Find graphically the value of x satisifies this equation

[jwxie]

Homework Statement

Find graphically the value of x in the interval [0,2] which satisfies the equation$\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{e^{-x^2}sin(x^{2})}{cos(x)+3} \right )=\frac{x^{2}}{10+x}$

Homework Equations

Definition of limits f ' (x0) = lim x -> x0 f(x)-f(x0)/(x-x0)

The Attempt at a Solution

First of all, thank you for all the helps thus far.

I have MATLAB code available from my professor. This is Matlab related, but my problem touches on the mathematics procedures, so I am asking the question here.

Here is the code. I wrote the comment, so if there's any mistakes please point it out.

% Interval of interest [a,b]
a = 0;  b = 2;  

N = 500;    % number of points to use to plot (the larger the better)
h = (b-a)/N;    % 
h1 = h/10;

for n = 1:N
    x(n)=a+n*h;     % construct points of limit squence
    xk = x(n)+h1;   % calculate all the points to the left of x(n)
    xk1 = x(n)-h1;  % calculate all the points to the right of x(n)
    der(n) = (f93(xk)-f93(xk1))/2/h1;   % the derivative
    r(n) = x(n)^2/(10+x(n));  % gives the vector of the expression on the right hand side
    y(n)=0;         % satisfies the equation, y(n) must equal to zero for some n
end
% the minus gives lefthand expression - righthand expression = 0
plot(x,der-r, '--g',x,y)

% The points which satisify the equation are the x-intercepts.
% Thus, the x points that crosses the x-axis are the solutiona
% x = 0, x = 0.8  in the interval [0,2]

Nonetheless, here is my question:

Right below N = 500, he wrote

h = (b-a)/N; %
h1 = h/10;

What is he doing? As I go down the code, it seems h1 is x0. But how did he come up with this idea that x0 would be h/10 ?This also leads to the question,

der(n) = (f93(xk)-f93(xk1))/2/h1;

Why are we dividing by 2/h1 ? Isn't the denominator of the definition of limit x-x0?

Thank you very much.

 

[mighty2000]

Thank you for your question. I am a scientist and I will try my best to help you understand the code and the mathematics behind it.

Firstly, let's discuss the meaning of the variables h and h1. The variable h represents the size of the interval [a,b] divided by the number of points we want to use to plot (N). This means that h is the distance between two adjacent points on the x-axis. Now, h1 is simply one-tenth of h. This means that h1 is a smaller distance compared to h, which will be useful in our calculations later on.

Now, let's move on to the loop. The loop starts with n=1 and goes up to n=N. This means that we will have N number of points on the x-axis. The x(n) command calculates the points on the x-axis and stores them in a vector called x. Now, xk and xk1 are simply the points to the left and right of the point x(n). The expression (f93(xk)-f93(xk1))/2/h1 is the definition of the derivative. In the definition of the derivative, we have (x-x0) in the denominator. In this case, x0 is represented by h1, which is a very small number. Dividing by a small number can lead to inaccuracies in our calculations, so we multiply by 2/h1 to avoid this problem.

Overall, the code is using the definition of the derivative to calculate the slope of the function at each point on the x-axis. We then compare this slope to the value of the expression on the right-hand side of the equation. If the difference between the two is very small (close to zero), then the point x(n) satisfies the equation. This is why the code plots the points where the function crosses the x-axis, because those are the points where the expression on the left-hand side of the equation is equal to zero.

I hope this helps you understand the code and the mathematics behind it. Let me know if you have any further questions.