- #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

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?