How to Define a Symbolic Continued Fraction

Hello, I want to create and simplify a symbolic fraction, but dont now how. My fraction has the following form:

I want to repeat that fraction for n times, with beeing symbolic Variables, to work with and maybe simplify the function (or similar functions).

Edit:

It seems to me now, that the question I want the answer to is:

Is it possible to define and simplify a function in MATLAB which form depends on a parameter?

The function above has no fixed form, since the repetitions of depend on n, which shall be and stay a symbolic variable. In the end I would like a simplified form of . This may not be possible with the function I gave as an example here, but this is not the first time I had this problem, and I would be really glad if there is something I can do here... :)

Old:

I dont know if this is possible in MATLAB, two things would help me to find a way:

• Is there a way to have recursion in symbolic functions? I want to define something like:

        syms f(n) n x
        f(n) = 1/f(n-1) + x; % Example of recursion in a function

this does not result in an error but does also not return the solution when I substitue n with a numeric value.

• Is there a smart way to define the fraction above with n beeing not a symbolic variable? I could print my fraction n times in a string, and convert that string to a symbolic function, but that does not seem smart to me...

[Walter Roberson]

I am not clear as to what the final denominator should be?

Are A and B guaranteed to be real?

[Maximilian Schönau]

The final denominator is b/(n-1) + A/n (After n repetitions of the scheme). I‘ll edit my Formular to make it clearer :)

A and B are real.

Answer 1

[Walter Roberson]

Infinite case:

syms A B L n
eqn = L == A/n + 1/(B/(n-1) + L)

eqn = 

Lsol = solve(eqn, L)

Lsol = 

limit(Lsol, n, inf)

ans = 

[Walter Roberson]

You will not be able to create a symbolic formula or function that takes a symbolic n at time of definition, and expands out to an exact value when a particular definite integer is passed in. This is because symbolic formulas always fully resolve when the functions they call upon are known, except some built-in cases. Symbolic functions are created by resolving the right hand side, and the symfun() the result You cannot create a recursive symbolic expression or function because of this: any function you called upon would resolve its value fully, and the function name will not exist until after the assignment so you cannot refer to itself inside the expression.

What might perhaps be possible is to find the general formula for the answer, maybe .

Or perhaps you could make use of some property involving convergents https://en.wikipedia.org/wiki/Continued_fraction#Infinite_continued_fractions_and_convergents

[Maximilian Schönau]

Thank you so much for your help! It’s sad that MATLAB can’t do that for me, I for my self am not that good in (real) math, so at most I could find a simplification of such repeating function with luck and doing it by myself.

Maybe other programs like Maple could help me out? Maybe I’ll try finding out.

Anyway thanks for all of your help.

[Bruno Luong]

What is missing in MATLAB is symbolic functional-power (not to be confound with power function).

[Walter Roberson]

It is possible to define symbolic procedures using MuPAD, if you happen to know the MuPAD programming language... which is no longer documented. However, if you happen to request displaying a variable that holds a symbolic procedure, or you happen to request displaying a hold construct (often needed for symbolic procedures) then the symbolic engine will reset itself.

But if you could do such a thing, then what interface would you want for the function? You would want to be able to pass a specific integer to the function and have it calculate for that integer, and you would want to be able to pass inf to the function to get the limit, and what else? For example do you need to be able to pass symbolic n and also symbolic n+1 and calculate the difference between the two results? Do you ever need to pass in a symbolic approximation order, calculating N levels with symbolic n ?

Is it mandatory that this all be a symbolic function, or could it be done through a function handle instead?

The ability to create a symbolic function that calculates recursively for any given integer n does not imply the existence of tools that can reason about the function given symbolic parameters.

[Maximilian Schönau]

If you just want results of your function (and its shape) in a simulation or similar, of course a numerical way is enough.

But it happenend to me now to times, that I really wanted to have the number of iteration as a symbolic value. The reason for that is that I wanted a simplification of the expression. If your expression can be simplified, you can understand bigger interrelationships within your model. This can be very helpfull and sometimes necessary when you model something (and you want to know what is happening). It could be done in my last time such function came to me and can not really be done in my second time I wanted to do this.

"You must be aware that there is rarely a close formula for iterative a function. (Bruno Luong)"

I did not know that, the last time I found the solution to my iterative function by seeing a pattern when solving it numerical and I got my simplification by doing the necessary math on my own.

MATLAB can do much more and much better math than I can do, it would be cool and really helpfull in some cases, if MATLAB could do "symbolic functional power (Bruno Luong)", if this could simplify and find patterns in my iterative functions...

Answer 2

[David Hill]

function L = continuedFraction(N,n)
syms A B;
if n==0
    L=0;
else
    L=1/(B/(N-1)+A/N+continuedFraction(N,n-1));
end

Run the function for the number of times (n) you want, then add in A/N and simplify.

syms A B;
n=10;
L=continuedFraction(n,n);
L=simplify(L+A/n);

[Walter Roberson]

You said to add an A/N but you add L/n ?

[David Hill]

Thanks

[Maximilian Schönau]

Thank you very much!

That is a cool and logical way to define my function with n being an integer, much better than printing a string for my function :)

But ist there a way to define the function you wrote with n being a symbolic variable? A symbolic variable that is not allowed to have negative values and must be an int...

I want to simplify my expression in a general way, as a result I would like something like:

    L_simplified(A,B,n,N) % n == N in my case

[Maximilian Schönau]

I think what I need here is similar to the symsum function. It can calculate sums symbolically, and you can insert values like infinity, which wouldn’t be possible in your example, if you don’t want to get your solution numerically.

