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I' just so stumped right now. I want to get $x^{n}$ to equal $x^{2n+1}$. Right now I have that: $$(\sqrt{x})^{2n} = x^n$$ But I don't know what to do to x to get: $$x^n = \{\text{something done to $x$}\}^{2n+1}$$

MRT
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    \begin{eqnarray} \left( x^{\frac{n}{2n+1}} \right)^{2n+1} \end{eqnarray} – Donald Splutterwit Mar 06 '17 at 20:07
  • $\left(x^{\frac n{2n+1}}\right)^{2n+1}=x^n$ and this is pretty much as simple as it gets. –  Mar 06 '17 at 20:10
  • @G is there a way to have x not involving x like $$x^{\frac{n}{2n+1}}$$ – MRT Mar 06 '17 at 20:12
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    @M.Turner To what end? It's that thing, regardless of how you write it. Would, I don't know, $\exp\left(\frac n{2n+1}\ln x\right)$ better suit your taste? –  Mar 06 '17 at 20:21
  • @G no no. See my comment below and you'll see what I'm trying to do – MRT Mar 06 '17 at 20:24
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    You should have written it in the question to begin with, instead of playing the guessing game with strangers. Multiply the series by $\dfrac{\lvert x\rvert^{1/2}}{\lvert x\rvert^{1/2}}$ and set $y=\lvert x\rvert^{1/2}$. Be careful of the sign, though. –  Mar 06 '17 at 20:25
  • @G I'm still not getting it. I want to know what $\sin(\text{something})$ to get the series below? – MRT Mar 06 '17 at 20:37

2 Answers2

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Recall the following exponent law:

$$x^{ab} = (x^a)^b$$

In your case, you have this:

$$x^n = (x^a)^{2n+1}$$

Note that this is just a more mathematical way of stating exactly what you have in your question:

But I don't know what to do to x to get: $$x^n = \{\text{something done to $x$}\}^{2n+1}$$

So for your particular problem, you have $ab = n$ and $b = 2n+1$, and you want to find the value of $a$ that makes all of this work. Well, since you know $ab = n$, then $a = n/b$, and since you know $b = 2n+1$, then we can conclude $a = \dfrac n{2n+1}$.

$$ x^n = \left(x^{\frac n{2n+1}}\right)^{2n+1} $$

  • Damn. I need to express $$\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{(2n+1)!}$$ In terms of sine? and sine has $x^{2n+1}$ not $x^n$ so I'm super stuck – MRT Mar 06 '17 at 20:18
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    @M.Turner, I think you should update your original question to reflect exactly what it is you were asked to do at the very beginning of whatever led you to this requirement. –  Mar 06 '17 at 21:25
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$(x^n)=x^{\frac{n}{2n+1}^{2n+1}}$ $\forall n\in\Bbb R/\{-\frac{1}{2}$}