Maclaurin Series for ln(1+e^x) and \frac{x}{(e^x-1)}

[Bucky]

obtain the maclaurin series expansions of the following:
$$ln(1+e^x)$$
ok I am quite lost..i assume you set it equal to f(x) then differentiate..but what happens when you differentiate that?
also question (b) is
$$\frac{x}{(e^x-1)}$$
does that work down to
(e^x - 1)^-x ?

and if so..would you then differentiate the whole of the bracket? but then the chain rule comes into use doesn't it? so it would be -x times e^x?

 

[finchie_88]

I've had a go at the first part of your question, and I think I might have got somewhere with it but I'm not sure, this is what I did...
the maclaurin series for,
$$ln(1+x)=x-(0.5x^2)+(0.333x^3)-(0.25x^4)+...$$
$$ln(1+y) = y-(0.5y^2)+(0.333y^3)-(0.25y^4)+...$$
$$e^x = 1+x+((x^2)/2!)+((x^3)/3!)+...$$
y = e^x
when we combine them, we get something very complicated, but is something like...
Generalised result:
$$\ln(1+e^x) = kx+\frac{\(n^2x^2)(-1)^(n-1)}{2!}+\frac{\(n^3x^3)(-1)^(n-1)}{3!}+...$$

At the moment, I can't work out what k is, but I think I might have the rest of it (emphasis on the word think).
btw, I am crap at this maths symbol stuff, so if it comes out dodgy, make the most of it.

 

[Integral]

obtain the maclaurin series expansions of the following:
$$ln(1+e^x)$$
ok I am quite lost..i assume you set it equal to f(x) then differentiate..but what happens when you differentiate that?

Do it and find out, follow the defintion of a Maclaurin's series. See what you get.

also question (b) is
$$\frac{x}{(e^x-1)}$$
does that work down to
(e^x - 1)^-x ?

and if so..would you then differentiate the whole of the bracket? but then the chain rule comes into use doesn't it? so it would be -x times e^x?

No,
$$\frac{x}{(e^x-1)}= x (e^x-1)^{-1}$$

use the quotient or product rule, should come out the same.

 

[Hurkyl]

obtain the maclaurin series expansions of the following:
$$ln(1+e^x)$$
ok I am quite lost..i assume you set it equal to f(x) then differentiate..but what happens when you differentiate that?

You could say that f(x) = ln(1+e^x) and then try to find f'(x), f''(x), et cetera. That would be an equivalent problem.

How far have you gotten in trying to differentiate it? Do you remember your rules of differentiation?

 

[Bucky]

ok here's how far i got:

f(x) = ln(1+e^x)

f'(x) = (1+e^x)^(-1)

f''(x) = -(1+e^x)^(-2)

f'''(x) = 2(1 + e^x)^(-3)

f^(iv)(x) = -6(1 + e^x)^(-4)

f^(v)(x) = 24(1+e^x)^(-5)

f(0) = ln(2)
f'(0) = 1/2
f''(0) = 4
f'''(0) = 16
f^(iv)(0) = 96
f^(v)(0) = 768but now I am not sure what to do. the notes get a bit hazy at this point

 

[HallsofIvy]

ok here's how far i got:

f(x) = ln(1+e^x)

f'(x) = (1+e^x)^(-1)

No, you haven't used the chain rule: now multiply by the derivative of 1+ e^x which is e^x:
$$f'(x)= \frac{e^x}{1+e^x}$$

f''(x) = -(1+e^x)^(-2)

Since your first derivative was wrong, this is wrong (and it's not even the derivative of (1+e^x)^-1: again you have not multiplied by the derivative of (1+ e^x).
Use the quotient rule: the derivative of $\frac{e^x}{1+e^x}$ is
$$\frac{e^x(1+e^x)- e^{2x}}{(1+e^x)^2}= \frac{e^x}{(1+e^x)^2}$$
You should be able to see a pattern.

f'''(x) = 2(1 + e^x)^(-3)

f^(iv)(x) = -6(1 + e^x)^(-4)

f^(v)(x) = 24(1+e^x)^(-5)

f(0) = ln(2)
f'(0) = 1/2

These are correct

f''(0) = 4
f'''(0) = 16
f^(iv)(0) = 96
f^(v)(0) = 768

These are incorrect

but now I am not sure what to do. the notes get a bit hazy at this point

If your notes have the basic formula:
$$\Sigma_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$
it should not be hazy at all.

 

[Bucky]

ok I am just rubbish at math it seems..
quotient rule is bottom times differentiated top minus top times differentiated bottom. all divided by bottom squared. right?
so
f'''(x) = (1+e^x)^2 e^x - e^x over
2(1+e^x) + e^x

f'''(x) = e^3x-e^x over
r2+3e^x(latex ftw ¬.¬ )
i get the feeling I am deeply wrong

i read on and i see how this works now, its just differentiating e to the power of stuff (which in practical terms i doubt ill need to know but ho hum..) and applying chain/quotient rules that's getting me.