Taylor series helpWatch

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Thread starter 11 years ago
#1
Can you pls help me w/the following question?

Taylor Series for ln(x^2) ??
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11 years ago
#2
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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Thread starter 11 years ago
#3
(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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11 years ago
#4 - right? (Integrate -1/(1-u) term by term)

now, use the substitution I said, and it should give the result?
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Thread starter 11 years ago
#5
(Original post by nota bene) - 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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11 years ago
#6
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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Thread starter 11 years ago
#7
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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11 years ago
#8
Yeah, that's the same thing.
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Thread starter 11 years ago
#9
(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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11 years ago
#10
(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: for So by substitute , we have: for 0
Thread starter 11 years ago
#11
I think expanding about the origin...

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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11 years ago
#12
You can cheat on both expansions by using the double angle formulae.
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11 years ago
#13
(Original post by Edamame)
I think expanding about the origin...

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. for all real x. 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}    (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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11 years ago
#14
Well, you could have just used the fact that ...
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11 years ago
#15
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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