Differentiate expression involving an integral

Question

How can I minimize this expression with respect to $c_i$ i.e. differentiate this expression, equate to zero and solve for $c_i$. The expression is $\sqrt{\int_a^b \left| f(x) - \sum_{i=1}^k c_i \phi_i(x) \right|^2 dx}$. I need to obtain $c_i$ in terms of the other terms.

Answer 1

1. Start by taking the derivative of the expression with respect to $c_i$. Let's call the expression: $$ E(c_i) = \sqrt{\int_a^b \left| f(x) - \sum_{i=1}^k c_i \phi_i(x) \right|^2 dx} \ $$ Then, the derivative of $E(c_i)$ with respect to $c_i$ is:

$$ \frac{dE}{dc_i} = \frac{1}{2}\left(\int_a^b \frac{2}{2} \frac{d}{dc_i}\left| f(x) - \sum_{i=1}^k c_i \phi_i(x) \right|^2 dx\right)^{-\frac{1}{2}} \cdot \frac{d}{dc_i}\left| f(x) - \sum_{i=1}^k c_i \phi_i(x) \right|^2 $$

2. Now, you need to compute the derivative of the absolute value term: $$ \ \frac{d}{dc_i}\left| f(x) - \sum_{i=1}^k c_i \phi_i(x) \right|^2 \ $$ You can use the chain rule here. Let's denote the absolute value term as $g(x) = \left| f(x) - \sum_{i=1}^k c_i \phi_i(x) \right|^2$. Then, $$ \ \frac{d}{dc_i}g(x) = \frac{d}{dg}g(x) \cdot \frac{d}{dc_i}\left(f(x) - \sum_{i=1}^k c_i \phi_i(x)\right)^2 \ $$ Since the derivative of $g(x)$ with respect to $g$ is simply $2g(x)$, you have: $$ \ \frac{d}{dc_i}g(x) = 2\left| f(x) - \sum_{i=1}^k c_i \phi_i(x) \right|\cdot \frac{d}{dc_i}\left(f(x) - \sum_{i=1}^k c_i \phi_i(x)\right) \ $$
3. Now, calculate the derivative of $f(x) - \sum_{i=1}^k c_i \phi_i(x)$ with respect to $c_i$: $$ \ \frac{d}{dc_i}\left(f(x) - \sum_{i=1}^k c_i \phi_i(x)\right) = -\phi_i(x) \ $$
4. Plug the result from step 3 into step 2: $$ \ \frac{d}{dc_i}g(x) = -2\left| f(x) - \sum_{i=1}^k c_i \phi_i(x) \right|\cdot \phi_i(x) \ $$
5. Now, plug the result from step 4 into step 1: $$ \ \frac{dE}{dc_i} = \frac{1}{2}\left(\int_a^b \frac{2}{2} \left(-2\left| f(x) - \sum_{i=1}^k c_i \phi_i(x) \right|\cdot \phi_i(x)\right) dx\right)^{-\frac{1}{2}} \ $$
6. Simplify and solve for $c_i$: $$ \ \frac{dE}{dc_i} = -\frac{\left(\int_a^b \left| f(x) - \sum_{i=1}^k c_i \phi_i(x) \right|\cdot \phi_i(x) dx\right)}{E(c_i)} \ $$
7. Set $\frac{dE}{dc_i}$ equal to zero and solve for $c_i$: $$ \ 0 = \int_a^b \left| f(x) - \sum_{i=1}^k c_i \phi_i(x) \right|\cdot \phi_i(x) dx \ $$
8. Solving the equation can be a good challenge.