Power Series Question Picture on inside

Question

anonymous · Answer

attachment

anonymous · Answer

My process was

anonymous · Answer

$$\frac{ 1 }{ 1-4x }= \sum_{0}^{\infty} (4x)^n $$

anonymous · Answer

I then differentiated both sides

anonymous · Answer

$$\frac{ 4 }{ (1-4x)^2 }= \sum_{n=1}^{\infty} n(4x)^{n-1} $$

anonymous · Answer

$$\frac{ 8 }{ (1-4x)^2 }= 2 \sum_{n=1}^{\infty} n(4x)^{n-1}$$

anonymous · Answer

but this isn't giving me the proper coefficients did i make a mistake somewhere?

amistre64 · Answer

what is your derivative of 8(1-4x)^(-2) ?

anonymous · Answer

Did i need to take the derivative of that?

amistre64 · Answer

$$f(x)=0!c_0+\sum_{1}c_nx^n$$
$$f'(x)=1!c_1+\sum_{2}c_nx^n$$
$$f''(x)=2!c_2+\sum_{3}c_nx^n$$
$$f'''(x)=3!c_3+\sum_{4}c_nx^n$$
etc ....

amistre64 · Answer

at x=0, or whatever you center it as,  the summation part goes to 0, and we are left with a solution for the coeffs

amistre64 · Answer

$$c_n=\frac{f^{(n)}(a)}{n!}$$

anonymous · Answer

Oh.. I thought that was only for Taylor series. I'll try it out thank you.

amistre64 · Answer

taylor series are power series ...

amistre64 · Answer

sometimes a power series can be generated by simply working the division ...

amistre64 · Answer

spose we integrate your function up to 2/(1-4x)


            2+2^3x+2^5x^3
           -------------
1-4x   | 2
            (2-2^3x)
                 8x
                (8x-2^5x^2)
                       32x^2
                       (32x^2-2^7x^3)

F(x) = sum 2^(2n+1) x^n

take the derivative

f(x) = sum n 2^(2n+1) x^(n-1)

amistre64 · Answer

attachment

anonymous · Answer

Thank you for all your help I understand where I went wrong now and I got the answer.

amistre64 · Answer

good :)