Math Genius: Find the maximum and minimum value of $f(x)$

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$$f(x) = sin x + int_{-frac pi 2}^{frac pi 2} (sin x + tcos x)f(t)dt$$
Find the minimum and maximum value of $f(x)$.

My attempt:

Rewrite the functional equation as
$$f(x) = sin x left( 1 + int_{-frac pi 2}^{frac pi 2} f(t)dtright) + cos x int_{-frac pi 2}^{frac pi 2} tf(t)dt$$
Then differentiate both sides
$$f'(x) = cos x left( 1 + int_{-frac pi 2}^{frac pi 2} f(t)dtright) – sin x int_{-frac pi 2}^{frac pi 2} tf(t)dt$$
for maxima/minima, $f'(x)$ = 0
$$cos x left( 1 + int_{-frac pi 2}^{frac pi 2} f(t)dtright) = sin x int_{-frac pi 2}^{frac pi 2} tf(t)dt$$
I got stuck at this point.

I found a solution. Since
$$f(x) = sin x left( 1 + int_{-frac pi 2}^{frac pi 2} f(t)dtright) + cos x int_{-frac pi 2}^{frac pi 2} tf(t)dt$$
We can rewrite it as $$f(x) = Asin x + Bcos x$$
This gives us the equations
$$begin{gather}
A = 1 + int_{-frac pi 2}^{frac pi 2}f(t)dt tag{1} \
B = int_{-frac pi 2}^{frac pi 2} tf(t)dt tag{2}
end{gather}$$

Using some integration properties, it is easy to see that $int_{-frac pi 2}^{frac pi 2}f(t) = int_{-frac pi 2}^{frac pi 2}Bcos t dt$ and $int_{-frac pi 2}^{frac pi 2}tf(t)dt = int_{-frac pi 2}^{frac pi 2}Atsin t dt$. Evaluating these integrals and substituting in the equations, they simplify to
$$begin{gather}
A = 1 + 2B tag{1} \
B = 2A tag{2}
end{gather}$$

Solving this system gives $A = -frac 13$ and $B = -frac 23$. Thus
$$f(x) = -frac 13 sin x – frac 23 cos x$$

The maximum and minimum values of $f(x)$ are $frac{sqrt{5}}{3}$ and $-frac{sqrt{5}}{3}$ respectively.

The solution is a bit convoluted, but here goes

Step 1: Differentiate the given equation by $x$

$$f'(x) = cos x + int_{frac{-pi}{2}}^{frac{pi}{2}} (cos x – tsin x)f(t)dt$$

Step 2: Consider the following sum

$$f'(x)cos x + f(x) sin x = 1 + int_{frac{-pi}{2}}^{frac{pi}{2}}f(t)dt = f(frac{pi}{2})$$

Now, if you solve this differential equation in $f(x)$, you would get the general solution to be

$$f(x) = Asin x + Bcos x$$

Now, to satisfy the given functional equation, the constants need to satisfy the following

$$f(0) = int_{frac{-pi}{2}}^{frac{pi}{2}}tf(t)dt$$
$$f(frac{pi}{2}) = 1 + int_{frac{-pi}{2}}^{frac{pi}{2}}f(t)dt$$

This should give you two equations in $A, B$, and you can solve to get

$$A = -frac{1}{3}, B = -frac{2}{3}$$

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