What is the point in the Taylor Series expansion?

I understand how to use the Taylor (and Maclaurin) Series, but I don't understand what the point in using it in the first place is...

So it is used to approximate the value of a function at x = a ...

but hey, instead of doing all that working out and just approximating the function value at a, I could just do f(a) instead, and then I'm directly lead to the right answer, it's not even an approximation! :O

Why would you use the Taylor Series in a real life situation instead of just doing f(a)?

Reply 1

jameswhughes

There's more to it than that, e.g if you were asked to draw f(x)= (sin x)/x the expansion would be very useful for finding what happens to the function as x approaches 0, however if you just try finding f(0) you won't be able to do it.

Reply 2

Mr M

Study Forum Helper

What would you do if you didn't know how to find f(a)?
.
What would you do if you needed to integrate f but didn't know how to? Perhaps you could expand it and integrate term by term ...

Reply 3

FireGarden

Perhaps f(a) is not what you require, but is something you know; along with derivatives at that point. Then you can approximate f(x) to however accurate you can, and can then use it for more values of x, which may be of more interest. They have way more uses than that, though; not an expert with them, but I've used them to solve differential equations before.

Reply 4

Fing4

Prove that

\displaystyle\frac{d(sin(x))}{dx}=cos(x)

and other derivatives without them

(edited 12 years ago)

Reply 5

JoMo1

1) how do you calculate sin(a)? Note that "put it into a calculator" just steps the problem back to a problem of "How does your calculator figure out sin(a)?

2) Expansions have a habit of allowing nice elegant proofs of lots of different things, especially with regard to limits, e.g. sin(x)/f(x), where f(x) is a polynomial.

3) Rather fundamental to complex analysis and holomorphic functions (wikipedia it) which are in turn beautiful and horrifyingly useful.

Reply 6

nuodai

How do you think calculators evaluate

\sin(x)

when the only operations their hardware can cope with are addition and multiplication? What they do is use enough terms in the Taylor expansion

\sin(x) = \displaystyle \sum_{n=0}^{\infty} \dfrac{(-1)^nx^{2n+1}}{(2n+1)!}

to display the answer accurate to enough decimal places.

Reply 7

Mr M

Study Forum Helper

Original post by JoMo1

How does your calculator figure out sin(a)?

Original post by nuodai

How do you think calculators evaluate

\sin(x)

when the only operations their hardware can cope with are addition and multiplication? What they do is use enough terms in the Taylor expansion

\sin(x) = \displaystyle \sum_{n=0}^{\infty} \dfrac{(-1)^nx^{2n+1}}{(2n+1)!}

to display the answer accurate to enough decimal places.

It is a bit of a myth that calculators use power series for trigonometry.

Google Cordic Algorithm.

(edited 12 years ago)

Reply 8

JoMo1

Original post by Mr M

It is a bit of a myth that calculators use power series for trigonometry.

Google Cordic Algorithm.

I do indeed know this, but felt confusing the issue with proper reality was mostly unnecessary, and taking the broadest definition of "calculator", it wasn't entirely false.

Reply 9

poohat

Often linear functions are easier to work with than nonlinear functions. The easiest way to approximate f(x) by a linear function is to take its Taylor series, and then throw away all the higher order terms (ie you get f(x+a) ~= f(a) + f'(a)(x-a) which is linear in x), and using that form can make things easier. You can then add in higher order terms if you need more accuracy.

This kind of thing comes up all the time, both in applications, and in proofs. See: http://en.wikipedia.org/wiki/Linearization

(edited 12 years ago)

Reply 10

the bear

Original post by JoMo1

I do indeed know this, but felt confusing the issue with proper reality was mostly unnecessary, and taking the broadest definition of "calculator", it wasn't entirely false.