Appendix C Solutions¶

\[ a_n = \frac{2}{L}\left[\frac{L}{2\pi n}\sin\!\left(\frac{2\pi n}{L}x\right)\right]_0^L = \frac{2}{L} \cdot \frac{L}{2\pi n}\left[\sin(2\pi n) - \sin(0)\right] = 0 \]

Calculation of \(b_n\) (\(n \geq 1\)):

\[ b_n = \frac{2}{L}\int_0^L 1 \cdot \sin\!\left(\frac{2\pi n}{L}x\right) dx = \frac{2}{L}\left[-\frac{L}{2\pi n}\cos\!\left(\frac{2\pi n}{L}x\right)\right]_0^L \]

\[ = \frac{2}{L} \cdot \left(-\frac{L}{2\pi n}\right)\left[\cos(2\pi n) - \cos(0)\right] = -\frac{1}{\pi n}[1 - 1] = 0 \]

Final Answer¶

\[ \boxed{a_0 = 2, \quad a_n = 0 \;(n \geq 1), \quad b_n = 0 \;(n \geq 1)} \]

Verification¶

Substituting into the Fourier series gives \(f(x) = \frac{a_0}{2} = \frac{2}{2} = 1\). The constant function is correctly reproduced. ✓

B-2. Using the orthogonality of complex exponentials (Eq. (C.12)), evaluate the following integral¶

→ Back to problem

Solution Strategy¶

Combine the integrand into the form \(e^{i\frac{2\pi(m-n)}{L}x}\) and apply equation (C.12).

Detailed Calculation¶

\[ \int_0^L e^{i \frac{2\pi \cdot 3}{L} x}\, e^{-i \frac{2\pi \cdot 5}{L} x}\, dx = \int_0^L e^{i \frac{2\pi(3-5)}{L} x}\, dx = \int_0^L e^{-i \frac{4\pi}{L} x}\, dx \]

From equation (C.12), with \(m = 3\), \(n = 5\) and \(m \neq n\):

\[ \int_0^L e^{i \frac{2\pi(m-n)}{L} x}\, dx = L\,\delta_{mn} = L \cdot 0 = 0 \]

Final Answer¶

\[ \boxed{0} \]

Verification¶

Confirmed by direct calculation:

\[ \int_0^L e^{-i \frac{4\pi}{L} x}\, dx = \left[\frac{e^{-i \frac{4\pi}{L} x}}{-i \frac{4\pi}{L}}\right]_0^L = \frac{L}{-4\pi i}\left(e^{-4\pi i} - 1\right) = \frac{L}{-4\pi i}(1 - 1) = 0 \quad \checkmark \]

B-3. Using Euler's formula, show the following¶

→ Back to problem

Part 1: Complex Exponential Representation of \(\cos\)¶

Solution strategy: Add \(e^{i\theta}\) and \(e^{-i\theta}\) using Euler's formula.

From Euler's formula:

\[ e^{i\theta} = \cos\theta + i\sin\theta \]

\[ e^{-i\theta} = \cos\theta - i\sin\theta \]

Adding both expressions:

\[ e^{i\theta} + e^{-i\theta} = 2\cos\theta \]

Therefore:

\[ \cos\theta = \frac{1}{2}\left(e^{i\theta} + e^{-i\theta}\right) \]

Substituting \(\theta = \frac{2\pi n}{L}x\):

\[ \boxed{\cos\!\left(\frac{2\pi n}{L}x\right) = \frac{1}{2}\left(e^{i\frac{2\pi n}{L}x} + e^{-i\frac{2\pi n}{L}x}\right)} \]

Part 2: Relationship Between Real and Complex Fourier Coefficients¶

Solution strategy: Rewrite \(\cos\) and \(\sin\) in Eq. (C.5) as complex exponentials and compare with Eq. (C.10).

Rewriting Eq. (C.5) using Eq. (C.9):

\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left[a_n \cdot \frac{e^{ik_n x} + e^{-ik_n x}}{2} + b_n \cdot \frac{e^{ik_n x} - e^{-ik_n x}}{2i}\right] \]

where \(k_n = \frac{2\pi n}{L}\). Using \(\frac{1}{2i} = -\frac{i}{2}\) and rearranging:

\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left[\frac{a_n}{2} e^{ik_n x} + \frac{a_n}{2} e^{-ik_n x} - \frac{ib_n}{2} e^{ik_n x} + \frac{ib_n}{2} e^{-ik_n x}\right] \]

\[ = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left[\frac{a_n - ib_n}{2} e^{ik_n x} + \frac{a_n + ib_n}{2} e^{-ik_n x}\right] \]

On the other hand, Eq. (C.10) gives:

\[ f(x) = \sum_{n=-\infty}^{\infty} c_n \, e^{ik_n x} = c_0 + \sum_{n=1}^{\infty} c_n \, e^{ik_n x} + \sum_{n=1}^{\infty} c_{-n} \, e^{-ik_n x} \]

Comparing both expressions:

\[ \boxed{c_0 = \frac{a_0}{2}, \qquad c_n = \frac{a_n - ib_n}{2} \;(n \geq 1), \qquad c_{-n} = \frac{a_n + ib_n}{2} \;(n \geq 1)} \]

Verification¶

When \(f(x)\) is real, we should have \(c_{-n} = c_n^*\). Indeed:

\[ c_n^* = \left(\frac{a_n - ib_n}{2}\right)^* = \frac{a_n + ib_n}{2} = c_{-n} \quad \checkmark \]

(since \(a_n, b_n\) are real.)

