# Thread: Taylor and Maclaurin series questions

1. ## Taylor and Maclaurin series questions

1. Find the Taylor series for f(x) = cos (3x) centered at a = (Pi/2). Find its radius of convergence.

2. Let f(x) = (1+x)^(-1/2)
a) Find the Maclaurin series for f.
b) Use the series found in 2a to find the maclaurin series for (1+x^2)^(-1/2) and use that result to find a Maclaurin series for sinh^-1 (x).

I cannot figure out how to do these. For 1, I substituted the 3x for x in the cos(x) series... what do I do from there? I know I have to put (x-a) at the end for the center.

For 2, I have no idea what to do. It says to use the binomial series, which we didn't really cover. It's a bonus on a homework.

2. For the binomial series of $\displaystyle \frac{1}{\sqrt{1+x}}$

If we let $\displaystyle m=\frac{-1}{2}$ in the binomial series,

$\displaystyle (1+x)^{m}=1+mx+\frac{m(m-1)}{2!}x^{2}+.......+\frac{m(m-1)...(m-k+1)}{k!}x^{k}$, if |x|<1.

$\displaystyle \frac{1}{\sqrt{1+x}}=1-\frac{1}{2}x+\frac{(\frac{-1}{2})(\frac{-1}{2}-1)}{2!}x^{2}$$\displaystyle +\frac{(\frac{-1}{2})(\frac{-1}{2}-1)(\frac{-1}{2}-2)}{3!}x^{3}+.....+\frac{(\frac{-1}{2})(3/2)(5/2)...(-1/2-k+1)}{k!}x^{k}+...$

$\displaystyle =1-\frac{1}{2}x+\frac{3}{2^{2}2!}x^{2}-\frac{15}{2^{3}3!}x^{3}+....+(-1)^{k}\frac{1*3*5....(2k-1)}{2^{k}k!}x^{k}+....$

$\displaystyle 1-\frac{x}{2}+\frac{3x^{2}}{8}-\frac{5}{16}x^{3}+..........$

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

The derivative of $\displaystyle sinh^{-1}(x)=\frac{1}{\sqrt{1+x^{2}}}$

3. Originally Posted by galactus
$\displaystyle =1-\frac{1}{2}x+\frac{3}{2^{2}2!}x^{2}-\frac{15}{2^{3}3!}x^{3}+....+(-1)^{k}\frac{1*3*5....(2k-1)}{2^{k}k!}x^{k}+....$

$\displaystyle 1-\frac{x^{2}}{2}+\frac{3x^{4}}{8}-\frac{5}{16}x^{6}+..........$

How did you go from x to x^2, from x^2 to x^4, from x^3 to x^6, etc.?

4. That's a stupid typo. Here is what it should be.

$\displaystyle 1-\frac{x}{2}+\frac{3x^{2}}{8}-\frac{5x^{3}}{16}+\frac{35x^{4}}{128}-\frac{63x^{5}}{256}+..............$

5. So how would I do part b? Would I substitute x^2 in for every x?

6. Yes.

Here is something to go on to know what to look for:

The series for $\displaystyle sinh^{-1}(x)=x-\frac{x^{3}}{6}+\frac{3x^{5}}{40}-\frac{5x^{7}}{112}+\frac{35x^{9}}{1152}-................$

7. Originally Posted by galactus
$\displaystyle (1+x)^{\frac{-1}{2}}=1+\sum_{k=1}^{\infty}\frac{(\frac{-1}{2})(\frac{-3}{2})....(\frac{-1}{2}-k+1)}{k!}x^{k}$
I'm still pretty confused with everything. Is that the answer?

And I'm very much lost with the (1+x^2)^(-1/2) series.

8. Can someone help with #1? I did a table of derivatives and evaluated them at Pi/2:

derivative, evaluated at Pi/2

f(x) = cos3x ,0
f '(x) = -3sin3x ,3
f ''(x) = -9cos3x , 0
f '''(x) = 27sin3x ,-27
f ''''(x) = 81cos3x ,0
f^5 (x) = -243sin3x ,243

The nth derivative would be (-1)^n * [3^(2n)], right? SO then I just used the nth derivative evaluated at (pi/2) / n! formula to get c sub n.

Summation n=0 to infinity [(nth derivative evaluated at Pi/2) / n!] * (x-Pi/2)^n

Then I substituted the actual nth derivative values to get:
0 + ([3(x-Pi/2)] / 1!) - ([27(x-Pi/2)^3] / 3!) + 0 + ([243(x-Pi/2)^5] / 5!) +...

So then I got the Taylor series of cos(3x) to be:

(-1)^n * 3^(2n+1) * (x-Pi/2)^(2n+1)
(2n+1)!

I'm really confused with this stuff. It's been awhile since I've done it and it's final time.