# Taylor Polynomial Approximations (Apostol Section 7.8 #7)

1. Sep 9, 2011

### process91

1. The problem statement, all variables and given/known data
Prove that $0.493948<\int_0^{1/2} \frac{1}{1+x^4} dx<0.493958$

2. Relevant equations
This chapter is about Taylor Polynomials, and specifically this section deals with Taylor's formula with remainder:
$$f(x)=\sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k + E_n(x)$$
The general formula for $E_n(x)$ is given in integral form:
$$E_n(x)=\frac{1}{n!}\int_a^x (x-t)^n f^{(n+1)} (t) dt$$

There is a theorem which gives bounds for the error:
If the (n+1)st derivative of f satisfies the inequalities
$$m \le f^{(n+1)} (t) \le M$$

for all t in some interval containing a, then for every x in this interval we have the following estimates:
$$m \frac{(x-a)^{n+1}}{(n+1)!} \le E_n(x) \le M \frac{(x-a)^{n+1}}{(n+1)!} \qquad \text{if} \quad x>a$$

(and a similar one if x<a, which I don't think we'll need since 1/2 > 0)

3. The attempt at a solution
The only example similar to this is approximating $\int_0^{1/2} e^{-t^2}dt$, which is much easier since we know that the nth derivative of $e^x$ is always $e^x$.

I've tried a number of methods, but all seem unlikely to be what Apostol had in mind, given their complexity. The most straightforward, I suppose, would be to use a method similar to the book's, however there's no telling what the n+1st derivative is in general, and to get close to the book's bounds we would need to get to at least the sixth derivative, and then figure out where it is equal to zero on (0,1/2) so that we could go back and put bounds on the fifth derivative (yuck!).

Alternatively, I had found that in the work for #3 (a lot of these problems build on previous ones) I had come up with the formula

$$\frac{1}{1+x^2} - \sum_{k=0}^{n-1}(-1)^k x^{2k} -x^{2n} \le 0$$

Which, if we replace x by x^2, gives an integrable formula for the upper bound. There was also an analogous formula for the lower bound. The problem is, in order to approach the bounds given in the problem, I had to take it to n=4, which yields some serious arithmetic for a book written before calculators. Also, I don't get the lower bound exactly, I get a much higher lower bound.

Is this really what Apostol had in mind? I feel like I'm missing some crucial simplifying shortcut.

Last edited: Sep 9, 2011
2. Sep 9, 2011

### Dick

Actually, it's pretty easy to expand 1/(1+x^4) in a taylor series without taking any derivatives if you remember the geometric sum formula 1+u+u^2+...=1/(1-u) for |u|<1. What should u be?

Actually, I've realized I'm probably not answering your question here. I would integrate that sum term by term and use that it's an alternating series. There is another way to estimate the sum of an alternating series without using a remainder term. I'm not sure what Apostol had in mind, but even before calculators approximating something like that to six decimal places was not unheard of.

Last edited: Sep 9, 2011
3. Sep 9, 2011

### process91

I have the expansion. Even through long division, it comes out to be

$$\frac{1}{1+x^4}=1-x^4+x^8-x^{12}+....$$

The book hasn't covered infinite series yet, however. This is similar to another method I tried, however. I expanded it out to be:

$$\frac{1}{1+x^4}=\sum_{k=0}^{n-1} (-1)^k x^{4k} + E_{2n+3}(x)$$

Where $E_{2n+3}(x)=(-1)^{n+1} \frac{x^{4n+2}}{1+x^2}$.

4. Sep 9, 2011

### Dick

I added to my last post. How about integrating term by term and using an alternating series estimate?

5. Sep 9, 2011

### process91

We haven't covered any series stuff yet, that's all in two chapters. Specifically, alternating series estimates have not been discussed.

6. Sep 9, 2011

### Dick

Ok, funny you would do taylor series before doing series. Then I guess you have to use a remainder term. Integrating the series f(x)=1-x^4+x^8-x^16+... gives you something easy enough to sum even up to six, seven or even eight terms. You just have to make sure the remainder term is small enough. If you've got a remainder term like x^(4n+2)/(1+x^2) that should be easy enough to estimate.

