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shouldn't the n=1 be n=2 at the highlighted part? as it was differentiated twice?

enter image description here

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  • $\begingroup$ There's no differentiation going on between those two expressions. $\endgroup$ Nov 23 '16 at 23:15
  • $\begingroup$ They simplified the expression by canceling factorial terms. Look after the $=$ sign that a factor $(2n)$ has disappeared from the numerator. After this , see Pat's answer $\endgroup$
    – imranfat
    Nov 23 '16 at 23:16
  • $\begingroup$ Your question is why does it not go to $n=2$. The answer is that a term only disappears when it is $x^0$. The expression for $f(x)$ has such a term (when $n=0$), but the expression for $f'(x)$ does not, since the smallest term (when $n=1$) has an $x^1$. $\endgroup$
    – Dr Xorile
    Nov 23 '16 at 23:19
  • $\begingroup$ @ Dr Xorile In another word, the change of the index all depends on if any of the terms have been cancelled? $\endgroup$
    – DSL
    Nov 23 '16 at 23:48
  • $\begingroup$ @ Dr Xorile In another word, the change of the index all depends on if any of the terms have been cancelled? for example, if I were to find $f(x)'''$, then another first turn would drop, because it is 1. right? so if it is $f(x)^{n} $and n is even, then there is no need to change n, if it is odd,n would need to change? $\endgroup$
    – DSL
    Nov 24 '16 at 0:08
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They didn't do anything with that summation. All they did was cancel out the inter-term. In the next part of the equation, that's when they changed the indices for the summation.

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Normally, when we differentiate a power series, we lose a term off the front because it's constant. But in this case, the powers are all even. Since there is no $x^1$ term in $f(x)$, there is no constant term in $f'(x)$. We're counting by $2$'s here, so only every other differentiation erases a term.

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  • $\begingroup$ @ B Goddard In another word, the change of the index all depends on if any of the terms have been cancelled? for example, if I were to find $f(x)'''$, then another first turn would drop, because it is 1. right? so if it is $f(x)^{n} $and n is even, then there is no need to change n, if it is odd,n would need to change? $\endgroup$
    – DSL
    Nov 24 '16 at 0:08
  • $\begingroup$ @Tmm Yes. Because the index is really $2n$. $\endgroup$
    – B. Goddard
    Nov 24 '16 at 3:29
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No this is correct. The $(2n)(2n-1)$ in the numerator simplifies the $(2n)!$ in the denominator, and the exponent is not changed.

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${(2n)(2n-1)\over (2n)!}={1\over (2(n-1))!}$ . . .

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