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Solving a recurrence equation

  • Thread starter yoran
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118
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Hi, I'm trying to solve this recurrence equation.

1. Homework Statement
Solve [tex]y[k+2]+y[k]=sin(k)[/tex].
The answer is already given, it's
[tex]y[k]=c_1sin(\frac{\pi}{2}k) + c_2cos(\frac{\pi}{2}k) + \frac{sin(k)+sin(k-2)}{2(1+cos(2))}[/tex]

2. Homework Equations


3. The Attempt at a Solution
First I solve the homogeneous equation.
[tex]y[k+2]+y[k]=0[/tex].
The characteristic function is
[tex]r^2+1=0[/tex]
This has two complex roots [tex]r=i[/tex] and [tex]r=-i[/tex].
Thus the general solution is
[tex]y[k]=c_1(i)^k+c_2(-i)^k[/tex]
If you change the notation for the complex numbers, then it gives
[tex]y[k]=c_1(cos(\frac{\pi}{2}k)+isin(\frac{\pi}{2}k)) + c_2(cos(\frac{-\pi}{2}k)+isin(\frac{-\pi}{2}k))[/tex]
After grouping together it gives
[tex](c_1+c_2)cos(\frac{\pi}{2}k) + (c_1-c_2)isin(\frac{\pi}{2}k)[/tex]
Apparently, the answer is without complex numbers so I guess they only keep the real part. Then it must be so that
[tex]c_1-c_2=0\Leftrightarrow c_1=c_2[/tex]
Then I get as solution
[tex]y[k]=2c_1cos(\frac{\pi}{2}k)[/tex]
which is not the same solution as the general solution in the answer. What did I do wrong?

I also tried to find a particular solution for the non-homogeneous equation. I tried as a solution
[tex]y[k]=\alpha sin(k)[/tex]
with [tex]\alpha[/tex] the parameter to determine. I don't this is the right solution. Can anyone help me with this problem?

Thanks,

Yoran
 

Answers and Replies

334
0
They don't "only keep the real part". Plug the solution back into the homogeneous equation; the sine part falls out without imposing any constraint on c1 and c2.
 
118
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Hi,

Thanks for your quick reply. I tried to do what you told me but when I plug it into the homogeneous equation, everything cancels out! I get 0=0. This is what I did:
We have that
[tex]y[k]=(c_1+c_2)cos(\frac{\pi}{2}k)+(c_1-c_2)isin(\frac{\pi}{2}k)[/tex]
I plug it into
[tex]y[k+2]+y[k]=0[/tex]
I get
[tex](c_1+c_2)cos(\frac{\pi}{2}(k+2))+(c_1-c_2)isin(\frac{\pi}{2}(k+2))+(c_1+c_2)cos(\frac{\pi}{2}k)+(c_1-c_2)isin(\frac{\pi}{2}k)=0[/tex]
Because
[tex]cos(\theta + \pi)=-cos(\theta)[/tex] and
[tex]sin(\theta + \pi)=-sin(\theta)[/tex]
I get that
[tex]-(c_1+c_2)cos(\frac{\pi}{2}k)-(c_1-c_2)isin(\frac{\pi}{2}k)+(c_1+c_2)cos(\frac{\pi}{2}k)+(c_1-c_2)isin(\frac{\pi}{2}k)=0[/tex]
which makes
[tex]0=0[/tex]
How come?
 
334
0
Actually, you are correct. That was a silly suggestion of me, because 0=0 simply confirms your solution satisfies the equation from which you derived the solution. I will edit this post if I come up with something.

Edit: Wow I had been reading it wrong for a good 45 minutes, sorry! Consider complex coefficients and make the substitution d1=c1+c2 and d2=i(c1-c2).

Can you find the particular solution from here?
 
Last edited:
118
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Hi,

Yeah of course... just give different names as they are just parameters... Thanks a lot.
I tried the substitution [tex]\alpha sin(k)[/tex] because there is [tex]sin(k)[/tex] on the right side of the equation. However, I can't find the right answer. This is what I have done:
Try
[tex]y[k]=\alpha sin(k)[/tex]
Plug it into the equation:
[tex]\alpha sin(k+2) + \alpha sin(k) = sin(k)[/tex]
Then we get
[tex]\alpha = \frac{sin (k)}{2sin(k+1)cos(1)}[/tex]
by using the identity
[tex]sin(\theta) + sin(\alpha) = 2sin(\frac{\theta + \alpha}{2})cos(\frac{\theta - \alpha}{2})[/tex]
The answer for [tex]\alpha[/tex] is obviously not the right answer.
Did I make the wrong substitution?
 
118
0
Anyone?
 
tiny-tim
Science Advisor
Homework Helper
25,789
249
Try
[tex]y[k]=\alpha sin(k)[/tex]
Hi yoran! :smile:

Try [tex]y[k]=\alpha\,sin(k\,-\,1)[/tex] . :smile:
 
118
0
Ok, I tried
[tex]y[k]=\alpha \sin(k-1)[/tex].
Then I get that
[tex]\alpha = \frac{1}{2\cos 1}[/tex]
So given the solution for the homogeneous equation and this solution, I have that
[tex]y[k]=c_1\sin{\frac{\pi}{2}k} + c_2\cos{\frac{\pi}{2}k} + \frac{1}{2\cos{1}}\sin{k-1}[/tex]
But the answer should be
[tex]y[k]=c_1\sin{\frac{\pi}{2}k} + c_2\cos{\frac{\pi}{2}k} + \frac{\sin{k}+\sin(k-2)}{2(1+\cos{2})}[/tex]
Where does it go wrong in my solution?
 
tiny-tim
Science Advisor
Homework Helper
25,789
249
Hi yoran! :smile:
…I have that
[tex]y[k]=c_1\sin{\frac{\pi}{2}k} + c_2\cos{\frac{\pi}{2}k} + \frac{1}{2\cos{1}}\sin{k-1}[/tex]
But the answer should be
[tex]y[k]=c_1\sin{\frac{\pi}{2}k} + c_2\cos{\frac{\pi}{2}k} + \frac{\sin{k}+\sin(k-2)}{2(1+\cos{2})}[/tex]
Where does it go wrong in my solution?
:rofl: But they're the same! :rofl:​

Standard trig equation: 1 + cos(2) = 2cos²(1).

So [tex]\frac{\sin{k}+\sin(k-2)}{2(1+\cos{2})}[/tex] = … ? :smile:
 
118
0
Hehe, of course... Thanks a lot!
[tex]\frac{\sin{k}+\sin(k-2)}{2(1+\cos{2})}=\frac{2\sin(k-1)\cos(1)}{4\cos^2(1)}=\frac{\sin(k-1)}{2\cos 1}[/tex]
 

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