B-4. Find all Fourier coefficients and on the interval¶

→ Back to problem

Solution Strategy¶

Directly apply the orthogonality relations (C.3) and (C.4).

Detailed Calculation¶

Calculation of \(a_n\):

\[ a_n = \frac{2}{L}\int_0^L \sin\!\left(\frac{2\pi}{L}x\right)\cos\!\left(\frac{2\pi n}{L}x\right) dx \]

From equation (C.4), the integral of the product of \(\sin\) and \(\cos\) is always zero:

\[ a_n = 0 \quad \text{(for all } n = 0, 1, 2, \ldots \text{)} \]

Calculation of \(b_n\):

\[ b_n = \frac{2}{L}\int_0^L \sin\!\left(\frac{2\pi}{L}x\right)\sin\!\left(\frac{2\pi n}{L}x\right) dx \]

From equation (C.3):

\[ \int_0^L \sin\!\left(\frac{2\pi \cdot 1}{L}x\right)\sin\!\left(\frac{2\pi n}{L}x\right) dx = \frac{L}{2}\,\delta_{1n} \]

Therefore:

\[ b_n = \frac{2}{L} \cdot \frac{L}{2}\,\delta_{1n} = \delta_{1n} \]

Final Answer¶

\[ \boxed{a_n = 0 \;(\text{for all } n), \qquad b_1 = 1, \qquad b_n = 0 \;(n \geq 2)} \]

Verification¶

Substituting into the Fourier series gives \(f(x) = b_1 \sin\!\left(\frac{2\pi}{L}x\right) = \sin\!\left(\frac{2\pi}{L}x\right)\). The original function is reproduced. ✓

B-5. Gaussian Integral Formula¶

→ Back to problem

Solution Strategy¶

Using convention (b), we complete the square in \(\tilde{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-3x^2} e^{-ikx}\,dx\) and reduce it to a Gaussian integral.

Detailed Calculation¶

\[ \tilde{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-3x^2 - ikx}\,dx \]

Completing the square in the exponent:

\[ -3x^2 - ikx = -3\left(x^2 + \frac{ik}{3}x\right) = -3\left(x + \frac{ik}{6}\right)^2 - \frac{k^2}{12} \]

Verification: \(-3\left(x + \frac{ik}{6}\right)^2 = -3\left(x^2 + \frac{ik}{3}x + \frac{(ik)^2}{36}\right) = -3x^2 - ikx + \frac{k^2}{12}\)

Therefore \(-3\left(x + \frac{ik}{6}\right)^2 - \frac{k^2}{12} = -3x^2 - ikx\). ✓

\[ \tilde{f}(k) = \frac{1}{\sqrt{2\pi}}\,e^{-k^2/12}\int_{-\infty}^{\infty} e^{-3\left(x + \frac{ik}{6}\right)^2}\,dx \]

Substituting \(t = x + \frac{ik}{6}\) gives \(dt = dx\). The integration path is shifted in the complex plane, but since the integrand is Gaussian, it yields the same value as integration along the real axis (by Jordan's lemma):

\[ \int_{-\infty}^{\infty} e^{-3t^2}\,dt = \sqrt{\frac{\pi}{3}} \]

Therefore:

\[ \tilde{f}(k) = \frac{1}{\sqrt{2\pi}}\,e^{-k^2/12} \cdot \sqrt{\frac{\pi}{3}} = \frac{1}{\sqrt{2\pi}} \cdot \frac{\sqrt{\pi}}{\sqrt{3}} \cdot e^{-k^2/12} \]

\[ = \frac{1}{\sqrt{6}}\,e^{-k^2/12} \]

Final Answer¶

\[ \boxed{\tilde{f}(k) = \frac{1}{\sqrt{6}}\,e^{-k^2/12}} \]

Verification¶

For the general formula \(f(x) = e^{-ax^2}\), convention (b) gives \(\tilde{f}(k) = \frac{1}{\sqrt{2a}}\,e^{-k^2/(4a)}\). Substituting \(a = 3\) yields \(\tilde{f}(k) = \frac{1}{\sqrt{6}}\,e^{-k^2/12}\). This agrees. ✓

B-6. Following the definition of convolution (Eq. (C.21)), compute the convolution of and ( is a constant). Here is D¶

→ Back to problem

Solution Strategy¶

Apply the sifting property of the δ function to the definition of convolution.

