I'm trying to solve a recurrence relation by generating terms and looking for a pattern. I've learned that i can't stop

Maple's autosimplification process and the best I can do is use Parse from the InertForm package. The hand drawn picture below is what I'm trying to replicate. I know I can use rsolve but I'm trying to do the steps I would with pencil and paper.

Hi, friends!

I'm not a math =) but it is interesting

How can i solve this equation like the gambler's ruin with Maple's function rsolve

f(n)=0.5*f(n-1)+0.5*f(n+1), f(0)=1, f(6)=0

rsolve({f(n)=0.5*f(n-1)+0.5*f(n+1), f(0)=1, f(6)=0}, {f});

it returns this

{f(n) = 7 f(5) - f(5) (n + 1)}

I don't understand :(

for example Wolfram Alpha return the true result

How to compute the recurrence relation and I find the problem when the summation of U because appear noise term self-canceling and I can not find the nth component of U?Mixed_volterra_-Fredholm_(278)_Ex(8.17).mw

Download Mixed_volterra_-Fredholm_(278)_Ex(8.17).mwMixed_volterra_-Fredholm_(278)_Ex(8.17).mw

thank you for helping:)

Hello,

I want to solve and plot a multitime recurrence of the Samuelson Hicks Model (http://www.mathem.pub.ro/proc/bsgp-22/K22-gh-A84.pdf).

Feel free to share any tips that could help.

Thank you.

How do I solve this in Maple:

M(k,n)=M(k,n-1)+M(k-1,n-1)+M(k-2,n-2) and

M(k,1)=1 and M(1,n)=n

?

Thank you, Jan

I'm trying to create and populate the Array `A` with a sequence of lower triangular matrices based on the following relation:

A[k+1][i,j] = a (1+k) A[k][i,j] + b ((1+j) A[k][i-1,j] - j A[k][i-1,j-1])

The rows and columns of each matrix are given by i and j; a and b are scalars. The structure of A is not important provided a matrix can be accessed for a given k. The highest k value will not be more than about 100.

The starting matrix for k=1 is

[-1 0][ 1 -1]

Any guidance on where to start with setting up the appropriate procedure would be greatly appreciated.

Thanks.

Greetings to all.

At the following Math.Stackexchange Discussion a certain constant was computed in relation to a Master Theorem Type recurrence being solved. This prompted me to try to identify it by the use of the eponymous command. What follows is the content of the Maple session. You may want to read the post in order to get an understanding of what the constant means and how its exact value is calculated.

> fsolve(2/2^a+1/4^a=1, a); 1.271553303 > identify(%); 1.271553303 > identify(%,all); 1/2 1/2 2 2 3 arcsech(------ + ---- - 1/6 Zeta(5)) 7 6 > evalf(log[2](1+sqrt(2))); 1.271553303

My questions/observations are:

In concluding I would like to say, why the complicated formula and not the simple one? Let me congratulate you just the same on providing this very useful command. I have worked on pandigital approximations which are slightly related and I understand that adding an operation like the logarithm to an integer base up to some max base value can dramatically increase the search space and may not always be feasible.

Best regards,

Marko Riedel

C(n,k)=((2*(k-2-n))/(k*(k-1)))*C(n,k-2)if n in {even} then C(n,0)=((-1)^(n/2))*(factorial(n)/factorial(n/2)) and C(n,1)=0if n in {odd} then C(n,0)=0 and C(n,1)=(2*(-1)^((n-1)/2))*(factorial(n)/factorial((n-1)/2))

i've been trying for hours to get a procedure involving this to work :(

Thanks

Having a recurrence relation of the form:

f(n)*c_{n+1} + g(n)*c_n + h(n)*c_{n-1}=0

with coefficient functions f,g,h is it possible to convert this into a matrix of dimension N+1

thus if the sequences c_n are the coefficients of a series sum_{n=0}^{infinity} c_n x^n I'm only interested up to order N of the series

e.g. N=2

n=0: f(0)*c_{1} + g(0)*c_0 =0

n=1: f(1)*c_{2} + g(1)*c_1 + h(1)*c_{0}=0

n=2: g(2)*c_2 + h(2)*c_{1}=0

I initialise a matrix by recurrence, the dimension of the matrix and its elements depend on the (N1,N2) given. then, I create a vector of the same length of the matrix, its dimension is also defined by (N1,N2). then, I would like to linearsolve the matrix by the vector for a general solution with (N1,N2) as parameters. would this be possible?here is the definition of the matrix and the vector:

-------------------------------------t:...

Is it possible when solving a differential equation to get the corresponding recurrence relation of the series expansion instead of the actual solution?

e.g.

ode:=y''=omega^2*y(x)

the solution is obviously exp(\pm omega*x)

But I want

a_(n+2) = omega^2/(n+1)(n+2) * a_n

or something like that

Hi,

I'm using Maple 14 and I'm trying to solve a system of recurrence relations with initial conditions (a Vector Autoregressive model). The system is

const:=Vector[column]([0.009421681, -0.0005441856]):lagY:=Vector[column]([0.796519372, 0.0179147112]):lagZ:=Vector[column]([0.133049059, 0.5240695764]):A:=<<const>|<lagY>|<lagZ>>;eq1:=y(t) = A[1, 1] + A[1, 2]*y(t - 1) + A[1, 3]*z(t - 1);eq2:=z(t) = A[2, 1] + A[2, 2]*y(t - 1) + A[2, 3]*z(t - 1);

I've got a recurrence equation system like:

fn:={f(x)=sum(f(k)*f(x-k-1),k=0..x-1),f(0)=1}

appearently

rsolve(fn,f(k))

doesn't work.

Is there a way to solve this with maple?

Actually I also would like to solve this step by step, by hand. Could you refer me to something that could help me, solve my problem?

Thanks a lot

Let N(1, 1)=1 and N(1,k)=0 if k<>1. Let N(s+1,k) = N(s,k) + N(s,k-1) + N(s,k+1).Acting in such a way, we obtain the triangle 1 1,1,1 1,2,3,2,1 1,3,6,7,6,3,1.................,where only the nonzero terms are shown.What is the explicit formula for N(s,k)? How to find it with Maple?

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