trouble understanding this problem

A>0 Assuming a function

f(x)=Sum(0 to infinity) c_{n}x^{n}

is the solution to the differential equation:

f''(x)=af(x)

with the conditions f(0)=1 f'(0)=0 find a formula for the coefficients of f. what is its radius of convergence?

completely lost and do not understand the problem someone point me in the right direction??

Re: trouble understanding this problem

Start by writing down your series for f(x), differentiate it twice for f''(x), substitute them into your DE, compare the coefficients of the terms that have like powers of x.

Re: trouble understanding this problem

Quote:

Originally Posted by

**Prove It** Start by writing down your series for f(x), differentiate it twice for f''(x), substitute them into your DE, compare the coefficients of the terms that have like powers of x.

thanks very much for your reply

but I believe i understood up to the "compare the coefficents..." line

I attempted the second derivative (i believe its right)

f''(x) = SUM_{n to infinity} n(n-1)c_{n}x^{n-2}

Substituting this into the equation

SUM_{n to infinity} n(n-1)c_{n}x^{n-2} = a(SUM_{n to infinity} c_{n}x^{n})

and then i don't understand what you mean by compare the coefficients that have like powers of x

do you mean that bring a into the sum SUM_{n to infinity} ac_{n}x^{n}) in which case a=n(n-1)???

I also am not sure how to deal with the f(0)=1 and f'(0)=0 conditions

in the case of f(x)=SUM_{n to infinity} c_{n}x^{n} if x = 0 then the series = 0 meaning i have to add +1 to equation to make the condition correct

Re: trouble understanding this problem

Quote:

Originally Posted by

**jvang1222** thanks very much for your reply

but I believe i understood up to the "compare the coefficents..." line

I attempted the second derivative (i believe its right)

f''(x) = SUM_{n to infinity} n(n-1)c_{n}x^{n-2}

Substituting this into the equation

SUM_{n to infinity} n(n-1)c_{n}x^{n-2} = a(SUM_{n to infinity} c_{n}x^{n})

and then i don't understand what you mean by compare the coefficients that have like powers of x

do you mean that bring a into the sum SUM_{n to infinity} ac_{n}x^{n}) in which case a=n(n-1)???

I also am not sure how to deal with the f(0)=1 and f'(0)=0 conditions

in the case of f(x)=SUM_{n to infinity} c_{n}x^{n} if x = 0 then the series = 0 meaning i have to add +1 to equation to make the condition correct

f(x) = $c_0+c_1 x + c_2 x^2 \dots$

f''(x) = $2 c_2 + 6c_3 x+12 c_4 x^2 \dots = a f(x) = a\left(c_0+c_1 x + c_2 x^2 \dots\right)$

now compare like powers of $x$ and equate their coefficients

$c_2 = a c_0, 6c_3=a c_1, 12c_4=a c_2$, etc.

two initial conditions will completely specify a 2nd order linear differential equation so plug in your conditions and solve for the resulting coefficients.

Re: trouble understanding this problem

Quote:

Originally Posted by

**romsek** f(x) = $c_0+c_1 x + c_2 x^2 \dots$

f''(x) = $2 c_2 + 6c_3 x+12 c_4 x^2 \dots = a f(x) = a\left(c_0+c_1 x + c_2 x^2 \dots\right)$

now compare like powers of $x$ and equate their coefficients

$c_2 = a c_0, 6c_3=a c_1, 12c_4=a c_2$, etc.

two initial conditions will completely specify a 2nd order linear differential equation so plug in your conditions and solve for the resulting coefficients.

okay you clarified the cofficents idea but now in terms of deriving an equation for f would it be something like

c_{0}=2c_{2}/a and then a general equation of c_{n}=2c_{n+2}/a

the deriving equation line has me confused. am I solving for a?? and then substituing that into the f(x) formula? ex.

a = 2c_n+2 / c_n and then substituting into f(x)=\sum_{0}{infinity} 2c_{n+2}c_{n}/c_{n}x^{n}

and upon figuring this all out i have to find the radius of convergence of the formula?!

Re: trouble understanding this problem

The initial conditions give you $\displaystyle c_0$ and $\displaystyle c_1$, and then you figure out what the rest of the c's are using the recurrence relation ($\displaystyle c_{n+2}$ in terms of $\displaystyle c_n$).

Once you have an expression for $\displaystyle c_n$, you should be able to use the ratio test to get the radius of convergence (it's almost always the ratio test).

- Hollywood