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

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SUMMARY

The discussion focuses on deriving the Taylor series for the function ln[(x - h²) / (x + h²)] using Taylor's Theorem. Participants emphasize the necessity of knowing the center of expansion, denoted as 'a', to properly formulate the series. The standard Taylor series formula is referenced, highlighting the relationship between the function's derivatives and the expansion terms. The conversation concludes with a suggestion to clarify the problem statement with the professor for further guidance.

PREREQUISITES
  • Understanding of Taylor's Theorem and its applications
  • Familiarity with logarithmic functions and their properties
  • Knowledge of series expansions and convergence
  • Basic calculus concepts, including derivatives and limits
NEXT STEPS
  • Study the derivation of Taylor series for various functions
  • Learn about the significance of the center of expansion in Taylor series
  • Explore the error term in Taylor's Theorem and its implications
  • Review examples of logarithmic series expansions for better understanding
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Students studying calculus, mathematicians interested in series expansions, and educators seeking to clarify Taylor's Theorem applications.

Alec11
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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]
 
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Alec11 said:

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]
 
Mark44 said:
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.
 
Alec11 said:
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.
 
Mark44 said:
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.
 
Alec11 said:
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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