# Taylors polynomials

1. Aug 11, 2010

### Telemachus

1. The problem statement, all variables and given/known data
Find an approximated value for $$\sqrt[ ]{9.03}$$ using a Taylors polynomial of third degree and estimate the error.

2. Relevant equations

3. The attempt at a solution
I thought of solving it by using

$$f(x)=\sqrt[]{x}$$ centered at $$x_0=9$$

So

$$P_n(x)=3+\dysplaystyle\frac{(x-9)}{6}-\dysplaystyle\frac{(x-9)^2}{216}+\dysplaystyle\frac{3(x-9)^3}{3888}$$

Then I evaluated it at x=9.03, so I get:

$$P_n(x)=3+\dysplaystyle\frac{(0.3)}{6}-\dysplaystyle\frac{(0.3)^2}{216}+\dysplaystyle\frac{3(0.3)^3}{3888}\approx{3.049604167}$$

I don't know if this is right, I've tried with the calculator and it gives 3.00500.... Now, how do I estimate the error? just by resting to the value the calculator gives the one I get?

2. Aug 11, 2010

### vela

Staff Emeritus
9.03-9 = 0.03, not 0.3. :)

3. Aug 11, 2010

### Telemachus

****, thanks haha

4. Aug 11, 2010

### Staff: Mentor

For the error estimate, there's a formula for the error in approximating a Taylor series by the first n terms (terms up through degree n). If my memory is correct, it looks something like this.

$$R_n(x)= \frac{f^{n + 1}(c)}{(n + 1)!}(x - a)^{n + 1}$$

Since it is usually impossible to determine c (which is between x and a), in practice, one finds an upper bound on |f(n+1)(x)|.

5. Aug 12, 2010

### Telemachus

Thank you Mark. I think I get it.

Here it is:

$$R_{n+1}(x)=\displaystyle\frac{-15(x-9)^4}{4!16\sqrt[ ]{\alpha}}$$

$$\alpha=9+0.03\theta$$, $$\theta\in{(0,1)}$$

Last edited: Aug 12, 2010
6. Aug 12, 2010

### jegues

I could be wrong, but shouldn't this be an inequality? That's how I remember it.

7. Aug 12, 2010

### Staff: Mentor

No, it shouldn't be an inequality. Checking with a calculus text, the formula for the remainder is correct.

8. Aug 12, 2010

### Telemachus

It can be thought as inequality like this:

$$|R_n(x)|= |\frac{f^{n + 1}(c)}{(n + 1)!}(x - a)^{n + 1}|\leq{}\displaystyle\frac{(x-a)^{n+1}}{(n+1)!} k$$
Where $$k=max{f(b) \textsf{ such that b} \in{[x,a]}$$

Last edited: Aug 12, 2010
9. Aug 12, 2010

### Staff: Mentor

The remainder is an equality. Usually it suffices to get an upper bound on that derivative - then you have an inequality.

10. Aug 12, 2010

### Hurkyl

Staff Emeritus
The various forms of the Taylor remainder are generalizations of the mean value theorem to higher derivatives and higher degree polynomials -- the remainder is exactly equal to the given expression for some point c in a known interval.

Typical practical applications find the maximum over the entire interval, and then use that as an upper bound on the true remainder.

11. Aug 12, 2010

### jegues

Thank you both, Mark and Hurkyl for clearing things up!