Where, T is the temperature in Kelvin, Beta (or β) is the thermal expansion coefficient (1/K), and Alpha (or α) is the thermal diffusivity (m2/s). Using MATLAB find the properties for T= 321, 440, and 571, respectively.

clc 
clear all 

T=[300 400 500 600]; 
Beta=[1.33e3 1.5e3 0.86e3 0.65e3]; 
Alpha=[0.2128e4 0.4625e4 0.6384e4 0.7190e4];

coeff_Beta=polyfit(T,Beta,1); 
Beta =@(T) coeff_Beta(1)*(T)+coeff_Beta(2);

coeff_Alpha=polyfit(T,Alpha,1); 
Alpha =@(T) coeff_Alpha(1)*(T)+coeff_Alpha(2);

T=[321 440 571]; 
Beta=Beta(T); 
Alpha=Alpha(T);

table(T',Beta',Alpha','variableNames',{'T','Beta','Alpha'})

There is an error occuring in the last line, kindly tell me the problem in the syntax!

Answer 1

[Alan Stevens]

I get no error when running your script exactly as it is:

ans =

3×3 table

T Beta Alpha

___ ______ ______

321 1430.7 2895.8

440 1111.8 4912.3

571 760.72 7132.1

However, I suggest you don't use identical names for a variable and a function. For example, I'd change

Beta =@(T) coeff_Beta(1)*(T)+coeff_Beta(2);

to something like

Betafn =@(T) coeff_Beta(1)*(T)+coeff_Beta(2);

etc.

[Ameer Hamza]

% this function computes the interpolating polynomials
% for the given data, x and y, using Newton's interpolatory
% divided-difference formula. The polynomials of degree
% 1, 2, ..., n are computed, where n is one less than the
% number of data points. The polynomials are then evaluated
% at x0.
function newt_div(x,y,x0)
% n is the maximum degree possible with this data
n=length(y)-1;
% initialize divided-difference table
d=y;
for i=1:n,
% compute divided differences, overwriting those
% we don't need
for j=n+1:-1:i+1,
d(j)=(d(j)-d(j-1))/(x(j)-x(j-i));
end
% use nested multiplication to evaluate the
% polynomial at x0
y0=d(i+1);
for j=i:-1:1,
y0=d(j)+(x0-x(j))*y0;
end
% display coefficients of Newton form of interpolating
% polynomial, and the value at x0
disp(sprintf('Degree %d:',i))
Coefficients=d(1:i+1)
Value=y0
end

[Ameer Hamza]

How we will implement the above mentioned programming to find the values for T??

[Alan Stevens]

Do you mean you want to find the value of T that corresponds to a particular value of Beta, say? If so, then use [1.33e3 1.5e3 0.86e3 0.65e3] for x, [300 400 500 600] for y, and the value of Beta of interest for x0.

[Ameer Hamza]

No! I want to find the value of T (the same question above, that you have answered) by Newtons Divided Difference Mehod as I have mentioned its coding above...

[Ameer Hamza]

This Coding:

% this function computes the interpolating polynomials
% for the given data, x and y, using Newton's interpolatory
% divided-difference formula. The polynomials of degree
% 1, 2, ..., n are computed, where n is one less than the
% number of data points. The polynomials are then evaluated
% at x0.
function newt_div(x,y,x0)
% n is the maximum degree possible with this data
n=length(y)-1;
% initialize divided-difference table
d=y;
for i=1:n,
% compute divided differences, overwriting those
% we don't need
for j=n+1:-1:i+1,
d(j)=(d(j)-d(j-1))/(x(j)-x(j-i));
end
% use nested multiplication to evaluate the
% polynomial at x0
y0=d(i+1);
for j=i:-1:1,
y0=d(j)+(x0-x(j))*y0;
end
% display coefficients of Newton form of interpolating
% polynomial, and the value at x0
disp(sprintf('Degree %d:',i))
Coefficients=d(1:i+1)
Value=y0
end

[Alan Stevens]

" ... I want to find the value of T (the same question above, that you have answered) ..."

The initial question did not find the value of T! The values of T were specified and the routine found the corresponding values of Alpha and Beta. If that's what you want from the interpolating polynomial, then use [300 400 500 600] for x, [1.33e3 1.5e3 0.86e3 0.65e3] for y (to find the corresponding Beta values) and the values [321 440 571] for x0.

[Ameer Hamza]

Yes I need to find the values of Alpha and Beta for T = 321 and 571 respectively by Newtons Divided Difference Mehod