# Expansion of a power series

## Homework Statement

Suppose we are given the power series expansion ##f(x) = -\sum_{n=1}^{\infty} \frac{x^{n}}{n} ## which converges for |z|<1.

What is the radius of convergence?

Sum this serie and derive a power series expansion for the resulting function around -1/2, 1/2, 3/4 and 2.

## The Attempt at a Solution

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I can let ##f(x) = \sum_{n=0}^{\infty} \frac{x^{n}+1}{n+1}## so ##f'(x) = x^n = \frac{1}{1-x}## Integrating back gives ##f(x) = -ln(\frac{1}{1-x})=-\sum_{n=1}^{\infty} \frac{x^n}{n} ##

Now I can just plug in the values -1/2, 1/2, 3/4 and 2 right?

My question is, why do they ask me to do this with many values if all I need to do is plug it in in the final equation? I think I probably need to do something else/more...

Also, how do I get the radius of convergence? I recall it is 1 in this case, but not how? I've heard I can take the inverse of the results of the ratio test, but thats gives me an expression, ##\frac{x_n}{n+1}## and not a number?

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Suppose we are given the power series expansion ##f(x) = -\sum_{n=1}^{\infty} \frac{x^{n}}{n} ## which converges for |z|<1.

What is the radius of convergence?

Sum this serie and derive a power series expansion for the resulting function around -1/2, 1/2, 3/4 and 2.

## The Attempt at a Solution

[/B]
I can let ##f(x) = \sum_{n=0}^{\infty} \frac{x^{n}+1}{n+1}## so ##f'(x) = x^n = \frac{1}{1-x}## Integrating back gives ##f(x) = -ln(\frac{1}{1-x})=-\sum_{n=1}^{\infty} \frac{x^n}{n} ##

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NO: ##f'(x) ## is a sum, not a single term. However, ##f'(x) = 1/(1-x)## is correct.

Then ##f(x) = -\ln (1-x),## NOT ##-\ln (1/(1-x)).##

Also note: do not type "ln w" in LaTeX; it produces the ugly result ##ln w##; instead, use LaTeX the way it was designed to be used: type "\ln w" instead. That produces the clear and pleasing result ##\ln w.##

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Now I can just plug in the values -1/2, 1/2, 3/4 and 2 right?

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No: the question did not ask for ##f(-1/2)## or ##f(1/2)##, etc. It asked for the series expansions of ##f(x)## about the points ##x = -1/2## or ##x = 1/2##, etc. These would be power series in ##(x+1/2)## or ##(x-1/2)##, etc.

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My question is, why do they ask me to do this with many values if all I need to do is plug it in in the final equation? I think I probably need to do something else/more...

Also, how do I get the radius of convergence? I recall it is 1 in this case, but not how? I've heard I can take the inverse of the results of the ratio test, but thats gives me an expression, ##\frac{x_n}{n+1}## and not a number?

You really do need to read up about and clarify your understanding of "radius of convergence". What does it MEAN if we say that the radius of convergence of a series ##\sum c_n x^n## is ##r > 0?##

Finally: why would you post such a thread in the "Precalculus" forum? It is very definitely a "Calculus and Beyond" question.

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vela
Staff Emeritus
Homework Helper
Now I can just plug in the values -1/2, 1/2, 3/4 and 2 right?

My question is, why do they ask me to do this with many values if all I need to do is plug it in in the final equation? I think I probably need to do something else/more...
The answer to your question is in your original post. You wrote, "derive a power series expansion for the resulting function around -1/2, 1/2, 3/4 and 2."

I've heard I can take the inverse of the results of the ratio test, but thats gives me an expression, ##\frac{x_n}{n+1}## and not a number?
It sounds like you heard some vague notion that you can use the ratio test to find the radius of convergence, but that's about it. That's a good starting point, but you should read your textbook or notes to sharpen up that notion into what you actually need to do instead of just assuming or guessing.

It seems to me that if you simply slow down and think a bit more carefully, you'll be able to figure this problem out.

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NO: ##f'(x) ## is a sum, not a single term. However, ##f'(x) = 1/(1-x)## is correct.

Then ##f(x) = -\ln (1-x),## NOT ##-\ln (1/(1-x)).##

Also note: do not type "ln w" in LaTeX; it produces the ugly result ##ln w##; instead, use LaTeX the way it was designed to be used: type "\ln w" instead. That produces the clear and pleasing result ##\ln w.##

*********************************************************

**********************************************************

No: the question did not ask for ##f(-1/2)## or ##f(1/2)##, etc. It asked for the series expansions of ##f(x)## about the points ##x = -1/2## or ##x = 1/2##, etc. These would be power series in ##(x+1/2)## or ##(x-1/2)##, etc.

***********************************************************

You really do need to read up about and clarify your understanding of "radius of convergence". What does it MEAN if we say that the radius of convergence of a series ##\sum c_n x^n## is ##r > 0?##

Finally: why would you post such a thread in the "Precalculus" forum? It is very definitely a "Calculus and Beyond" question.
You have clearly misunderstood the question. (yes my bad on the misstakes, that should not loose your focus on the key question here).

You pointed out what I don't understand, well why do you think I posted in the forum? Because I understand everything? lol

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The answer to your question is in your original post. You wrote, "derive a power series expansion for the resulting function around -1/2, 1/2, 3/4 and 2."

It sounds like you heard some vague notion that you can use the ratio test to find the radius of convergence, but that's about it. That's a good starting point, but you should read your textbook or notes to sharpen up that notion into what you actually need to do instead of just assuming or guessing.

It seems to me that if you simply slow down and think a bit more carefully, you'll be able to figure this problem out.
thanks!

Yes but it's the "expand around a point" part I don't understand. I can derive a general expansion, as done above (with mistakes) but I can't see how that result may differ for different points. I'm new to this and it's not in our book. Is it a straight forward way to explain this?

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