Insert symbolic equation in another symbolic equation

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Mirko Rizzi
Mirko Rizzi am 17 Apr. 2023
Kommentiert: Mirko Rizzi am 17 Apr. 2023
Hello i have the equation eq5 that has been solved and is now only a function of y. I want now to insert this function (r) inside eq6 and solve that, but it gives me an error. How can i solve this?
eq5 = hw2*(q-r)==y;
r = solve(eq5,r);
eq6 = @(y) sigma*eps*((r^4)-T_inf^4)-y;
y0=0;
sol = fsolve(eq6,y0);
  8 Kommentare
Mirko Rizzi
Mirko Rizzi am 17 Apr. 2023
Bearbeitet: Mirko Rizzi am 17 Apr. 2023
if i use this:
% eq6 = sigma*eps*((r^4)-T_inf^4)==y;
% sol = solve(eq6,y)
it gives me a vector of 4 same elements:
root(z^4 - (148549786974450477987*z^3)/27487790694400 + (66201117630463850684060566839278314716507*z^2)/6044629098073145873530880000 - (25694677813792110401306682574502596589303194435085805665710978025618652510443*z)/2420199345095688424498763567867944239211734959652864000000 + 3804255296569717375581777916032849021112749976946199317897911767146631227587687/1141798154164767904846628775559596109106197299200000000, z, 1)
root(z^4 - (148549786974450477987*z^3)/27487790694400 + (66201117630463850684060566839278314716507*z^2)/6044629098073145873530880000 - (25694677813792110401306682574502596589303194435085805665710978025618652510443*z)/2420199345095688424498763567867944239211734959652864000000 + 3804255296569717375581777916032849021112749976946199317897911767146631227587687/1141798154164767904846628775559596109106197299200000000, z, 2)
root(z^4 - (148549786974450477987*z^3)/27487790694400 + (66201117630463850684060566839278314716507*z^2)/6044629098073145873530880000 - (25694677813792110401306682574502596589303194435085805665710978025618652510443*z)/2420199345095688424498763567867944239211734959652864000000 + 3804255296569717375581777916032849021112749976946199317897911767146631227587687/1141798154164767904846628775559596109106197299200000000, z, 3)
root(z^4 - (148549786974450477987*z^3)/27487790694400 + (66201117630463850684060566839278314716507*z^2)/6044629098073145873530880000 - (25694677813792110401306682574502596589303194435085805665710978025618652510443*z)/2420199345095688424498763567867944239211734959652864000000 + 3804255296569717375581777916032849021112749976946199317897911767146631227587687/1141798154164767904846628775559596109106197299200000000, z, 4)
Mirko Rizzi
Mirko Rizzi am 17 Apr. 2023
if i view the entire function r, copy what it is written and substitute it inside works fine, but as it is in the cycle i can't copy and paste manually at every passage

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Akzeptierte Antwort

Dyuman Joshi
Dyuman Joshi am 17 Apr. 2023
Bearbeitet: Dyuman Joshi am 17 Apr. 2023
Use vpa to get numerical values. Convert them to double() if you need the values to be in numeric data type.
syms z
%first root, copied from above
sol1=vpa(root(z^4 - (148549786974450477987*z^3)/27487790694400 + (66201117630463850684060566839278314716507*z^2)/6044629098073145873530880000 - (25694677813792110401306682574502596589303194435085805665710978025618652510443*z)/2420199345095688424498763567867944239211734959652864000000 + 3804255296569717375581777916032849021112749976946199317897911767146631227587687/1141798154164767904846628775559596109106197299200000000, z, 1))
sol1 = 
549277.87721495737433337758873113
double(sol1)
ans = 5.4928e+05
  3 Kommentare
Dyuman Joshi
Dyuman Joshi am 17 Apr. 2023
They look same because the syntax of the output above is same except for the number of root.
Since solve() was unable to find the explicit value, it returns the solution as -
root(equation,variable,1)
root(equation,variable,2)
root(equation,variable,3)
root(equation,variable,4)
You can see at the end of the each expression there's a number, denoting which root it refers to. The value of roots will, of course, depend upon the coefficients.s
Mirko Rizzi
Mirko Rizzi am 17 Apr. 2023
thanks for explaining me!

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