Find a Maclaurin series expansion for each function.

f(x) = cos x

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

f(x) = x sin x

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- June 4th 2008, 01:25 PMRabbit016Maclaurin series
Find a Maclaurin series expansion for each function.

f(x) = cos x

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

f(x) = x sin x - June 4th 2008, 01:55 PMo_O
These are pretty standard but if the intent is to "discover" these on your own, just directly apply the formula for the Maclaurin series (a taylor series centred around a = 0):

Taking a = 0:

For the f(x) = cos x, you would need to find:

, , ..

Then generalize a pattern and that should be your Maclaurin series (Of course, if you wanted to prove it you can do it with induction to verify your series).

Also, a google search for maclaurin series would probably give you these results. - June 4th 2008, 03:49 PMMathstud28
Using o_O's above formula we see that

this is applicable

Because applying the use of the root test for finding intervals of convergence we get

Therefore we run into an identity verifying that cos(x) is analytic across the reals

for

Either recognize this as the converse of the sum of an infinite geometric seires or notice that the temrs are given by

Using the Root test again we get

Now seeing that at x=1 and x=-1 we have a divergent series this function is analytic only on

For the third I will leave it up to you to show the termiwise derivation but

and applying the ratio or root test again we see this is also analytic across the reals