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For many recurrence relations, spotting a pattern in the first few terms allows us to conjecture a possible closed form formula solution for the general term which can be verified by induction. I have no problem with that.

Often, however, solving first order recurrence relations with non-linear coefficients proves to be more difficult; eg when no simple pattern is immediately evident. In these cases, eg if we wanted a closed form formula for $t_n$, a good method is to use a substitution of the form $t_n=f(n)u_n+g(n)$ and solve the simpler relation in terms of $u_n$ and then substitute the solution for $u_n$ back into $t_n$ which is what we are aiming for.

The following recurrence relations can be solved by different substitution, as I write explicitly below: $$t_{n+1}=2t_n+1,~t_0=0~~\text{solved with}~t_n=u_n+k$$ $$t_{n+1}=2t_n+n,~t_1=2~~\text{solved with}~t_n=u_n+an+b$$ $$t_{n+1}=2t_n+2^n,~t_1=1~~\text{solved with}~t_n=a\times2^nu_n$$ $$t_{n+1}=(n+2)t_n+n!,~t_1=1~~\text{solved with}~t_n=a\times n!\times u_n$$ and in each case a choice of $a$, $b$ or $k$ is made to obtain an easy recurrence relation in terms of $u_n$ to work with.

My question is, assuming the recurrence relation in question can be solved by substitution, is there a general method that immediately tells us which sort of substitution to use, like when $f(n)=1$ or when $g(n)=0$ or any other choice for that matter? For example, how would I know to use the substitution $t_n=a\times n!\times u_n$ as in the final example?

Thank you for your help.

A-Level Student
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  • No, there is no general method for solving 1st order nonlinear difference equations. See, e.g., https://math.stackexchange.com/questions/147075/why-is-solving-non-linear-recurrence-relations-hopeless – Gerry Myerson Oct 08 '20 at 12:07
  • @GerryMyerson ok but what suggests the type of substitution we should make? – A-Level Student Oct 08 '20 at 12:09
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    That question presumes there is always a substitution that works. There isn't. – Gerry Myerson Oct 08 '20 at 12:10
  • @GerryMyerson thanks, edited. – A-Level Student Oct 08 '20 at 12:11
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    The way you know what substitution to use, in those few cases where there actually is a substitution that works, is: by experience. If you've solved enough of these equations (or seen enough of them solved in textbooks), you develop an idea of what's going to work, and when. It's much like doing indefinite integrals. The more of them you've done, the more likely you are to recognize what to do the next time. – Gerry Myerson Oct 08 '20 at 12:15
  • @GerryMyerson do you have any source of relevant good recurrence relations questions that I can have a look at? – A-Level Student Oct 08 '20 at 21:00
  • https://books.google.com.au/books/about/Finite_Difference_Equations.html?id=qEH6RF3txPsC&redir_esc=y – Gerry Myerson Oct 08 '20 at 21:47
  • @GerryMyerson thanks. – A-Level Student Oct 09 '20 at 10:26

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