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anonymous

  • one year ago

Power Series Question Picture on inside

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    My process was

  3. anonymous
    • one year ago
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    \[\frac{ 1 }{ 1-4x }= \sum_{0}^{\infty} (4x)^n \]

  4. anonymous
    • one year ago
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    I then differentiated both sides

  5. anonymous
    • one year ago
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    \[\frac{ 4 }{ (1-4x)^2 }= \sum_{n=1}^{\infty} n(4x)^{n-1} \]

  6. anonymous
    • one year ago
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    \[\frac{ 8 }{ (1-4x)^2 }= 2 \sum_{n=1}^{\infty} n(4x)^{n-1}\]

  7. anonymous
    • one year ago
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    but this isn't giving me the proper coefficients did i make a mistake somewhere?

  8. amistre64
    • one year ago
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    what is your derivative of 8(1-4x)^(-2) ?

  9. anonymous
    • one year ago
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    Did i need to take the derivative of that?

  10. amistre64
    • one year ago
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    \[f(x)=0!c_0+\sum_{1}c_nx^n\] \[f'(x)=1!c_1+\sum_{2}c_nx^n\] \[f''(x)=2!c_2+\sum_{3}c_nx^n\] \[f'''(x)=3!c_3+\sum_{4}c_nx^n\] etc ....

  11. amistre64
    • one year ago
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    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
    • one year ago
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    \[c_n=\frac{f^{(n)}(a)}{n!}\]

  13. anonymous
    • one year ago
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    Oh.. I thought that was only for Taylor series. I'll try it out thank you.

  14. amistre64
    • one year ago
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    taylor series are power series ...

  15. amistre64
    • one year ago
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    sometimes a power series can be generated by simply working the division ...

  16. amistre64
    • one year ago
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    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
    • one year ago
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  18. anonymous
    • one year ago
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    Thank you for all your help I understand where I went wrong now and I got the answer.

  19. amistre64
    • one year ago
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    good :)

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