# Taylor/Maclaurin series of a function

• EEristavi
In summary: It's only going to work when one of the functions is ##1/x## or ##1/x^2##; or, if one is a polynomial, then that it already the Taylor series. In general, you have to expand both.Thank you :)
EEristavi

## Homework Statement

Obtain Maclaurin Series for:
f(x) = sin(x2)/x

## Homework Equations

f(x) = ∑f(n)(c) (x-c)n / n! (for Maclaurin c = 0)

## The Attempt at a Solution

I know that sin(x2) = x2 - (x2*3/3! +...

from the final answer I see, that this is just multiplied to 1/x.

This bothers me,
doesn't we have to expand 1/x also? Moreover, I see that every term for P(1/x) at x=0 is undefined => range must be (-∞, 0)∪(0, ∞)

Last edited:
EEristavi said:

## Homework Statement

Obtain Maclaurin Series for:
f(x) = sin(x2)/2

## Homework Equations

f(x) = ∑f(n)(c) (x-c)n / n! (for Maclaurin c = 0)

## The Attempt at a Solution

I know that sin(x2) = x2 - (x2*3/3! +...

from the final answer I see, that this is just multiplied to 1/x.

This bothers me,
doesn't we have to expand 1/x also? Moreover, I see that every term for P(1/x) at x=0 is undefined => range must be (-∞, 0)∪(0, ∞)
I don't see what you are talking about. The series you have given
$$\sum_{n=1}^\infty \frac{(x^2)^{2n-1}}{(2n-1)!}=\sum_{n=1}^\infty \frac{x^{4n-2}}{(2n-1)!}$$ converges absolutely for all ##x## and there are no ##\frac 1 x## terms anywhere.

LCKurtz said:
there are no 1x1x\frac 1 x terms anywhere.

Aaah... sorry about that :( I accidentally divided by 2 - not x. I've edited now.
sorry again

EEristavi said:
from the final answer I see, that this is just multiplied to 1/x.

This bothers me,
doesn't we have to expand 1/x also?

Taylor series are unique. When you multiply by ##1/x## you get a power series. There's no need to generate a power series for ##1/x## to do this.

EEristavi
So, if I understand correctly:
If I have complex function [e.g. f(x)g(x)] I can expand one of them and multiply to another

EEristavi said:
So, if I understand correctly:
If I have complex function [e.g. f(x)g(x)] I can expand one of them and multiply to another

It's only going to work when one of the functions is ##1/x## or ##1/x^2##; or, if one is a polynomial, then that it already the Taylor series. In general, you have to expand both.

EEristavi
Thank you :)

## 1. What is a Taylor/Maclaurin series of a function?

A Taylor/Maclaurin series of a function is a representation of a function as an infinite sum of terms, where each term is a polynomial of increasing degree. It is used to approximate a function at a specific point by using the function's derivatives at that point.

## 2. How is a Taylor/Maclaurin series of a function calculated?

A Taylor/Maclaurin series is calculated by taking the derivatives of the function at a specific point and plugging them into the formula for the series, which is: f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...

## 3. What is the difference between a Taylor series and a Maclaurin series?

A Taylor series is a more general form of the Maclaurin series, where the point of expansion (a) can be any value. A Maclaurin series is a specific case of a Taylor series, where the point of expansion is 0.

## 4. What are the applications of Taylor/Maclaurin series?

Taylor/Maclaurin series are used in calculus to approximate functions, to find the values of functions at points where the function is difficult to evaluate, and to solve differential equations. They are also used in physics, engineering, and other fields to model and analyze real-world phenomena.

## 5. How accurate is a Taylor/Maclaurin series approximation?

The accuracy of a Taylor/Maclaurin series approximation depends on the number of terms used in the series. The more terms included, the closer the approximation will be to the actual value of the function. However, for some functions, the series may not converge or may converge very slowly, resulting in a less accurate approximation.

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