Find the area out of the functional

Calculus Level 3

If $f$ is a real-valued function such that $f(x) \neq 0$ for all real $x$ , and satisfies the functional equation $\large f(x)=x\sqrt{x\sqrt{x\sqrt{x \sqrt{f(x)}}}}$ for all $x>0$ . Then find the area under the graph of $f$ between $x=1$ to $x=4$ .

Note: This is my original problem. For more of my problems check the set Questions I've Made

The answer is 21.

3 solutions

Brian Charlesworth
Dec 27, 2014

Simplifying, we have that

$f(x) = x * x^{\frac{1}{2}} * x^{\frac{1}{4}} * x^{\frac{1}{8}} * (f(x))^{\frac{1}{16}}$

$\Longrightarrow (f(x))^{\frac{15}{16}} = x^{\frac{15}{8}} \Longrightarrow f(x) = x^{2},$

the integral of which from $x = 1$ to $x = 4$ is $\frac{64}{3} - \frac{1}{3} = \boxed{21}$ .

There is 1 more solution $f(x) = 0$ then the answer could be 0

Krishna Sharma - 6 years, 5 months ago

Good point. I guess that the question should start with "If $f(x)$ is a positive, real-valued function ...." to avoid the ambiguity.

Brian Charlesworth - 6 years, 5 months ago

I've already mentioned it in my question that $x>0$ .

Sanjeet Raria - 6 years, 5 months ago

But we can still have the function $f(x) = 0$ for all $x \gt 0$ satisfying the functional equation. Only if it is specified that $f(x)$ is a positive real-valued function does $f(x) = x^{2}$ become the unique solution function.

P.S.. I didn't report the problem, but I can see now that, as presently worded, an ambiguity exists.

Brian Charlesworth - 6 years, 5 months ago

@Brian Charlesworth – Yeah... you are right. I've changed the wordings now. Thank you.

Sanjeet Raria - 6 years, 5 months ago

@Sanjeet Raria – FYI The zero function is the function such that $f(x) = 0 \forall x$ . Saying that it is not the zero function only means that there is some value such that $f( x^* ) \neq 0$ . It does not mean that $f(x) \neq 0 \forall x$ .

Calvin Lin Staff - 6 years, 5 months ago

I don't know what is wrong with my eyes. I kept reading this problem repeatedly as the "x" was associated with the radical. I think the only way to eliminate this bizarre ambiguity that seemed to have only affected me is to explicitly add the * for multiplication after each x. Then it is a simple, even charming problem. Sorry my brain is not working like others.

Bradley Slavik - 5 years, 11 months ago

Prakhar Gupta
Jan 19, 2015

Here is my solution for this problem. $f(x) = x\sqrt{x\sqrt{x\sqrt{x\sqrt{f(x)}}}}$ Here we can replace $f(x)$ in the square root by the one defined in the question. Replacing $f(x)$ every time it will become an infinite series of square roots.

In that case it will become like this. $f(x) = x\sqrt{x\sqrt{x \ldots}}$ Hence we can write $f(x)$ as $f(x) = x\sqrt{f(x)}$ Hence $f(x) = x^{2}$ Hence the area under the curve is $21$

Add a step at the end stating the area under the curve is $\int_1^4 f(x)dx$ . :)

Hem Shailabh Sahu - 6 years, 2 months ago

Bill Bell
Oct 9, 2015

Some of you may find this peculiar. I found that ${ y }_{ k+1 }=x\sqrt { x\sqrt { x\sqrt { x\sqrt { { y }_{ k } } } } }$ converges quickly for values in the given interval making it easy to discern that $f\left( x \right) ={ x }^{ 2 }$ is a solution.

Find the area out of the functional

Note: This is my original problem. For more of my problems check the set Questions I've Made

The answer is 21.

3 solutions

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