# Power series expansion

## Homework Statement

"Use power series to evaluate the function at the given point"
## ln (x+ \sqrt{1+{x^2}}) - sin x ##

at ## x = 0.001 ##

## Homework Equations

Relevant power series:
A: ## ln (1+x) = \Big( \sum_{n=0}^\infty\frac{({(-1)^{n+1}}{x^n})}{n} \Big) ##

B: ## {(1+x)^p} = \sum_{n=0}^{\infty }\binom{p}{n}{x^n} ##

C: ## sin x = \Big ( \sum_{n=0}^\infty\frac{({(-1)^{n}{x}^{2n+1}})}{(2n+1)!} \Big) ##

3. The Attempt at a Solution

Starting out with the logarithmic expression, there are two power series.

If ## \sqrt{1+{x^2}} = y ##

then ## ln (x+y) = \Big( \sum_{n=0}^\infty\frac{({(-x)^{n+1}}{y^n})}{n} \Big) ##

and ## y = \sum_{n=0}^{\infty }\binom{(\frac{1}{2})}{n}{({(x^2)^n)}} ##

I hope I am correct this far? Of course, I can now expand the series and multiply term by term, but this is quite tedious - especially since (using wolfram alpha), the first terms are in the power of 5 and 7. I also tried making a general expression, but then I end up with something like..
## ln (x+y) = \Big( \sum_{n=0}^\infty\frac{({(-x)^{n+1}})}{n} \Big) * (\sum_{n=0}^{\infty }\binom{(\frac{1}{2})}{n}{({(x^2)^n)}}) ##

.. and I don't know how to simplify this.

There must be some obvious mistake, or a simpler way to do this?

mfb
Mentor
If ## \sqrt{1+{x^2}} = y ##

then ## ln (x+y) = \Big( \sum_{n=0}^\infty\frac{({(-x)^{n+1}}{y^n})}{n} \Big) ##
That does not look right and I don't see where this comes from. You cannot simply replace occurences of 1 by a variable. In particular, the two occurences of "1" in the expansion given above have a completely different meaning.

and ## y = \sum_{n=0}^{\infty }\binom{(\frac{1}{2})}{n}{({(x^2)^n)}} ##
What is ##\binom{(\frac{1}{2})}{n}##?
There is a power series for ##\sqrt{1+x}## that you can use.

With the correct power series, it is sufficient to consider the first few terms - I guess you just need the first non-vanishing order of x.

"In particular, the two occurrences of "1" in the expansion given above have a completely different meaning."
Yes, I was unsure of this.

And maybe I was unclear - I used the binomial series to expand ## \sqrt{1+{x^2}} ##
## \sqrt{1+{x^2}} = \sum_{0}^{\infty }\binom{\frac{1}{2}}{n}{(x^2)^n} = 1 + \frac{x^2}{2} - \frac{x^4}{8} + \frac{x^6}{16} ... ## where ## \binom{\frac{1}{2}}{n} ## is the binomial expression 0.5 choose n.
I think this is correct?

I don't fully understand, however - how do I make a power series of:
## ln (1 + x + \frac{x^2}{2} - \frac{x^4}{8} + \frac{x^6}{16}...) ## ... ?

mfb
Mentor
Okay, I found the appropriate generalization of the standard binomial coefficients needed for that.

##\ln(1+x+\frac{x^2}{2}+...) = \ln(1+c)= ...## with c coming from the sum. You won't need many terms to find one that does not cancel.

ngc2024
Of course! Thanks!

Ray Vickson
Homework Helper
Dearly Missed
"In particular, the two occurrences of "1" in the expansion given above have a completely different meaning."
Yes, I was unsure of this.

And maybe I was unclear - I used the binomial series to expand ## \sqrt{1+{x^2}} ##
## \sqrt{1+{x^2}} = \sum_{0}^{\infty }\binom{\frac{1}{2}}{n}{(x^2)^n} = 1 + \frac{x^2}{2} - \frac{x^4}{8} + \frac{x^6}{16} ... ## where ## \binom{\frac{1}{2}}{n} ## is the binomial expression 0.5 choose n.
I think this is correct?

I don't fully understand, however - how do I make a power series of:
## ln (1 + x + \frac{x^2}{2} - \frac{x^4}{8} + \frac{x^6}{16}...) ## ... ?

Write ##1 + x + \frac{x^2}{2} - \frac{x^4}{8} + \frac{x^6}{16} + \cdots## as ##1+z##, and use the series for ##\ln(1+z)## in powers of ##z##.

ngc2024