Using Power Series to Evaluate ln and sin at a Given Point

[ngc2024]

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]

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.

 

[ngc2024]

"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]

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]

"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 $z$.