n}=5_{5n+1}+\color{red}2\color{black}F_{5n}$ and we should have $$F_{5(n+1)}\ =\ 5F_{5n+1}+3F_{5n}\ =\ 5\left(F_{5n+1}+3m \right)$$ For example, $$F_{10}\ =\ 55\ =\ 40+15\ =\ 5F_6+3F_5$$ $$F_{15}\ =\ 610\ =\ 445+165\ =\ 5F_{11}+3F_{10}$$
Yes, I have corrected my post.

 April 3rd, 2014, 10:15 AM #5 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,182 Thanks: 181 Excellent , MarkFL is the winner! Olinguito gets a bonus point for proofreading. MarkFL , you have the next question. Thanks from MarkFL and Olinguito
 April 3rd, 2014, 10:21 AM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Okay...here is a fun exercise: Problem: The double integral $\int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy$ is an improper integral and could be defined as the limit of double integrals over the rectangle: $[0,t]\times[0,t]$ as $t\to 1^{-}$. 1.) Expand the integrand as a geometric series to show that $\int_0^1 \int_0^1 \frac{1}{1-xy}\,dx\,dy = \sum_{n=1}^{\infty}\frac{1}{n^2}$ 2.) Leonhard Euler proved that $\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$ Prove this fact by evaluating the integral found in (1). Thanks from agentredlum, mathbalarka and Olinguito
 April 4th, 2014, 10:41 AM #7 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,182 Thanks: 181 This proof is from Brendan W. Sullivan who gives Tom Apostol credit for inventing it. http://docs.google.com/viewer?url=ht...3Rznk0-l89V8rw I will add some details. For MarkFL's 1) We observe $0 \le xy \le 1$ So we let $xy = r$ without fear in order to use the formula for a geometric series. $$\dfrac{1}{1-xy} = \dfrac{1}{1-r} = 1 + r + r^2 + r^3 + ... = \\ 1 + xy + (xy)^2 + (xy)^3 + ... = \sum_{n = 1}^{\infty}(xy)^{n-1}$$ Now you can do this $$\int_{0}^1 \int_{0}^1 ( 1 + xy + x^2y^2 + x^3y^3 + ... ) dxdy$$ Work from the inside out , integrate term by term with respect to x keeping y constant. $$\int_{0}^1 (x + \dfrac{x^2y}{2} + \dfrac{x^3y^2}{3} + \dfrac{x^4y^3}{4} + ...) |_{0}^1)dy$$ Plug in the upper limit x = 1 (the lower limit x = 0 vanishes making no contribution) $$\int_{0}^1 (1 + \dfrac{y}{2} + \dfrac{y^2}{3} + \dfrac{y^3}{4} + ... ) dy$$ Integrate with respect to y $$(y + \dfrac{y^2}{2 \cdot 2} + \dfrac{y^3}{3 \cdot 3} + \dfrac{y^4}{4 \cdot 4} + ...) |_{0}^1$$ Plug in the upper limit y = 1 (the lower limit y = 0 vanishes), $$1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + ... + \dfrac{1 }{n^2} + ... = \sum_{n = 1}^{\infty } \dfrac{1 }{n^2}$$ This confirms MarkFL's 1) For MarkFL's 2) I refer the reader to the link given above because giving details here would turn this into a very long post.
 April 4th, 2014, 10:58 AM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs This problem was given as a university level problem of the week at MHB, and this is the solution I posted: 1.) Expanding the integrand as a geometric series, we may write: $\displaystyle \frac{1}{1-xy}= \sum_{n=0}^{\infty}(xy)^n$ Hence the integral becomes: $\displaystyle \int_0^1\int_0^1 \sum_{n=0}^{\infty}(xy)^n\,dx\,dy=\int_0^1\left[\sum_{n=0}^{\infty}\frac{x^{n+1}y^n}{n+1} \right]_0^1\,dy=$ $\displaystyle \int_0^{1}\sum_{n=1}^{\infty}\frac{y^{n-1}}{n}\,dy=\left[\sum_{n=1}^{\infty}\frac{y^{n}}{n^2} \right]_0^1=\sum_{n=1}^{\infty}\frac{1}{n^2}$ 2.) Using the change of variables: $\displaystyle (x,y)=(u-v,u+v)$ we obtain: $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\iint_{R}\frac{1} {1-u^2+v^2}\left|\frac{\partial (x,y)}{\partial (u,v)} \right|\,du\,dv$ Calculating the Jacobian matrix, we find: $\displaystyle \frac{\partial (x,y)}{\partial (u,v)}=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\\\end{vmatrix}=\begin{vmatrix}1&-1\\1&1\\\end{vmatrix}=1(1)-(-1)(1)=2$ Thus, we have: $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=2\iint_{R}\frac{1 }{1-u^2+v^2}\,du\,dv$ Remapping the boundaries in terms of the new variables, we find $R$ is a square in the $uv$-plane with vertices: $\displaystyle (0,0),\,\left(\frac{1}{2},-\frac{1}{2} \right),\,\left(\frac{1}{2},\frac{1}{2} \right),\,(1,0)$ Reversing the order of integration and using the symmetry of the square, we obtain: $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\int_0^{ \frac{1}{2}}\int_0^u \frac{dv\,du}{1-u^2+v^2}+\int_{ \frac{1}{2}}^1\int_0^{1-u} \frac{dv\,du}{1-u^2+v^2} \right)$ Next, we may compute: $\displaystyle \int_0^u\frac{dv}{1-u^2+v^2}=\left[\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{v}{\sqrt{1-u^2}} \right) \right]_0^u=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}} \right)=\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)$ $\displaystyle \int_0^{1-u}\frac{dv}{1-u^2+v^2}=\left[\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{v}{\sqrt{1-u^2}} \right) \right]_0^{1-u}=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}} \right)=\frac{1}{\sqrt{1-u^2}}\tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)$ If we let: $\displaystyle \tan(\theta)=\sqrt{\frac{1-u}{1+u}}$ Squaring, we obtain: $\displaystyle \tan^2(\theta)=\frac{1-u}{1+u}$ Add through by 1: $\displaystyle \tan^2(\theta)+1=\frac{1-u}{1+u}+1$ Apply a Pythagorean identity on the left and combine terms on the right: $\displaystyle \sec^2(\theta)=\frac{2}{1+u}$ Invert both sides: $\displaystyle \cos^2(\theta)=\frac{u+1}{2}$ Solving for $u$, we find: $\displaystyle u=2\cos^2(\theta)-1=\cos(2\theta)$ Hence, we find: $\displaystyle \tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)=\theta=\frac{1}{2}\cos^{-1}(u)$ Using the identity $\displaystyle \sin^{-1}(u)+\cos^{-1}(u)=\frac{\pi}{2}$ we finally have: $\displaystyle \tan^{-1}\left(\sqrt{\frac{1-u}{1+u}} \right)=\frac{1}{2}\left(\frac{\pi}{2}-\sin^{-1}(u) \right)$ Utilizing these results, we now may state: $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\int_0^{ \frac{1}{2}}\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du+\frac{1}{2}\int_{ \frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\left(\frac{\pi}{2}-\sin^{-1}(u) \right)\,du \right)$ Now, let's look at the first integral: $\displaystyle \int_0^{ \frac{1}{2}}\frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du$ Using the substitution: $\displaystyle \alpha=\sin^{-1}(u)\,\therefore\,d\alpha=\frac{1}{\sqrt{1-u^2}}\,du$ we now have: $\displaystyle \int_0^{\frac{\pi}{6}}\alpha\,d\alpha=\frac{1}{2} \left[\alpha^2 \right]_0^{\frac{\pi}{6}}= \frac{1}{2}\left(\frac{\pi}{6} \right)^2=\frac{\pi^2}{72}$ Next, let's break the second integral into two parts: i) $\displaystyle \frac{\pi}{4}\int_{\frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\,du=\frac{\pi}{4}\left[\sin^{-1}(u) \right]_{\frac{1}{2}}^1=\frac{\pi}{4}\left(\frac{\pi}{2}-\frac{\pi}{6} \right)=\frac{\pi^2}{12}$ ii) $\displaystyle -\frac{1}{2}\int_{\frac{1}{2}}^1 \frac{1}{\sqrt{1-u^2}}\sin^{-1}(u)\,du$ Using the substitution: $\displaystyle \alpha=\sin^{-1}(u)\,\therefore\,d\alpha=\frac{1}{\sqrt{1-u^2}}\,du$ we now have: $\displaystyle -\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \alpha\,d\alpha=-\frac{1}{4}\left[\alpha^2 \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}=-\frac{1}{4}\left(\left(\frac{\pi}{2} \right)^2-\left(\frac{\pi}{6} \right)^2 \right)=-\frac{\pi^2}{18}$ Thus, putting these results together, there results: $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=4\left(\frac{\pi^ 2}{72}+\frac{\pi^2}{12}-\frac{\pi^2}{18} \right)=4\left(\frac{\pi^2}{24} \right)$ And, we may then state: $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ Shown as desired. So...agentredlum, you now have the floor. Thanks from agentredlum
 April 4th, 2014, 11:39 AM #9 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,182 Thanks: 181 Thanx , it would have taken me 2 hours + to write in rhe details using mathjax with my PS3 controller. Are you concerned at all that when $$x = y = 1$$ we get $$\dfrac{1}{1-1}$$ and the Geometric series does not apply? Thanks from MarkFL
April 4th, 2014, 11:48 AM   #10
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Quote:
 Originally Posted by agentredlum Thanx , it would have taken me 2 hours + to write in rhe details using mathjax with my PS3 controller. Are you concerned at all that when $$x = y = 1$$ we get $$\dfrac{1}{1-1}$$ and the Geometric series does not apply?
I see $x$ and $y$ as "approaching" 1 from beneath as given by the definition of the rectangle.

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