You would make a good start by simplifying your equations (by collecting all the constant terms for example)can u pls solve this? i tried all what i know.

the elationship between the resistance R of the thermistor and the temperature is given by

(1/T)-1.129241*10^(-3)=2.341077*10^(-4)*log(R)+8.775468*10^(-8)*(log(R))^(3)

where T is in Kelvin and R is in ohms.

A thermistor error of no more than 0.01C is acceptable. To find the range of the

resistance that is within this acceptable limit at 19C , we need to solve

1/19.01+ 273.15=1.129241×10^(-3) + 2.341077×10^(−4)*ln(R) + 8.775468 × 10^(−8)* {ln(R)}^(3)

and

1/18.99 + 273.15= 1.129241× 10^(−3) + 2.341077 × 10^(−4) ln(R) + 8.775468 × 10^(−8) {ln(R)}^3

Use Newton’s method to find the resistance at 18.99C. Find the absolute relative

approximate error at the end of each iteration, and plot relative approximate error versus

iteration number.

clear all

clc

R=solve('(1/(18.99+273.15))-1.129241*10^(-3)=2.341077*10^(-4)*log(R)+8.775468*10^(-8)*(log(R))^(3)'); %gercek R degeri

Ri=3; tol=273.15+0.01; error=273.15+10;

k=1; a=1;

while error>=tol

f=1.129241*10^(-3)+2.341077*10^(-4)*log(Ri)+8.775468*10^(-8)*(log(Ri))^(3)-(1/(18.99+273.15));

df=((2.341077*10^(-4))/(Ri))+(8.775468*10^(-8)*3*(log(Ri))^2)/Ri;

Rnew=Ri-(f/df);

fnew=1.129241*10^(-3)+2.341077*10^(-4)*log(Rnew)+8.775468*10^(-8)*(log(Rnew))^(3)-(1/(18.99+273.15));

error=abs(fnew);

relativeerror=(error/R);

Ri=Rnew

k=k+1;

end

plot(k,relativeerror)

i wrote program this but i cant take the plot of all iteration what do u sugges??

CB