taylor polynomial of order 'n' help needed!

[johnq2k7]

Give the Taylor polynomial of order n about x=x_0

a.) sin(Pi*x); x_0=1/2

b.)ln x; x_0=e

work shown:

for taylor series sin(x)= x-(x^3)/3! + (x^5)/5! -..... for all x

do i simply sub. in (pi*x) for x to solve a.) and how do i go about doing so?

for b.) what do i do here.. what is the taylor' s series for ln x is it simply ln e=1...

please help

 

[fasteddie65]

Give the Taylor polynomial of order n about x=x_0

a.) sin(Pi*x); x_0=1/2

sin x = x - x^3/3! + x^5/5! - ...

sin (?x) = ?x - (?x)^3/3! + (?x)^5/5! - ...

To generate a Taylor polynomial for sin x about ?/2, make a table of derivatives and evaluate each at x = ?/2:

f(x) = sin x f(?/2) = sin (?/2) = 1
f'(x) = cos x f'(?/2) = cos (?/2) = 0
f"(x) = - sin x f"(x) = - sin (?/2) = -1
etc.

So sin x = (x - ?/2) - (x - ?/2)^3/3! + (x - ?/2)^5/5! - ...

b.)ln x; x_0=e

Similarly for f(x) = ln x

f(x) = ln x f(e) = ln e = 1
f'(x) = 1/x f'(e) = 1/e
f"(x) = -1/x^2 f"(e) = -1/e^2

So ln x = 1 + (x - e) + (1/e)(x - e)^2/2! - (1/e^2)(x - 3)^3/3! + ...