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I finally figured out that you could differentiate $x^n$ and get $nx^{n-1}$ using the derivative quotient, but that required doing binomial expansion for non-integer values.

The most I can find with binomial expansion is the first, second, last, and second to last terms.

So how do I find something like $(x+a)^{\pi}$? When differentiating in calculus, I didn't need to find terms after the second because I knew they would all cancel out, but how do you find these terms?

Do they work for negative exponents as well?

And does this work for complex exponents?

Which came first, Euler's method for complex exponents or binomial expansion for complex exponents?

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  • $\begingroup$ How did you define $\pi$? $\endgroup$ – flawr Dec 21 '15 at 14:54
  • $\begingroup$ @flawr what do you mean? I can tell you the second term is $a\pi x^{\pi-1}$. $\endgroup$ – Simply Beautiful Art Dec 21 '15 at 14:56
  • $\begingroup$ supply x,n,a and the expression has an answer $\endgroup$ – JMP Dec 21 '15 at 14:59
  • $\begingroup$ @JonMarkPerry No, but could you give me full expansion formula for $(x+a)^{\pi}$? And some arguments about negative/complex exponentials. $\endgroup$ – Simply Beautiful Art Dec 21 '15 at 15:11
  • $\begingroup$ you need to define 'choose 3.14 objects from 76+3i objects' and it gets technical $\endgroup$ – JMP Dec 21 '15 at 15:13
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The Binomial theorem for any index $n\in\mathbb{R}$ with $|x|<1,$ is

$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\ldots$

For $(x+a)^\pi$ one could take $x$ or $a$ common according as if $|a|<|x|$ or $|a|<|x|$ and use Binomial theorem for any index. i.e., $x^\pi(1+a/x)^\pi$ in case $|a|<|x|.$

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    $\begingroup$ I'm confused. How does the whole $n(n-1)(n-2)...$ work for non-integer $n$? $\endgroup$ – Simply Beautiful Art Dec 21 '15 at 15:20
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    $\begingroup$ For eg. $(1+x)^{-1}=1-x+x^2-x^3+\ldots$ if $|x|<1.$ i.e., the series in the right hand sides converges to $(1+x)^{-1}$ if $|x|<1.$ $\endgroup$ – Suhail Dec 21 '15 at 15:24
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    $\begingroup$ No, but how does non-integer $n$ work for the numerator of the n-th term? I know that series, but I can't define $n(n-1)(n-2)\cdots(n-\pi)$. $\endgroup$ – Simply Beautiful Art Dec 21 '15 at 15:45
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    $\begingroup$ @SimpleArt: You just substitute. So if $n$ is equal to pi, then $\pi(\pi-1)(\pi-2)\dots$. $\endgroup$ – Tito Piezas III Dec 21 '15 at 15:50
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    $\begingroup$ Oh... my bad. I'm terribly sorry. @TitoPiezasIII thanks. $\endgroup$ – Simply Beautiful Art Dec 21 '15 at 15:52

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