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

• Alec11
In summary, the Taylor series for ln[(x - h2) / (x + h2)] can be written as follows: ln[(x - h2) / (x + h2)] = ln(x-h2) - ln(x + h2) if a = 0, and ln[(x - h2) / (x + h2)] = ln(x-h2) if a = x.
Alec11

## 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]

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...

Last edited:

## 1. What is Taylor's Theorem and how is it used to establish a Taylor series in terms of h?

Taylor's Theorem is a mathematical theorem that states that any infinitely differentiable function can be approximated by a polynomial of degree n, where n is the number of derivatives used in the approximation. To establish a Taylor series in terms of h, we can use the formula: f(h) = f(0) + f'(0)h + (f''(0)h^2)/2! + ... + (f^(n)(0)h^n)/n!, where f^(n)(0) represents the nth derivative of the function evaluated at 0.

## 2. What is the significance of h in Taylor's Theorem and Taylor series?

h represents the variable used to express the distance between the point of approximation and the point where the Taylor series is centered (usually 0). It is important because it allows us to generalize the Taylor series and make it applicable to a wider range of functions.

## 3. How do you determine the accuracy of a Taylor series in terms of h?

The accuracy of a Taylor series in terms of h depends on the remainder term, R_n(h), which is given by the formula: R_n(h) = (f^(n+1)(c)h^(n+1))/(n+1)!, where c is a value between 0 and h. The smaller the value of R_n(h), the more accurate the Taylor series will be.

## 4. Can Taylor's Theorem be applied to non-polynomial functions?

Yes, Taylor's Theorem can be applied to any infinitely differentiable function, regardless of whether it is a polynomial or not. However, the accuracy of the Taylor series approximation may vary depending on the function and the choice of h.

## 5. How is Taylor's Theorem related to the concept of a Maclaurin series?

A Maclaurin series is a special case of a Taylor series, where the series is centered at 0. This means that h=0, and the formula for the Maclaurin series becomes: f(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + ... + (f^(n)(0)x^n)/n!. Therefore, Taylor's Theorem is essential in establishing the Maclaurin series for a function by providing a general formula for the coefficients of the series in terms of derivatives evaluated at 0.

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