For exp(x), separate the integer from the fractional component of the argument.

Question

For exp(x), separate the integer from the fractional component of the argument. For example if the function is called as exp(3.123), separate the 3 from the fractional component 0.123. Tip: you can use x-floor(x) to do this. Recognizing that exp(x+y)=exp(x)*exp(y), compute the integer portion of the argument using a simple loop that performs the multiplication and calculate the fractional component using the series expansion.

Additional Information:

You may use the following code fragments for the sin(x), cos(x), and exp(x) functions. Note, however, that these routines calculate a fixed number of terms, based on the value of N. you will have to modify the routines to continue to calculate terms until the accuracy criterion has been established. Tips: do NOT try to write the complete, fully-functional routine all at once! Better is to work on the individual pieces, test them independently, then assemble together for the final functional version.

% calculate exp(x), given x and N (number of terms)

s= 1; % this is the sum of terms and our final answer

xt= 1; % first term is 1, so let's just put it in to start

fact= 1; % factorial of first term is also 1, so put it in here

for i=1:N % now lets add N terms, as requested by user

xt= xt*x; % calculate numerator of next term (x^i)

fact= fact*i; % calculate denominator of next term (i!)

s=s + xt/fact; % add term to total sum, and go back for more

end

Explanation / Answer

function [s] = mexp(x)

n = floor(x); % get integer part of x
x = x - floor(x); % get the decimal part

p = 1; % initialize the p to 1 to store e^n
for i = 1:n
p = p * exp(1); % p will store the product of n e's i.e e^n
end

epsilon = 0.000005; % for accuracy
ps = 0; % initialize previous term sum to zero
i = 1; % to iterate in while loop
s = 1; % this is the sum of terms and our final answer
xt= 1; % first term is 1, so let's just put it in to start
fact= 1; % factorial of first term is also 1, so put it in here
while abs(s - ps) > epsilon % check if difference of ps and s is less then epsilon
xt= xt*x; % calculate numerator of next term (x^i)
fact= fact*i; % calculate denominator of next term (i!)
ps = s; % store previous sum
s=s + xt/fact; % add term to total sum, and go back for more
i = i + 1; % increment i
end

s = s * p; % s is the final answer that is returned by function

end

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