Taylor Series Help....?

Question

We've never done any sort of problem like this, and it's the beginning of the year and I'm getting extremly confused...

1) Look up the Taylor (or Maclaurin) series for the following functions:

e^x

cos x

sin x

This "series" is an infinitely long polynomial that ("strangely") approximates a function that is not a polynomial. For now, you can just copy them down. Try graphing the first 4 or 5 terms of the polynomial if you want to see how they behave.


So now I've tried looking up a bunch of stuff, but I keep getting extremely confused!!

Answer 1

People have already stated what the Taylor series for these functions are, so instead of addressing that directly, I'm hoping I can clear up a little bit of the confusion about why this works...

Well, a Taylor Series is just an infinite representation of Taylor's Theorem:

f(t) = f(t0) + f'(t0)*(t-t0) + f''(t0)*(t-t0)^2/2! + f'''(t0)*(t-t0)^3/3!...

Maclaurin series are just Taylor series with t0 = 0, or in mathematical terms, expansions around 0.

An intuitive way to see why a Taylor series comes about is the following:

You have a function f. You want to know its value at t, but you only have it at t0. So you take a stab at the value of f(t) and say maybe it's f(t0).

To make corrections for the fact that f is not constant and you're trying to model the function's behavior at t, then you multiply f'(t0) by (t-t0) to account for a linear change in f over (t-t0) and add that to f(t0)...

now you have f(t) =~ f(t0) + f'(t0)*(t-t0)

As a second correction for possibilities of the function slope changing, you correct with the curvature, and consider f''(t0) and multiply it by (t-t0)/2! and add it to your previous result...

Then you have to consider the 3rd derivative at t0, multiplying it by (t-t0)/3! and adding it, and you keep doing this, and the pattern continues until you have infinitely many of these things (or finitely many and a remainder term)...

For Taylor Series, you suppose that f can be expressed in terms of an infinite power series in (t-t0), where t0 is just the point about which you're expanding the thing, and for Maclaurin series, t0 = 0.

The basic idea of a Taylor series, in terms of deriving it, is:

suppose f can be modeled by a power series in (t-t0)...

Write it as Sum[c_k*(t-t0)^k] from k = 0 to infinity, and c_k being the kth coefficient in the series...

the first thing to see is that if you set t = t0, all terms but the first drop, so the constant coefficient, c_0) = f(t0).

differentiate the series just like you would any polynomial, and plug in t=t0. You'll see that all terms go away except c_1, which is f'(t0).

repeat differentiation and plugging t=t0, and you'll see that all terms disappear except for 2*c_2...so you see that f''(t0)/2! is c_2

keep doing this and the pattern becomes immediate...
c_k is the kth derivative of f, evaluated at t0, multiplied by (t-t0)^k/k!

Answer 2

Follow the wikipedia link, the Taylor series you are looking for are at the bottom of the page

Answer 3

f(x) = e^x so f(0) = 1
f'(x) = e^x so f'(0) = 1
f"(x) = e^x so f"(0) = 1
....................................
f^n(x) = e^x so f^n(0) =1
Thus the Maclaurin expansion is:
e^x = 1+x+x^2/2! +x^3/3! +....+ x^n/n! +...
By the ratio test, this series converges for all values of x

f(x) = sin x so f(0) = 0
f'(x) = cosx so f'(0) = 1
f"(x) = -sinx so f"(0) = 0
f"'(x) = -cosx so f"'(0) = -1
f""(x) = sinx so f""(0)= 0
..........................................
f^n(x) = sin(x + 1/2npi) so F^n(0) = sin 1/2npi
So Maclaurin expansion is:
sinx = x-x^3/3! +x^5/5! - ... +(-1)^m+1 x^(2m-1)/(2m-1)! +...
To find the Taylor series about the point pi/3, substitute pi /3 for all the derivatives instead of 0.

cos x similar to sin x

Answer 4

you could not evaluate a guy to a woman. each and every is distinctive. yet a guy, and a woman, supplement one yet another. In different words, one guy and one woman (at the same time) is one unit. they're a complementary pair.

Answer 5

this is all available in
http://mathworld.wolfram.com/TaylorSeries.html