Establish Taylor series using Taylor's Theorem in terms of h

  • #1
7
0

Homework Statement


Find the Taylor series for:

ln[(x - h2) / (x + h2)]

Homework Equations


f(x+h) =∑nk=0 f(k)(x) * hk / k! + En + 1

where En + 1 = f(n + 1)(ξ) * hn + 1 / (n + 1)!

The Attempt at a Solution


ln[(x - h2) / (x + h2)] = ln(x-h2) - ln(x + h2)

This is as far as I have been able to get. I don't understand what to set x and h to, and without that I don't think I can do much else.
[/B]
 

Answers and Replies

  • #2
34,533
6,229

Homework Statement


Find the Taylor series for:

ln[(x - h2) / (x + h2)]
Isn't there any more to this question? Usually, if they want a Taylor's series, they'll give information like "expanded around a = 1" or the like. This would mean that the Taylor's series would look like ##a_0 + a_1(x - 1) + a_2(x - 1)^2 + \dots + a_n(x - 1)^n + \dots##.

If a = 0, the series is a Maclaurin series.
Alec11 said:

Homework Equations


f(x+h) =∑nk=0 f(k)(x) * hk / k! + En + 1

where En + 1 = f(n + 1)(ξ) * hn + 1 / (n + 1)!

The Attempt at a Solution


ln[(x - h2) / (x + h2)] = ln(x-h2) - ln(x + h2)

This is as far as I have been able to get. I don't understand what to set x and h to, and without that I don't think I can do much else.[/B]
 
  • #3
7
0
Isn't there any more to this question? Usually, if they want a Taylor's series, they'll give information like "expanded around a = 1" or the like. This would mean that the Taylor's series would look like ##a_0 + a_1(x - 1) + a_2(x - 1)^2 + \dots + a_n(x - 1)^n + \dots##.

If a = 0, the series is a Maclaurin series.
No, I gave you all of the information that is given. I assume they just want me to write out the first few terms of the series without evaluating it.
 
  • #4
34,533
6,229
No, I gave you all of the information that is given. I assume they just want me to write out the first few terms of the series without evaluating it.
My point is that you can't write the first few terms without knowing what the "center" of the expansion is. IOW, without knowing what a is in all of the x - a terms. My advice is to contact your professor and ask him/her to clarify this question.
 
  • #5
7
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My point is that you can't write the first few terms without knowing what the "center" of the expansion is. IOW, without knowing what a is in all of the x - a terms. My advice is to contact your professor and ask him/her to clarify this question.
It is possible if you're using Taylor's Theorem in terms of h that I specified in the OP.
 
  • #6
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6,229
It is possible if you're using Taylor's Theorem in terms of h that I specified in the OP.
This is a different form than I have seen Taylor's Theorem. The usual way the Taylor Formula is presented is
##f(x) = \sum_{n = 0}^\infty \frac{f^{(n)}(a)}{n!} (x - a)^n##. (There's also another form that has a degree-n polynomial and an error term.) See https://en.wikipedia.org/wiki/Taylor's_theorem

From the above, ##f(x + h) = \sum_{n = 0}^\infty \frac{f^{(n)}(a)}{n!} (x + h - a)^n##.

If you let a = x, the right side becomes ##\sum_{n = 0}^\infty \frac{f^{(n)}(x)}{n!} h^n##, which agrees with your formula.

I need to run right now, but will look in on this later this evening...
 
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