# Fall 2016, problem 29

Find at least one function $f: \mathbb R \rightarrow \mathbb R$ such that $f(0)=0$ and $f(2x+1) = 3f(x) + 5$ for any real $x$.

Nelix
5 years ago

This one works:

$$f(x) := \begin{cases} \frac{5}{2} \left(\exp\left({\frac{\log(3)}{\log(2)}\log(|x+1|)}\right) -1\right) & \textrm{ if } x \neq -1 \\ -\frac{5}{2} & \textrm{ otherwise} \end{cases}$$

This is easily checked and it's kind of nice to see how everything falls into place, when you do the algebra.

I arrived at this expression something like this:

First I tried to solve the recursive equations

• $a_{n+1} = 2 a_n + 1$
• $b_{n+1} = 3 b_n + 5$

After trying around a little, I found the particular solutions

• $a_n = 2^n - 1$
• $b_n = \frac{5}{2} (3^n -1)$

Now I let: $$f(a_n) := b_n \textrm{ (Eq.1)}$$

Then $f(0) = 0$, anyway, by sheer luck. Otherwise, I would have tried to shift the index $a_n \rightarrow a_{n+k}$ appropriately.

Now you just reinterpret $n$ as a real number and write $$\begin{eqnarray*} x &:=& a(n) = 2^n - 1 \\ \implies b(n) &=& \frac{5}{2}(3^n -1) = \frac{5}{2}(3^{\frac{\log(x+1)}{\log(2)}} -1) \end{eqnarray*}$$ Plug these expressions into (Eq.1) and you arrive at: $$f(x) := \frac{5}{2} \left(\exp\left({\frac{\log(3)}{\log(2)}\log(x+1)}\right) -1\right)$$ There is just one thing left to fix: $\log(x+1)$ is not defined for $x \leq -1$. But remember, that in some respects the function $y \rightarrow \log(|y|)$ reasonably generalizes the $\log$ function to all reals $y \neq 0$. For example in the sense, that $\log|y|$ is an antiderivative of $\frac{1}{y}$ in all of $\mathbb{R} \setminus \{0\}$. This motivated the final form of the result as given in the beginning. Observe also, how the function $x \rightarrow 2x + 1$ maps each of the sets $]-\infty;-1[$, $\{-1\}$ and $]-1; \infty[$ bijectively onto itself, which fact corresponds to the three cases in the final result.

This board has become somewhat of a depressing affair and I feel, I've had a part in that development. I should have stopped posting here every week a long time ago. So let this be my last post here. Bye.

Nelix 5 years ago

Could you explain your decision? Without you, thiis board will probably disappear. The interactive formula of Purdue POW is not a complete success .... The problems are often very interesting but: It's very difficult for many people to use Mathjax. Perhaps Purdue could accept solutions in pdf format. It would be useful that people of Purdue University who invent all these beautiful problems present their solutions after a certain period of time (as in the ancient version of POW).

francois 5 years ago

I think the only problem is that problems are now a bit too hard,, Purdue staff should change that in order that at least two or three good solutions appear each week,,

Hubert 5 years ago

Still sticking with that decision. But, for what it's worth, here's my thoughts:

I don't think MathJax /Tex is the problem. To me it's really not a nuisance as much as a big convenience. You type something as simple and natural as \sum{i=0}^\infty enclosed in dollar signs and have a beautifully typeset sum sign magically appear right away. The Tex language is natural like this, there's not a huge amount of technicalities you have to wade through, in order to produce something simple. Also it is the definitive standard for writing mathematics using a keyboard and making people learn it in a nonserious setting such as this board is not unreasonable. And there's like tons of excellent tutorials. Maybe all that's missing is a clearly visible message in the answering dialog, that all you really have to do is type dollarsign some tex code dollarsign to display mathematics in your answer.

Bottomline: Tex is your friend. MathJax makes it magically work in your post. It's not an arcane weirdness associated with this particular forum, at all.

