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idku

  • one year ago

tell me if I am doing this correctly. I haven't touched math for a while now.

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  1. idku
    • one year ago
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    (I am just reviewing some concepts from the past) I want to find the power series representation for \(\displaystyle\large\color{black}{\tan^{-1}(x)}\) now I will show my work....

  2. idku
    • one year ago
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    \(\displaystyle\large\color{black}{\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n}\) \(\displaystyle\large\color{black}{\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n}\) \(\displaystyle\large\color{black}{\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n}}\) \(\displaystyle\large\color{black}{\int \frac{1}{1+x^2}~dx~=~\int~\sum_{n=0}^{\infty}(-1)^nx^{2n}~dx}\) \(\displaystyle\large\color{black}{\tan^{-1}(x)~=~\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2n+1}}\)

  3. idku
    • one year ago
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    and I am not placing +C, because I am looking for specific function - representation and not a family of functions.

  4. idku
    • one year ago
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    oh my power is off

  5. idku
    • one year ago
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    \(\displaystyle\large\color{black}{\tan^{-1}(x)~=~\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}}\)

  6. idku
    • one year ago
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    this should be correct, if I didn't err anywhere once again.

  7. geerky42
    • one year ago
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    What's the problem here? You already represented \(\tan^{-1}(x)\) as series in your other question here.

  8. geerky42
    • one year ago
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    http://openstudy.com/users/idku#/updates/554a6664e4b048e824ff3507

  9. idku
    • one year ago
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    yeah forgot about that..... my apologies.

  10. idku
    • one year ago
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    tnx for the confirmation, and bye:)

  11. geerky42
    • one year ago
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    Ok no problem.

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