Question

Derive the first three (non-zero) terms of Taylor's series expansion for the function

(a) f(x)=sin(x)f(x)=sin⁡(x)  about the origin and thereby estimate sin(0.2)=​

Answer 1

We want to find the first 3 terms of the Taylor's series expansion for f(x) = sin(x) around x = 0.

Remember that a Taylor's series expansion of a function f(x) around the point x₀ is given by:

`f(x) = f(x_0) + (1)/(2!)f'(x_0)*(x - x_0) + (1)/(3!)*f''(x_0)*(x - x0)^2 + ...`

Where in the formula we have the first 3 terms of the expansion (but there are a lot more).

So, if:

f(x) = sin(x)

x₀ = 0

The terms are:

`f(x_0) = sin(0) = 0`

`(1)/(2!)*f(x_0)'*(x - x_0) = (1)/(2) cos(0)*(x - 0) = x/2`

`(1)/(3!)*f(x_0)''*(x - x_0)^2 = (1)/(6)*-sin(0)*(x - 0)^2 = 0`

`(1)/(4!)*f(x_0)'''*(x - x_0)^3 = (1)/(24)*-cos(0)*(x- 0)^3 = -(x^3)/(24)`

We already can see that the next term is zero (because when we derive the cos part, we will get a sin() that is zero when evaluated in x = 0), then the next non zero term is:

`(1)/(6!)*f(x_0)''''*(x - x_0)^5 = (1)/(2*3*4*5*6) *(x - 0)^5 = (x^5)/(720)`

Then we can write:

`sin(x) = (x)/(2) - (x^3)/(24) + (x^5)/(720)`

Evaluating this in x = 0.2, we get:

`sin(0.2) = (0.2)/(2) - (0.2^3)/(24) + (0.2^5)/(720) = 0.099667`