interpolation for 4cos(x) − ex = 0

theres only one positive root for the equation 4cos(x) − ex = 0 for [0,1] domain. use a parabolic function calculated through interpolation to find the approximation for this root. you can choose whatever points you want.

any idea on how to solve this? thanks yall in advance!

[Matt J]

what points have you chosen?

[xbhax]

y = [-4.71 -1.51]; but i might be missing one bc its parabolic i guess? any suggestions?

[James Tursa]

Are you supposed to pick any three points you want, fit a parabola to those points, and then find the root of that parabola in the [0,1] domain? Can you use the MATLAB polyfit( ) and roots( ) functions for this? Or are you supposed to be using hand-written code?

[xbhax]

no, i cant use polyfit nor roots for this... and yes to the 1st question :)

Answer 1

[James Tursa]

Without seeing the text of your assignment, I would presume that this is the outline of what you are supposed to do:

1) Define the function y(x) = 4*cos(x) − exp(x)

2) Pick three points of this function near the desired root in interval [0,1], e.g. (0,f(0)), (0.5, f(0.5)), (1,f(1))

3) Fit a parabola to those three points

4) Find the root of that parabola in the [0,1] interval

Steps 1 and 2 are very easy. Step 3 would have been very easy with polyfit( ), but it is still easy to solve for arbitrary parabola coefficients using a hand written algorithm. Just ask yourself if you have three points (x1,y1,), (x2,y2), and (x3,y3), how would you write the equation of a parabola that passed through these three points? That would give you three equations in three unknowns (the parabola coefficients a, b, c). Then step 4 is easy with your own quadratic formula or other quadratic root code.

See if you can figure out step 3 and come back with any problems you might have.

[James Tursa]

Start with a generic parabola expression

y = a*x^2 + b*x + c

You have three points on this parabola, (x1,y1), (x2,y2), and (x3,y3). So plug them in to get three equations:

y1 = a*x1^2 + b*x1 + c

y2 = a*x2^2 + b*x2 + c

y3 = a*x3^2 + b*x3 + c

You know all the x and y values, so rewrite this as a matrix equation for the unknowns a, b, c:

Y = [some matrix] * [a;b;c]

Now you've got a matrix expression and you are trying to find the vector [a;b;c]. Do you know how to solve a linear system using MATLAB? This is the approach I was steering you towards.

But if you were directed to use a different approach or if this isn't really the problem you are supposed to be solving, you will need to let us know. Maybe post the text of your assignment.

[xbhax]

Heyy, thank you very much for your reply! followed step by step and i think i got it...

This is what ive got so far

syms x 
f = @(x) (4.*cos(x) - exp(x)); 
x = [0,0.5,1]; 
y = [f(0),f(0.5),f(1)];
a = f(x(1))
b = f(x(2))
c = f(x(3))

% polyfit(x,y,2) -> checking

p = fLagrange(x,y);
p = simplify(p);
p = vpa(p,5)


P = [0,1]; 
vp = subs(p,P)

one last question is how do i find the approximation for the root if i dont know the x value? thank you!

[James Tursa]

I was using a, b, and c to denote the coefficients of the quadratic function. This doesn't match what you have for a, b, and c above.

To solve for the roots of a quadratic function, you could simply use the quadratic formula. That will give you the x value approximation of the root to the original function.

[xbhax]

got it! thanks a lot! :)