1. ## Taylor problem #2

Explain the expression that sqrt(1+x) is "close" to 1+(1/2)*x for "small" x.

I tried to solved with epsilon-delta:
Let e>0, so there is delta(e)>0 so that
|x|<delta ==> |sqrt(1+x)-{1+(1/2)*x}|<e

2. The function can be written as an Taylor polynome. Doing this for the first two terms leads to:

(1+0)^2*x^0/0! + 0.5*(1+0)*x^1/1! = 1 + 0.5*x

3. Originally Posted by Jeroentje
The function can be written as an Taylor polynome. Doing this for the first two terms leads to:

(1+0)^2*x^0/0! + 0.5*(1+0)*x^1/1! = 1 + 0.5*x

Also explain what close means here by considering the form of the remainder for this truncated Taylor series.

CB

4. You are right CB.

For most functions it can be stated that the for small x the first terms of the Taylor series will become dominant over the higher terms. This is simular to a normal polynome.

5. Looking to all three factors of the Taylor serie

$\displaystyle f^{(n)}(x=0)$ will become positive and negative alternatly, but the absolute value will become lower for higher n.

$\displaystyle x^n$ will become lower for higher n (for small x)

$\displaystyle 1/n!$ will become lower for higher n.

As you can see, all the three terms becomes lower for higher n. This means that the lower n terms will become dominant. The smaller x is, the more dominant the first terms will be.