How do I calculate:

$\displaystyle y_{n+1}-y_{n}=\sin(n) $

$\displaystyle n>=0, y_0=0$

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- Nov 30th 2006, 11:10 AMa4sweDifference equation
How do I calculate:

$\displaystyle y_{n+1}-y_{n}=\sin(n) $

$\displaystyle n>=0, y_0=0$ - Nov 30th 2006, 11:34 AMtopsquark
- Nov 30th 2006, 11:41 AMa4swe
Yes, I would agree to that.

But what then? - Nov 30th 2006, 12:05 PMgalactus
You could start by building your iterations.

$\displaystyle y_{n+1}-y_{n}=sin(n)\Rightarrow{y_{1}+y_{0}=sin(0)}\Righta rrow{y_{1}=0}$

$\displaystyle y_{2}-y_{1}=sin(1)\Rightarrow{y_{2}=sin(1)}$

$\displaystyle y_{3}-y_{2}=sin(2)\Rightarrow{y_{3}-sin(1)=sin(2)}$

$\displaystyle y_{4}-y_{3}=sin(3)\Rightarrow{y_{4}-(sin(2)+sin(1))=sin(3)}$

And so on and so on. - Nov 30th 2006, 12:15 PMa4swe
I want to have a formula for ANY n, that method won't help me.

I know there is one, and getting to that is my problem.

$\displaystyle \frac{\sin(\frac{n}{2})*\sin(\frac{n-1}{2})}{\sin(\frac{1}{2})}$

Is what I want it to be. - Nov 30th 2006, 03:07 PMtopsquark
(Hmmm...It seems I was off by a term in my earlier post. Galactus has it right.)

Have you tried an induction proof?

For n = 1: $\displaystyle \frac{\sin(\frac{1}{2})*\sin(\frac{1-1}{2})}{\sin(\frac{1}{2})} = 0$

Since $\displaystyle y_1 = 0$ this is good. (The formula also works for $\displaystyle y_0 = 0$.)

So it works for some n. Let's see if it works for n + 1.

Now, $\displaystyle y_{n+1} = \sum_{i = 0}^{n+1} sin(i) =(?) \frac{\sin(\frac{n+1}{2})*\sin(\frac{n}{2})}{\sin( \frac{1}{2})}$

By hypothesis $\displaystyle y_{n+1} = y_n + sin(n)$ so

$\displaystyle \frac{\sin(\frac{n+1}{2})*\sin(\frac{n}{2})}{\sin( \frac{1}{2})} = \frac{\sin(\frac{n}{2})*\sin(\frac{n-1}{2})}{\sin(\frac{1}{2})} + sin(n)$

should be true.

$\displaystyle \sin(\frac{n+1}{2})*\sin(\frac{n}{2}) = \sin(\frac{n}{2})*\sin(\frac{n-1}{2}) + sin(\frac{1}{2})sin(n)$

On the LHS:

$\displaystyle \sin(\frac{n+1}{2})*\sin(\frac{n}{2}) = sin(\frac{n}{2} + \frac{1}{2}) * sin(\frac{n}{2})$

= $\displaystyle \left [ sin(\frac{n}{2}) cos(\frac{1}{2}) + sin(\frac{1}{2}) cos(\frac{n}{2}) \right ] sin(\frac{n}{2})$

= $\displaystyle sin^2(\frac{n}{2}) cos(\frac{1}{2}) + sin(\frac{1}{2}) sin(\frac{n}{2}) cos(\frac{n}{2})$

= $\displaystyle sin^2(\frac{n}{2}) cos(\frac{1}{2}) + \frac{1}{2} sin(\frac{1}{2}) sin(n)$

Similarly we may show that the RHS becomes

$\displaystyle \sin(\frac{n}{2})*\sin(\frac{n-1}{2}) + sin(\frac{1}{2})sin(n) = $ $\displaystyle sin^2(\frac{n}{2}) cos(\frac{1}{2}) - \frac{1}{2} sin(\frac{1}{2}) sin(n) + sin(\frac{1}{2})sin(n)$

Which again gives:

$\displaystyle sin^2(\frac{n}{2}) cos(\frac{1}{2}) + \frac{1}{2} sin(\frac{1}{2}) sin(n)$

So the two sides are equal.

So the theorem is true for n and n + 1 etc.

-Dan - Dec 1st 2006, 01:15 AMa4swe
Ahhh, yes!

Thank you tops.