Detailed Calculation¶

From the definition of convolution in Eq. (C.21):

\[ (f * g)(x) = \int_{-\infty}^{\infty} f(x')\,g(x - x')\,dx' \]

Since \(g(x) = \delta(x - a)\), we have \(g(x - x') = \delta((x - x') - a) = \delta(x - x' - a)\):

\[ (f * g)(x) = \int_{-\infty}^{\infty} f(x')\,\delta(x - x' - a)\,dx' \]

Using the sifting property of the δ function \(\int h(x')\,\delta(x' - c)\,dx' = h(c)\). Here, \(\delta(x - x' - a) = \delta(-(x' - (x-a))) = \delta(x' - (x-a))\) (since the δ function is an even function):

\[ (f * g)(x) = f(x - a) \]

Substituting \(f(x) = e^{-|x|}\):

Final Answer¶

\[ \boxed{(f * g)(x) = e^{-|x - a|}} \]

Verification¶

This is consistent with the general property that convolution with a δ function simply translates the function. When \(a = 0\), we get \((f * \delta)(x) = f(x) = e^{-|x|}\), confirming that the δ function is the identity element for convolution. ✓

B-7. Evaluate the following integral using the Fourier integral representation of the δ function (Eq. (C.19))¶

→ Back to problem

Solution Strategy¶

Set \(k' = 0\) in equation (C.19).

Detailed Calculation¶

Equation (C.19) is:

\[ \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i(k - k')x}\,dx = \delta(k - k') \]

Setting \(k = 7\), \(k' = 0\):

\[ \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i \cdot 7 x}\,dx = \delta(7 - 0) = \delta(7) \]

Since \(\delta(7) = 0\) (the \(\delta(x)\) function is zero for \(x \neq 0\)):

Final Answer¶

\[ \boxed{\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i \cdot 7 x}\,dx = \delta(7) = 0} \]

Verification¶

From a physical perspective, \(e^{i7x}\) is an oscillating function, and when integrated over the entire real line, the positive and negative contributions cancel each other out, yielding zero. This is consistent with \(\delta(7) = 0\). ✓

B-8. Verify the following using Parseval's equality (Eq. (C.18)) for the case¶

→ Back to problem

Solution Strategy¶

Compute the left-hand side and right-hand side independently, then confirm they are equal.

Detailed Calculation¶

Left-hand side:

\[ \int_{-\infty}^{\infty} |f(x)|^2\,dx = \int_{-\infty}^{\infty} e^{-2|x|}\,dx = 2\int_0^{\infty} e^{-2x}\,dx = 2 \cdot \left[-\frac{1}{2}e^{-2x}\right]_0^{\infty} = 2 \cdot \frac{1}{2} = 1 \]

Right-hand side:

From \(\tilde{f}(k) = \sqrt{\frac{2}{\pi}}\,\frac{1}{1+k^2}\):

\[ \int_{-\infty}^{\infty} |\tilde{f}(k)|^2\,dk = \int_{-\infty}^{\infty} \frac{2}{\pi}\,\frac{1}{(1+k^2)^2}\,dk = \frac{2}{\pi}\int_{-\infty}^{\infty} \frac{dk}{(1+k^2)^2} \]

Using the formula \(\int_{-\infty}^{\infty}\frac{dk}{(1+k^2)^2} = \frac{\pi}{2}\):

\[ = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1 \]

Final Answer¶

\[ \boxed{\text{LHS} = 1 = \text{RHS}} \]

Parseval's theorem is confirmed to hold.

Verification¶

We verify the derivation of \(\tilde{f}(k)\). The Fourier transform of \(f(x) = e^{-|x|}\) (convention (b)):

\[ \tilde{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-|x|} e^{-ikx}\,dx = \frac{1}{\sqrt{2\pi}}\left[\int_0^{\infty} e^{-(1+ik)x}\,dx + \int_{-\infty}^0 e^{(1-ik)x}\,dx\right] \]

\[ = \frac{1}{\sqrt{2\pi}}\left[\frac{1}{1+ik} + \frac{1}{1-ik}\right] = \frac{1}{\sqrt{2\pi}} \cdot \frac{2}{1+k^2} = \sqrt{\frac{2}{\pi}}\,\frac{1}{1+k^2} \quad \checkmark \]

Medium¶

M-1. Find the Fourier coefficients and of the function defined on the interval , and write down the Fourier series (Equation (C.5¶

→ Back to problem

Solution Strategy¶

Substitute \(f(x) = x\) into equations (C.6) and (C.7), and compute using integration by parts.

