Very Easy Taylor Series Approximation Help

[Farzan]

Homework Statement

Approximate f by a Taylor polynomial with degree n at the number a.

f(x) = x^(1/2)

a=4
n=2

4<x<4.2
(This information may not be needed for this, there are two parts but I only need help on the first)

Homework Equations

Summation f^(i) (a) * (x-a)^i / i!

The Attempt at a Solution

Derivatives...

x^(1/2)<br /> 0.5 * x^(-1/2)<br /> -0.25 * x^(-3/2)

So the Taylor series of order n=2 should be...

2 + 0.25(x-4) - (1/64)(x-4)(x-4)

Now, my question...

To find the approximation of square root of x, do I just plug in 4 to that?

That would make most of the terms zero, leaving me with 2

Does this mean that whenever the Taylor series is centered at a number, the first term is the approximation?Ugh, sorry for the failure using tex tags :( It's my first post here

 

[Dick]

Yes. Exactly. If x=a the first term IS the approximation. The usual game is to approximate something like 4.1 using the series.

 

[HallsofIvy]

I think Dick is interpreting the question differently than I am.

What you have: 2+ (1/4)(x-4)- (1/64)(x- 4)² alread is the quadratic approximation to \sqrt{x} around x= 4.

Putting x= 4 would just give you the obvious \sqrt{4}= 2.