# Thread: Series of function problem

1. ## Series of function problem

Find $\displaystyle y=\Sigma c_n x^n$, with the next conditions:

1. $\displaystyle y=1$ when $\displaystyle x=1$

2. $\displaystyle y'=1$ when $\displaystyle x=1$

3. $\displaystyle y''+y=0$

I need an answer without using the fact that the function: $\displaystyle y=sinx+cosx$ satisfies all conditions above.

What I'm asking is, how can I solve it strictly from those conditions...(without guessing)

Thank you!

2. Note that $\displaystyle y = sinx + cosx$ only satisfies the ODE, not the initial conditions (1, 2).

To generally solve this, consider

$\displaystyle y = \sum_{n=0}^{\infty}c_nx^n$ so
$\displaystyle y' = \sum_{n=1}^{\infty}nc_nx^{n-1}$
and $\displaystyle y'' = \sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}$

Now put those two in the ODE ($\displaystyle y'' + y =0$) and put them together under the same summation (consider the change of variable k = n-2 for y'')
You'll get a recursive formula for $\displaystyle c_n$, from which you will be able to derive a general solution. After that, plug in the initial conditions to get the specific solution.

3. Originally Posted by Also sprach Zarathustra
Find $\displaystyle y=\Sigma c_n x^n$, with the next conditions:

1. $\displaystyle y=1$ when $\displaystyle x=1$

2. $\displaystyle y'=1$ when $\displaystyle x=1$

3. $\displaystyle y''+y=0$

I need an answer without using the fact that the function: $\displaystyle y=sinx+cosx$ satisfies all conditions above.

What I'm asking is, how can I solve it strictly from those conditions...(without guessing)

Thank you!
First take two derivatives of the series

$\displaystyle y''=\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}$

and plug this into the ODE to get

$\displaystyle y''+y=\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}\sum_{n=0}^{\infty}c_{n}x^{n}=0$

Now re-index the the first sereis to start at 0.

$\displaystyle \sum_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^{n}+\sum_{n= 0}^{\infty}c_{n}x^{n}=0$

$\displaystyle \sum_{n=0}^{\infty}\left[(n+2)(n+1)c_{n+2}+c_{n}\right]x^{n}=0$

This gives us a relationship between the coeffients of the power series.

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

Using this we get

$\displaystyle c_2=-\frac{c_{0}}{1\cdot 2}$
$\displaystyle c_3=-\frac{c_{1}}{2\cdot 3}$
$\displaystyle c_4=\frac{c_{2}}{3\cdot 4}=\frac{c_{0}}{4!}$
$\displaystyle c_5=\frac{c_{1}}{4\cdot 5}=\frac{c_{1}}{5!}$
$\displaystyle c_6=-\frac{c_{2}}{5\cdot 6}=\frac{c_{0}}{6!}$
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Notice the even $\displaystyle c_n$'s give

$\displaystyle y=c_{0}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=c_{0}\cos(x)$

Notice the odd $\displaystyle c_n$'s give

$\displaystyle y=c_{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=c_{1}\sin(x)$

4. A Taylor expansion around x=1 is easy to find considering that is $\displaystyle y(1)=1$ , $\displaystyle y^{'} (1) = 1$ and that $\displaystyle y^{''} (x) = - y(x)$ , so that is...

$\displaystyle y^{''}(1) = -y(1) = -1$

$\displaystyle y^{(3)} (1) = - y'(1) = -1$

$\displaystyle y^{(4)} (1) = - y^{''} (1) = 1$

$\displaystyle y^{(5)} (1) = - y^{(3)} (1) = 1$

$\displaystyle \dots$

... that brings to the Taylor expansion...

$\displaystyle y(x) = 1 + (x-1) - \frac{(x-1)^{2}}{2!} - \frac{(x-1)^{3}}{3!} + \frac{(x-1)^{4}}{4!} + \frac{(x-1)^{5}}{5!} - \dots$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$