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**1. Homework Statement**

Show that:

[tex]\varphi_{0}(x) = f_{0}(x)[/tex]

and

[tex]\varphi_{1}(x) = f_{1}(x) - \frac{\left\langle\right\varphi_{0},f_{1}\rangle}{\left\|\varphi_{0}\right\|^{2}}\varphi_{0}(x)[/tex]

are orthogonal on the interval [a,b].

**2. Homework Equations**

Orthogonal functions satisfy:

[tex]\left\langle\right\varphi_{m},\varphi_{n}\rangle = \int^{b}_{a}\varphi_{m}(x)\varphi_{n}(x)dx = g(m)\delta_{mn}[/tex]

Where, [tex]\delta_{mn}[/tex] is the Delta Kronecker.

Also:

[tex]\left\langle\right\varphi_{m},\varphi_{m}\rangle = \left\|\varphi_{m}\right\|^{2}[/tex]

**3. The Attempt at a Solution**

Since m and n (0 and 1) are not equal, the Delta Kronecker is zero and therefore the proof is a matter of proving that:

[tex]\left\langle\right\varphi_{0},\varphi_{1}\rangle = 0[/tex]

Having substituted the functions into the inner product formula in 2:

[tex]\left\langle\right\varphi_{0},\varphi_{1}\rangle = \int^{b}_{a}\varphi_{0}(x)\varphi_{1}(x)dx =

\int^{b}_{a}f_{0}(x)\left[f_{1}(x) - \frac{\left\langle\right\varphi_{0},f_{1}\rangle}{\left\|\varphi_{0}\right\|^{2}}\varphi_{0}(x)\right]dx[/tex]

Because the Delta Kronecker is zero, all I have to do is show that:

[tex]\int^{b}_{a}f_{0}(x)\left[f_{1}(x) - \frac{\left\langle\right\varphi_{0},f_{1}\rangle}{\left\|\varphi_{0}\right\|^{2}}\varphi_{0}(x)\right]dx = 0[/tex]

I'm unsure as to whether I should use integration by parts to do the resulting integral because there is another integral embedded in the [tex]\varphi_{1}(x)[/tex] function; which (because it is a definite integral) would be tricky to differentiate or integrate.