Check this link: https://in.mathworks.com/help/matlab/ref/meshgrid.html

Alternatively, PFA a crude code (debugging required). Use the above method as a better practice.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

clear all

clc

x = 0.1:1/10:10;

y = 0.1:1/10:10;

lenX = length(x);

lenY = length(y);

[X,Y] = meshgrid(x,y);

z = zeros(lenX,lenY);

for i = 1:length(x)

for j = 1:length(y)

temp = 0;

for n = 0:100

temp = temp + (x(i).^(2-n))*(y(j).^(n/2));

end

z(i,j) = temp;

end

end

surf(X,Y,z);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Here is a (numerical) solution of your problem:

n = 0:1:100;

x = 0:0.1:10; % generate x coordinates

y = 0:0.1:10; % generate y coordinates

[X,Y] = meshgrid(x,y); % generate mesh

temp = zeros(length(y), length(x), length(n)); % preallocate array for temporary results

for i = 1:length(n)

temp(:, :, i) = X.^(2-n(i)).*Y.^(n(i)/2); % evaluate values at given n

end

fxy = sum(temp, 3); % summation along the 3rd dimension

surf(X, Y, fxy);

The two previous answers are a great start, BUT ...not sure we can assume that the variation for n from 0 to 100 is the same as over the grid of x and y ?? clarification please

x=linspace(0,10,50);

y=linspace(0,10,100);

n = 0:1:100;

z=zeros(size(x,2),size(y,2));

for i=1:size(n,2)

z= z + (x.^(2-n(i)))'*(y.^(n(i)/2));

end

surf(y,x,z)

my solution.

Dear Prof. Samanta and Somolik,

Thank you for your kind suggestions. Comparing the both your suggestions the idea diverges each other. Although, the shape of the of plots are converging but the directions are opposite to each other.

Let me do a quick check of an arbitrary function of a known shape.

Let me choose the following function:

f(x,y) = \sum_(n=0)^4 (x - 10)^n + y^n, where 0

My previous concern was based on an erroneous reading -- I see now what the function looks like, and that it is to be evaluated at a grid of X,Y values.