# Thread: Find the minimum of an integral

1. ## Find the minimum of an integral

Find $\displaystyle \min_{a,b \in \mathbb{R}}\int_{0}^{2\pi }(1+x-(a+b\cos x))^{2}dx$ without using mathematical analysis or calculus.

Do you have any ideea how can I do this?

2. ## Re: Find the minimum of an integral

Originally Posted by cristi92
Find $\displaystyle \min_{a,b \in \mathbb{R}}\int_{0}^{2\pi }(1+x-(a+b\cos x))^{2}dx$ without using mathematical analysis or calculus. Do you have any ideea how can I do this?
Hint On the space $\displaystyle \mathcal{C}[0,2\pi]$ consider the inner product $\displaystyle <f,g>=\int_0^{2\pi}f(x)g(x)dx$ . Then, the integral is the square of the norm of $\displaystyle 1+x-(a+b\cos x)$ .

Thank you!

4. ## Re: Find the minimum of an integral

Hmm, I don't get this.

I just verified that the result is $\displaystyle {2\pi^3 \over 3}$.
But how do you get that from the norm?

5. ## Re: Find the minimum of an integral

Originally Posted by ILikeSerena
Hmm, I don't get this. I just verified that the result is $\displaystyle {2\pi^3 \over 3}$. But how do you get that from the norm?
The given integral is $\displaystyle I(a,b)=\left\|{1+x-(a+b\cos x)}\right\|^2=d^2(1+x,a+b\cos x)$ , so the minimum of $\displaystyle I(a,b)$ is the square root of the distance of $\displaystyle 1+x$ to its orthogonal projection onto $\displaystyle \textrm{Span}[1,\cos x]$ .

6. ## Re: Find the minimum of an integral

Nice!

I'm still wondering though how to arrive at the result....

The only way I can think of, is calculating the integrals of (1+x) with 1 and with cosx (after norming them) to find the projection, which is (1+pi).
And then calculate the integral of ((1+x)-(1+pi))^2, which is indeed $\displaystyle 2\pi^3 \over 3$.

But isn't this analysis and calculus?
Is there an easier way?

7. ## Re: Find the minimum of an integral

Originally Posted by ILikeSerena
But isn't this analysis and calculus? Is there an easier way?
Well, there is no strict boundary between Calculus and Algebra, but if you use the inner product and the concept of orthogonal projection we can consider more an algebraic than analytic method. Besides, if we compute the Gram matrix $\displaystyle G$ with respect to the basis $\displaystyle B=\{1,x\cos x\}$ of $\displaystyle V=\textrm{span}[1,x,\cos x]$ we can express $\displaystyle <f(x),g(x)>=\sqrt{[f]_B^t\;G\;[g]_B}$ and from this point, the problem is purely algebraic.

I see.
Thanks!