J4James

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8 years, 344 days

MaplePrimes Activity


These are replies submitted by J4James

@dharr No more recursive error, I dont know why it was happening in the first place. Anyways I am still facing the large expression and Rootof. eta(H)sheet2.mw

@dharr If I follow your approach then I get Error, recursive assignment and I am unable to use the expression for Th and Tc in eta?

@dharr 

If assign value to H then I easily get a numerical value. What I am looking for is an expression interms of H.

 

@Kitonum 

Is it possible to smooth this curve?

@tomleslie 

The value for parameter "M" is missing. 

@Carl Love 

It just procduces zero repeatedly.  

Vector[column](13, [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])

@tomleslie 

Thank you for your efforts!

@Carl Love Thanks for the concise answer. 

If I just want to extract values of P1 then which syntax will work? 

Vector([seq(eval(diff(J(y),y), F2[Q]),Q=-3..3,ec)])


 

@Muhammad Usman 

You need to watch your language. This is community based site. People here are not paid (volunteers) to answer your wishes. 

You should present your problem along with your attempt. If someone wants to answer then they will otherwise you will get nothing? 

We should be appreciative of the responders for their valuable time and answers.

I think, you need to apologize to @nm for your bad behaviour.  

@DSkoog 

 

Thank you for your efforts. But I am still waiting for you to follow it.

@Christopher2222 BTW, I like SE format. Any ways, If you follow the proposed site, I will really, appreciate? 

@tomleslie 

What if we are suppose to use different intervals for a, b and c. say a is in (0,1), b is in (0.1, 0.5) and c is in (1,2)

along with a small change in the expression Expr:=f(x)+b*g'(x)+c*f'(x)*g(x) #

@tomleslie 

I really appreciate your patience. You were very close. I have modified your code to make it the way I wanted. But the problem is now that I have three loops, which I assume may cause problems if I have a more complicated system.

odeExpr2.mw

@tomleslie Once you used a = b = 0.5, for solving the system and then varied it for the expression. Doesn't it effect the results?

What I am trying to say is that a and b should vary simultaneously for both the system and Expr.  

e.g.,  Expr:=(a=0.1)*f(1)+(b=0.1)*g'(1)+c*f'(1)*g(1) # at x=1: Here f(1) etc has been found for a=0.5, which should be continuously updated for each value of a and b then used in Expr subsequently. 

I think the confusion arise, becuase I used a=b=0.5 in dsolve. I was just trying to show that the system has a solution.

 

@Thomas Richard 

Thanks. 

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