## anonymous one year ago Power Series Question Picture on inside

1. anonymous

2. anonymous

My process was

3. anonymous

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

4. anonymous

I then differentiated both sides

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. amistre64

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

9. anonymous

Did i need to take the derivative of that?

10. amistre64

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

11. amistre64

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

12. amistre64

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

13. anonymous

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

14. amistre64

taylor series are power series ...

15. amistre64

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

16. amistre64

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)

17. amistre64

18. anonymous

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

19. amistre64

good :)