1. ## Inner Products.

For V = C[0,1] with inner product <f,g> = ∫ f(x)g(x)dx [For limits 0 to 1] if f(0) = 0 = f(1) show that f'(x) and f(x) are orthogonal.

2. $\displaystyle \left\langle {f,f'} \right\rangle = \int_0^1 {f\left( x \right) \cdot f'\left( x \right)dx}$

Since: $\displaystyle {\textstyle{1 \over 2}} \cdot \left( {f^2 \left( x \right)} \right)^\prime = f\left( x \right)f'\left( x \right)$ it follows that: $\displaystyle \int_0^1 {f\left( x \right) \cdot f'\left( x \right)dx} = \left. {{\textstyle{{f^2 \left( x \right)} \over 2}}} \right|_0^1 = {\textstyle{{f^2 \left( 1 \right) - f^2 \left( 0 \right)} \over 2}} = 0$

Thus: $\displaystyle \left\langle {f,f'} \right\rangle =0$

3. May I know how did you come to this part :

4. Originally Posted by pearlyc
May I know how did you come to this part :

$\displaystyle f^2 \left( x \right) = f\left( x \right) \cdot f\left( x \right)$

and now we apply the product rule: $\displaystyle \left[ {f\left( x \right) \cdot f\left( x \right)} \right]^\prime = f'\left( x \right) \cdot f\left( x \right) + f\left( x \right) \cdot f'\left( x \right)$

5. Oh right!

Thank you.