Generalisation of terms in a series

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jackiepollock
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Homework Statement
I'm stuck at understanding the generalisation of the terms in a series
Relevant Equations
Maclaurin series
Hello. I'm not sure how the generalisation comes about (where I circle).

I assume that r means the the rth derivative of f(x). If that's the case, as I plug 3 = r into this generalisation, the third derivative term should equal to (-1)^3x^7 /7!, but the third derivative term is -1x^3/3!.

What's the problem?
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##\frac{(-1)^r}{(2r+1)!}x^{2r+1}## is the ##(2r+1)## term of the series, which means that for r=3 you get the ##2r+1=2\cdot 3+1=7## that is the 7th term of the series. The 3rd term of the series is for the ##r'## such that ##2r'+1=3## or ##r'=1##.
In case you wonder what happens to the ##2r## terms of the series, they are all zero because the corresponding derivative at 0 ##f^{(2r)}(0)=0## is equal to zero

Essentially what I am telling you is that the even terms of the series are all zeros and the odd terms are ##\frac{(-1)^r}{(2r+1)!}x^{2r+1}##
 
jackiepollock said:
I assume that r means the the rth derivative of f(x). If that's the case, as I plug 3 = r into this generalisation, the third derivative term should equal to (-1)^3x^7 /7!, but the third derivative term is -1x^3/3!.

What's the problem?
Your assumption is incorrect. The ##r## in the expansion for ##\sin x## and the ##r## in the general formula for the Maclaurin series are different variables. Note in the general formula, the exponent of ##x## is equal to the order of the derivative, so the ##x^7## term is matched to the seventh derivative of ##f##.
 
vela said:
Your assumption is incorrect. The ##r## in the expansion for ##\sin x## and the ##r## in the general formula for the Maclaurin series are different variables. Note in the general formula, the exponent of ##x## is equal to the order of the derivative, so the ##x^7## term is matched to the seventh derivative of ##f##.
Yes I think what confuses him is that the even terms of the series are zero for the specific expansion of ##\sin x## and that they use the same letter r for the generic series formula and for the specific series of sinx.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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