🌊 Symmetry makes Fourier series much simpler.
If a function is even or odd, many Fourier coefficients automatically disappear. This means the series contains only cosine terms or only sine terms, making calculations faster and the structure of the series easier to understand.
When working with Fourier series, calculations become much easier if the function has symmetry.
There are two important types of symmetry.
Even function
- Symmetric with respect to the y-axis
- Example: \( f(x)=x^{2} \)
- \( f(-x)=f(x) \)
Odd function
- Symmetric with respect to the origin
- Example: \( f(x)=x^{3} \)
- \( f(-x)=-f(x) \)
Theorem 1
If \( f \) is even and periodic with period \( 2\pi \), then its Fourier series contains only cosine terms.
Proof
The sine coefficients are
\[ b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}
\underbrace{
\underbrace{f(x)}_{\text{even}}
\underbrace{\sin(nx)}_{\text{odd}}
}_{\text{odd}}
dx=0\quad\quad n=1,2,\cdots \]
An even function multiplied by an odd function results in an odd function.
Therefore the integrand is an odd function.
A key property of integrals states:
- The integral of an odd function over symmetric limits is always zero.
Thus the Fourier series contains only cosine terms.
Theorem 2
If \( f \) is odd and periodic with period \( 2\pi \), then its Fourier series contains only sine terms.
Proof
The cosine coefficients are
\[ a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}
\underbrace{
\underbrace{f(x)}_{\text{odd}}
\underbrace{\cos(nx)}_{\text{even}}
}_{\text{odd}}
dx=0\quad\quad n=0,1,2,\cdots \]
An odd function multiplied by an even function results in an odd function.
Again the integrand is odd.
A key property of integrals states:
- The integral of an odd function over symmetric limits is always zero.
So the Fourier series contains only sine terms.
Exercises
Exercise 1
If \( f \) is even and periodic with period \( 2\pi \), then the Fourier coefficients become:
\[ a_{n}=\frac{2}{\pi}\int_{0}^{\pi}f(x)\cos(nx)dx\quad\quad n=0,1,2,\cdots \]
Show that this formula is correct!
Proof
Start from the general formula:
\[ a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx \]
Split the integral:
\[ a_n=\frac{1}{\pi}
\left[
\int_{-\pi}^{0}f(x)\cos(nx)dx
+
\int_{0}^{\pi}f(x)\cos(nx)dx
\right] \]
Substitute x = −t in the first integral:
Because:
- \( f \) is even, we have \(f(-t)=f(t) \).
- cosinus is even, we have \( \cos(n(-t))= \cos(nt) \)
- d(-t) = −dt
- The integral lower limit: \( x=-\pi\Rightarrow t=\pi \)
- The integral upper limit: \( x=0\Rightarrow t=0 \)
\[ =\frac{1}{\pi}
\left[
\int_{\pi}^{0}f(-t)\cos(n(-t))d(-t)
+
\int_{0}^{\pi}f(x)\cos(nx)dx
\right] \]
After cleaning up:
\[ =\frac{1}{\pi}
\left[
-\int_{\pi}^{0}f(t)\cos(nt)dt
+
\int_{0}^{\pi}f(x)\cos(nx)dx
\right] \]
Reversing the limits of the first integral gives:
\[ =\frac{1}{\pi}
\left[
\int_{0}^{\pi}f(x)\cos(nx)dx
+
\int_{0}^{\pi}f(x)\cos(nx)dx
\right] \]
Because:
- t and x are dummy variables
- they only live inside the integral
- renaming them changes nothing.
\[ \int_{0}^{\pi} f(x)\,dx
=
\int_{0}^{\pi} f(t)\,dt \]
Conclusion:
\[ a_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\cos(nx)dx \]
Exercise 2
If \( f \) is odd and periodic with period \( 2\pi \), then the Fourier coefficients become:
\[ b_{n}=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx\quad\quad n=1,2,\cdots \]
Show that this formula holds!
Proof
Start from the general formula:
\[ b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx \]
Split the integral:
\[ =\frac{1}{\pi}
\left[
\int_{-\pi}^{0}f(x)\sin(nx)dx
+
\int_{0}^{\pi}f(x)\sin(nx)dx
\right] \]
Substitute x = −t in the first integral:
Because:
- \( f \) is odd, we have \(f(-t)=-f(t) \)
- sinus is odd, we have \( \sin(n(-t))=-\sin(nt) \)
- d(-t) = −dt
- The integral lower limit: \( x=-\pi\Rightarrow t=\pi \)
- The integral upper limit: \( x=0\Rightarrow t=0 \)
\[ =\frac{1}{\pi}
\left[
\int_{\pi}^{0}f(-t)\sin(n(-t))d(-t)
+
\int_{0}^{\pi}f(t)\sin(nt)dt
\right] \]
After cleaning up:
\[ =\frac{1}{\pi}
\left[
-\int_{\pi}^{0}f(t)\sin(nt)dt
+
\int_{0}^{\pi}f(t)\sin(nt)dt
\right] \]
After reversing limits:
\[ =\frac{1}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx +\frac{1}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx \]
The switch from t to x in a definite integral is allowed because the letter used for the variable has no meaning outside the integral.
Therefore:
\[ b_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx \]
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