function f(x)=xe^x
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Write a function my_fun.m which returns the value of f(x)=xe^x given an input x. This
function should be a function of x only.
3 个评论
Steven Lord
2019-8-24
This sounds like a homework assignment. If it is, show us what you've done to try to solve the problem and ask a specific question about where you're having difficulty and we may be able to offer some guidance.
Deanna Jaramillo
2019-8-27
yes it was a homework problem. I do not understand much code and I am wanting to learn. The instructor showed us some 'for' loops and 'while' loops and a function program and I am not sure when to use what. Here is the question for the first part.
a) Considering the function
f(x) = xe^x (1)
(a) Write a function my_fun.m which returns the value of (1) given an input x. This function should be a function of x only.
I came up with:
function y=my_fun(x)
y=x*exp(x);
end
Then the next part:
b) Write a function my_mean.m which takes four arguments: 1) a function (my_fun.m), 2) a number a, 3) another number b satisfying a ≤ b, and 4) a positive integer N. This function should return an approximate value for:
(2)Where f(x) represents (1). The approximate value of (2) should be calculated via:
where:
h = (b − a)/ N , xj = a + (j − 1)h
HW #1 For example, a call to the function would resemble:
my_mean(@my_fun, 0, 5, 100)
(c) Use my_mean.m and my_fun.m to compute an approximation to
In this case the exact solution for (4) can be computed as:
1/ e
Plot (using semilogy) the absolute error between M and your approximations versus N for:
N(n) = 2n , 1 ≤ n ≤ 13
So, for the codes, I have:
function y= my_mean(my_fun,a,b,N)
h=(b-a)/N;
x=linspace(a,b,N);
y=0;
for j=1:length(x)
y = (1/N)*(a+((j-1)*h));
end
end
and the final for the output plot:
close all
clear
clc
format longg
n=1:13;
N=2.^n;
M=0;
for j=1:length(N)
M(j)=abs(my_mean(@my_fun,-1,1,N(j)));
end
semilogy(N,M,'-ob');
xlabel('N', 'Fontsize', 10);
ylabel('M', 'Fontsize', 10);
title('Absolute Error', 'Fontsize', 15)
I got the plot to show up but the y-axis goes from 10^-4 to 10^-1, when the instructor's goes from 10^-4 to 10^0.
How do I know what to use when?
-Deanna
Steven Lord
2019-8-27
James already discussed your parts b and c, so I just want to offer a little extra information. This is slightly more advanced than you may have learned so far, but it's something that can come in very useful when working with MATLAB.
For part a your function will work as long as the x input to my_fun is a scalar (just one number.) You could modify it ever so slightly to allow it to work on any sized array (a scalar, a vector with ten elements, a 5-by-5 matrix, a 2-by-3-by-4-by-5-by-6 array, etc.) by using the array multiplication operator .* instead of the matrix multiplication operator *. [For scalars they behave the same; for non-scalars they don't.] See the Array vs. Matrix Operations documentation page for more information.
回答（2 个）
James Tursa
2019-8-27
编辑：James Tursa
2019-8-27
Take a look at this loop from your code:
for j=1:length(x)
y = (1/N)*(a+((j-1)*h)); % <-- This replaces y at each step ... it doesn't sum anything up!
end
And compare it to the summation (3) in the instructions. Two problems: You are not calling your function, and you are not summing anything up. You should be summing up f(xj) values (N of them) and then dividing that sum by N according to the (3) formula. So the loop should be something like this instead:
for j=1:N % <-- Use the limits in the (3) formula
y = y + my_fun(a+(j-1)*h); % <-- Sums up individual calculations into y
end
y = y/N;
You don't need (or want) that x-linspace(etc) call.
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