## 48 Reputation

17 years, 18 days

## Hi, acer,...

Hi, acer, Yes, you misunderstood what I wrote. But from your suggestion I have worked out, it is g := proc(n) local f,F,i; f := (t,n) -> n+1/t; F := n; for i from n-1 to 1 by -1 do F := f(F,i); od; F; end proc; which is what I wanted. Use the function cfrac suggested by Tarr to verify: a := g(9); cfrac(a); David

## Hi, acer,...

Hi, acer, Yes, you misunderstood what I wrote. But from your suggestion I have worked out, it is g := proc(n) local f,F,i; f := (t,n) -> n+1/t; F := n; for i from n-1 to 1 by -1 do F := f(F,i); od; F; end proc; which is what I wanted. Use the function cfrac suggested by Tarr to verify: a := g(9); cfrac(a); David

## I see. Thanks a lot. David ...

I see. Thanks a lot. David

## Prof. Douglas,...

Prof. Douglas, Thank you for your excellent explanation and code, esp. the two modifications in my code. I have tried it out. BTW, when this line is included, > with( plots ): Maple returns a warning message Warning, the name changecoords has been redefined What does this mean? Another question is I did try to avoid using df := 2*x; by using df := diff(f(x),x); but it doesn't work, while yours df := D(f); works, why? So it seems that in Maple plotting pairs of data is quite different from Matlab, in the latter it is much more simple, just plot(a,b). David

## Prof. Douglas,...

Prof. Douglas, Thank you for your excellent explanation and code, esp. the two modifications in my code. I have tried it out. BTW, when this line is included, > with( plots ): Maple returns a warning message Warning, the name changecoords has been redefined What does this mean? Another question is I did try to avoid using df := 2*x; by using df := diff(f(x),x); but it doesn't work, while yours df := D(f); works, why? So it seems that in Maple plotting pairs of data is quite different from Matlab, in the latter it is much more simple, just plot(a,b). David

## thank you, alec....

thank you, alec. I agree the writing f(x=0) is not standard, I just want exppress the function value of f when its independent variable x=0, or x=x1. x1 is the same as x0 in the code, just some given value of x. In my code which I followed you did (below the ===== line) there is no x1. If dverk78 method fails for my problem, you are suggesting me to use other method like rkf45? Or any other methods that I use. Maybe I should try it myself first. David

## thank you, alec....

thank you, alec. I agree the writing f(x=0) is not standard, I just want exppress the function value of f when its independent variable x=0, or x=x1. x1 is the same as x0 in the code, just some given value of x. In my code which I followed you did (below the ===== line) there is no x1. If dverk78 method fails for my problem, you are suggesting me to use other method like rkf45? Or any other methods that I use. Maybe I should try it myself first. David

## Yeah, I also got this...

Yeah, I also got this trivial solution too. There is definitely other non-trivial solution, which I am looking for.

## Yeah, I also got this...

Yeah, I also got this trivial solution too. There is definitely other non-trivial solution, which I am looking for.

## well, this is one of the...

well, this is one of the solutions, but not the one I am interested. I am trying to get the other nontrival solution.

## well, this is one of the...

well, this is one of the solutions, but not the one I am interested. I am trying to get the other nontrival solution.

## Thank alec and Doug. For...

Thank alec and Doug. For alec's suggestion, I tried to follow, but I could not get a solution for my problem, which is (to answer Doug's question) DEq := diff(f(x),x) = c1*sqrt( g(f(x)) - g(f(x=x0)) ); Eq := ug - f(x=0) = c2*sqrt( g(f(x=0)) - g(f(x=x0)) ); where g(f) := exp(f-xn) +exp(-f) +(1-exp(-xn))*f; c1,c2,xn,ug are parameters, and x0 is some specific given value of x. Actually, I am solving a Poisson equation which reads DEq2 := diff(f(x),x\$2) = c *( exp(f-xn) -exp(-f) +(1-exp(-xn)) ); constrains (mixed boundary conditions): 1. ug - f(x=0) = c2*sqrt( g(f(x=0)) - g(f(x=x0)) ); (g(f) defined above) 2. diff(f(x),x) = 0 when x=x0; The above first-order DEq is just a first-integration of DEq2 with constrain 2. alec or Doug, do you have any suggestion on the above problem? Thanks again for your time. BTW, can you let me know how to post Maple workseet contents as in Doug's lastet post? David

## Thank alec and Doug. For...

Thank alec and Doug. For alec's suggestion, I tried to follow, but I could not get a solution for my problem, which is (to answer Doug's question) DEq := diff(f(x),x) = c1*sqrt( g(f(x)) - g(f(x=x0)) ); Eq := ug - f(x=0) = c2*sqrt( g(f(x=0)) - g(f(x=x0)) ); where g(f) := exp(f-xn) +exp(-f) +(1-exp(-xn))*f; c1,c2,xn,ug are parameters, and x0 is some specific given value of x. Actually, I am solving a Poisson equation which reads DEq2 := diff(f(x),x\$2) = c *( exp(f-xn) -exp(-f) +(1-exp(-xn)) ); constrains (mixed boundary conditions): 1. ug - f(x=0) = c2*sqrt( g(f(x=0)) - g(f(x=x0)) ); (g(f) defined above) 2. diff(f(x),x) = 0 when x=x0; The above first-order DEq is just a first-integration of DEq2 with constrain 2. alec or Doug, do you have any suggestion on the above problem? Thanks again for your time. BTW, can you let me know how to post Maple workseet contents as in Doug's lastet post? David

## Thank you. I tried, it...

Thank you. I tried, it works.

## Thank you. I tried, it...

Thank you. I tried, it works.
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