Detailed Calculation¶

Calculation of \(a_0\):

\[ a_0 = \frac{2}{L}\int_0^L x\,dx = \frac{2}{L} \cdot \frac{L^2}{2} = L \]

Calculation of \(a_n\) (\(n \geq 1\)):

\[ a_n = \frac{2}{L}\int_0^L x\cos\!\left(\frac{2\pi n}{L}x\right) dx \]

Integration by parts (\(u = x\), \(dv = \cos\!\left(\frac{2\pi n}{L}x\right)dx\)):

\[ = \frac{2}{L}\left[\frac{L}{2\pi n}x\sin\!\left(\frac{2\pi n}{L}x\right)\bigg|_0^L - \frac{L}{2\pi n}\int_0^L \sin\!\left(\frac{2\pi n}{L}x\right)dx\right] \]

First term: At \(x = L\), \(\sin(2\pi n) = 0\); at \(x = 0\), it is \(0\). Therefore the first term \(= 0\).

Second term:

\[ -\frac{L}{2\pi n}\int_0^L \sin\!\left(\frac{2\pi n}{L}x\right)dx = -\frac{L}{2\pi n}\left[-\frac{L}{2\pi n}\cos\!\left(\frac{2\pi n}{L}x\right)\right]_0^L = \frac{L^2}{(2\pi n)^2}[\cos(2\pi n) - \cos(0)] = 0 \]

Therefore:

\[ a_n = 0 \quad (n \geq 1) \]

Calculation of \(b_n\) (\(n \geq 1\)):

\[ b_n = \frac{2}{L}\int_0^L x\sin\!\left(\frac{2\pi n}{L}x\right) dx \]

Integration by parts (\(u = x\), \(dv = \sin\!\left(\frac{2\pi n}{L}x\right)dx\)):

\[ = \frac{2}{L}\left[-\frac{L}{2\pi n}x\cos\!\left(\frac{2\pi n}{L}x\right)\bigg|_0^L + \frac{L}{2\pi n}\int_0^L \cos\!\left(\frac{2\pi n}{L}x\right)dx\right] \]

First term: \(-\frac{L}{2\pi n}\left[L\cos(2\pi n) - 0\right] = -\frac{L^2}{2\pi n}\)

Second term:

\[ \frac{L}{2\pi n}\left[\frac{L}{2\pi n}\sin\!\left(\frac{2\pi n}{L}x\right)\right]_0^L = \frac{L^2}{(2\pi n)^2}[\sin(2\pi n) - 0] = 0 \]

Therefore:

\[ b_n = \frac{2}{L}\left(-\frac{L^2}{2\pi n}\right) = -\frac{L}{\pi n} \]

Fourier series:

\[ \boxed{f(x) = \frac{L}{2} - \frac{L}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}\sin\!\left(\frac{2\pi n}{L}x\right)} \]

Verification at \(x = L/2\):

\[ f(L/2) = \frac{L}{2} - \frac{L}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}\sin(\pi n) \]

Since \(\sin(\pi n) = 0\) for all integers \(n\):

\[ f(L/2) = \frac{L}{2} - 0 = \frac{L}{2} \]

This is consistent with \(f(L/2) = L/2\). ✓

Verification¶

Check at \(x = 0\): \(f(0) = \frac{L}{2} - \frac{L}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}\sin(0) = \frac{L}{2}\). However, \(f(0) = 0\) is expected. This is because the Fourier series represents a periodic function, so at \(x = 0\) it converges to the average of \(f(0^+) = 0\) and \(f(L^-) = L\), which is \(L/2\) (Dirichlet's theorem). This is the correct behavior of a Fourier series at a point of discontinuity. ✓

M-2. Fourier Transform of a Gaussian Function and Parseval's Theorem¶

→ Back to problem

Solution Strategy¶

As in D5, we perform completing the square and obtain the result for general \(a\).

Detailed Calculation¶

Fourier Transform Calculation:

\[ \tilde{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ax^2} e^{-ikx}\,dx \]

Completing the square in the exponent:

\[ -ax^2 - ikx = -a\left(x + \frac{ik}{2a}\right)^2 - \frac{k^2}{4a} \]

Therefore:

\[ \tilde{f}(k) = \frac{1}{\sqrt{2\pi}}\,e^{-k^2/(4a)}\int_{-\infty}^{\infty} e^{-a\left(x + \frac{ik}{2a}\right)^2}\,dx = \frac{1}{\sqrt{2\pi}}\,e^{-k^2/(4a)} \cdot \sqrt{\frac{\pi}{a}} \]

\[ \boxed{\tilde{f}(k) = \frac{1}{\sqrt{2a}}\,e^{-k^2/(4a)}} \]

The result is again a Gaussian function. ✓

Verification of Parseval's Theorem:

Left-hand side:

\[ \int_{-\infty}^{\infty} |f(x)|^2\,dx = \int_{-\infty}^{\infty} e^{-2ax^2}\,dx = \sqrt{\frac{\pi}{2a}} \]

Right-hand side:

\[ \int_{-\infty}^{\infty} |\tilde{f}(k)|^2\,dk = \int_{-\infty}^{\infty} \frac{1}{2a}\,e^{-k^2/(2a)}\,dk = \frac{1}{2a}\sqrt{\frac{\pi}{1/(2a)}} = \frac{1}{2a}\sqrt{2\pi a} = \sqrt{\frac{\pi}{2a}} \]

Final Answer¶

\[ \boxed{\int_{-\infty}^{\infty} e^{-2ax^2}\,dx = \sqrt{\frac{\pi}{2a}} = \int_{-\infty}^{\infty} \frac{1}{2a}\,e^{-k^2/(2a)}\,dk} \]

Both sides equal \(\sqrt{\frac{\pi}{2a}}\), confirming Parseval's theorem. ✓

M-3. Solve the following problem using the convolution theorem (Eq. (C.22))¶

→ Back to problem

Solution Strategy¶

Find the Fourier transforms of \(f(x) = g(x) = e^{-x^2}\), apply the convolution theorem, and then take the inverse transform.

