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I want to show the "simple" relation:

$$\sin' x=\cos x$$

by using power series. I know that:

$$\sin x=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{(2n+1)!}$$

$$\cos x=\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}$$

Differentiating $\sin x$:

$$\sum_{n=1}^{\infty}(-1)^n(2n+1) \frac{x^{2n}}{(2n+1)!}=\sum_{n=1}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!}$$

But now my sum starts at $n=1$ while the power series for $\cos x$ starts at $n=0$. Am I not missing the constant term "$1$" in my $\sin' x \space \space $power series?

Ken
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qmd
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    The sum in the sine starts at n=0, why the sum in the derivative of the sine starts at n=1? – Did Jun 02 '15 at 15:21
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    I saw in a book that when you differentiate the sum starts at $1$ and not at $0$ but I guess that was because there was a constant term that became $0$ when differentiated which is not the case here. – qmd Jun 02 '15 at 15:28

2 Answers2

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You've just missed that the $n=0$ term of the sine series is $x$, so it doesn't disappear when you differentiate it.

Chappers
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loock at the firsts terms of the $\sin$ series: $$ \sin x= x-\dfrac{x^3}{3!}+\cdots $$ whan you derive you have:

$$ (\sin x)'= 1-\dfrac{x^2}{2!}+\cdots $$

so the series have allways a first term for $n=0$.

Emilio Novati
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  • I misunderstood something in my textbook. The author started the sum at $n=1$ after differentiating but as I now realize and as the other answers have pointed out, that only works when the first term is constant and disappears after differentiating. – qmd Jun 02 '15 at 15:30