Interpolation Formula in MATLAB | Solving for y(ti) with Code Example

[sara_87]

Homework Statement

If i have an interpolation formula, say:

$$y(t_i)=3+h\sum^i_{j=1}jy(t_j)$$

where i=0 to n
h=(a+b)/n
t_i=i*h
and
when i=0, we have: y(t₀)=0.

How would i write a code to find y(t_i)?

Homework Equations

The Attempt at a Solution

function y=examp(a,b,n)
h=(a+b)/n
for i=1:n+1
t(i)=(i-1)*h
for j=1:i+1

and now I am stuck because i don't know how write a code that uses the previous value of y to compute the next value of y.

 

[badphysicist]

Before the first loop, I would set up the y's(actually, it would be faster for your code to init the t's to zeros to, but unless n is very large it shouldn't matter):

y = zeros(n+1,1)

Then in the loop it would be something like
y(j) = 3 + sum( y(1:j-1).*(1:j-1) )

the .* does an element-wise product

 

[Mene]

Can anyone help me with this?



There are a few ways you can write a code to solve for y(ti) using the interpolation formula provided. Here is one possible approach:

function y = interpolate(a, b, n)

% Initialize h and ti
h = (a + b) / n;
ti = 0;

% Create an array to store y values
y = zeros(1, n+1);

% Use a for loop to iterate through each value of ti
for i = 1:n+1
% Use the interpolation formula to calculate y(ti)
y(i) = 3 + h * sum(1:i .* y(1:i));

% Update ti for the next iteration
ti = ti + h;
end

% Set y(t0) = 0
y(1) = 0;

% Plot the results
plot(t, y);
xlabel('t');
ylabel('y(t)');
title('Interpolation of y(t)');

This code creates a function called "interpolate" that takes in the values of a, b, and n as inputs. It then calculates h and initializes ti to 0. Next, it creates an array to store the values of y and uses a for loop to iterate through each value of ti, calculating y(ti) using the interpolation formula provided. The code also sets y(t0) = 0 and plots the results.

Keep in mind that this is just one possible approach and there may be other ways to write the code. It's always a good idea to test your code with different inputs to make sure it is working correctly.