Alternating partial sums of a series

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    Partial Series Sums
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Discussion Overview

The discussion centers on the properties of the Taylor series expansion of the function ##e^{-x}##, specifically regarding the behavior of its alternating partial sums for ##x > 0##. Participants are exploring how these partial sums bound the function from above and below and are seeking a comprehensive proof that encompasses all cases of these inequalities.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the partial sums of the Taylor series for ##e^{-x}## alternate in bounding the function, suggesting that ##1 > e^{-x}## and ##1 - x < e^{-x}##.
  • Others express a need for a proof that addresses all cases of these inequalities simultaneously, indicating a desire for a more general approach.
  • A participant notes that the inequalities imply ##1 - e^{-x} > 0## and ##1 - x - e^{-x} < 0##, which suggests a relationship between the partial sums and the function.
  • There is a discussion about the complexity of proving two separate cases for even and odd partial sums, indicating uncertainty about the best method to approach the proof.
  • One participant describes the behavior of alternating series, noting that partial sums ending in a plus sign are greater than those ending in a minus sign, which supports the bounding behavior of the series.
  • A later reply questions whether Taylor's theorem could be utilized to formulate a proof, indicating an exploration of different mathematical tools for the problem.

Areas of Agreement / Disagreement

Participants generally agree on the bounding nature of the partial sums but express differing views on how to prove this behavior comprehensively. The discussion remains unresolved regarding the best approach to the proof and whether to treat even and odd cases separately.

Contextual Notes

Participants acknowledge the complexity of the proof and the need for clarity in defining the cases being considered. There is also a mention of a qualitative argument that supports the bounding behavior, but no consensus on a formal proof method has been reached.

spaghetti3451
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Consider the Taylor series expansion of ##e^{-x}## as follows:

##\displaystyle{e^{-x}=1-x+\frac{x^{2}}{2}-\frac{x^{3}}{6}+\dots}##

For ##x>0##, the partial sums ##1##, ##1-x##, ##\displaystyle{1-x+\frac{x^{2}}{2}}## bound ##e^{-x}## from above and from below alternately.

How do I prove this?
 
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failexam said:
Consider the Taylor series expansion of ##e^{-x}## as follows:

##\displaystyle{e^{-x}=1-x+\frac{x^{2}}{2}-\frac{x^{3}}{6}+\dots}##

For ##x>0##, the partial sums ##1##, ##1-x##, ##\displaystyle{1-x+\frac{x^{2}}{2}}## bound ##e^{-x}## from above and from below alternately.

How do I prove this?
Be clear what you are trying to prove: that
1 > e-x, and
1-x < e-x, etc.
 
Yes.

I am looking for a proof that covers all the cases ##1>e^{-x}, 1-x<e^{-x}, \dots## at once.
 
failexam said:
Yes.

I am looking for a proof that covers all the cases ##1>e^{-x}, 1-x<e^{-x}. \dots## at once.
Ok, so what does it say about the sum of the rest of the terms in the expansion in each case?
 
Well, it say that ##1-e^{-x}>0, 1-x-e^{-x}<0, \dots##.
 
failexam said:
I am looking for a proof that covers all the cases 1>e−x,1−x<e−x,…1>e−x,1−x<e−x,…1>e^{-x}, 1-x
Ok, that edit happened while I was replying.
That was not clear from the OP. It looked like you were only asking for those three specific partial expansions.
Anyway, my hint in post #4 still applies.
 
But, it's difficult to start the general proof since there are two separate cases.

Are we supposed to prove two inequalities, one for even partial sums and one for odd partial sums?
 
this is a typical alternating series. each term is smaller than the previous one and the signs alternate. It follows that any partial sum ending in a plus sign is larger than any partial sum ending in a minus sign, hence also larger than the full sum, etc...I.e. the partial sums are jumping back and forth across the full sum, or final limit. draw a picture, as this is essentially obvious.

i.e. start from home and go a block north. then go back a 1/2 block south. you are obviously somewhere between home and the first block. then go back 1/4 block north. you are obviously somewhere between the 1/2 block point and the one block point...

in answer to your question, yes there do seem to be two cases.the best discussion i know of for series including alternating ones is in courant's calculus.

here is a free copy:

https://archive.org/details/DifferentialIntegralCalculusVolI
 
The qualititative argument works well for me.

But, just for further info, do you think Taylor's theorem can be used to write a proof?
 

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