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MathSofiya

  • 3 years ago

Unsure.... \[f(x)=x^2tan^{-1}(x^3)\]

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  1. MathSofiya
    • 3 years ago
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    I'll post the remainder of the question in just a second...bear with me

  2. MathSofiya
    • 3 years ago
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    From the MacLaurin Table: \[tan^{-1}(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\]

  3. MathSofiya
    • 3 years ago
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    This is what I've come up with so far: \[x^2\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+1}}{2n+1}\]

  4. MathSofiya
    • 3 years ago
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    Hmmmm....could it be: \[x^2\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}\]?

  5. MathSofiya
    • 3 years ago
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    I meant: \[\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}\]

  6. KingGeorge
    • 3 years ago
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    I think both the first and third solutions you have are correct.

  7. colorful
    • 3 years ago
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    I am inclined to agree

  8. MathSofiya
    • 3 years ago
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    Yes! So this would be correct? \[\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}\]

  9. KingGeorge
    • 3 years ago
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    Looks good to me. If you wanted to, you could also say\[(x^3)^{2n+3}=x^{6n+9}\]

  10. MathSofiya
    • 3 years ago
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    ok Thanks!

  11. KingGeorge
    • 3 years ago
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    You're welcome.

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