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Power series expansion

  • Thread starter ngc2024
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
34,056
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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.
 
  • #3
15
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"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}...) ## ... ?
 
  • #4
34,056
9,922
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.
 
  • #5
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Of course! Thanks!
 
  • #6
Ray Vickson
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Homework Helper
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"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##.
 

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