Taylor & MacLaurin Series

[jetter2]

My class is getting into these two types of series, and I have the process down for for representing Functions AS a series, and finding the radius of convergence, but my issue is how can I apply this method to Taylor and MacLaurin Series?


For Example;

I know that [x/(4x+1)] can be represented by the summation from 0->inf of (-4x)^n with a radius of -.25<x<.25...

But how can I apply this to the Taylor, Maclaurin, or Binomial Series?

For example.

(1)Find the Taylor Series centered @ A.

f(x) = sin x, a= pi/2


(2) Obtain the Maclaurin Series for;

f(x) = x cos 2x


Moreover, can someone help me understand the difference between THESE methods, and expanding a binomial series?

How would I use the binomial series to expand;

1/(x+4)^4

If the denominator wasn't raised to the 4th power I could attack this as a normal Function to Power series conversion..but that power is throwing me for a loop.

Another toughie I'm killing myself over with the Binomial Expansion;

(1-x)^(2/3)


Am I just over thinking the Binomial series, or does it truly differ from the other methods?

 

[royhaas]

A binomial series with an positive integer exponent will terminate. This is related to taking the derivative of a monomial. In your case, \(\displaystyle (x+4)^{-4}\) will be an infinite series, since the exponent is a negative integer. One way to evaluate it is to write out the first several terms of \(\displaystyle 1/(1+x/4)\) and take the derivative three times. This may help you see the relationship between a MacLaurin series and a binomial series.

 

[jetter2]

A binomial series with an positive integer exponent will terminate. This is related to taking the derivative of a monomial. In your case, \(\displaystyle (x+4)^{-4}\) will be an infinite series, since the exponent is a negative integer. One way to evaluate it is to write out the first several terms of \(\displaystyle 1/(1+x/4)\) and take the derivative three times. This may help you see the relationship between a MacLaurin series and a binomial series.

I'm still not following what you mean, can you show me a few steps so that I my follow along in your explanation?

 

[lookagain]

How would I use the binomial series to expand;

1/(x+4)^4

If the denominator wasn't raised to the 4th power I could attack this as a normal Function to Power series conversion..
but that power is throwing me for a loop.

Another toughie I'm killing myself over with the Binomial Expansion;

(1-x)^(2/3)

\(\displaystyle 1/(x + 4)^4 \ = \ (x + 4)^{-4}\)


If you use the form of \(\displaystyle (a + b)^n\)for the Binomial Expansion, \(\displaystyle a^n + \frac{na^{n - 1}}{1!} + \frac{(n - 1)na^{n - 2}}{2!} \ + \ . . . ,\)


then the first three terms (on the right side of the equals sign) for the infinite series would be:


\(\displaystyle (x + 4)^{-4} \ = \ x^{-4} \ + \ \frac{(-4)x^{-5}(4)^1}{1!} \ + \ \frac{(-5)(-4)x^{-6}(4)^2}{2!}\)


------------------------------------------------------------------------------------------------------------------


For (1 - x)^(2/3), you have the exponent as a rational noninteger.


The first three terms of this infinite series are shown here:



\(\displaystyle (1)^{\frac{2}{3}} \ + \ \frac{\frac{2}{3}(1)^\frac{-1}{3}(-x)^1}{1!} \ + \ \frac{(\frac{-1}{3})(\frac{2}{3})(1)^\frac{-4}{3}(-x)^2}{2!}\)

 

[jetter2]

THANK YOU!!

This makes perfect sense =]

I got the taylor and Mac series down for most my practice problems, but I'm still having a problem with representing these fucntions as a series and finding the convergence radius;


f(x) = x^2 / ((1-2x)^2)

I know here that I have to integrate due to the denominator having a power or a log, but I'm not really sure on the mechanics of it.

Same with f(x) = arctan(x/3)

How do you approach this one, do you have to use the inverse trig Identity and then multiply through or something?

What about a Macluren Series for the following;

f(x) = sin 2(x)

or

f(x) = xcos 2x


This is the last chunk of this chapter I am still getting hung up on, everything else you guys have said makes perfect sense =]

 

[galactus]

THANK YOU!!

This makes perfect sense =]

I got the taylor and Mac series down for most my practice problems, but I'm still having a problem with representing these fucntions as a series and finding the convergence radius;

\(\displaystyle f(x) = \frac{x^{2}}{(1-2x)^{2}}dx\)

You have to integrate this one?. If so,

You can make a sub \(\displaystyle u=1-2x, \;\ \frac{-du}{2}=dx\)

Making the subs gives:

\(\displaystyle \frac{-1}{8}\int\frac{(u-1)^{2}}{u^{2}}du=\frac{1}{4}\int\frac{1}{u}du-\frac{1}{8}\int\frac{1}{u^{2}}du-\frac{1}{8}\int du\)

integrate and resub. They are rather straightforward now.

Same with \(\displaystyle f(x) = arctan(x/3)\)

How do you approach this one, do you have to use the inverse trig Identity and then multiply through or something?

[/quote]

Use parts to integrate. Let \(\displaystyle u=tan^{-1}(x/3), \;\ dv=dx, \;\ du=\frac{3}{x^{2}+9}dx, \;\ v=x\)

\(\displaystyle xtan^{-1}(x/3)-3\int\frac{x}{x^{2}+9}dx\)

MacLaurin series?

\(\displaystyle f(x) = sin (2x)\)

Use the series for sin and then sub in 2x for x.

\(\displaystyle f(x) = xcos 2x\)

[/quote]

Use the series for cos, sub in 2x for x, then multiply by x.

The series for cos is \(\displaystyle \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!}\)

Make the x a 2x, then multiply by x:

\(\displaystyle \sum_{k=0}^{\infty}(-1)^{k}\frac{4^{k}(x)^{2k+1}}{(2k)!}\)

 

[jetter2]

One last thing...


So I was presented with the Series n goes 1-> inf. sigma [[(-2)^n]/[n^.5]](x+3)^n

Now When I see this problem I imediatly want to try the alternating series test. My only problem with that is I end up getting to a form where I do the Ratio Test as well. But I do some algebra and end up with 2 which = divergance.

If I just directly attack this as a ratio test problem, I find that it converges at zero with an interval of (-7/2, 5/2]

Should I ONLY use the alternating series test if (-1)^n is in the numerator? Why does the alternating series test NOT work here?

 

[galactus]

It would appear, at first glance, that the Alternating Series test can not be applied because the series does not pass the second condition.

That is, it does not pass:

\(\displaystyle a_{n+1}\leq a_{n}, \;\ \forall \;\ n\)