7. Sep 9, 2011

### process91

That's kind of what my original work resulted in, however I don't think I would have been able to accurately calculate that. Here's the calculation:

http://www.wolframalpha.com/input/?i=1%2F2+-+%281%2F2%29^5+%2F5+%2B%281%2F2%29^9+%2F9+-%281%2F2%29^13%2F13+%2B%281%2F2%29^17%2F17

I would be sure to make errors in doing $\frac{(1/2)^{17}}{17}$, and it just seems out of place to me that he would ask such a difficult computational question when, in a question above it, he gives you that $2^{16} = 65536$, presumably because he felt it would be too difficult to calculate. The other reason that I suspect he's hinting at a different method is that the estimates I get are so much more precise than his, but stopping any shorter gives estimates well outside his.

8. Sep 9, 2011

### process91

I know it seems like I'm just whining about doing some calculations, but what concerns me is that I think there may be a different method that he wants to emphasize here that I am unaware of. The question above it asks to prove that

$\int_0^1 \frac{1+x^{30}}{1+x^{60}} dx = 1 + \frac{x}{31}, \qquad where \quad 0<x<1$

Again, I can brute force this to a smaller range than he specifies above using some difficult calculations, however I feel there is a more elegant solution which is eluding me. For the problem above, I'm sure it has something to do with equating the standard integral form for the error (above) with the Lagrange form of the error:

$E_n (x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1},\qquad \text{where} \quad a<c<x$

with a=0 and x=1, however I'm still messing with that one - it doesn't come out quite cleanly. I don't know if the method that I'm missing is related to both of them, or if there really isn't one. I probably shouldn't waste too much time on it, at any rate.

Last edited: Sep 9, 2011
9. Sep 9, 2011

### Dick

Oh, I don't know. I got 2^16=65536 on my second try by doing it long hand from 256*256. I muffed the first one because I'm apparently not very good at lining up columns any more. That the integral of (1+x^30)/(1+x^60) is 1/31 to any reasonable precision is pretty easy, the powers are just so LARGE. You can ignore everything after the first two terms. If there are any secrets here, I'm not aware of them.

10. Sep 9, 2011

### process91

Yeah, I was surprised to see $2^{16}$ given, since so many other calculations were much more tedious.

I probably shouldn't derail my own thread, but with regard to the x^30 and x^60 equation, here was my approach:

We have that $E_n (x) = \frac {1}{n!} \int_a^x (x-t)^n f^{(n+1)}(t)dt$ and also that $E_n (x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1},\qquad \text{where} \quad a<c<x$. Setting these equal to each other, with the following substitutions:

$$a=0, x=1, n=30$$
$$f^{(n+1)}(t) = \frac{1+t^{30}}{(1+t^{60})(1-t)^{30}}$$

We have
$$\int_0^1 \frac{1+t^{30}}{1+t^{60}} dt = \frac{1}{n+1} \frac{1+c^{30}}{(1+c^{60})(1-c)^{30}} \qquad \text {where} \quad 0 < c < 1$$
which is so remarkably similar to the original question. All that is left is to prove that $31<f^{(n+1)}(c) <32 \qquad \forall c \in (0,1)$, but this is not true.

Another form of the remainder given is $E_n(x) = \frac{f^{(n+1)}(c)}{n!}(x-c)^n(x-a),\qquad \text{where} \quad a<c<x$, and with the same substitutions we arrive at

$$\int_0^1 \frac{1+t^{30}}{1+t^{60}} dt = \frac{1+c^{30}}{1+c^{60}} \qquad \text{where} \quad 0<c<1$$

Now, all that is left is to prove $1 \le \frac{1+c^{30}}{1+c^{60}} \le \frac{32}{31}\qquad \forall c\in(0,1)$. This gives a nice lower bound of 1, as desired, but the upper bound is too lenient.

11. Sep 10, 2011

### awkward

Here is an approach that I think may work.

$$\frac{1}{1+t} = 1 - t + t^2 - t^3 + E_3(t)$$
and develop a bound on the error term $$E_3(t)$$
Then make the substitution $$t = x^4$$ and integrate from 0 to 1/2. Use your bound on E_3 to develop a bound on the error in the integral.