# Even Symmetry

A function $f(t)$ is even if its plot is symmetrical about the vertical axis; that is,
$$f(t)=f(-t) \tag{1}$$
Examples of even functions are $t^{2}, t^{4}$, and $\cos t$. Figure $1$ shows more examples of periodic even functions.
Fig. 1: Typical examples of even periodic functions.
Note that each of these examples satisfies Eq. (1). A main property of an even function $f_{e}(t)$ is that:
$$\int_{-T / 2}^{T / 2} f_{e}(t) d t=2 \int_{0}^{T / 2} f_{e}(t) d t \tag{2}$$
because integrating from $-T / 2$ to 0 is the same as integrating from 0 to $T / 2$. Utilizing this property, the Fourier coefficients for an even function become
$$\begin{array}{l}a_{0}=\frac{2}{T} \int_{0}^{T / 2} f(t) d t \\a_{n}=\frac{4}{T} \int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t \\b_{n}=0\end{array} \tag{3}$$
Since $b_{n}=0$, This makes sense because the cosine function is itself even. It also makes intuitive sense that an even function contains no sine terms since the sine function is odd. To confirm Eq. (3) quantitatively, we apply the property of an even function in Eq. (2) in evaluating the Fourier coefficients in Eqs. (6), (8), and (9) of the PAGE (Trigonometric Fourier Series). It is convenient in each case to integrate over the interval $-T / 2 < t < T / 2$, which is symmetrical about the origin. Thus,
$$a_{0}=\frac{1}{T} \int_{-T / 2}^{T / 2} f(t) d t=\frac{1}{T}\left[\int_{-T / 2}^{0} f(t) d t+\int_{0}^{T / 2} f(t) d t\right] \tag{4}$$
We change variables for the integral over the interval $-T / 2 < t < 0$ by letting $t=-x$, so that $d t=-d x, f(t)=f(-t)=f(x)$, since $f(t)$ is an even function, and when $t=-T / 2, x=T / 2$. Then,
\begin{aligned}a_{0} &=\frac{1}{T}\left[\int_{T / 2}^{0} f(x)(-d x)+\int_{0}^{T / 2} f(t) d t\right] \\&=\frac{1}{T}\left[\int_{0}^{T / 2} f(x) d x+\int_{0}^{T / 2} f(t) d t\right]\end{aligned} \tag{5}
showing that the two integrals are identical. Hence,
$$a_{0}=\frac{2}{T} \int_{0}^{T / 2} f(t) d t \tag{6}$$
as expected.
$$a_{n}=\frac{2}{T}\left[\int_{-T / 2}^{0} f(t) \cos n \omega_{0} t d t+\int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t\right] \tag{7}$$
We make the same change of variables that led to Eq. (5) and note that both $f(t)$ and $\cos n \omega_{0} t$ are even functions, implying that $f(-t)=f(t)$ and $\cos \left(-n \omega_{0} t\right)=\cos n \omega_{0} t$. Equation (7) becomes
\begin{aligned}a_{n} &=\frac{2}{T}\left[\int_{T / 2}^{0} f(-x) \cos \left(-n \omega_{0} x\right)(-d x)+\int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t\right] \\&=\frac{2}{T}\left[\int_{T / 2}^{0} f(x) \cos \left(n \omega_{0} x\right)(-d x)+\int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t\right] \\&=\frac{2}{T}\left[\int_{0}^{T / 2} f(x) \cos \left(n \omega_{0} x\right) d x+\int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t\right]\end{aligned} \tag{8.a}
or
$$a_{n}=\frac{4}{T} \int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t \tag{8.b}$$
as expected. For $b_{n}$,
$$b_{n}=\frac{2}{T}\left[\int_{-T / 2}^{0} f(t) \sin n \omega_{0} t d t+\int_{0}^{T / 2} f(t) \sin n \omega_{0} t d t\right] \tag{9}$$
We make the same change of variables but keep in mind that $f(-t)=$ $f(t)$ but $\sin \left(-n \omega_{0} t\right)=-\sin n \omega_{0} t$. Equation (9) yields
\begin{aligned}b_{n} &=\frac{2}{T}\left[\int_{T / 2}^{0} f(-x) \sin \left(-n \omega_{0} x\right)(-d x)+\int_{0}^{T / 2} f(t) \sin n \omega_{0} t d t\right] \\&=\frac{2}{T}\left[\int_{T / 2}^{0} f(x) \sin n \omega_{0} x d x+\int_{0}^{T / 2} f(t) \sin n \omega_{0} t d t\right] \\&=\frac{2}{T}\left[-\int_{0}^{T / 2} f(x) \sin \left(n \omega_{0} x\right) d x+\int_{0}^{T / 2} f(t) \sin n \omega_{0} t d t\right] \\&=0\end{aligned} \tag{10}
confirming Eq. (3).