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Let's say: $${\int_{n}^{n+\frac{1}{n}}f(x) }\space\text {d}x=C$$

I am looking for some function $f$ that would create, for all $n$ values inputted, a constant $C$ to be created. What is $f(x)$? What is the $C$ value?

Another way of looking at this problem (I am bad at explanations) $$\int_{2}^{2+\frac{1}{2}}f(x)\space\text{d}x=C$$ $$n=2$$ $$\int_{3}^{3+\frac{1}{3}}f(x)\space\text{d}x=C$$ $$n=3$$ $$\vdots$$ For all values of $n$, a constant comes out. And that constant is $C$. What is the function and the constant? Please non-trivial solutions are preferred.

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  • $\begingroup$ Is $n$ restricted to being an integer, or can $n$ be any real number? Also, I assume you are looking for solutions other than the trivial solution $f(x) = 0$ and $C = 0$, right? $\endgroup$ – JimmyK4542 Jul 17 '15 at 4:06
  • $\begingroup$ Well, you could take the derivative of both sides to get $\left(1-\frac{1}{n^2}\right)f\left(n+\frac{1}{n}\right)=f(n)$ $\endgroup$ – msinghal Jul 17 '15 at 4:28
  • $\begingroup$ @JimmyK4542 Yeah, I was looking for more than that answer. It is a little bit lazy in my book. :) $\endgroup$ – user253055 Jul 17 '15 at 6:17
  • $\begingroup$ $C$ can be any value. If $f$ yields $C$, $C'f$ yields $C'C$. $\endgroup$ – Yves Daoust Jul 17 '15 at 7:00
  • $\begingroup$ If you don't set any particular condition on $f$ (like being infinitely derivable or with a closed form formula...), the answer is trivial. Where is this question coming from ? $\endgroup$ – Yves Daoust Jul 17 '15 at 7:24
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If you are interested in finding a nice function that extrapolates the given condition, then you may be interested in the following construction of an analytic solution.


Construction. Let $\mathcal{D} = \{ z \in \Bbb{C} : | \arg z | < \pi / 4 \}$ and for $z \in \mathcal{D}$ define

$$ a_0(z) = z \quad \text{and} \quad a_{n+1}(z) = a_n(z) + \frac{1}{a_n(z)}. $$

Then introduce the following function

$$ F(z) = z^2 + \sum_{k=0}^{\infty} \left( \frac{1}{a_k(z)^2} - \frac{1}{2(k+1)} \right). \tag{1} $$

Claim. $F(z)$ is well-defined by $\text{(1)}$ and analytic on $\mathcal{D}$. Moreover, $F(z)$ satisfies $$ F(z+\tfrac{1}{z}) = F(z) + 2. $$

Assuming this claim, our choice is obvious: let $f = F'$. Then the condition is satisfied with the choice $C = 2$.

Proof. To this end, introduce a new sequence $b_n(z) = a_n(z)^2$ and write $b_n = x_n + iy_n$ for $x_n, y_n \in \Bbb{R}$. Then it is easy to observe that

$$ x_{n+1} = x_n + 2 + \frac{x_n}{|b_n|^2}. $$

Now fix a compact subset $K$ of $\mathcal{D}$. Then the image of $K$ under squaring is again a compact subset of the (open) right-half plane. Now we focus only for $z \in K$. Then it follows easily from induction that for any $n$,

$$ x_n > x_{n-1} + 2 > \cdots > x_0 + 2n. $$

So we have the following uniform estimate

$$ \left| \sum_{k=0}^{n-1} \frac{1}{b_k} \right| \leq \sum_{k=0}^{n-1} \frac{1}{|b_k|} \leq \sum_{k=0}^{n-1} \frac{1}{x_0 + 2k} = \mathcal{O}_K(\log n). $$

(The big-Oh notation $\mathcal{O}_K$ means that generic bounds depend only on the set $K$.) This tells us that

$$ b_n = b_0 + \sum_{k=0}^{n-1} (b_{k+1} - b_k) = z^2 + 2n + \sum_{k=0}^{n-1} \frac{1}{b_k} = z^2 + 2n + \mathcal{O}_K(\log n). $$

From this we find that

$$ \left| \frac{1}{a_k^2} - \frac{1}{2(k+1)} \right| = \frac{|2k+2-b_k|}{2(k+1)|b_k|} \leq \frac{2+|z|^2+\mathcal{O}_K(\log n)}{2(k+1)(2k+\delta)} = \mathcal{O}_K \left( \frac{\log k}{k^2} \right). $$

Therefore the series defining $F(z)$ converges uniformly on $K$ and hence is analytic on $K$ by the Weierstrass M-test. Since $K$ was arbitrary, $F$ remains analytic on the set $\mathcal{D}$ as well.

