The most scary things that I find in mathematics are functional equations, as there is no general method for solving them as far as I know. The purpose of this note is to gather some information about how to solve these type of equations, and most importantly, whether every functional equation has a solution or not. I know that some of my fellow brilliant members know about this, so help me! As for an example, I encountered this one when I was working on something confidential.

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TopNewestI can't solve it either. I suppose you don't know f(0) or f(1)? I find that the behaviour of this equation depends strongly on the boundary value. If the first value is 0 then all subsequent values are (as you have doubtless noticed). If the first is just \( \pi \) then subsequent terms are \( \pi \). If the first term is a fairly small fraction of \( \pi \) then subsequent terms drop off to zero quickly. If the first term is 1.25 times \( \pi \) then subsequent terms rise quickly to twice \( \pi \). Incidentally I used this code to do my little investigation.

– Bill Bell · 2 years, 1 month agoLog in to reply

– Danny Kills · 2 years, 1 month ago

Only if i knew programming! Anyways, as I have written in the post, im not going to tell what this equation is about, but it would be great help if you could do some analysis on it.... I'll give u credit i find a major breakthrough! (Although thats highly unlikely!)Log in to reply

– Danny Kills · 2 years, 1 month ago

Did you graph it???Log in to reply

plot f(n+1)=f(n)-arcsin(sin(f(n))/sqrt(2)),f(1)=pi/4

You can see the initial condition at the right.

I'll have to admit at the outset that my analysis skills are limited. I like numerical methods; I'm not at all skilled or knowledgeable in algebra. – Bill Bell · 2 years, 1 month ago

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– Danny Kills · 2 years, 1 month ago

My actual aim with this problem is to find the general term of this sequence. I can plot the graph to see how it varies but it will not help me solve it unless i superimpose it with thousands of graphs to see which fits! I can tell you what this is about but not on brilliant....Log in to reply

– Bill Bell · 2 years, 1 month ago

Have you considered asking for a solution on quora.com? There are people on there orders of magnitude more able than I am.Log in to reply

– Danny Kills · 2 years, 1 month ago

Thanks.. will surely try!Log in to reply