Generalisation of terms in a series

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Homework Help Overview

The discussion revolves around the generalization of terms in a series, specifically in the context of the Maclaurin series for the sine function. Participants are exploring the relationship between the derivatives of a function and the terms in the series expansion.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the interpretation of the variable r in the context of derivatives and series terms. There is confusion regarding the matching of derivative orders to the corresponding terms in the series.

Discussion Status

Some participants have provided clarifications regarding the distinction between the variable r in the general formula and its use in the specific case of the sine function. The discussion is actively addressing misunderstandings about the series structure and the nature of even and odd terms.

Contextual Notes

There is an emphasis on the fact that even terms in the series expansion for sine are zero, which may contribute to the confusion about the variable r and its implications in the series generalization.

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?
Screenshot 2021-08-04 at 11.34.27.png


Screenshot 2021-08-04 at 11.45.13.png
 
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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}##
 
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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##.
 
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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.
 

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