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

1. Dec 13, 2011

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

2. Dec 13, 2011

### 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!!

3. Dec 13, 2011

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?

4. Dec 13, 2011

### 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.

5. Dec 13, 2011

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

6. Dec 13, 2011

### micromass

Yes, that's exactly what your book did!

7. Dec 13, 2011

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?

8. Dec 13, 2011

### micromass

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

9. Dec 13, 2011