Solve 3rd Degree Taylor Polynomial for \sqrt[ ]{9.03}

[Telemachus]

Homework Statement

Find an approximated value for $$\sqrt[ ]{9.03}$$ using a Taylors polynomial of third degree and estimate the error.

Homework Equations

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?

 

[vela]

9.03-9 = 0.03, not 0.3. :)

 

[Telemachus]

****, thanks haha

 

[Mark44]

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)|.

 

[Telemachus]

Thank you Mark. I think I get it.

Here it is:

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

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

 

[jegues]

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)|.

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

 

[Mark44]

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

 

[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 $$k=max{f(b) \textsf{ such that b} \in{[x,a]}$$

 

[Mark44]

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

 

[Hurkyl]

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

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.

 

[jegues]

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.

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