How to use a loop to find a maximum value

The equation 1+x+((x^2)/2)+((x^3)/6) is often used to approximate e^x. Find to two decimal places the largest value of x before the error exceeds 1 percent. The calculation for error is: e^x-(1+x+((x^2)/2)+((x^3)/6))/e^x.

I know i need to use a loop to find this. However, I am not sure how to format this one or whether to use a for or while loop.

Any suggestions?

[lvn]

I would first make a plot of both e^x and the approximation, so you get an idea where that largest x will be.

Then I would simply do a for loop to find the first x (e.g from x=-10 to 10 in steps of 0.01 (two decimal places) - and put a break a soon as your error exceeds 1%).

PS. For the error, do not forget to take the absolute value.

[Matthew Quinones]

Ok so i put the e^x formula to begin. Then I have

for x=-10:.01:10

x=((ex)-(1+x+((x^2)/2)+((x^3)/6))/ex

if x<=.01

end

After this, what do I do?

Answer 1

[Image Analyst]

You don't need to use a loop. You can simply do

x = 0: 0.01 : 1; 
theErrorPercentage = exp(x)-(1+x+((x.^2)/2)+((x.^3)/6)) ./ exp(x);
plot(x, theErrorPercentage);
xlabel('x');
ylabel('Percent Error');
grid on;

% Find where it's closest to 1
[theDifference, indexAtMin] = min(abs(theErrorPercentage - 1));
xAt1Percent = x(indexAtMin)

At least that's one way. Do you want to use a loop?

[Matthew Quinones]

I think I am supposed to use a loop with a break in it. However I am not sure how to format it

[Matthew Quinones]

Ok so far i have:

x=0;

for k=0:.01:10

    x=1+x+((x^2)/2)+((x^3)/6);

    y=(x-(1+x+((x^2)/2)+((x^3)/6))/x);

    if y<=0.01

        break

    end
end
x

The output I get x=1 and y=-1.6667

not sure if that sounds right? Can't there be another value that would be closer to 0.01% error?

Answer 2

[Rohan Sanjay Patel]

error = 0;
x = 0;
while error < 1
x = x + 0.01;
approx = 1 + x + x^2/2 + x^3/6;
error = 100*(exp(x) - approx)/exp(x);
end
disp(x)