## 28 Reputation

14 years, 294 days

## mathematica...

@Markiyan Hirnyk

Can you explain more about what Mathematica has missed?

## mathematica...

@Markiyan Hirnyk

Can you explain more about what Mathematica has missed?

## conjunction again...

Thank you but I still find it hard to understand the solution space maple has given. Is there a way to get maple to plot it?

## conjunction again...

Thank you but I still find it hard to understand the solution space maple has given. Is there a way to get maple to plot it?

## symbolic=false...

Thank you. How do you interpret the  "And(2 argument([...]" mathematically?

## symbolic=false...

Thank you. How do you interpret the  "And(2 argument([...]" mathematically?

## rsolve...

Hi,

I am really just trying to plot the solution to this recurrence for a(n-1) as n grows.

rsolve({a(1) = 1+2/n, a(p) = 1+(p-1)*a(p-1)/n+2*b(p-1)/n, b(0) = 1, b(p) = 1+p*b(p-1)/n}, {a(p), b(p)});

This makes sense for non-integer n but it happens that in my case I am only interested in the case where n is a positive integers.  I am however slightly surprised that the graph isn't smooth when n takes non-integer values. Maybe it is obvious why this is?

## rsolve...

Hi,

I am really just trying to plot the solution to this recurrence for a(n-1) as n grows.

rsolve({a(1) = 1+2/n, a(p) = 1+(p-1)*a(p-1)/n+2*b(p-1)/n, b(0) = 1, b(p) = 1+p*b(p-1)/n}, {a(p), b(p)});

This makes sense for non-integer n but it happens that in my case I am only interested in the case where n is a positive integers.  I am however slightly surprised that the graph isn't smooth when n takes non-integer values. Maybe it is obvious why this is?

## better but still broken...

'add' produces a better but still basically broken graph. You get all kinds of weird effects at the non-integer positions.

## better but still broken...

'add' produces a better but still basically broken graph. You get all kinds of weird effects at the non-integer positions.

## Thanks...

Thanks very much!

## Thanks...

Thanks very much!

## General solution...

Thanks. If you just take the second recurrence alone, it seems that you should be able to get a closed form solution for B(k) in terms of A(0) and n but I can't get rsolve to do it. How would you do that?

## General solution...

Thanks. If you just take the second recurrence alone, it seems that you should be able to get a closed form solution for B(k) in terms of A(0) and n but I can't get rsolve to do it. How would you do that?

## Pick out A(0)...

How do I just output A(0) using your code? My crude method is

solutions:= evalf(solve({SA,SB},{seq(A(k),k=0..n-2),seq(B(k),k=1..n-1)}));

eval(A(0), solutions);

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