How to get Numerical value from a Symbol?
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zongquan
am 29 Jan. 2012
Kommentiert: Torsten
am 19 Apr. 2023
Hi, everyone,
I was working on extracting the numerical value of a symbol, however, it does't work for me.
x=2*y+5*y^3+3*y^5 and I have a series values of 'x'. What I want is to get the numerical values of 'y' which contains only pure real values.
When I wrote "solve('x=2*y+5*y^3+3*y^5','y')", it gave me 5 numerical answers on screen. However, it showed "5*1 syms" in the Workspace. Moreover, when I put the "solve" in the 'for' loop, it only showed 1*1 sym.
I am wondering who could help me this. Your suggestion is highly appreciated!
Thank you
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Walter Roberson
am 29 Jan. 2012
Use double() to convert a symbolic number to a double precision value.
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Vani Madhavi Tukkapuram
am 19 Apr. 2023
D = 1000;
Co = 2000;
Cs = 200;
Cp = 100;
Ch = 0.20;
Cd = 0.15;
R = 20;
t1 = 5;
syms D Co Cs Cp Ch Cd R t1 T
%theta(t) = 1 / (1 + R - t);
%I1(t) = L - D * t;
%I2(t) = D(1 + R - t) * log((1 + R - t1)/(1 + R - T));
L = D * (t1 + ((1 + R - t1) * log((1 + R - t1)/(1 + R - T))));
OC = Co;
HC_1 = (L * t1)-((D * t1^2)/2) + D *(1 + R);
HC_2 = (-T + t1);
HC_3 = (1 + R)/2;
HC_4 = (3/2) * (t1^2/(1+R));
HC_5 = log((1 + R - t1)/(1 + R - T));
HC_6 = (T + 3*t1)/2;
HC_7 = t1*log(1 + R-t1)-T*log(1+R-T);
HC = Ch*(HC_1*((HC_2-HC_3-HC_4)*(HC_5+HC_6+HC_7)));
DC_1 = D*Cd;
DC_2 = (T-t1);
DC_3 = (1+R-t1)*log((1 + R - t1)/(1 + R - T));
DC = DC_1*(-DC_2+DC_3);
PC_1 = D*Cp;
PC_2 = t1;
PC_3 = (1+R-t1)*log((1 + R - t1)/(1 + R - T));
PC = PC_1 * (PC_2 + PC_3);
SRC = D*Cs*T;
TP = (SRC - OC - HC - DC - PC)./T;
f = TP==0;
diff(f,T);
disp('---------------------------------');
2 Kommentare
Torsten
am 19 Apr. 2023
Your expression is too difficult to be solved for T explicitly.
Give values to the other parameters and use "vpasolve".
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