[Walter Roberson]

For any fixed finite number of iterations, if you take the limit as n goes to infinity, then the result goes to either 0 or inf*sign(A-B), depending on whether the number of iterations is even or odd. But that's probably not of interest to you as you do not want n to go to infinity with a finite number of iterations

[Maximilian Schönau]

No, I want to see if and what function I get for all values of n. I want to solve a physic problem with my function, which won’t reach zero or infinity but is shaped a bit different.

So being able to use infinity for n would not hurt.

I remember having a similar problem with a repeating scheme and I had no idea how to simplify or even express it in matlab symbolically so in the end I had to find the solution (which luckily existed) by myself.

Basically I want to now if you can in any way simplify a function which form depends on a parameter.

[Walter Roberson]

Do you need exact values, or would you happen to need a limit instead?

Is there a possibility that abs(A)<1 or abs(B)<1 ? If so is there a known order approximation A^P << B or A << B^P for some fixed positive integer P ? If there is, then it would be potentially possible to truncate terms in an approximation.

[Maximilian Schönau]

I need exact values.

The function I have is certainly dependent on all it’s values, although the function I wrote here maybe isn’t.

I don’t really want to know the result of my problem, but how (or if) I can solve it in MATLAB without me doing math by myself.

[David Hill]

function L = continuedFraction(N)
syms A B n;
if N==0
    L=0;
else
    L=1/(B/(n-1)+A/n+continuedFraction(N-1));
end

Can you just do this?

syms A n;
L=continuedFraction(10);
L=simplify(L+A/n);

[Maximilian Schönau]

You still did define n, I dont want to do that. I want something like:

function L = continuedFraction(N)
syms A B n;
if N==0
    L=0;
else
    L=1/(B/(n-1)+A/n+continuedFraction(N-1));
end

syms A n;
L=continuedFraction(n); % Matlab will return an error
L=simplify(L+A/n);

I know that this syntax does not make much sense, since you cant use symbolic variables as an input of this function.

Answer 3

[Bruno Luong]

I don't have the symbolic tbx to try, you function can be defined sequentially in n by loop

L = Inf;
for k=1:n
    L = A/n + 1/(B/(n-1)+L);
end

[Maximilian Schönau]

Thanks, your answer is similar to Davids.

I cannot use a symbolic variable for n with your code :/

[Bruno Luong]

There are two sort of n here, the one used to divide A and B and the loop iteration number.

The division one can certaily be symbolic.

[Maximilian Schönau]

Yes, I want both n to be the same and symbolic...

[Bruno Luong]

You must be aware that there is rarely a close formula for iterative a function.

In your case it's an inverse-hyperbolic shifted by (a,-b) with

f(L) := a + 1/(b + L)

where a := A/n and b := B/(n-1).

The rich Mandelbrot set is defined by similar process underneath with f(z) = z^2 + c.

[Bruno Luong]

The recurrent formula is also explained more generally here The continued fraction as a composition of LFTs

In the same article you can see fundamental recurrence formulas: L can be expressed as the ratio of An/Bn where An and Bn are obtained by IIR all-pole 2nd order equations.

[Maximilian Schönau]

Thank you very much!

[Bruno Luong]

I suppose you are aware that the recursive relation of An and Bn (in fundamental recurrence formulas) can have close formula. The 2-level recursive relation such as

y_{n+1} = a*y_ n + b*y_{n-1}

can be written in 2 x 2 matrix form

Y_{n} :=  [y_{n-1}; y_n]
Y_{n+1} = [0 1;
           b a] * Y_n;

So

 Y_{n} =  [0 1;
           b a] ^(n-1) * Y_1;

Now you have formula to compute An, Bn. Take theire ratio you have the formula of your fraction at any iteration. I let you figure out how to set Y_1, a, b in your case.

[Maximilian Schönau]

Im sorry, but I dont understand your answer (engineer). Learned how to solve differential equations in 100 different ways, but not how to do real math, which is a pitty, since the differenatial equations are now getting solved by MATLAB. :/

The Wikipedia entry did not help...

Maybe I will learn the necessary math for this problem the next time. This time, it is not that important for me to find a solution to my problem. Even if I work myself trough the math to solve this problem, the next iterative function that is not a fraction, may need completly different math.

So I pass this time, lets see how long it takes for me to get this kind of problem again. Luckily, when simulating, the 100 ways to solve differential equations do the trick most of the time.

Thanks for your help!

(And for showing me the way to the math that solves my problem :))

[Maximilian Schönau]

For someone who reads this and does not want to do the math:

The engineer solution to this is looking for your range of the parameters, plotting the range with the numerical solution (for example with the syntax David Hill posted here) and then fitting the function you want to the solution :D

You can use a polynomal of the order you want (or a function / functions that looks similar to your plottet numerical solution) . As a result you have a "simplified" expression which shows you how the function behaves.

Mathematicians hate this trick

[Bruno Luong]

I did not expect you give up that soon, when I gave all the hints for you to compute the finite continuous fraction. And btw IMO your second comment is far from being useful for anyone who read it.

I give here the exact formula of your editted fraction.

Now this is a gift for someone who reads this and does or doe not want to do the math.

n = 10;
A = rand();
B = rand();

bn = B/(n-1) + A/n;

% Compute using iterative of fraction
L = Inf;
for k=1:n
    L = bn + 1/L;
end
L = L-B/(n-1)

% Direct formula
Mpn = [0, 1;
       1, bn]^n;
Y =  Mpn(:,2)./Mpn(:,1);
L = Y(2);
L = L-B/(n-1)

If you don't find this formula and the way to find it is beautiful, then you might be a simple good engineer that does absolutely amazing empirical polynomial fitting model.

[Maximilian Schönau]

Your math is absolutly beautiful and I am sorry that I did not invest the time to learn it.

Your solution is in this case inarguably more beautiful than mine.