Detailed Calculation¶

Step 1: Computing the Fourier Transforms

Setting \(a = 1\) in the result from S2:

\[ \tilde{f}(k) = \tilde{g}(k) = \frac{1}{\sqrt{2}}\,e^{-k^2/4} \]

Step 2: Applying the Convolution Theorem

The convolution theorem from equation (C.22) (convention (b)):

\[ \widetilde{(f*g)}(k) = \sqrt{2\pi}\,\tilde{f}(k)\,\tilde{g}(k) \]

\[ \widetilde{(f*g)}(k) = \sqrt{2\pi} \cdot \frac{1}{\sqrt{2}}\,e^{-k^2/4} \cdot \frac{1}{\sqrt{2}}\,e^{-k^2/4} = \sqrt{2\pi} \cdot \frac{1}{2}\,e^{-k^2/2} = \frac{\sqrt{2\pi}}{2}\,e^{-k^2/2} \]

Step 3: Inverse Fourier Transform

\[ (f*g)(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \frac{\sqrt{2\pi}}{2}\,e^{-k^2/2}\,e^{ikx}\,dk = \frac{1}{2}\int_{-\infty}^{\infty} e^{-k^2/2 + ikx}\,dk \]

Completing the square in the exponent:

\[ -\frac{k^2}{2} + ikx = -\frac{1}{2}(k - ix)^2 - \frac{x^2}{2} \]

\[ (f*g)(x) = \frac{1}{2}\,e^{-x^2/2}\int_{-\infty}^{\infty} e^{-\frac{1}{2}(k-ix)^2}\,dk = \frac{1}{2}\,e^{-x^2/2} \cdot \sqrt{2\pi} = \frac{\sqrt{2\pi}}{2}\,e^{-x^2/2} \]

Final Answer¶

\[ \boxed{(f*g)(x) = \sqrt{\frac{\pi}{2}}\,e^{-x^2/2}} \]

(We simplified \(\frac{\sqrt{2\pi}}{2} = \sqrt{\frac{\pi}{2}}\).)

Verification¶

We confirm by direct calculation. \((f*g)(x) = \int_{-\infty}^{\infty} e^{-x'^2}\,e^{-(x-x')^2}\,dx'\).

The exponent: \(-x'^2 - (x-x')^2 = -2x'^2 + 2xx' - x^2 = -2(x' - x/2)^2 - x^2/2\)

\[ (f*g)(x) = e^{-x^2/2}\int_{-\infty}^{\infty} e^{-2(x'-x/2)^2}\,dx' = e^{-x^2/2}\sqrt{\frac{\pi}{2}} = \sqrt{\frac{\pi}{2}}\,e^{-x^2/2} \quad \checkmark \]

M-4. Fourier Integral Representation of the δ Function¶

→ Back to problem

(a) \(\delta(-x) = \delta(x)\) (Even Function)¶

Starting from the Fourier integral representation:

\[ \delta(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ikx}\,dk \]

Substituting \(x \to -x\):

\[ \delta(-x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-ikx}\,dk \]

Changing the integration variable \(k \to -k\) (\(dk \to -dk\), the limits of integration reverse and return to the original):

\[ = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ikx}\,dk = \delta(x) \]

\[ \boxed{\delta(-x) = \delta(x)} \]

(b) Scaling Property \(\delta(\alpha x) = \frac{1}{|\alpha|}\,\delta(x)\)¶

From the Fourier integral representation:

\[ \delta(\alpha x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ik(\alpha x)}\,dk \]

Substituting \(k' = \alpha k\) gives \(dk = dk'/\alpha\):

Case 1: \(\alpha > 0\)

\[ \delta(\alpha x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ik'x}\,\frac{dk'}{\alpha} = \frac{1}{\alpha}\,\delta(x) \]

Case 2: \(\alpha < 0\)

When \(k' = \alpha k\) with \(\alpha < 0\), the limits of integration are reversed:

\[ \delta(\alpha x) = \frac{1}{2\pi}\int_{\infty}^{-\infty} e^{ik'x}\,\frac{dk'}{\alpha} = \frac{1}{2\pi} \cdot \frac{1}{\alpha}\int_{\infty}^{-\infty} e^{ik'x}\,dk' = \frac{1}{2\pi} \cdot \frac{-1}{\alpha}\int_{-\infty}^{\infty} e^{ik'x}\,dk' = \frac{1}{|\alpha|}\,\delta(x) \]

(Since \(\alpha < 0\), we have \(-1/\alpha = 1/|\alpha|\))

\[ \boxed{\delta(\alpha x) = \frac{1}{|\alpha|}\,\delta(x)} \]

(c) \(x\,\delta(x) = 0\)¶

For any test function \(\phi(x)\):

\[ \int_{-\infty}^{\infty} x\,\delta(x)\,\phi(x)\,dx = \int_{-\infty}^{\infty} \delta(x)\,[x\,\phi(x)]\,dx \]

By the sifting property of the δ function, letting \(h(x) = x\,\phi(x)\):

\[ = [x\,\phi(x)]_{x=0} = 0 \cdot \phi(0) = 0 \]

Since the integral is zero for any test function, in the distributional sense:

\[ \boxed{x\,\delta(x) = 0} \]

Verification¶

Special case of (b): when \(\alpha = -1\), we get \(\delta(-x) = \delta(x)\). This is consistent with (a). ✓

Alternative confirmation of (c): \(\delta(x)\) is nonzero only at \(x = 0\). At \(x = 0\), the factor of \(x\) is zero, so the product is zero everywhere. ✓

M-5. Differentiation Property of the Fourier Transform¶

→ Back to problem

(a) Proof that the Fourier transform of \(f'(x)\) is \(ik\,\tilde{f}(k)\)¶

Calculation using convention (b):

\[ \mathcal{F}[f'](appendix_c/k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f'(x)\,e^{-ikx}\,dx \]

Integration by parts (\(u = e^{-ikx}\), \(dv = f'(x)\,dx\)):

\[ = \frac{1}{\sqrt{2\pi}}\left[f(x)\,e^{-ikx}\right]_{-\infty}^{\infty} - \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)\,(-ik)\,e^{-ikx}\,dx \]

Assuming \(f(x) \to 0\) as \(x \to \pm\infty\), the boundary terms vanish:

\[ = ik \cdot \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)\,e^{-ikx}\,dx = ik\,\tilde{f}(k) \]

\[ \boxed{\mathcal{F}[f'](appendix_c/k) = ik\,\tilde{f}(k)} \]

(b) Solution of the differential equation \(f'(x) + \beta f(x) = 0\)¶

Method: Directly compute the Fourier transform of \(f(x) = Ce^{-\beta x}\theta(x)\).

Let \(f(x) = Ce^{-\beta x}\) for \(x > 0\) and \(f(x) = 0\) for \(x < 0\). That is, \(f(x) = Ce^{-\beta x}\theta(x)\).

Fourier transform:

\[ \tilde{f}(k) = \frac{1}{\sqrt{2\pi}}\int_0^{\infty} Ce^{-\beta x}\,e^{-ikx}\,dx = \frac{C}{\sqrt{2\pi}}\int_0^{\infty} e^{-(\beta + ik)x}\,dx = \frac{C}{\sqrt{2\pi}} \cdot \frac{1}{\beta + ik} \]

Verification: Examining \(f'(x) + \beta f(x) = 0\) (for \(x > 0\)) in Fourier space.

\(f'(x) = -\beta Ce^{-\beta x}\theta(x) + C\delta(x)\) (the delta function arises from differentiating \(\theta\))

Taking the Fourier transform:

\[ \mathcal{F}[f'] = ik\,\tilde{f}(k) = \frac{ikC}{\sqrt{2\pi}(\beta + ik)} \]

\[ \beta\,\tilde{f}(k) = \frac{\beta C}{\sqrt{2\pi}(\beta + ik)} \]

\[ ik\,\tilde{f}(k) + \beta\,\tilde{f}(k) = \frac{(ik + \beta)C}{\sqrt{2\pi}(\beta + ik)} = \frac{C}{\sqrt{2\pi}} \]

This equals \(\mathcal{F}[C\delta(x)] = \frac{C}{\sqrt{2\pi}}\). Thus \(f' + \beta f = C\delta(x)\) holds, and since the right-hand side is zero for \(x > 0\), the equation \(f' + \beta f = 0\) is satisfied.

Alternative solution (more direct): The general solution of the differential equation \(f' + \beta f = 0\) is \(f(x) = Ce^{-\beta x}\). When \(\beta > 0\), a physically decaying solution is obtained for \(x > 0\).

\[ \boxed{f(x) = Ce^{-\beta x} \quad (x > 0)} \]

Verification¶

Substituting \(f(x) = Ce^{-\beta x}\) into the differential equation: \(f' = -\beta Ce^{-\beta x}\), so \(f' + \beta f = -\beta Ce^{-\beta x} + \beta Ce^{-\beta x} = 0\). ✓

Advanced¶

A-1. Fourier-Analytic Proof of the Uncertainty Relation¶

→ Back to problem

(a) Derivation of \(\Delta x \cdot \Delta k \geq \frac{1}{2}\)¶

Solution strategy: Substitute \(u(x) = xf(x)\), \(v(x) = f'(x)\) into the Cauchy–Schwarz inequality, and evaluate the left-hand side using integration by parts.

