Methods for Finding Taylor Polynomials

Aside from the "standard" way of listing all the derivatives and applying the Taylor Polynomial formula, we have some other ways to find Taylor Polynomials as well.

Arithmetic

Example

Find the Taylor Polynomial of:

(see cardinal sine)

I.e, what is for ?

solution

Substitution

Example

Find the Taylor polynomial for at .

solution
We can try the standard way of listing all the derivatives:

But if we keep trying to differentiate, it becomes more and more of a pain in the ass.

Instead, let . Then the Taylor polynomial at is . Thus:

Shifting the Point of Tangency

Example

Find the Taylor polynomial of , .

solution
We can re-write our function so we're still centred at , somehow.

(see symmetry identities).

Let . Then, when we have , the equality still holds, so we can evaluate the Taylor polynomial of at (nice):

which is slightly easier than finding the derivatives of cosine and seeing a pattern.

Example

Approximate using Taylor Polynomials.

solution

Note that we chose to make the Taylor polynomial nicer, and then we evaluated it at .

(see Newton's binomial theorem)

Shortcut

As an example, we take the linear approximation to at :

If we let , we find that:

which is actually the quadratic approximation of the function .

To be sure, lets let and , and differentiate them:

Evaluating everything at 0:

The values of , , and are matched by the corresponding values of the polynomial at . That means that is the second-order Maclaurin polynomial of .

Info

Letting be the order Maclaurin polynomial, where and , will always yield the -order Maclaurin polynomial of the function , as long as exists.

Remark

In other words, the same changes that apply to the original functions must be applied to the Maclaurin polynomials as well.

Example

Determine the third order Maclaurin polynomial for . Then use the shortcut method to find:

  1. The third order Maclaurin polynomial for
  2. The second order Maclaurin polynomial for
  3. The fourth order Maclaurin polynomial for

solution
First, finding the Maclaurin polynomial for .

which gives:

Next, we find the Maclaurin polynomial for . Note that , so we sub in :

Next, we find the Maclaurin polynomial for . Note that , so we can differentiate and multiply the result by :

Finally, we find the Maclaurin polynomial for . Note that , so we find the antiderivative of :

To determine , remember that our polynomial is centred at , so . Thus:

Since , we conclude that . Thus the desired polynomial is: