130_notes.dvi

so that we can easily compute the coefficients.

an=

1

2 L

∫L

−L

f(x)e

−inπx

L dx

In summary, the Fourier series equations we will use are

f(x) =

∑∞

n=−∞

ane

inπxL

and

an=

1

2 L

∫L

−L

f(x)e

−inπx

L dx.

We will expand the interval to infinity.

5.6.2 Fourier Transform*

To allow wavefunctions to extend to infinity, we will expand the interval used

L→∞.

As we do this we will use the wave number

k=

nπ

L

.

AsL→∞.,kcan take on any value, implying we will have a continuous distribution ofk. Our sum
overnbecomes an integral overk.

dk=

π

L

dn.

If we defineA(k) =

√

2

πLan, we can make the transform come out with the constants we want.

f(x) =

∑∞

n=−∞

ane

inπxL

Standard Fourier Series

an= 21 L

∫L

−L

f(x)e

−inπx

L dx Standard Fourier Series

An=

√

2

πL

1

2 L

∫L

−L

f(x)e

−inπx

L dx redefine coefficient

An=√^12 π

∫L

−L

f(x)e

−inπx

L dx

f(x) =

√π

2

1

L

∑∞

n=−∞

Ane

inπxL

f stays the same

f(x) =

√π

√

2 L

∞∫

−∞

A(k)eikxLπdk but is rewritten in new A and dk

f(x) =√^12 π

∞∫

−∞

A(k)eikxdk result

A(k) =√^12 π

∫∞

−∞

f(x)e−ikxdx result