# Using Taylor's Theorem to approximate

1. May 9, 2005

### Shameel

Hi Guys,

Is there any whay I can use the following theorem to approximate the value of e^x at x=0?

$$f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + ...$$

If the above function is not used for approximation, then what is it used to do?

Thanks heaps

2. May 9, 2005

### arildno

Why bother approximating $$e^{0}????$$
You know what it is: $$e^{0}=1$$

However, since you know that, you are now in a position to approximate with only a couple of terms in the Taylor series function values like $$e^{0.01}, e^{0.17671234}$$

3. May 9, 2005

### Shameel

I am a n00b. Could you give me an example how Taylor approximations work?

4. May 9, 2005

### dextercioby

Sure.

$$e^{0.001}\simeq e^{0}+1(0.001-0)+\frac{1}{2}(0.001-0)^{2}$$

Daniel.

5. May 9, 2005

### arildno

Let us try to approximate $$e^{0.01}$$ by forming the Taylor series about the origin:
Then, we have:
$$e^{x}=f(x)\approx{f}(0)+f'(0)(x-0)+\frac{1}{2}f''(0)(x-0)^{2}$$
where I've retained the 3 first terms in the Taylor series.
But, now, we have:
$$f(0)=e^{0}=1,f'(0)=e^{0}=1,f''(0)=e^{0}=1$$
since the derivative of the exponential function is itself
Thus, we have:
$$e^{0.01}\approx{1}+1*(0.01-0)+\frac{1}{2}*1*(0.01-0)^{2}=1+\frac{1}{100}+\frac{5}{100000}=1.01005$$