Edamame
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Can you pls help me w/the following question?

Taylor Series for ln(x^2) ??
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nota bene
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Do you know the Taylor/Maclaurin series for ln(1+x) ?

edit: To find the series of ln(1-u) consider the geometric series 1/(1-u) and write that as a power series. Then, letting u=x^2+1 you'll have your desired expansion.
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Edamame
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(Original post by nota bene)
Do you know the Taylor/Maclaurin series for ln(1+x) ?

edit: To find the series of ln(1-u) consider the geometric series 1/(1-u) and write that as a power series. Then, letting u=x^2+1 you'll have your desired expansion.
do you mean u = x^2 - 1

I am confused.

Well ln(1 + y) = y - y^2/2 + y^3/3 - ....

If I let y = x^2 - 1, then

log (x^2) = x^2 - 1 - ... very complicated expansions... can somebody help - am I doing something wrong??
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nota bene
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\ln(1-u)= - \displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n} - right? (Integrate -1/(1-u) term by term)

now, use the substitution I said, and it should give the result?
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Edamame
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(Original post by nota bene)
\ln(1-u)= - \displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n} - right? (Integrate -1/(1-u) term by term)

now, use the substitution I said, and it should give the result?
Um I want ln(x^2)

what you suggest is u = x^2 + 1 substitution in ln(1 - u)

then I'd get ln(1 - x^2 + 1) = ln(2 - x^2) - not what I wanted!!
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nota bene
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Keep track of your brackets, it's ln(1-(x^2+1))=ln(x^2)...

edit: Although you'll need the substitution to be u=-x^2+1 sorry about that, but it's an easy fix.
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Edamame
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um why do you have to use ln(1 - u)

can I use ln(1 + u) and then use the substitution of u = x^2 - 1
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nota bene
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Yeah, that's the same thing.
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Edamame
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(Original post by nota bene)
Yeah, that's the same thing.
But the thing is - it doesn't really work. Try it out.

Unless, of course I don't expand the brackets.
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student271828183
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(Original post by Edamame)
Can you pls help me w/the following question?

Taylor Series for ln(x^2) ??
Edamame, when you talk about Taylor series of a function, it is necessary to specify the point at which the function is expanding about.

The Taylor series of ln(x) about x=1 is given by:
ln(x)=\sum_{n=1}^{\infty}{\frac{  (-1)^{n+1}}{n}(x-1)^n} for \mid{x-1}\mid< 1

So by substitute x=u^2, we have:
ln(u^2)=\sum_{n=1}^{\infty}{\fra  c{(-1)^{n+1}}{n}(u^2-1)^n} for \mid{u^2-1}\mid< 1
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Edamame
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I think expanding about the origin...

Anyways how about this:

given the expansions of sinx and cosx about the origin,

what is the expansion of (sinx)^2 and sinxcosx. I tried mulitplying the expansions, but the expressions are so complicated. So many factorials!!
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Zhen Lin
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You can cheat on both expansions by using the double angle formulae.
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unizon
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(Original post by Edamame)
I think expanding about the origin...

Anyways how about this:

given the expansions of sinx and cosx about the origin,

what is the expansion of (sinx)^2 and sinxcosx. I tried mulitplying the expansions, but the expressions are so complicated. So many factorials!!
In order to find such expressions, you need to be familiar with the product of power series, which is beyond the scope of A-level Maths.

sin(x)=\sum_{n=0}^{\infty}{\frac  {(-1)^{n}x^{2n+1}}{(2n+1)!}} for all real x.


sin^2(x)=( \sum_{n=0}^{\infty}{\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}})( \sum_{n=0}^{\infty}{\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}})
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
=\sum_{n=0}^{\infty}\{\sum_{j=0} ^{n}{\frac{(-1)^{j}(-1)^{n-j}}{(2j+1)!(2n-2j+1)!}\}x^{2n+1}


Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
=\sum_{n=0}^{\infty}\{\sum_{j=0} ^{n}{\frac{(-1)^n\dbinom{2n+2}{2j+1}}{(2n+2)! }\}x^{2n+1}


=\sum_{n=0}^{\infty}{\frac{(-1)^{n}2^{2n+1}x^{2n+1}}{(2n+2)!}  }


sin(x)cos(x)=\frac{sin(2x)}{2}
=\sum_{n=0}^{\infty}{\frac{(-1)^{n}(2x)^{2n+1}}{2(2n+1)!}}
=\sum_{n=0}^{\infty}{\frac{(-1)^{n}4^{n}x^{2n+1}}{(2n+1)!}}

(ps. you may need to check my answer as I am not absolutely sure, I hope you can get the idea)
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Zhen Lin
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Well, you could have just used the fact that (\sin x)^2 = \frac{1}{2}\left( 1 - \cos 2x \right)...
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unizon
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(Original post by Zhen Lin)
Well, you could have just used the fact that (\sin x)^2 = \frac{1}{2}\left( 1 - \cos 2x \right)...
Absolutely, this indeed saves a lot of works. I did the multiplication of power series just because Edamame has mentioned about it.

(Original post by Edamame)
what is the expansion of (sinx)^2 and sinxcosx. I tried mulitplying the expansions, but the expressions are so complicated. So many factorials!!
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