Question: Expand $\displaystyle \frac{1}{(1+x)^\frac{1}{4}} $ as a power series and use it to estimate $\displaystyle \frac{1}{1.1^\frac{1}{4}} $ correct to three decimal places.

I am clueless!

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- Mar 5th 2010, 10:40 PMEm Yeu AnhFinding a power series
Question: Expand $\displaystyle \frac{1}{(1+x)^\frac{1}{4}} $ as a power series and use it to estimate $\displaystyle \frac{1}{1.1^\frac{1}{4}} $ correct to three decimal places.

I am clueless! - Mar 6th 2010, 04:14 AMCaptainBlack
- Mar 6th 2010, 04:15 AMHallsofIvy
That is $\displaystyle f(x)= (1+ x)^{-1/4}$

$\displaystyle f'(x)= -\frac{1}{4}(1+ x)^{-5/4}$

$\displaystyle f"(x)= \frac{5}{16}(1+ x)^{-9/4}$

$\displaystyle f'''(x)= -\frac{45}{64}(1+ x)^{-13/4}$

The nth derivative is, I believe, $\displaystyle \frac{(-1)^n}{1(3)(9)\cdot\cdot\cdot(4n+1)}(1+ x)^{-(4n+1)/4}$.

Write it as a Taylor's series about x= 1. Since the sign is alternating, the error is never larger than the**next**term in the series. Expand it until the next term is less than .001.

(Well, yeah, the generalized binomial series is probably easier!) - Mar 6th 2010, 07:41 AMshawsend
Also check out "Binomial Theorem" on Wikipedia. Rising factorial, falling factorial, I'm confused. How about I show you this Mathematica code and then you figure out how I got it and what it means.

Code:`n[858]:=`

myFunctions[x_] := 1/(1 + x)^(1/4)

mySeries[x_, n_] :=

Sum[(FactorialPower[-4^(-1), k]/k!)*

x^k, {k, 0, n}]

mySeries[0.1, 20]

myFunctions[0.1]

Out[860]=

0.9764540896763106

Out[861]=

0.9764540896763105