JasonWhat is the Taylor series for ln(x+2) about x = 0?

[zzmanzz]

Homework Statement

Using power series, expand ln(x + 2) about a = 0 (Taylor series)

Homework Equations

The Attempt at a Solution

Is this appropriate?

ln(x+2) = ln((x+1)+1)

x' = x+1

ln(x'+1) = \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x')^{n+1}

or

ln(x+2) = ln(\frac{x}{2}+1)

x' = \frac{1}{2}x

ln(x'+1) = \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x')^{n+1}

 

[vanhees71]

Well, you rather have
\ln(2+x)=\ln[2(1+x/2)].
You can, however still use your trick with this correction made!

 

[zzmanzz]

So would your formula then be:

ln2 + ln(1+(x/2)) = ln2 + \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x/2)^{n+1}

or

ln(2(x/2+1)) = 2*\sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x/2)^{n+1}

 

[vanhees71]

This you should be able to answer yourself by giving the reasoning behind each of the two claims you made. Of course only one of them can be right!

 

[zzmanzz]

I believe that the first one is right because

ln(AB) = lnA + lnB

A = 2, B = (1+x/2)

 

[Chestermiller]

I think you want to do the expansion around x = -1, so that a = 0.

So, the first few terms are: ln(x+2) = 0 + (x-1)f'(-1)+...=(x-1)+...

 

[vanhees71]

It was asked to do the expansion around 0, and of course the first equation is correct, i.e.,
\ln(x+2)=\ln 2 + \ln(1+(x/2)) = \ln 2 + \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}(x/2)^{n+1}.
Finally you should think about the domain, where the series is convergent!

 

[Ray Vickson]

Homework Statement

Using power series, expand ln(x + 2) about a = 0 (Taylor series)

Homework Equations

The Attempt at a Solution

Is this appropriate?

ln(x+2) = ln((x+1)+1)

x' = x+1

ln(x'+1) = \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x')^{n+1}

or

ln(x+2) = ln(\frac{x}{2}+1)

x' = \frac{1}{2}x

ln(x'+1) = \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x')^{n+1}

Why not just use the Maclaurin series formula
f(x) = f(0) + f'(0)\,x + f''(0)\, \frac{x^2}{2!} + \cdots \:\:?
It is really very simple to use in this case.

 

[zzmanzz]

Yes, that's what I had in mind. For the x terms in your expansion. Can I do:
f(x) = f(0) + f'(0)(\frac{x}{2}) + f''(0)\frac{1}{2!}(\frac{x}{2})^2 + ...

 

[Ray Vickson]

Yes, that's what I had in mind. For the x terms in your expansion. Can I do:
f(x) = f(0) + f'(0)(\frac{x}{2}) + f''(0)\frac{1}{2!}(\frac{x}{2})^2 + ...

Your could write that, but you would be WRONG. The Correct form of the expansion is exactly as I wrote it. Your series above is for f(x/2), not for f(x).

 

[zzmanzz]

ok I think I get it now.

I use your mclaurin series to expand the formula..

Then based on the series outcome, we build the sum. So the following sum is still correct:

ln2 + \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x/2)^{n+1}

 

[Chestermiller]

It was asked to do the expansion around 0

Is that what it means when it says "expand ln(x + 2) about a = 0?" I'm not familiar with this type of terminology. What the heck is "a"?

Chet

 

[Ray Vickson]

Is that what it means when it says "expand ln(x + 2) about a = 0?" I'm not familiar with this type of terminology. What the heck is "a"?

Chet

My guess would be that it was a typo, and should have said "...about x = 0".