# 3rd order Polynomial approximation to `e^x ` , `x in [0,1] ` using Calculus. Let `V ` be the vector space of all continuous functions `f:[0,1] -> RR ` with the standard inner product...

3rd order Polynomial approximation to `e^x ` , `x in [0,1] ` using Calculus.

Let `V ` be the vector space of all continuous functions

`f:[0,1] -> RR ` with the standard inner product

`(f,g) = int_0^1 f(x)g(x) dx `

Let `W ` = Span {`1,x,x^2,x^3 ` }

Find the best approximation to `f(x) = e^x ` that exists in `W `, that is, find the

function `g in W ` such that the Euclidean distance between the functions `f ` and `g ` , ||f-g||, is minimized.` `

### 2 Answers | Add Yours

The equation after 'I will rewrite this as' didn't display for some reason.

Anyway, it should have been

`||f-g||= int_0^1 (e^x - a_0-a_1x-a_2x^2-a_3x^3)^2 dx`

You need to square the term under the integral sign before applying the formula they give at the end of the question.

The problem is to find the best approximation to the function

`f(x)=e^x ` where `x in [0,1] `

using a function `g ` that is in W, where W is the span of `{1,x,x^2,x^3} ` ie the first 4 powers in `x ` (including the zeroeth power).

A span of a set of elements is the set of all possible linear combinations of those elements.

A potential near solution that should come easily to your mind is the Taylor series approximation to the function `f `, about `x_0=0` for example:

`g_T = 1 + x + x^2/2 +x^3/6 `

This is a good benchmark to work from, and noticing this near solution would gain some marks I think.

Now, the measure we are told to use to judge the quality of the approximation is the Euclidean distance, `||f-g|| `. That measure is equivalent to the standard inner product of `(f-g) ` with itself (its Euclidean 'norm'), which we are nearly given in the question.

Adapting the formula they give a bit we have that

`||f-g|| = int_0^1 (f(x)-g(x))^2dx`

I will rewrite this as``

``

We are then interested in solving this equation to find `{a_0,a_1,a_2,a_3} `

Using the integral formula given in the question for each of the terms in `g ` you should arrive at

`||f-g|| = 1/2(e^2-1)-2a_0(e-1)-2a_1-2a_2(e-2)-2a_3(6-2e) `

` ``+a_0^2+a_0a_1+2/3a_0a_2+a_1^2/3+1/2a_0a_3+1/2a_1a_2+2/5a_1a_3+1/5a_2^2`

`+1/3a_2a_3 + a_3^2/7 `

To minimize this with calculus, we collect the four partial derivative equations in `{a_0,a_1,a_2,a_3} ` which leaves us with a 4D linear system to solve. The method is to reduce the following matrix to row reduced Echelon Form:

2 1 2/3 1/2 | 2(e-1)

1 2/3 1/2 2/5 | 2

2/3 1/2 2/5 1/3 | 2(e-2)

1/2 2/5 1/3 2/7 | 2(6-2e)

After 8 row operations (!) I get (math editor broken now for some reason)

a_3 = (1/20)*(311-117*e)*(-1400/1749) = 0.2817199

a_2 = 135*e - 345 -(153/2)*a_3 = 0.4164714

a_1 = 363 - 141*e + (756/10)*a_3 = 1.02029

a_0 = (53/2)*e - 135/2 - (251/20)*a_3 = 0.9988832

Plugging this solution into the distance measure ||f-g|| gives a value of

1.132644e-07

Whereas, plugging in the Taylor series approx about 0 discussed above gives a value of

0.0002831481

which is about a 2500 times larger error! Remember the Taylor series approximation is only good in the region close to where it is centered (x_0).

**Best approximation obtained according to Euclidean distance measure:**

**g(x) = a_0 + a_1*x + a_2*x^2 + a_3*x_3**

**where {a_0,a_1,a_2,a_3} as given above.**

**Sources:**