The maximum value assumed by the function on the interval

What is the maximum value assumed by the function f(x) = sin(5*cos(x))*cos(5*sin(x)) on the interval [0,1] ?

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

Answer 1

[James Browne]

Greetings,

I wrote an example for you which finds the maximum value of f(x) over the specified interval and plots the resuts. Here you go, hope this helps!

%Define the interval over which f(x) will be evaluated
xInterval = [0 1];

%specify resolution of the evaluation
dx = 0.0001;

%Create a vector of x values based on the interval of evaluateion and
%resolution specifications
x = xInterval(1): dx : xInterval(2);

%Calculate f(x) values
for i = 1:length(x)
fx(i) = sin(5*cos(x(i)))*cos(5*sin(x(i)));
end

%Find the magnitude and index of the maximum f(x) value and print the
%maximum magnitude to the command window
[M,idx] = max(fx);
fprintf('The maximum value in the given range of x is: %5.4f\n',M)

%Use the index of the maximum f(x) value to find the x value which produced
%the maximum f(x) value
xForMaxFx = x(idx);

%Determine title based on interval of evaluation parameters
titleName = strcat('Maximum f(x) Value over the interval [',num2str(xInterval(1)),',',num2str(xInterval(2)),']');

%Plot f(x) and highlight the maximum value
plot(x,fx,xForMaxFx,M,'kd')
title(titleName)
xlabel('x (units)')
ylabel('f(x) (units)')
ylim([-1,1])
legend('f(x)','Maximum f(x) value','location','southeast')

[Aidan Welch]

Just FYI this code does not work generally. It only worked because the original function had no powers. When you attempt this with a polynomial, it reads as incorreect implementation of matrix powers since x is a vector. Not sure how to fix this without fundementally re-approaching how this code is structured.

[Walter Roberson]

@Aidan Welch

Which part of the code would fail with polynomials? The user posted

for i = 1:length(x)
fx(i) = sin(5*cos(x(i)))*cos(5*sin(x(i)));
end

which shows evaluating the function scalar by scalar, and scalar^power is the same as scalar.^power

There would be a potential issue if the function evaluation was vectorized, such as

fx = sin(5*cos(x))*cos(5*sin(x));   %WOULD FAIL

and would need to be written

fx = sin(5*cos(x)).*cos(5*sin(x));   %WOULD FAIL

but the issue for ^ needing to be .^ in vector calculations would not be different than the issue for * needing to be .* in vector calculations.

[Aidan Welch]

The line which defines the function fails under power conditions and under dot power conditions. It fails on 2020a with a polynomial via the aforementioned error with the default power function. When the dot power function is applied, I get an error concerning "incompatible Matrix sizes"

[Walter Roberson]

The line that defines the function? You mean

fx(i) = sin(5*cos(x(i)))*cos(5*sin(x(i)));

??

But that line only uses scalar values.

[Walter Roberson]

Example of the function changed to use ^ without using .^

No failure.

%Define the interval over which f(x) will be evaluated
xInterval = [0 1];

%specify resolution of the evaluation
dx = 0.0001;

%Create a vector of x values based on the interval of evaluateion and
%resolution specifications
x = xInterval(1): dx : xInterval(2);

%Calculate f(x) values
for i = 1:length(x)
fx(i) = sin(5*cos(x(i)))^2*cos(5*sin(x(i)))^3;
end

%Find the magnitude and index of the maximum f(x) value and print the
%maximum magnitude to the command window
[M,idx] = max(fx);
fprintf('The maximum value in the given range of x is: %5.4f\n',M)

The maximum value in the given range of x is: 0.9195

%Use the index of the maximum f(x) value to find the x value which produced
%the maximum f(x) value
xForMaxFx = x(idx);

%Determine title based on interval of evaluation parameters
titleName = strcat('Maximum f(x) Value over the interval [',num2str(xInterval(1)),',',num2str(xInterval(2)),']');

%Plot f(x) and highlight the maximum value
plot(x,fx,xForMaxFx,M,'kd')
title(titleName)
xlabel('x (units)')
ylabel('f(x) (units)')
ylim([-1,1])
legend('f(x)','Maximum f(x) value','location','southeast')

Answer 2

[Walter Roberson]

Step 1: Plot the function to see visually the approximate peak. Find an approximate interval for the peak; it does not have to be precise at all.

Step 2: Differentiate the function. Solve for a zero of that over the approximate interval that you identified. That gives the location of the peak.

Step 3: substitute the location of the peak into the formula to get the value of the peak.