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jlee-21-Visitor
1-Visitor
November 24, 2014
Solved

Can I calculate differential by using "solve block"?

  • November 24, 2014
  • 6 replies
  • 2724 views

캡처7.PNG

I saw prime 3.0 book and then I got some question

In this picture, T has three "variables". it is mean " i , j , k "

I tried to calculate differential about " r " but I got error.

So I want to calculate in "solve block"

But I am worried that I saw that "solve block" can calculate only "one variable"

IS is right?

and Now t has variables " i, j, k"

" r " include " i " . so when " i " change, " r " is also changed.

so the thing that I want to say is possible to get values about " r " when I calculate differential about " i "

Best answer by Werner_E

1) t is not set up to be dependent on rc, so you sure should not differentiate with respect to rc (unless you want zero as result).

2) You can only integrate and differentiate continuous functions. Your function T uses the argument i as an index to a vector rc. This makes T a discrete, non-continuos function which cannot be integrated or differtiated. Think over it - what should happen if i has the value 1.5?

You use a lot of nested functions in your sheet which makes it very hard to track. Maybe you have success if you make rc a function of i rather than a vector lookup. If ORIGIN is set to default 0, create a vector idx:=stack(0,1,2,....last(cr)) and you may also use Stuarts routines to make it easier to create that vector. Then create a function rc_f(i):=linterp(idx, rc,i) and use rc_f(i) instead of rc[i. Maybe this helps but I haven't looked at the other functions involved so there may be more problems. Your functions are too much nested and complicated for me to look at.

BTW, of course you can solve for more than one variable using a solve block. You sure will find examples for solving a system of equations in the help as well as in your book.

6 replies

Werner_E
Werner_E25-Diamond IAnswer
25-Diamond I
November 24, 2014

1) t is not set up to be dependent on rc, so you sure should not differentiate with respect to rc (unless you want zero as result).

2) You can only integrate and differentiate continuous functions. Your function T uses the argument i as an index to a vector rc. This makes T a discrete, non-continuos function which cannot be integrated or differtiated. Think over it - what should happen if i has the value 1.5?

You use a lot of nested functions in your sheet which makes it very hard to track. Maybe you have success if you make rc a function of i rather than a vector lookup. If ORIGIN is set to default 0, create a vector idx:=stack(0,1,2,....last(cr)) and you may also use Stuarts routines to make it easier to create that vector. Then create a function rc_f(i):=linterp(idx, rc,i) and use rc_f(i) instead of rc[i. Maybe this helps but I haven't looked at the other functions involved so there may be more problems. Your functions are too much nested and complicated for me to look at.

BTW, of course you can solve for more than one variable using a solve block. You sure will find examples for solving a system of equations in the help as well as in your book.

J
jlee-21-VisitorAuthor
1-Visitor
November 24, 2014

Thank you for helping

I understood a half.

I have a question about your talk

Maybe this helps but I haven't looked at the other functions involved so there may be more problems.

you said this sentence, but i can't understand well.

could you explain for me?

and then you said I need to think if "i" is 1.5 .

I learned it is possible to integrate if i have each section data.

so Is it possible to integrate because I has value i=1,2,3..17

Werner_E
Werner_E25-Diamond I
25-Diamond I
November 24, 2014

jinsuk Lee wrote:

Thank you for helping

I understood a half.

I have a question about your talk

Maybe this helps but I haven't looked at the other functions involved so there may be more problems.

you said this sentence, but i can't understand well.

could you explain for me?

Your function T calls other functions - vr, c and cn. I have not looked ath the definition of those functions, so it may be that there are additional problems there. The two problems I pointed you to are the fact that you should not differentiate with respect to a variable the function is not dependent on and the fact that T is not a continuous function in i and so cannot be differentiated or integrated with respect to i.

and then you said I need to think if "i" is 1.5 .

I learned it is possible to integrate if i have each section data.

so Is it possible to integrate because I has value i=1,2,3..17

Not, that would not be integration but a simple summation! Integration uses an infinitesimal small di and thats not possible when i can only be a number of discrete values!

F
Fred_Kohlhepp23-Emerald I
23-Emerald I
November 24, 2014

Attached is a short discussion about range variables, vectors, functions, integration and differentiation, couched (hopefully) in examples that you can relate to.

J
jlee-21-VisitorAuthor
1-Visitor
November 24, 2014

thank you very much

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