Asked by Max Bernstein
on 18 Oct 2011

Hello,

I have the following equation that I'm trying to solve. I need to solve for T for a very long array of known RT

function RT=RTD(T) a=0.003925; b=0.110361; d=1.492690; R0=100; RT=R0*(1+a*(1+d/100)*T-a*d/100^2*T^2+a*b/100^3*T^3-a*b/100^4*T^4);

I dont think I have fsolve or symbolic toolbox installed. Is there anyway to automate this without using those?

Thanks

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Answer by Naz
on 19 Oct 2011

Assuming that each RT will yield four roots:

a=0.003925;

b=0.110361;

d=1.492690;

R0=100;

T=zeros(4,length(RT));

for n=1:length(RT)

T(:,n)=roots(-a*b/100^4*R0 a*b/100^3*R0 -a*d/100^2*R0 a*(1+d/100)*R0 (1-RT(n))*R0);

end

If you want to create an actual function, go to file->new->script and paste the following:

function [T]=RTD(RT)

a=0.003925;

b=0.110361;

d=1.492690;

R0=100;

T=zeros(4,length(RT));

for n=1:length(RT)

T(:,n)=roots(-a*b/100^4*R0 a*b/100^3*R0 -a*d/100^2*R0 a*(1+d/100)*R0 (1-RT(n))*R0);

end

end

Then you can call this function and pass your array of RT. The function will return you an array of T.

T=RTD(RT)

P.S. Make sure there is enough parenthesis in the equation so the coefficients are calculated properly. I recommend to calculate the coefficients prior sending them to roots function. This also will save some time, since the coefficients will be calculated only once instead of each loop.

Walter Roberson
on 19 Oct 2011

You left out the [] in that roots() call, so this code will end up taking roots() of a scalar.

Answer by Andrei Bobrov
on 19 Oct 2011

a=0.003925; b=0.110361; d=1.492690; R0=100; syms k T cf = coeffs(k-R0*(1+a*(1+d/100)*T-a*d/100^2*T^2+a*b/100^3*T^3-a*b/100^4*T^4),T) fcf = matlabFunction(cf(end:-1:1)) RTD = @(RT)cell2mat(arrayfun(@(x)roots(fcf(x)),RT,'un',0))

use function `RTD`

T = RTD(RT)

Answer by Walter Roberson
on 18 Oct 2011

roots( [R0*a*b, -100*R0*a*b, 10000*R0*a*d, -100000000*R0*a-1000000*R0*a*d, -100000000*R0+100000000*RT] )

Note that roots() can only process one set of coefficients at a time: it is not vectorized.

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