<p>Please help me with the following problem:</p>
<p>Consider a function f(x) whose magnitude never exceeds 1 and whose integral on [0,1] equals 0. Find the maximum of the integral of [f(x)]^3 over [0,1].</p>
<p>(I posted this in a subforum, but the question was never answered, so I assume that it was buried a little too deep).</p>
<p>I believe it’s against forum rules to ask for help with homework problems…</p>
<p>I think it’s pretty obvious that this isn’t homework.</p>
<p>But it’s also not standardized test prep. They don’t test calculus.</p>
<p>But surely this isn’t a SAT problem, or a problem that will likely appear on any high school math course or most introductory college math courses.</p>
<p>Anyway I’ll bite and hopefully the moderator won’t close the thread.</p>
<p>Intuitively the function can be thought of as being positive on some segment of the interval [0,1] and negative on what’s left over. Suppose then that the interval on which it’s positive is [0,x] and the interval on which it’s negative is [x,1].</p>
<p>Rather than trying to pick from all possible functions I’ll assume a step function where the function is “1” on [0,x]. Because of the constraint that the integral of the function over [0,1] is 0 the function (assumed here to be a step function) must equal -x/(1-x) on the interval [1-x].</p>
<p>For this step function, compute the integral of the (function^3) over [0,1]. Then maximize that integral. A bit of algebra shows that the integral is I(x) = x(1-2*x^2)/((1-x)^2). Find the maximum of I(x) (i.e. find where the derivative of I(x) is 0). That occurs at x = 1/3. At x = 1/3 the integral of [function^3] over [0,1] is 1/4. That’s the maximum under the assumption that the function is a step function. Note that the function I(x) has critical points other than x=1/3. They are easily eliminated as not relevant to the problem.</p>
<p>
Isn’t this a forum for general test preparation? That could include competitive exams like the AIME or the Putnam, where this sort of stuff is tested. </p>
<p>Fogcity, thank you for your solution. I was trying something along the lines of factoring an expression involving [f(x)]^3 and f(x) (and then integrating), but wasn’t able to actually find a suitable one.
I do have a question, involving
How do you know this? Namely, how do you know that if the function is unity on [0,x], then it is -x/(1-x) on [0, 1-x]?</p>
<p>To answer your question:
</p>
<p>I’m computing that integral.</p>
<p>(1) Assume that the function is “1” on [0,x]. The contribution to the integral on this part of the interval of [0,1] is x.
(2) Then the function (if a step function) must equal -x/(1-x) on the interval [x,1]. The contribution to the integral on this part of the interval of [0,1] is - (x/(1-x))*(1-x) or -x.</p>
<p>Note that I’ve corrected the typo … the interval for (2) is <a href=“not%20%5B1-x%5D”>x,1</a></p>
<p>1<em>x - (x/(1-x))</em>(1-x) = 0 as required.</p>
<p>Draw a picture. x is a number (the position where the step function goes from positive to negative) somewhere between 0 and 1.</p>
<p>I wasn’t quite convinced with letting f(x) = 1 on (0,1/2) and f(x) = -1 elsewhere (what if f(x) = 2/3 on a larger interval maximizes it?).</p>
<p>(Note: The below expressions are in LaTeX, if you don’t know LaTeX, you can search TeXeR and copy/paste the below code).</p>
<p>Here’s what I did: Let $f(x) = c$ on $(0,a)$ and $f(x) = -1$ on $(a,1)$. By comparing areas, we have $ac = 1-a \Rightarrow c = \frac{1-a}{a}$. The integral we wish to maximize is</p>
<p>$$\int<em>0^a f(x)^3 , dx = \int</em>0^a \bigg( \frac{1-a}{a} \bigg)^3 , dx$$</p>
<p>The above integral can easily be evaluated. Then maximize it.</p>
<p>^^^^ (comment to rspense)</p>
<p>
</p>
<p>Please reread the analysis. Nothing of the sort is assumed or stated. Nowhere do I state that f is a function of x. Your a is my x. The intervals are [0,x] and [x,1]. (You have them as [0,a] and [a,1]).</p>
<p>The length of the interval is key to the proposed solution. That’s the point of the “x”. The function is assumed to be “1” on [0, x]. Then (if it’s a step function) it must be -x/(1-x) on the rest of the interval.</p>
<p>Can you finish your calculation? Conceptually this is a very difficult problem. It’s not fair to leave it as an exercise for the reader. In any case your equation seems identical to that used to get the proposed answer in #5. Did you review that post? Do you get a different answer? Do you agree with the approach?</p>
<p>I was able to find another solution, based on the idea that I outlined in my previous post.
Essentially, I thought that we could integrate</p>
<p>polynomial in f(x) in factored form = polynomial in f(x) in expanded form</p>
<p>over [0,1].
The advantage of this is that we can use the factored form to force the polynomial to always be less than 1 and thus yield a nonpositive integral.</p>
<p>The best factor is [f(x) - 1] because it is nonpositive and can equal 0, since f(x) can equal 1. To obtain a cubic in f(x), multiply by something like [f(x) + a]^2, which is always positive. However, we don’t want an [f(x)]^2 term, because we don’t know how to integrate that. The choice of a that forces the quadratic term to vanish is 1/2. So</p>
<p>[f(x) - 1][f(x) + 1/2]^2 = [f(x)]^3 - 0.75f(x) - 1/4
→ [f(x)]^3 = [f(x) - 1][f(x) + 1/2]^2 + 0.75[f(x)] + 1/4</p>
<p>Integrating over [0,1],</p>
<p>\int<em>{0}^{1} [f(x)]^3 dx = \int</em>{0}^{1} [f(x) - 1][f(x) + 1/2]^2 dx +. . . + 1/4</p>
<p>By the first constraint, the second integral vanishes, and the second integral is at most 0. It is 0 if f(x) = 1 or -1/2. The answer is therefore 1/4 as long as we can construct a piecewise f(x) from 1 and -1/2 that integrates to 0, which is easily doable. </p>
<p>f(x) = 1, 0 < x < 1/3
= -1/2, 1/3 < x < 1</p>
<p>The answer agrees with Fogcity’s.</p>
<p>Hmm, I think I may have misinterpreted the question. I’ll have to look at it later.</p>
<p>The flaw in the solution to #5 is that you assume f(x) is a step function. Fortunately, that assumption seems to be correct. I don’t really get what rspence did.</p>
<p>Yours is an elegant proof! Very impressive!</p>
<p>Can you apply your approach to the more general case?</p>
<p>Consider a function f whose magnitude never exceeds 1 and whose integral on [0,1] equals 0. Find the maximum of the integral of [f]^(2N + 1) over [0,1] where N is a positive integer.</p>
<p>The maximum increases as N increases. When N is very large the maximum approaches 1/2. What about when N = 2?</p>
<p>Interesting. I will start working on this now.</p>
<p>
</p>
<p>Thank you :D</p>
<p>It’s mostly boiled down to an algebra problem so far.</p>