# Making three equations orthonormal by forcing the constants

1. Make these three equations orthonormal to each other in the interval 1 less than or equal to x greater than or equal to 1 (this may be a typo as I think it should read 1 less than or equal to x less than or equal to one) by determining the appropriate values for the coefficients a b c d e f.

1) a
2) b+c*x
3) d+e*x+f*x^2

## Homework Equations

Definition of orthonormality is all we are given.

## The Attempt at a Solution

I think I understand how orthonormality works with vectors but this equation thing is throwing me off. Perhaps if someone could start by getting my on the right track then I can complete it (as it is a homework problem). Thanks!

Erik

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sorry. the paper says -1 less than or equal to x greater than or equal to 1 but I think it should be -1 less than or equal to x less than or equal to 1... as in [-1,1]

SammyS
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Homework Helper
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1. Make these three equations orthonormal to each other in the interval -1 less than or equal to x greater than or equal to 1 by determining the appropriate values for the coefficients a b c d e f.

1) a
2) b+c*x
3) d+e*x+f*x^2

## Homework Equations

Definition of orthonormality is all we are given.

## The Attempt at a Solution

I think I understand how orthonormality works with vectors but this equation thing is throwing me off. Perhaps if someone could start by getting my on the right track then I can complete it (as it is a homework problem). Thanks!

Erik

Hi Eric.

You could edit your original post to avoid confusion.

A set of functions, {fi(x) | i ∈ N}, is orthonormal on some interval [a, b] if:

$$\int_a^b f_i(x)\cdot f_j(x)\,dx=\left\{\begin{array}{cc}\ 1,&\mbox{ if } i=j\\ \ 0, & \mbox{ if } i\neq j\end{array}\right.$$

In this case, [a, b] = [-1, 1].

Therefore, do something like:

Let f1(x)=a.
Let f2(x)=b+c·x.
Let f3(x)=d+e·x+f·x2.

Clearly, a = 1/√(2).

$$\int_{-1}^1 (b+c\cdot x)^2 dx=1$$

$$b^2\int_{-1}^1 (1+\frac{2c}{b} x+\frac{c^2}{b^2}\,x^2) dx=1$$

$$\left(x+\frac{2c}{2b} x^2+\frac{c^2}{3b^2}\,x^3\right)_{x=-1}^{x=1} =\frac{1}{b^2}$$

$$2 + \frac{2c^2}{3b^2} =\frac{1}{b^2}$$

$$6\,b^2+2\,c^2 =3$$

Also,
$$\int_{-1}^1 (\frac{1}{\sqrt{2}})(b+c\cdot x) dx=0$$

$$\int_{-1}^1 (\frac{1}{\sqrt{2}})(b+c\cdot x) dx=0$$

$$\left(b\,x+\frac{c}{2} x^2\right)_{x=-1}^{x=1} =0$$

$$\displaystyle 2\,b=0\quad\to\quad b=0\,$$

Then:
$$c=\pm\sqrt{\frac{3}{2}}$$

Now, continue on with it.

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