A Why Question:Taylor Polynomial of e^x over x?

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    E^x Polynomial
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Discussion Overview

The discussion revolves around the use of Taylor and Laurent series for approximating the function \( \frac{e^x}{x} \) at \( x = 0 \). Participants explore the conditions under which these approximations can be applied, particularly focusing on the differences between Taylor and Laurent series and their applicability to various functions.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions why the Taylor polynomial for \( e^x \) can be used in the approximation of \( \frac{e^x}{x} \) at \( x = 0 \) and whether similar methods apply to \( \frac{e^x}{x^2} \) or \( \frac{e^x}{x^3} \).
  • Another participant asserts that a Taylor polynomial approximation cannot be used for \( \frac{e^x}{x} \) at \( x = 0 \) due to the function having a pole at that point, suggesting that a Laurent series should be used instead.
  • A participant provides a form of the Laurent series for \( \frac{e^x}{x} \) and states that the same approach applies to \( \frac{e^x}{x^2} \) or \( \frac{e^x}{x^3} \), but emphasizes that this should not be called a Taylor approximation.
  • Another participant references their textbook, which uses the Taylor polynomial \( T_5(x) \) of \( e^x \) in an integral approximation, questioning if this is equivalent to a Laurent series.
  • It is confirmed that while the textbook is correct in using the Taylor series for \( e^x \), it is incorrect to refer to \( \frac{T_5(x)}{x} \) as the Taylor series of \( \frac{e^x}{x} \), reiterating the need for a Laurent series.
  • Participants discuss the possibility of applying similar reasoning to other functions, such as \( \frac{x^2+x}{x} \), and confirm that it is valid to use the Taylor polynomial of the numerator.

Areas of Agreement / Disagreement

Participants generally agree on the distinction between Taylor and Laurent series, particularly regarding their applicability to functions with poles. However, there is some uncertainty about the implications of using Taylor series in specific contexts, such as in the textbook example.

Contextual Notes

Participants note that the Taylor series requires the function to be continuous and differentiable at the point of expansion, while Laurent series can accommodate poles. The discussion does not resolve the nuances of when to apply each series type in various contexts.

RadiantL
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So I was just wondering why when you approximate using the Taylor Polynomials for something like e^x/x at x = 0 you can just find the approximation for e^x and make it all over x, could you do the same for like e^x/x^2 or e^x/x^3?

I hope my question makes sense... thanks
 
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You can't find a Taylor polynomial approximation of [itex]\frac{e^x}{x}[/itex] at x=0. The function has a pole at 0. Finding a Taylor approximation at 0, requires the function to exist at that point (and be continuous, differentiable, etc. there).

However, you can make a Laurent series approximation. A Laurent series allows terms like [itex]\frac{1}{x}[/itex],[itex]\frac{1}{x^2}[/itex],... in its expansion.

The Laurent series is unique, so if you found one expression of your function in a Laurent series, then you found it. In our case, we can indeed do

[tex]\frac{e^x}{x}=\frac{1}{x}+1+\frac{x}{2}+...+\frac{x^n}{(n+1)!}+...[/tex]

The same thing will work for [itex]\frac{e^x}{x^2}[/itex] or [itex]\frac{e^x}{x^3}[/itex]. But don't call this a Taylor approximation!
 
Oh man, that's funky... my book under integration using the Taylor Polynomials it gives an example

∫e^x/x dx ≈∫ T5(x)/x dx

is that the same thing as the Laurent series?
 
Your book is correct. The [itex]T_5(x)[/itex] is indeed the Taylor series of [itex]e^x[/itex] (because [itex]e^x[/itex] exists at x=0 and is smooth there).

However, it would be incorrect to say that [itex]\frac{T_5(x)}{x}[/itex] is the Taylor series of [itex]\frac{e^x}{x}[/itex]. Here you have to use Laurent series.
 
Ah, so I'm wrong um so what my book did was just take an approximation for a part of that function? Specifically e^x
 
RadiantL said:
Ah, so I'm wrong um so what my book did was just take an approximation for a part of that function? Specifically e^x

Yes, that's exactly what your book did!
 
Intresting, so can I do that for everything else? for example...

x^2+x/x could I just take the Taylor polynomial of x^2+x and then make it over x?
 
RadiantL said:
Intresting, so can I do that for everything else? for example...

x^2+x/x could I just take the Taylor polynomial of x^2+x and then make it over x?

Yes. But the Taylor polynomial of [itex]x^2+x[/itex] will just be [itex]x^2+x[/itex]... (if you take the degree of the polynomial >1).
 
Haha of course, was trying to think up a random example :P Anyway, wow thanks so much your help was very much appreciated. I feel like you should be getting paid for this
 

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