Taylor/Maclaurin series of a function

[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 - (x^2*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, ∞)

 

[LCKurtz]

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 - (x^2*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.

 

[EEristavi]

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

 

[PeroK]

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

 

[PeroK]

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 :)