Step 1: Applying the Cauchy–Schwarz inequality

\[ \left|\int_{-\infty}^{\infty} [xf(x)]^*\,f'(x)\,dx\right|^2 \leq \int_{-\infty}^{\infty}|xf(x)|^2\,dx \cdot \int_{-\infty}^{\infty}|f'(x)|^2\,dx \]

The first factor on the right-hand side is \((\Delta x)^2\):

\[ \int_{-\infty}^{\infty}|xf(x)|^2\,dx = \int_{-\infty}^{\infty} x^2|f(x)|^2\,dx = (\Delta x)^2 \]

Step 2: Proof that \(\int |f'(x)|^2\,dx = (\Delta k)^2\)

From the result of S5(a), \(\mathcal{F}[f'](appendix_c/k) = ik\,\tilde{f}(k)\). Applying Parseval's theorem:

\[ \int_{-\infty}^{\infty}|f'(x)|^2\,dx = \int_{-\infty}^{\infty}|ik\,\tilde{f}(k)|^2\,dk = \int_{-\infty}^{\infty} k^2|\tilde{f}(k)|^2\,dk = (\Delta k)^2 \]

Step 3: Evaluating the left-hand side

\[ \int_{-\infty}^{\infty} x\,f(x)^*\,f'(x)\,dx \]

\(f(x)^*f'(x) = \frac{1}{2}\frac{d}{dx}|f(x)|^2 + \frac{i}{2}\frac{d}{dx}[\text{phase term}]\)...

More directly, we use integration by parts. Let \(I = \int_{-\infty}^{\infty} x\,f^*\,f'\,dx\).

From \(\frac{d}{dx}[x|f|^2] = |f|^2 + x\,f^*\,f' + x\,f\,(f^*)'\), we get:

\[ I + I^* = \int_{-\infty}^{\infty} x\,(f^*f' + f(f^*)') \,dx = \int_{-\infty}^{\infty} x\,\frac{d}{dx}|f|^2\,dx \]

Integration by parts:

\[ \int_{-\infty}^{\infty} x\,\frac{d}{dx}|f|^2\,dx = \left[x|f|^2\right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty}|f|^2\,dx = 0 - 1 = -1 \]

(\(f(x) \to 0\) sufficiently fast as \(x \to \pm\infty\), normalization condition \(\int|f|^2\,dx = 1\))

Therefore \(I + I^* = 2\,\text{Re}(I) = -1\), i.e., \(\text{Re}(I) = -\frac{1}{2}\).

Thus \(|I| \geq |\text{Re}(I)| = \frac{1}{2}\).

Step 4: Completing the inequality

From Cauchy–Schwarz:

\[ \frac{1}{4} \leq |I|^2 \leq (\Delta x)^2 \cdot (\Delta k)^2 \]

\[ \boxed{\Delta x \cdot \Delta k \geq \frac{1}{2}} \]

(b) Equality condition¶

The equality condition for Cauchy–Schwarz is \(v(x) = \lambda\,u(x)\) (\(\lambda\) is a complex constant):

\[ f'(x) = \lambda\,x\,f(x) \]

This is a separable differential equation:

\[ \frac{df}{f} = \lambda\,x\,dx \implies \ln f = \frac{\lambda}{2}x^2 + \text{const} \]

\[ f(x) = A\,e^{\lambda x^2/2} \]

For \(f(x)\) to converge to \(0\) as \(x \to \pm\infty\), we need \(\text{Re}(\lambda) < 0\).

Furthermore, examining the equality condition in detail: for \(|I| = |\text{Re}(I)|\) to hold, \(I\) must be real (and negative). Since \(I = \text{Re}(I) = -1/2\), \(I\) is indeed real.

\(I = \int x|f|^2 \cdot \frac{f'}{f}\,dx = \lambda\int x^2|f|^2\,dx = \lambda(\Delta x)^2\)

Since \(I\) is real, \(\lambda\) must also be real. Writing \(\lambda < 0\) as \(\lambda = -1/(2\sigma^2)\):

\[ f(x) = A\,e^{-x^2/(4\sigma^2)} \]

From the normalization condition \(\int|f|^2\,dx = 1\):

\[ |A|^2\int_{-\infty}^{\infty} e^{-x^2/(2\sigma^2)}\,dx = |A|^2\sqrt{2\pi\sigma^2} = 1 \]

\[ A = \left(\frac{1}{2\pi\sigma^2}\right)^{1/4} \]

\[ \boxed{f(x) = \left(\frac{1}{2\pi\sigma^2}\right)^{1/4} e^{-x^2/(4\sigma^2)}} \]

This is a Gaussian function (Gaussian wave packet).