I don't want to come across as preachy, it's just that I don't think MathJax/Tex poses as much of a hurdle to people as you think. Whatever hurdle it may pose can further be lowered, by linking to this tutorial, for example, in a prominent place. In my mind, encouraging pdfs and image files as answers is a bad solution for a problem that doesn't really exist.

I also don't think, the problems are too hard necessarily. Problem number 1 was the hardest yet and it had the best discussion. They should be reasonably hard in order to encourage partial solutions or simplifications to overcomplicated existing solutions, any activity at all.

The one problem that I can put a finger on is this: You've solved the problem, you're proud of your solution but some guy/gal has already posted a similiar solution. So you get zero recognition for posting your answer. There is no fun in answering an answered question and there's no fun in spoiling a problem to others. I have no solution to offer for this conundrum. Me continuing to play the role I've played here is no solution, at any rate. So no more answers from me, for now.

Nelix 5 years ago

When I said the board will stop without you, I think it's a kind of recognition.... And generally you can not hope for any recognition in a social network.

Some one find an idea , another read his solution, find an amelioration sometimes a small amelioration. It's life, it's normal, and often it's a good thing.

When I said, Purdue could accept pdfs, It's because the number of answers in the ancient version of the board was quite bigger. Here we are two or three to answer and I think it's not good.

francois 5 years ago

Most of the problems were by no means hard . At least as compared to problems that received lots of solutions in the former version of the board . And yes , non wysiwyg LaTeX editing is a nuisance.

rubs 5 years ago

@ francois: Thanks for the complement. I regret having used the much too strong word "recognition". What I was thinking of is something a lot milder: The ancient version of the board probably worked like this (I'm assuming here, but in any case other noninteractive formats work this way): You hand in your solution before deadline. If you got it right your (nick)name is mentioned. What happens is, you do something more or less hard correctly, you get "yes, correct". You do it wrong, you get "no, wrong" or no response at all. There is some basic fun in that process and that's all I meant when I awkwardly used the word "recognition": This basic action$\rightarrow$feedback mechanism.

Posting a solution on the current board is no fun, the way things are. Sure, thinking about a problem and arriving at a solution and even writing something up is. But going ahead and posting it here feels utterly pointless at best. At worst it feels like spoiling the problem to others.

The more I think about it, the more I lean towards some basically noninteractive system with maybe the option to comment ( after the deadline). In the end, recreational problems like these maybe just don't warrant lengthy discussions or stepwise ameliorations of solutions - and certainly no amount of praise for finding a solution (I am sorry for having made myself misunderstood in that direction). Getting a simple "yes correct" or "no wrong" is really the appropriate end result.

But then again, maybe just give it time and people will make this interactive format work, because it is a great idea in principle and the problem authors are doing a great job, I think.

Nelix 5 years ago
Hubert
5 years ago

$1)$ I think Nelix would here simplify by equivalent steps, that is

• find $a$ and $g(x)=f(x)+a$ to get a $g(2x+1)=3g(x)$,
• find $b$ and $h(x)=g(x+b)$ to get a $h(2x)=3h(x)$,
• find $c, d$ to get a $h(x)=c.x^d$ working

to get $a=\frac{\ln 3}{\ln 2}\Rightarrow~$

$f(x)=\frac 52(|1+x|^a-1)$ works

$2)$ Now he won't stop here and give all the solutions,, just the idea

first work on $x>0, \; h=0$ works, besides wlog $\exists~ h(x)>0$

$\varphi(x)=\ln(h(e^x))\Rightarrow~~\exists~u,v, \; \varphi(x+u)=\varphi(x)+v$

find $e$ and $\psi(x)=\varphi(x)+ex$ to get a $\psi(x+u)=\psi (x)$,

so $\psi$ is any $u-$ periodic map which gives - going backwards -

$\varphi \rightarrow h \rightarrow g$ and the general solution $\rightarrow f$;