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April 3rd, 2014, 12:31 AM   #1
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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

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April 3rd, 2014, 01:07 AM   #2
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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.
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Last edited by MarkFL; April 3rd, 2014 at 10:23 AM.
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April 3rd, 2014, 01:58 AM   #3
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Math Focus: Abstract algebra
Quote:
Originally Posted by MarkFL View Post
$\displaystyle F_{5(n+1)}-F_{5n}=3_{5n+2}+2F_{5n+1}-F_{5n}$

$\displaystyle F_{5(n+1)}-F_{5n}=5_{5n+1}+F_{5n}$
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,
$$F_{10}\ =\ 55\ =\ 40+15\ =\ 5F_6+3F_5$$
$$F_{15}\ =\ 610\ =\ 445+165\ =\ 5F_{11}+3F_{10}$$
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April 3rd, 2014, 08:24 AM   #4
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Quote:
Originally Posted by Olinguito View Post
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.
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April 3rd, 2014, 10:15 AM   #5
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Excellent , MarkFL is the winner!

Olinguito gets a bonus point for proofreading.

MarkFL , you have the next question.

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April 3rd, 2014, 10:21 AM   #6
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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).
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April 4th, 2014, 10:41 AM   #7
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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.

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April 4th, 2014, 10:58 AM   #8
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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
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April 4th, 2014, 11:39 AM   #9
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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
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April 4th, 2014, 11:48 AM   #10
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Quote:
Originally Posted by agentredlum View Post
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.
Thanks from agentredlum
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