Problem II: Taylor / Maclaurin Polynomial Estimate for An Important Integral from Probability

What is the area of the region in the plane bounded by the X-axis, X=0, X=1 and `Y = e^{-X^2}`? or

what is `int_0^1 e^{-x^2} dx` ?

What does your calculator give for an estimate?

What do you find when you integrate this with a TI 89?

What does Winplot give for an estimate?

The computer algebra system MAPLE gave an estimate of 0.7468241330 for this integral.

Winplot gave a midpoint estimate with n = 10 of 0.74713;with n = 1000 the estimate was 0.72682.

Problem 1: Find a fourth degree polynomial P(

Solution: `P(x) = 1 + x + {x^2}/2 + {x^3}/{2*3) + {x^4}/{4!}`

The key concept is to** use information about derivatives to design the polynomial.**

At x = 0, we find that *f*(0) = 1, *f *'(0) =1, and since
*f
*'(x)
= *f*(x) we have that *f* ^{(n)}(0) = 1 for all
natural numbers
*n*. We now consider the key result, typical of
the Taylor's theory approach to estimation using polynomials based on derivative
information.

**THEOREM**:

If
`P_n(x) = 1 + x + {x^2}/2 + {x^3}/{2*3) + {x^4}/{4!} + {x^5}/{5!} +...+ {x^n}/{n!} ` then P* _{n}*(

` P_n(0) = P_n '(0) = P_n''(0) =...= P_n^{(n)}(0) = 1`

and P* _{n}*(b) is approximately equal to e

In fact, if we let the error in this estimation be denoted by R

then for some c between 0 and b

`R_n = e^c {b^{n+1}}/{(n+1)!}`.

**Proof: **[Discuss this at the end of the presentation if time permits.]

To illustrate the key ideas
of
**Taylor's theory**, we start by estimating the number e with a polynomial**.**

**Estimating e:** We
**apply
the theorem using n = 5 to estimate the value of e** [= e

First, ` P_5(x) = 1 + x + {x^2}/2 + {x^3}/{2*3) + {x^4}/{4!} + {x^5}/{5!} = 1 + x + {x^2}/2 + {x^3}/6 + {x^4}/24 + {x^5}/120 `.

So according to the theorem, e is approximately equal to

` P_5(1) =1 + 1 + {1^2}/2 + {1^3}/6 + {1^4}/24 + {1^5}/120 ~~ 2.716667`

and R_{5 }= e - P* _{5}*(1) where `R_5 = e^c {1^6}/{6!} = e^c 1/720 `
for some c between 0 and 1.

Thus, e is approximately 2.716667 and the error
in using this estimate is R_{5}. Since c is between 0 and 1, e^{c}
is between 1 and e, so we can deduce that R_{5}, the difference
between e and the estimate, P* _{5}*(1), is no greater than
e/720. Using estimates of e from your earlier work we know
that e < 3, so our errror R

A more accurate estimate can be obtained by using a larger value for n. Try using n=6 and 7 to see the improvement. [This can be done progressively using a spreadsheet which you can examine now or later.]

We are now ready to estimate `int_0^1 e^{-t^2} dt`.

**Solution**: For each *t* between 0 and 1, -*t*^{2}
is between -1 and 0. Let *x* = -*t*^{2}. By our theorem with n = 4, **for each x between -1** and 0 we have

Since c < 0 , 0 < e^{c} < 1, and since x < 0 we
have that `0 > R_4(x) > {x^5}/120`.

Substituting -*t*^{2} for *x* ,we see that

` e^{-t^2} = 1 + (-t^2) + {(-t^2)^2}/2 + {(-t^2)^3}/6 + {(-t^2)^4}/24 + R_4(-t^2)`

`= 1 + (-t^2) + {t^4}/2 - {t^6}/6 + {t^8}/24 + R_4(-t^2)`

Now since `0 > R_4(-t^2) > {(-t^2)^5}/120 = {-t^10}/120` we have `int_0^1 P_4(-t^2) dt > int_0^1 e^{-t^2} dt > int_0^1 P_4(-t^2) - {t^10}/120 dt`.

By evaluating these integrals we obtain

`1 - 1/3 +1/10 - 1/42 + 1/216 > int _0^1 e^{-t^2} dt > 1 -1/3 + 1/10 - 1/42 + 1/216 - 1/1320` .

Therefore,

`int_0^1 e^{-x^2} dx ~~ 1 - 1/3 +1/10 - 1/42 + 1/216 ~~ 0.747486.`

and this is an overestimate by no more than `1/1320`.

**Comment**: We have been able to estimate
a definite integral involving the exponential function by using a Taylor
polynomial of degree 4. It should be apparent that by using a higher degree
for the estimating polynomial, the error term will become smaller and we
will obtain a more precise estimate. The systematic pattern in these polynomials
should allow you to find more precise estimates for `int_0^1 e^{-x^2} dx` without
much difficulty.

**Proof of Proposition**: We'll begin our proof by finding P* _{n}*'(

`P_n'(x) = 1 + x + {x^2}/2 + {x^3}/{2*3} + {x^4}/{4!} + ... +{nx^{n-1}}/{n!} `

`= 1 + x + {x^2}/2 + {x^3}/{2*3} + {x^4}/{4!} + ... + {x^{n-1}}/{(n-1)!} = P_{n-1}(x)`

From this it follows easily that ` P_n(0) = P_n '(0) = P_n''(0) =...= P_n^{(n)}(0) = 1`.

Now suppose that b is not 0 and let R* _{n}* = e

We need only justify the formula for evaluating R

`g(b) = e^b * P_n(b-b) + R* {(b-b)^{n+1}}/{b^{n+1}} = e^b*P_n(0) = e^b`, while `g(0) = e^0 * P_n(b-0) + R* {(b-0)^{n+1}}/{b^{n+1}} = P_n(b) + R = e^b` from the definition of R.

Furthermore `g'(t) = e^t*P_n(b-t) - e^t*P_n'(b-t)
- R* {(n+1)(b-t)^n}/{b^{n+1}}`

`= e^t* [P_n(b-t) - P_n'(b-t)] - R*{(n+1)(b-t)^n}/{b^{n
+ 1}}`

But `P_n(x) - P_n'(x) = {x^n}/{n!}, so

`g'(t) = e^t*{(b-t)^n}/{n!} - R* {(n+1)(b-t)^n}/{b^{n+1}}`.

Now we apply the Mean Value (or Rolle's) Theorem to the function *g*,
we can say that there is a number c between 0 and b where *g*'(c)=0.
Thus

`0 = g'(c) = e^c*{(b-t)^n}/{n!} - R* {(n+1)(b-t)^n}/{b^{n+1}}`

and

`e^t*{(b-t)^n}/{n!} = R* {(n+1)(b-t)^n}/{b^{n+1}}`.

Solving this last equation for R gives `R_n = R = e^c {b^{n+1}}/{(n+1)!}.