# Recursive sequence calculators

• luckis11
In summary, the conversation discusses the use of online calculators for solving differential equations and finding the values of a recursive sequence. The speaker also shares their method of solving differential equations through a relevant recursive sequence and expresses their belief that this is not a common approach. They also mention encountering issues with Wolfram Alpha and request advice on how to deal with imaginary solutions. The other person responds by suggesting a different approach for solving differential equations with recurrence relations and questioning the accuracy of Wolfram Alpha's calculations.

#### luckis11

Homework Statement
required for solving differential equations by elementary calculus
Relevant Equations
all differential equations
I know the http://www.calcul.com/ but it is offline some days. Do you know any other online free ones? I.e. that calculates the far values of a recursive sequence.

I need the values in order to test if the fit with my solution. And when the solution of the differential equation is unknown or weird, the only solution is the values of the recursive sequence! Usually I do not know what the differential equation is before I make up what the recursive sequence is. Every physicist and mathematician should know this, but I bet there are very few who can solve the differential equation through the relevant recursive sequence, like I do, which is what they did in the 1700's I guess.

Actually I find it strange that some differential equations cannot be solved through the relevant recursive sequence and they are solved otherwise, e.g. x"(t)=-kx'(t)^2, x(0)=0, x'(0)=U. Its recursive sequence is

u(n)=u(n-1)-ku(n-1)^2dt or u(n)=u(n-1)-ku(n)^2dt

where each next u(n) corresponds to a next dt. But this sequence cannot be solved. In other cases it can, e.g. at Stokes drag x''(t)=-kx'(t), x'(0)=U, x(0)=0, its recursive sequence is

u(n)=u(n-1)-ku(n-1)dt, u(0)=U=>u(n)=U(1 - dt k)^n

and you place n=t/dt (because 1 sec has 1/dt number of changes, therefore t secs have t*1/dt=n number of changes) and

lim dt->0 U(1 - dt k)^(t/dt)=Ue^(-k t)=u(t)=x'(t).

which is the solution of the differential equation. Partly, since this is only the x'(t), I haven't tried to solve for x(t).

Note that wolframalfa in some cases it gives semi-wrong and in some cases no answer for x"(t)=-kx'(t)^2, x(0)=0, x'(0)=U. If you plug it in like I wrote it, it gives an answer with i. What am I supposed to do with i?

Last edited:
This is all very difficult to read, you should learn to post in ## \LaTeX ## math (I see you have been told this before).

luckis11 said:
Every physicist and mathematician should know this, but I bet there are very few who can solve the differential equation through the relevant recursive sequence, like I do
This is a very strange thing to say. What makes you think that you know something very few other people know? How did you come by this knowledge?

luckis11 said:
and you place n=t/dt (because 1 sec has 1/dt number of changes, therefore t secs have t*1/dt=n number of changes) and

lim dt->0 U(1 - dt k)^(t/dt)=Ue^(-k t)=u(t)=x'(t).
This is not how you solve differential equations with recurrence relations, you do it by comparing coefficients.

luckis11 said:
which is the solution of the differential equation. Partly, since this is only the x'(t), I haven't tried to solve for x(t).
Well if you have ## x'(t) = U e^{-kt} ## then it should be easy for you to find ## x(t) ##.

luckis11 said:
Note that wolframalfa in some cases it gives semi-wrong and in some cases no answer for x"(t)=-kx'(t)^2, x(0)=0, x'(0)=U. If you plug it in like I wrote it, it gives an answer with i. What am I supposed to do with i?
Why do you think that Wolfram Alpha calculates this incorrectly? Have you checked it by differentiating? Note that it is easy to link to a solution on Wolfram Alpha by copying the contents of the address bar like this:
https://www.wolframalpha.com/input/?i=x''(t)=-kx'(t)^2,+x(0)=0,+x'(0)=U