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a) Find the first 5 terms of the Taylor series for $f(x)=1/\sqrt{x}$ centered at $a=4$

b) use the result from a) to estimate $1/\sqrt{3}$ and compare it to calculated value

My attempt at the solution:

first 5 terms:

$$f(x)=\frac{1}{\sqrt{x}},\hspace{0.5cm} f(4)=1/2$$

$$f'(x)=-\frac{1}{2x^{3/2}},\hspace{0.5cm} f'(4)=-1/16$$

$$f''(x)=\frac{3}{4x^{5/2}},\hspace{0.5cm} f''(4)=3/128$$

$$f'''(x)=-\frac{15}{8x^{7/2}},\hspace{0.5cm} f'''(4)=-15/1024$$

$$f^{4}(x)=\frac{105}{16x^{9/2}},\hspace{0.5cm} f^{4}(4)=105/8192$$

The Taylor series for this function will be difficult to produce because $f^n(a)$ will not be nice to find. So I will just try to find only the first 5 terms of the taylor series

General formula for the Taylor series is $\sum^\infty_{n=0}\frac{f^n(a)(x-a)^n}{n!}$

the first 5 terms of the Taylor series will be

$$\frac{1/2(x-4)^0}{0!}+\frac{-1/16(x-4)^1}{1!}+\frac{3/128(x-4)^2}{2!}+\frac{-15/1024(x-4)^3}{3!}+\frac{105/8192(x-4)^4}{4!}$$

$$=\frac{1}{2}+\frac{-1(x-4)}{16}+\frac{3(x-4)^2}{128*2}+\frac{-15(x-4)^3}{1024*6}+\frac{105(x-4)^4}{8192*24}$$

Here are my questions;

Firstly, is this valid? it seems cheating to not find a relation for $f^n(a)$ Secondly, is it as simple as substituting the required value for x, to approximate for that?

And otherwise, why does this work? why would i not center the series at a=3 in order to get a more accurate result?

Thanks

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  • $\begingroup$ Follow up question, what if i set my Taylor series to start at 1 instead of 0, would that change the solution? would i simply set n-1 for every instance of n in the Taylor series? $\endgroup$ – helpmeh Apr 2 '16 at 18:49
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Your work is correct. Generally, there might not be a nice relationship that allows you to write the series in summation notation, but I think this case does have a nice one.

$$\begin{align} f &= x^{-1/2} \\ f'&= (-1)\left(\frac12\right)x^{-3/2} \\ f''&=(-1)^2\left(\frac12\cdot\frac32\right)x^{-5/2} \\ f^{(3)}&=(-1)^3\left(\frac12\cdot\frac32\cdot\frac52\right)x^{-7/2} \\ \vdots\\ f^{(n)}&=\frac{(-1)^n(2n-1)!!}{2^n}x^{-(2n+1)/2} \\ \end{align}$$

Note the double factorial. Evaluating the derivative at $a=4$ leads to

$$\begin{align} \frac{1}{\sqrt x} &= \sum_{n=0}^\infty \frac{(-1)^n(2n-1)!!}{2^n}(4)^{-(2n+1)/2} \frac{(x-4)^n}{n!} \\ &=\sum_{n=0}^\infty \frac{(-1)^n(2n-1)!!}{2^n}(2)^{-(2n+1)} \frac{(x-4)^n}{n!} \\ &=\sum_{n=0}^\infty \frac{(-1)^n(2n-1)!!}{2^{3n+1}n!} (x-4)^n \end{align}$$

Anyway, the Taylor series is a representation of your function $f=\frac{1}{\sqrt x}$ about the point $x=4$ for some radius of convergence $\rho$. What that means is that this series is accurate only in some locality $|x-4|<\rho$. Outside that radius, this series fails to represent the function.

It just so happens that $x=3$ falls within the convergent region. So yeah, plug in $x=3$ and find the five term approximation.

Why not approximate about $x=3$?

Sure. Following the same logic, we find

$$\begin{align} \frac{1}{\sqrt x} &= \sum_{n=0}^\infty \frac{(-1)^n(2n-1)!!}{2^n3^{(2n+1)/2}} \frac{(x-3)^n}{n!} \\ \end{align}$$

Upon evaluating $f(3)$ every term vanishes but the $n=0$ term! You'll find that $\frac{1}{\sqrt 3} = \frac{1}{\sqrt 3}$. This doesn't really help us approximate the answer at all. On the other hand, the series about $x=4$ has rational coefficients, giving a very tractable polynomial. We can quickly compute an approximation to $\frac{1}{\sqrt 3}$ to some arbitrary degree of accuracy.

(EDIT) Additional explanation: There are other points of expansion that may be tempting. What about $x=1$ or $x=9$ or any other perfect square? These would give Taylor polynomials with rational coefficients (do you see why?). That's useful for computational purposes, but why $x=4$ specifically? Again, if we're interested in $f(3)$, then we need it to lie in the radius of convergence. Perhaps the other choices would not include $x=3$ in their radius of convergence.

What if I started the Taylor series at $n=1$?

A Taylor series is a power series $\sum_{n=0}^{\infty}a_nx^n$. It contains non-negative powers of your variable, $a_nx^n$, including a constant term $a_0x^0$. It is natural then to begin indexing your summation at $n=0$. You can reindex it. For this example, it'll go as follows. Let $k=n+1$ where $k$ is our new index. This is pretty much what you suggested. Substitute this relationship to find:

$$\sum_{k=1}^\infty \frac{(-1)^{k-1}(2(k-1)-1)!!}{2^{3(k-1)+1}(k-1)!} (x-4)^{k-1}$$ $$=\sum_{k=1}^\infty \frac{(-1)^{k-1}(2k-3)!!}{2^{3k-2}(k-1)!} (x-4)^{k-1}$$

Now rename $k\rightarrow n$ since it's merely an index:

$$\sum_{n=1}^\infty \frac{(-1)^{n-1}(2n-3)!!}{2^{3n-2}(n-1)!} (x-4)^{n-1}$$

It's useful to be able to reindex summations, but I don't think it serves a useful purpose here.

As a final bit of food for thought, you might be interested in an even more general notion: the Laurent series. $f(z) = \sum_{n=-\infty}^\infty a_n (z-c)^n$. It's like the Taylor series, but the index and thereby the powers include negatives. In fact, if $a_n = 0$ for $n<0$, then it is a Taylor series. Very useful in complex analysis.

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  • $\begingroup$ @helpmeh Sure thing. You caught me right after my morning coffee. The wheels were in motion hahah. $\endgroup$ – zahbaz Apr 2 '16 at 20:35

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