MATLAB: convert from euler (complex fourier) to sinus function (bn coefficients) - matlab

Question

I have the following script
clc; clear all; close all;
syms x n
f = x;
L = 1;
subplot(2,1,1)
h = ezplot(f,[-L,L])
set(h, 'Color','r','LineWidth',1)
a0 = (1/L) * int(f * cos(0* pi*x/L),-L,L)
an = (1/L) * int(f * cos(n* pi*x/L),-L,L)
bn = (1/L)* int(f* sin(n* pi*x/L),-L,L)
fx = a0/2 + symsum((an* cos(n*pi*x/L) + bn* sin(n*pi*x/L)),n,1,5)
% for n =5, the answer: fx = (2*sin(pi*x))/pi - sin(2*pi*x)/pi +
%(2*sin(3*pi*x))/(3*pi) - sin(4*pi*x)/(2*pi) + (2*sin(5*pi*x))/(5*pi)
hold on
h = ezplot(fx,[-L,L])
grid on
%Solution with complex Fourier
c0 = (1/(2*L))*int(f*exp(-j*0*pi*x/L),-L,L)
cn = (1/(2*L))*int(f*exp(-j*n*pi*x/L),-L,L)
subplot(2,1,2)
h = ezplot(f,[-L,L])
set(h, 'Color','r','LineWidth',1)
fx_c = c0 + symsum(cn*exp(j*n*pi*x/L),n,-5,-1) + ...
symsum(cn*exp(j*n*pi*x/L),n,1,5) % n for complex -5,5
hold on
h = ezplot(fx_c,[-L,L])
grid on
My question: Since the answer of fx should be equal to fx_c (complex fourier). We can see from the figures produced by these 2 functions. They are same. But
fx =
(2*sin(pi*x))/pi - sin(2*pi*x)/pi + (2*sin(3*pi*x))/(3*pi) - sin(4*pi*x)/(2*pi) + (2*sin(5*pi*x))/(5*pi)
and
fx_c =
exp(-pi*x*i)*((pi*i - 1)/(2*pi^2) + (pi*i + 1)/(2*pi^2)) - exp(pi*x*i)*((pi*i - 1)/(2*pi^2) + (pi*i + 1)/(2*pi^2)) - exp(-pi*x*2*i)*((pi*2*i - 1)/(8*pi^2) + (pi*2*i + 1)/(8*pi^2)) + exp(pi*x*2*i)*((pi*2*i - 1)/(8*pi^2) + (pi*2*i + 1)/(8*pi^2)) + exp(-pi*x*3*i)*((pi*3*i - 1)/(18*pi^2) + (pi*3*i + 1)/(18*pi^2)) - exp(pi*x*3*i)*((pi*3*i - 1)/(18*pi^2) + (pi*3*i + 1)/(18*pi^2)) - exp(-pi*x*4*i)*((pi*4*i - 1)/(32*pi^2) + (pi*4*i + 1)/(32*pi^2)) + exp(pi*x*4*i)*((pi*4*i - 1)/(32*pi^2) + (pi*4*i + 1)/(32*pi^2)) + exp(-pi*x*5*i)*((pi*5*i - 1)/(50*pi^2) + (pi*5*i + 1)/(50*pi^2)) - exp(pi*x*5*i)*((pi*5*i - 1)/(50*pi^2) + (pi*5*i + 1)/(50*pi^2))
How to convert fx_c to be fx?

Answer 1

They are related by Euler's formula. You can check it with rewrite command:
>> rewrite(exp(1i*x), 'cos')
ans =
cos(x) + sin(x)*1i
Applying it to your function and simplifying a bit, you can get to the same expression:
>> expand(rewrite(fx_c, 'cos'), 'ArithmeticOnly', true)
ans =
(2*sin(pi*x))/pi - sin(2*pi*x)/pi + (2*sin(3*pi*x))/(3*pi) - sin(4*pi*x)/(2*pi) + (2*sin(5*pi*x))/(5*pi)
>> fx
fx =
(2*sin(pi*x))/pi - sin(2*pi*x)/pi + (2*sin(3*pi*x))/(3*pi) - sin(4*pi*x)/(2*pi) + (2*sin(5*pi*x))/(5*pi)