# Very Easy Taylor Series Approximation Help

1. Apr 16, 2007

### Farzan

1. The problem statement, all variables and given/known data

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)

2. Relevant equations

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

3. The attempt at a solution

Derivatives...

$$x^(1/2) 0.5 * x^(-1/2) -0.25 * x^(-3/2)$$

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

2. Apr 16, 2007

### 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.

Last edited: Apr 17, 2007
3. Apr 17, 2007

### HallsofIvy

Staff Emeritus
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 $\sqrt{4}= 2$.