anonymous
  • anonymous
Unsure.... \[f(x)=x^2tan^{-1}(x^3)\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I'll post the remainder of the question in just a second...bear with me
anonymous
  • anonymous
From the MacLaurin Table: \[tan^{-1}(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\]
anonymous
  • anonymous
This is what I've come up with so far: \[x^2\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+1}}{2n+1}\]

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anonymous
  • anonymous
Hmmmm....could it be: \[x^2\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}\]?
anonymous
  • anonymous
I meant: \[\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}\]
KingGeorge
  • KingGeorge
I think both the first and third solutions you have are correct.
anonymous
  • anonymous
I am inclined to agree
anonymous
  • anonymous
Yes! So this would be correct? \[\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}\]
KingGeorge
  • KingGeorge
Looks good to me. If you wanted to, you could also say\[(x^3)^{2n+3}=x^{6n+9}\]
anonymous
  • anonymous
ok Thanks!
KingGeorge
  • KingGeorge
You're welcome.

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