anonymous
  • anonymous
Power Series Question Picture on inside
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
My process was
anonymous
  • anonymous
\[\frac{ 1 }{ 1-4x }= \sum_{0}^{\infty} (4x)^n \]

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anonymous
  • anonymous
I then differentiated both sides
anonymous
  • anonymous
\[\frac{ 4 }{ (1-4x)^2 }= \sum_{n=1}^{\infty} n(4x)^{n-1} \]
anonymous
  • anonymous
\[\frac{ 8 }{ (1-4x)^2 }= 2 \sum_{n=1}^{\infty} n(4x)^{n-1}\]
anonymous
  • anonymous
but this isn't giving me the proper coefficients did i make a mistake somewhere?
amistre64
  • amistre64
what is your derivative of 8(1-4x)^(-2) ?
anonymous
  • anonymous
Did i need to take the derivative of that?
amistre64
  • 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 ....
amistre64
  • 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
amistre64
  • amistre64
\[c_n=\frac{f^{(n)}(a)}{n!}\]
anonymous
  • anonymous
Oh.. I thought that was only for Taylor series. I'll try it out thank you.
amistre64
  • amistre64
taylor series are power series ...
amistre64
  • amistre64
sometimes a power series can be generated by simply working the division ...
amistre64
  • 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)
amistre64
  • amistre64
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anonymous
  • anonymous
Thank you for all your help I understand where I went wrong now and I got the answer.
amistre64
  • amistre64
good :)

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