Taylor Series

A Taylor Series is an expansion of some office into an infinite sum of terms, where each term has a larger exponent like x, tentwo, x3, etc.

Example: The Taylor Serial for due eastten

eastward x = one + x + ten2 2! + x3 3! + ten4 iv! + x5 five! + ...

says that the office: ex

is equal to the infinite sum of terms: one + x + xii/ii! + xiii/three! + ... etc

(Note: ! is the Factorial Function.)

Does it really work? Let's try it:

Example: eastx for ten=2

Well, we already know the answer is e2 = 2.71828... × ii.71828... = 7.389056...

But permit's endeavour more and more terms of our infinte series:

Terms Issue
ane+2 3
1+2+ ii2 ii! 5
1+2+ 22 2! + 2iii 3! vi.3333...
1+2+ 2two 2! + 23 3! + 24 4! 7
1+2+ 2ii ii! + 23 3! + twoiv 4! + 25 5! seven.2666...
i+two+ 22 2! + iiiii 3! + 24 four! + two5 5! + two6 6! vii.3555...
1+2+ ii2 2! + 2iii 3! + 24 4! + twov 5! + 2vi 6! + ii7 seven! 7.3809...
i+2+ 22 2! + 2three 3! + 2four 4! + ii5 v! + 26 6! + 27 7! + two8 eight! 7.3873...

It starts out really badly, but it then gets better and improve!

Try using "2^n/fact(n)" and north=0 to 20 in the Sigma Calculator and see what you get.

Here are some mutual Taylor Series:

Taylor Serial expansion As Sigma Annotation

e x = 1 + x + 10two 2! + x3 3! + ...

 Taylor: Sigma n=0 to infinity of (x^n)/n!

sin ten = x − 10three three! + 10five v! − ...

 Taylor: Sigma n=0 to infinity of [ (-1)^n / (2n+1)! ] times x^(2n+1)

cos x = 1 − x2 ii! + ten4 4! − ...

 Taylor: Sigma n=0 to infinity of [ (-1)^n / (2n)! ] times x^(2n)
Taylor 1/(1-x) = 1 + x + x^2 + x^3 + ... Taylor: Sigma n=0 to infinity of x^n

(At that place are many more.)

Approximations

We can utilize the start few terms of a Taylor Series to get an approximate value for a function.

Here nosotros bear witness better and better approximations for cos(x). The red line is cos(10), the blue is the approximation (try plotting it yourself) :

Y'all can also see the Taylor Series in action at Euler'due south Formula for Complex Numbers.

What is this Magic?

How can we plow a office into a serial of power terms like this?

Well, it isn't really magic. First we say we want to take this expansion:

f(x) = c0 + c1(ten-a) + cii(x-a)2 + c3(x-a)3 + ...

Then we cull a value "a", and piece of work out the values c0 , cane , c2 , ... etc

And it is done using derivatives (then nosotros must know the derivative of our function)

slope examples y=3, slope=0; y=2x, slope=2

Quick review: a derivative gives us the slope of a function at any point.

These basic derivative rules can assistance united states:

  • The derivative of a abiding is 0
  • The derivative of ax is a (example: the derivative of 2x is 2)
  • The derivative of tenn is nxnorthward-1 (example: the derivative of ten3 is 3x2 )

Nosotros will utilize the little marker ' to mean "derivative of".

OK, allow's start:

To get c0, cull ten=a and then all the (x-a) terms become zero, leaving us with:

f(a) = c0

So c0 = f(a)

To get c1, have the derivative of f(ten):

f'(x) = c1 + 2c2(10-a) + 3ciii(10-a)2 + ...

With x=a all the (x-a) terms become zero:

f'(a) = c1

So ci = f'(a)

To become cii, do the derivative once again:

f' '(x) = 2c2 + 3×2×c3(10-a) + ...

With x=a all the (10-a) terms get zero:

f' '(a) = 2c2

So ctwo = f' '(a)/ii

In fact, a design is emerging. Each term is

  • the next higher derivative ...
  • ... divided by all the exponents and then far multiplied together (for which we can utilize factorial annotation, for example three! = 3×ii×i)

And we get:

f(x) = f(a) + f'(a) 1! (x-a) + f''(a) two! (x-a)2 + f'''(a) three! (x-a)3 + ...

At present we have a fashion of finding our ain Taylor Series:

For each term: take the next derivative, divide by n!, multiply by (10-a)n

Example: Taylor Serial for cos(x)

Start with:

f(x) = f(a) + f'(a) 1! (x-a) + f''(a) 2! (x-a)2 + f'''(a) 3! (10-a)3 + ...

The derivative of cos is −sin, and the derivative of sin is cos, so:

  • f(x) = cos(x)
  • f'(x) = −sin(x)
  • f''(x) = −cos(10)
  • f'''(x) = sin(x)
  • etc...

And we get:

cos(ten) = cos(a) − sin(a) i! (10-a) − cos(a) 2! (x-a)2 + sin(a) 3! (x-a)3 + ...

Now put a=0, which is squeamish considering cos(0)=1 and sin(0)=0:

cos(x) = one − 0 one! (10-0) − 1 2! (x-0)ii + 0 3! (ten-0)3 + 1 4! (ten-0)four + ...

Simplify:

cos(10) = ane − x2/2! + x4/four! − ...

Attempt that for sin(x) yourself, information technology will help you to learn.

Or try it on another function of your choice.

The key thing is to know the derivatives of your role f(10).

Note: A Maclaurin Series is a Taylor Series where a=0, so all the examples we have been using and then far can also be called Maclaurin Series.