2. ## Re: Mathematical induction help

Originally Posted by Musawwir
Please pick out one of the problems, write down what you know such as a base step and any questions.

3. ## Re: Mathematical induction help

What do you mean by "help"- do it for you?

The second problem asks you to show, by induction, that $\displaystyle 10^n+ 3(4^{n+2})+ 5$ is divisible by 5 for n any positive integer.

Okay, "proof by induction" works on a pretty standard "recipe".

First show that the statement is true for n= 1. If n= 1, this is [math]10^1+ 3(4^3)+ 5= 10+ 3(64)+ 5= 207= 9(23) so is divisible by 9.

Now shoe that "if the statement is true for n= k (where k can be any number) then it is true for n= k+ 1.

If the statement is true for n= k then $\displaystyle 10^k+ 3(4^{k+ 2})+ 5$ is a multiple of 9- that is, $\displaystyle 10^k+ 3(4^{k+ 2})+ 5= 9m$ for some integer m.

Now take n= k+ 1. That is, look at $\displaystyle 10^{k+1}+ 3(4^{k+1+2})+ 5$ and try to go back to the previous case. We can write $\displaystyle 10^{k+1}= 10(10^k)$ and m[math]4^{k+ 1+ 2}= 4(4^{k+2}[math] so $\displaystyle 10^{k+1}+ 3(4^{k+1+2})+ 5= 10(10^k)+ 3(4(4^{k+2}))+ 5$. I notice that we can write that leading "10" as 9+ 1. And we can write that extra "4" in the middle term as 3+ 1. That is, $\displaystyle 10^{k+1}+ 3(4^{k+1+2})+ 5= 10(10^k)+ 3(4(4^{k+2}))+ 5= (9+1)(10^k)+ 3(3+ 1)(4^{k+2})+ 5= 10^k+ 3(4^{k+2})+ 5+ 9(10^k)+ 3(3)(4^{k+1})= 9m+ 9(10^k)+ 9(4^{k+ 2})= 9(m+ 10^k+ 4^{k+2})$ so is a multiple of 9.

5. ## Re: Mathematical induction help

if you could help for 3 and 5 and should you always start with n=1 n=K then n=k+1??

6. ## Re: Mathematical induction help

only one is tough

7. ## Re: Mathematical induction help

Yes, all of these depend upon some integer "n" so all are candidates for "proof by induction". And to prove a statement is true for all positive integers, yes, start by showing it is true for n= 1, then show that "if the statement is true for n= k it is also true for n= k+1". The idea is that you have proved it true for n= 1 so it is true for n= 1+ 1= 2. Since you now know it is true for n= 2, it is true for n= 2+ 1= 3, etc.

Number 3 asks you to show that $\displaystyle \sum_{r= 1}^n \frac{r^2+ r- 1}{(r+ 2)!}= \frac{1}{2}- \frac{n+1}{(n+2)!}$.

When n= 1 that is simply $\displaystyle \frac{1+ 1- 1}{(1+ 2)!}= \frac{1}{2}- \frac{1+ 1}{(1+ 2)!}$.

The left side is $\displaystyle \frac{1}{3!}= \frac{1}{6}$. The right side is $\displaystyle \frac{1}{2}- \frac{2}{3!}= \frac{1}{2}- \frac{2}{6}= \frac{3}{6}- \frac{2}{6}= \frac{1}{6}$.

Yes, they are the same.

Now, assume that $\displaystyle \sum_{r=1}^k\frac{r^2+ r- 1}{(r+2)!}= \frac{1}{2}- \frac{k+1}{(k+2)!}$ for some positive integer k.

We can write $\displaystyle \sum_{r=1}^{k+1} \frac{r^2+ r- 1}{(r+2)!}$ as $\displaystyle \sum_{r=1}^k\frac{r^2+ r- 1}{(r+2)!}+ \frac{k^2+ k- 1}{(k+2)!}$
and, using that "assumption" as $\displaystyle \frac{1}{2}- \frac{k+1}{(k+2)!}+ \frac{k^2+ k-1}{(k+ 2)!}$

Since those have the same denominator, add the fractions by adding the numerators:
$\displaystyle \frac{1}{2}- \frac{k+ 1- k^2+ k- 1}{(k+2)!}= \frac{1}{2}- \frac{-k^2+ 2k}{(k+2)!}$

All that is left is to show, by "algebraic manipulation" that the second fraction on the right can be written as
$\displaystyle \frac{k+ 1+ 1}{(k+1+ 2)!}$.

8. ## Re: Mathematical induction help

Number 5 is to show that $\displaystyle \frac{d^n}{dx^n}(e^x sin(x))= 2^{n/2} e^x sin(x+ \frac{n\pi}{4})$

Again. show this is true for n= 1. When n= 1 the left side is just $\displaystyle \frac{d}{dx}(e^x sin(x)$

By the "product rule" the derivative is $\displaystyle \frac{de^x}{dx}sin(x)+ e^x \frac{d sin(x)}{dx}$

I presume you know that $\displaystyle \frac{de^x}{dx}= e^x$ and that $\displaystyle \frac{d sin(x)}{dx}= cos(x)$.

So we have $\displaystyle e^x sin(x)+ e^xcos(x)$.

Now try to manipulate the right side, when n= 1, $\displaystyle 2^{1/2}e^x sin(x+ \frac{\pi}{4}$ to match that.

An obvious thing to try is the trig identity $\displaystyle sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)$ with $\displaystyle a= x$ and $\displaystyle b= \frac{\pi}{4}$ so have $\displaystyle sin(x+ \frac{\pi}{4})= sin(x)cos(\pi/4)+ cos(x)sin(\pi/4)$.

And, of course, $\displaystyle sin(\pi/4)= cos(\pi/4)= \frac{\sqrt{2}}{2}= 2^{-1/2}$.

9. ## Re: Mathematical induction help

(v)

\begin{align*} \dfrac{d}{dx} e^x\sin x & = e^x\cos x +e^x\sin x \\ & = \sqrt{2} e^x \left( \dfrac{ \cos x}{ \sqrt{2} }+\dfrac{ \sin x }{ \sqrt{2} } \right) \\ & = 2^{1/2}e^x\left( \cos x \sin \dfrac{ \pi }{4} + \sin x \cos \dfrac{ \pi }{4} \right) \\ & = 2^{1/2}e^x \sin \left( x+\dfrac{ \pi }{4} \right)\end{align*}

\begin{align*}\dfrac{d^{n+1}}{dx^{n+1}} e^x \sin x & = \dfrac{d}{dx} \left( \dfrac{d^n}{dx^n} e^x \sin x \right) \\ & = \dfrac{d}{dx} \left( 2^{n/2}e^x \sin \left( x+\dfrac{ \pi }{4} \right) \right) \end{align*}

Do the same trick that I did above where I factor out an extra factor of $\sqrt{2}$ and use the sum of angles formula for sine.

10. ## Re: Mathematical induction help

did the trick guys i have finally understand how to proceed and successfully did all the work
first test for n=1 assume statement is valid and now test for n=n+1 using the statement and i was a able to understand the h/w
thanks

11. ## Re: Mathematical induction help

I wouldn't phrase it that way. Yes, the first step is to show that the statement is true for n= 1. (Ocasionally you will have a statement that is only true for n greater than or equal to some positive integer other than 1. In that situation, use that value for the "base" case.)

Now, assuming that the statement is true for some n= k, prove it true for n= k+ 1 (I don't like to say "assume it is true for n and prove it is true for n+1. You are trying to prove the statement true for all positive integers, n, so shouldn't use n to represent a specific integer).