Cant find roots with fzero

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Jarl Bergström
Jarl Bergström 2015-6-8
评论： Walter Roberson 2017-3-15
This is pretty straight forward.
I have to following equation:
syms T
Qbalance(T) =(2585111364437669*T)/2251799813685248 + (7434051551537793*(T + 273)^4)/4835703278458516698824704 - 4489244199846279/70368744177664
And I'm trying to find one of the possible roots by using fzero: fzero(Qbalance,40) I know that one root is ~42 :-)
But it doesnt seem to work. Any idea what I can do to make it work anywhere between T = [-100 100] ?
Thanks!
  1 个评论
Torsten
Torsten 2015-6-8
fzero does not work with symbolic variables or expressions.
Use "solve" instead.
Best wishes
Torsten.

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回答（2 个）

Titus Edelhofer
Titus Edelhofer 2015-6-8
Hi,
either do what Torsten suggests, or create a function handle instead of using syms:
Qbalance = @(T) (2585111364437669*T)/2 ...
And yes, it's 42:
answerToEverything = round(fzero(Qbalance, 40))
answerToEverything =
42
;-)
Titus
  2 个评论
Jarl Bergström
Jarl Bergström 2015-6-8
Thank you very much for your replay!
The thing is, that doesn't seems to work if I have:
syms T
a=1+T
b=T
and then Q = @(T) a+b or Q=a+b
How can I put that equation into fzero?
The reason for why I dont want to use solve is because I get 4 roots and I don't know how to separate the correct one, because T can be both positive and negative.

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John D'Errico
John D'Errico 2015-6-8
编辑：John D'Errico 2015-6-8
No. It is NOT 42. Close, but no cigar. Unless of course, you round the result as did Titus. :)
If you are going to use syms, then why in the name of god and little green apples, why not solve? This is a 4th order polynomial after all.
syms T
Qbalance = (2585111364437669*T)/2251799813685248 + (7434051551537793*(T + 273)^4)/4835703278458516698824704 - 4489244199846279/70368744177664;
vpa(solve(Qbalance))
ans =
-1270.5696869775977378281084791948
42.330660289301496605774980024324
68.119513344148120611166749585241 + 814.64831383344987959986838110296i
68.119513344148120611166749585241 - 814.64831383344987959986838110296i
Looks like more like 42.33066... to me.
roots(sym2poly(Qbalance))
ans =
-1270.5696869776 + 0i
68.1195133441485 + 814.64831383345i
68.1195133441485 - 814.64831383345i
42.3306602893015 + 0i
Roots agrees.
Qbalance = @(T) (2585111364437669*T)/2251799813685248 + (7434051551537793*(T + 273).^4)/4835703278458516698824704 - 4489244199846279/70368744177664;
ezplot(Qbalance,[41,43])
grid on
Yep, the plot says so too. As does fzero, with absolutely no problems.
format long g
fzero(Qbalance,[0,100])
ans =
42.3306602893015
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Tewodros Bitaw
Tewodros Bitaw 2017-3-15
编辑：Tewodros Bitaw 2017-3-15
Hi Jarl I think this works better.
syms T
T0=298.15;
Qbalance(T) =(2585111364437669*T)/2251799813685248 + (7434051551537793*(T + 273)^4)/4835703278458516698824704 - 4489244199846279/70368744177664 Q=matlabFunction(Qbalance(T))
T=fzero(Q,T0)
T =
42.3307
Walter Roberson
Walter Roberson 2017-3-15
4835703278458516698824704 cannot be kept at full precision in the form shown.
syms T positive
Q = @(v) sym(v,'r');
T0 = Q(298.15);
Qbalance(T) =(sym('2585111364437669')*T)/sym('2251799813685248') + (sym('7434051551537793')*(T + sym(273))^4)/sym('4835703278458516698824704') - sym('4489244199846279')/sym('70368744177664')
solve(Qbalance(T))
The solution (in recent MATLAB) is
root(z^4 + 1092*z^3 + 447174*z^2 + (6156509634857202603287236*z)/7434051551537793 - 89068512980281706423859477/2478017183845931, z, 2)
which can be found to arbitrary precision using vpa(), or converted to double precision with double()

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