Cant find roots with fzero

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!

[Torsten]

fzero does not work with symbolic variables or expressions.

Use "solve" instead.

Best wishes

Torsten.

Answer 1

[Titus Edelhofer]

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

[Jarl Bergström]

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.

[Walter Roberson]

Q = matlabFunction(a+b);

Answer 2

[John D'Errico]

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

[Jarl Bergström]

Thank you very much for your replay!

The things is. I don't know how to separate the correct answer (in this case 42 :-) ) from -1270 which is also a "correct" solution. T can be both positive and negative.

[Titus Edelhofer]

Hi Jarl,

I'm not sure where the problem is: John's code shows using the symbolic toolbox both real results (-1270.56... and 42.33...). The call to fzero that I wrote gives the same 42.33... without problem (forget the fun about rounding). So what piece is missing? You need to tell us, why 42 is a better solution than -1270. I could imagine that "T" denotes temperature, so it should be larger than -273 I guess, is it this? So why not take the 42.33...?

Titus

[Tewodros Bitaw]

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]

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()