(c) The quantum mechanical uncertainty relation¶

Since \(p = \hbar k\), we have \(\Delta p = \hbar\,\Delta k\). Multiplying both sides of the result from (a), \(\Delta x \cdot \Delta k \geq \frac{1}{2}\), by \(\hbar\):

\[ \Delta x \cdot \hbar\,\Delta k \geq \frac{\hbar}{2} \]

\[ \boxed{\Delta x \cdot \Delta p \geq \frac{\hbar}{2}} \]

This is the Heisenberg uncertainty relation of quantum mechanics. Equality is achieved for Gaussian wave packets (coherent states).

Verification¶

For the Gaussian function \(f(x) = (2\pi\sigma^2)^{-1/4}e^{-x^2/(4\sigma^2)}\):

\((\Delta x)^2 = \int x^2|f|^2\,dx = \sigma^2\) (variance of the Gaussian distribution)
\(\tilde{f}(k) = (2\sigma^2/\pi)^{1/4}e^{-\sigma^2 k^2}\), so \((\Delta k)^2 = \frac{1}{4\sigma^2}\)
\(\Delta x \cdot \Delta k = \sigma \cdot \frac{1}{2\sigma} = \frac{1}{2}\)

Equality holds. ✓

A-2. From Fourier Series to Parseval's Identity: Derivation of \(\zeta(2) = \pi^2/6\)¶

→ Back to problem

(a) Derivation of Parseval's Equality for Fourier Series¶

Solution strategy: Integrate \(|f(x)|^2\) and use the orthogonality of the Fourier series.

From Eq. (C.5):

\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left[a_n \cos\!\left(\frac{2\pi n}{L}x\right) + b_n \sin\!\left(\frac{2\pi n}{L}x\right)\right] \]

Integrate \(|f(x)|^2 = f(x) \cdot f(x)\) from \(0\) to \(L\):

\[ \int_0^L |f(x)|^2\,dx = \int_0^L f(x) \cdot f(x)\,dx \]

Substituting the Fourier series twice into the right-hand side and expanding, the cross terms vanish by orthogonality (C.2)–(C.4). The surviving terms are:

Constant term with itself: \(\left(\frac{a_0}{2}\right)^2 \cdot L = \frac{a_0^2}{4} \cdot L\)
\(\cos\) terms with themselves (only \(n = m\)): \(a_n^2 \cdot \frac{L}{2}\)
\(\sin\) terms with themselves (only \(n = m\)): \(b_n^2 \cdot \frac{L}{2}\)

Therefore:

\[ \int_0^L |f(x)|^2\,dx = \frac{a_0^2}{4}\,L + \sum_{n=1}^{\infty}\left(a_n^2 + b_n^2\right)\frac{L}{2} \]

Dividing both sides by \(L\):

\[ \boxed{\frac{1}{L}\int_0^L |f(x)|^2\,dx = \frac{|a_0|^2}{4} + \frac{1}{2}\sum_{n=1}^{\infty}\left(|a_n|^2 + |b_n|^2\right)} \]

(b) Proof that \(\zeta(2) = \pi^2/6\)¶

Substitute the results from S1. For \(f(x) = x\):

\[ a_0 = L, \quad a_n = 0 \;(n \geq 1), \quad b_n = -\frac{L}{\pi n} \]

Left-hand side:

\[ \frac{1}{L}\int_0^L x^2\,dx = \frac{1}{L} \cdot \frac{L^3}{3} = \frac{L^2}{3} \]

Right-hand side:

\[ \frac{a_0^2}{4} + \frac{1}{2}\sum_{n=1}^{\infty}\left(a_n^2 + b_n^2\right) = \frac{L^2}{4} + \frac{1}{2}\sum_{n=1}^{\infty}\frac{L^2}{\pi^2 n^2} \]

Setting up the equality:

\[ \frac{L^2}{3} = \frac{L^2}{4} + \frac{L^2}{2\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2} \]

Dividing by \(L^2\):

\[ \frac{1}{3} = \frac{1}{4} + \frac{1}{2\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2} \]

\[ \frac{1}{3} - \frac{1}{4} = \frac{1}{2\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2} \]

\[ \frac{1}{12} = \frac{1}{2\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2} \]

\[ \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{2\pi^2}{12} = \frac{\pi^2}{6} \]

Final Answer¶

\[ \boxed{\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}} \]

Verification¶

Numerical check: \(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \cdots \approx 1.000 + 0.250 + 0.111 + 0.063 + 0.040 + \cdots \approx 1.55\)

\(\frac{\pi^2}{6} \approx \frac{9.870}{6} \approx 1.645\)

The partial sums converge to \(\pi^2/6\) from below, which is consistent. ✓

Also, the left-hand side of Parseval's equality \(\frac{1}{3}\) equals the right-hand side \(\frac{1}{4} + \frac{1}{2\pi^2} \cdot \frac{\pi^2}{6} = \frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{4}{12} = \frac{1}{3}\). ✓

Feedback on this page

Let us know if something was unclear, incorrect, or could be improved.