To derive the functional equation of $F$ we rephrase the definition $\text{(1)}$. To this end, observe that

$$ z^2 + \sum_{k=0}^{N-1} \left( \frac{1}{a_k(z)^2} - \frac{1}{2(k+1)} \right) = a_N(z)^2 - 2N - \sum_{k=1}^{N-1} \frac{1}{2(k+1)}. $$

Plugging this to $\text{(1)}$, we have

$$F(z) = \lim_{N\to\infty} \left( a_N(z)^2 - 2N - \sum_{k=1}^{N-1} \frac{1}{2(k+1)} \right). $$

Then it follows that

\begin{align*} F(z+\tfrac{1}{z}) &= \lim_{N\to\infty} \left( a_{N+1}(z)^2 - 2N - \sum_{k=1}^{N-1} \frac{1}{2(k+1)} \right)\\ &= 2 + \lim_{N\to\infty} \left( a_{N+1}(z)^2 - 2(N+1) - \sum_{k=1}^{N} \frac{1}{2(k+1)} \right) \\ &= 2 + F(z) \end{align*}

and the conclusion follows.


Remarks.

  1. $F$ is in fact a function of $z^2$, since it is constructed from $(b_n)$.

  2. Simulation suggests that $F$ is analytic on all of the right-half plane $\Re(z) > 0$, and has infinitely many poles along $\Re(z) = 0$. If we can prove that $\inf_{z \in K} \Re(b_n(z)) > 0$ for some $n = n(K)$ for any compact subset $K$ of the right-half plane, then the same proof works.

  3. if my calculation is correct, then $F$ satisfies the following asymptotics $$ F(x) = x^2 - \log x + \tfrac{1}{2}(\log 2 - \gamma) + \mathcal{O}(x^{-2}) $$ as $x \in \Bbb{R}$ and $x \to \infty$. The constant part is not important for our purpose, though.

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I can think of one example, though it is trivial:

$$f(x)=0=C.$$

I just thought I would point out that Mihir used Leibniz rule (or a slightly simpler version to compute the derivative https://en.wikipedia.org/wiki/Leibniz_integral_rule)

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  • $\begingroup$ This is a bit too trivial, sorry. $\endgroup$ – user253055 Jul 17 '15 at 4:16
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Consider the piecewise function $f: \displaystyle \bigcup_{n=1}^{M}\left(n,n+\frac 1n\right)\to \mathbb R$ with

$$f(x) = n, \quad x \in \left(n,n+\frac 1n\right).$$ In that case $C = 1$.

I tried to approach this problem in a geometric way. Since the proper integral $$\int\limits_n^{n+\frac 1n} f(x)\,dx$$ defines an area on the interval $\left(n, n+\dfrac 1n\right)$, this specific area must be constant for every $n \in \mathbb N$. Thus, I tried to create rectangles with base length $\dfrac 1n$ and height $n$, such that the area is constant and equal to $1$, for every $n = 1, \ldots, M$.

The graph of $f$ should look like:

enter image description here

In any case, we can set the height as $c_0 \cdot n, c_0 \in \mathbb R$, i.e. the function becomes $f(x) = c_0 \cdot n$, so we can get the constant $C = c_0$.

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If $n$ is restricted to be an integer and piecewise functions are allowed, the question is trivial.

Take any constant $C$ and any set of functions $g_n(t)$ that are integrable in a finite interval $[a_n,b_n]$ and yield non-zero $I_n$, then take

$$f:x\in[n,n+\frac1n]\mapsto C\frac{n(b_n-a_n)}{I_n}g_n\left((b_n-a_n)n(x-n)+a_n\right).$$

As there is no overlap between the intervals, the integrals are completely independent of each other.

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What you have here is a recurrence relation. If $F(x)$ is the antiderivative of $f$. It is easy to see that $F(x+1/x)-F(x)$=$C$. Imagine a $cx^0$ on the right side of this equation, what would happen if you substituted $1/x$ for $x$ on both sides is that you will end up with $F(x+1/x)-F(1/x)=C$ Looking back at our original equation and using the idea of transitive property we can easily see that $F(x)=F(1/x)$. Differentiating both sides we should get $f(x)=-1/x^2f(1/x)$. There are many solutions to this.

By the way $F(x)=c$ is the most obvious solution satisfies this if we look at our original equation so $f(x)=0$ could be a solution as the derivative of a constant is $0$.

From the formula we got its easy to see that $f(1)$ is restricted to $0$, in fact the graph of $f(x)$ when $f(x)$ is not zero every where is extremely similar to a logarithmic function. Both in its shape and in the fact the $log_a(1)=0$. It's not really exact but it's enough to see how the graph looks when it's not constantly at $0$.

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