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

[RadiantL]

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

 

[micromass]

You can't find a Taylor polynomial approximation of \frac{e^x}{x} 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 \frac{1}{x},\frac{1}{x^2},... 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

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

The same thing will work for \frac{e^x}{x^2} or \frac{e^x}{x^3}. But don't call this a Taylor approximation!

 

[RadiantL]

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?

 

[micromass]

Your book is correct. The T_5(x) is indeed the Taylor series of e^x (because e^x exists at x=0 and is smooth there).

However, it would be incorrect to say that \frac{T_5(x)}{x} is the Taylor series of \frac{e^x}{x}. Here you have to use Laurent series.

 

[RadiantL]

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

 

[micromass]

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!

 

[RadiantL]

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?

 

[micromass]

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 x^2+x will just be x^2+x... (if you take the degree of the polynomial >1).

 

[RadiantL]

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