Boas: Mathematical Methods for Phys Sci Pr.1.13.25

[kq6up]

Mclaurin Series with Division by Zero?

Boas: Mathematical Methods for Phys Sci Pr.1.13.25

Homework Statement

Using the methods of this section:
(a) Find the first few terms of the Maclaurin series for each of the following functions.
(b) Find the general term and write the series in summation form.
(c) Check your results in (a) by computer.
(d) Use a computer to plot the function and several approximating partial sums of the
series.

$$f(x)=\frac { 2x }{ { e }^{ 2x }-1 }$$

Homework Equations

$$f(x)=\sum_{n=0}^{\infty}f^{n}(0)x^{n}/n!$$

The Attempt at a Solution

Since f(0) is division by zero, how do you find a Maclaurin series for it?

Thanks,
Chris Maness

 

[HallsofIvy]

Yes, the function, as written, is not defined. But it is a "removable" discontinuity. I suspect they intend you to use the limit there. You can simplify by letting u= 2x. Then the limit is $\lim_{u\to 0}\frac{u}{e^u- 1}$. And now you can take that limit in a number of ways:
1) l'hospital's rule: differentiating both numerator and denominator gives $\frac{1}{e^u}$ which has limit 1 as u goes to 0.

2) Since you are dealing with MacLaurin series, the MacLaurin series of $e^u$ is $1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot$. So $e^u- 1= u+ u^2/2+u^3/6+ \cdot\cdot\cdot$ and $\frac{u}{e^u- 1}= \frac{u}{u+ u^2/2+ u^3/6+ \cdot\cdot\cdot}$$= \frac{1}{1+ u/2+ u^3/6+ \cdot\cdot\cdot}$ which again has limit 1 as u goes to 0.

(It would have been better if the problem had said "$f(x)= \frac{2x}{e^{2x}- 1}$ if $x \ne 0$, $f(0)= 1$".)

 

[vela]

Find the limit, if it exist, as x goes to 0.

 

[kq6up]

Yes, the function, as written, is not defined. But it is a "removable" discontinuity. I suspect they intend you to use the limit there. You can simplify by letting u= 2x. Then the limit is $\lim_{u\to 0}\frac{u}{e^u- 1}$. And now you can take that limit in a number of ways:
1) l'hospital's rule: differentiating both numerator and denominator gives $\frac{1}{e^u}$ which has limit 1 as u goes to 0.

2) Since you are dealing with MacLaurin series, the MacLaurin series of $e^u$ is $1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot$. So $e^u- 1= u+ u^2/2+u^3/6+ \cdot\cdot\cdot$ and $\frac{u}{e^u- 1}= \frac{u}{u+ u^2/2+ u^3/6+ \cdot\cdot\cdot}$$= \frac{1}{1+ u/2+ u^3/6+ \cdot\cdot\cdot}$ which again has limit 1 as u goes to 0.

(It would have been better if the problem had said "$f(x)= \frac{2x}{e^{2x}- 1}$ if $x \ne 0$, $f(0)= 1$".)

Yes, they don't mention using the limit if f(0)=D.N.E, and I should have use L'Hosp.

Thanks,
Chris