Power Series & interval of convergence

[Ashlander]

Hey all, I'm terribly stuck at trying to find the interval of convergence for this given function.

(SUM n=0 to Infinity) ((-1)^n * x^2n)/(n!) = 1 - x^2 + ((x^4)/2) - ...

I'm not sure if thats a little hard to read/understand, but so far I have reduced the given function to this:

|an+1 / an| = x^(2n+2) / ((n+1)*x^(2n))

and here is where I have no idea how to reduce the expression any more.. following the Ratio Method used in calculating the interval of convergence in a given power series is how I went along.

Also, can anyone explain to me what this means?
"b. Write out several terms of the series and verify that f'(x) = -2xf(x) [for all x in the interior of the interval of convergence].
c. Show that y = f(x) is a solution of the initial value problem y' = -2xy, y(0)=1."

(I'm not askin for answers, just a simple idea of what the question is asking for so I can go about doing it )

 

[soroban]

Hello, Ashlander!

I can help with the first two parts . . .

Find the interval of convergence:

. . \(\displaystyle \L f(x) \;= \:\sum^{\infty}_{n=0}(-1)^n\,\frac{x^{2n}}{n!} \;=\; 1 \,-\, x^2\,+\,\frac{x^4}{2!}\,-\,\frac{x^6}{3!}\,+\,\frac{x^8}{4!} \,-\,\cdots\)

The ratio is: \(\displaystyle \L\:R \:=\:\frac{x^{2n+2}}{(n+1)!}\,\cdot\,\frac{n!}{x^{2n}} \:=\:\frac{x^2}{n\,+\,1}\)

And: \(\displaystyle \L\:\lim_{n\to\infty}\frac{x^2}{n\,+\,1} \:=\:0\) . . . for all values of \(\displaystyle x.\)

Therefore, the interval of convergence is: \(\displaystyle \,(-\infty,\,\infty)\)

Also, can anyone explain to me what this means?

b. Write out several terms of the series and verify that \(\displaystyle f'(x)\:=\:-2x\cdot f(x)\)
[for all \(\displaystyle x\) in the interior of the interval of convergence.

The first few terms are: \(\displaystyle \L\: f(x)\;=\; 1 \,-\, x^2\,+\,\frac{x^4}{2!}\,-\,\frac{x^6}{3!}\,+\,\frac{x^8}{4!} \,-\,\frac{x^{10}}{5!}\,+\,\cdots\)

The derivative is: \(\displaystyle \L\:f'(x)\;=\;-2x\,+\,\frac{4x^3}{2!}\,-\,\frac{6x^5}{3!}\,+\,\frac{8x^7}{4!}\,-\,\frac{10x^9}{5!}\,+\,\cdots\)

Factor: \(\displaystyle \L\:f'(x)\;=\;-2x\left(1\,-\,\frac{2x^2}{2!}\,+\,\frac{3x^4}{3!}\,-\,\frac{4x^6}{4!}\,+\,\frac{5x^8}{5!}\,-\,\cdots\right)\)

Reduce: \(\displaystyle \L\:f'(x) \;= \;-2x\underbrace{\left(1\,-\,\frac{x^2}{2!}\,+\,\frac{x^4}{2!}\,-\,\frac{x^6}{3!}\,+\,\frac{x^8}{4!}\,-\,\cdots\right)}_{\text{This is }f(x)}\)

Therefore: \(\displaystyle \L\: f'(x)\;=\;-2\cdot f(x)\)

 

[galactus]

For the third portion, solve the DE.

\(\displaystyle \L\\y'=-2xy, \;\ y(0)=1\)

Separate variables:

\(\displaystyle \L\\\frac{y}{y'}=-2x\)

Integrate:

\(\displaystyle \L\\ln(y)=-x^{2}+C\)

\(\displaystyle \L\\y=C_{1}e^{-x^{2}}\)

Now, use the initial condition to find C1.

Then find the power series expansion. Does it look similar to your given f(x)?.

 

[Ashlander]

Oh man, you guys are so awesome!! Thanks for all the help