Fourier Transform Equation Solution

This problem requires us to find the value of \(f(0)\) by utilizing the given differential equation for its Fourier transform \(F(ω)\), an initial condition, and the properties of Fourier transforms.

Solving the Differential Equation for \(F(ω)\)

We are provided with the following first-order differential equation:

\(\dfrac{dF(ω)}{dω} = -ω F(ω)\)

This is a separable differential equation. We can rearrange the terms to group \(F\) with \(dF\) and \(\omega\) with \(dω\):

\(\dfrac{dF}{F} = -ω dω\)

Now, we integrate both sides of the equation:

\(\displaystyle\int \dfrac{dF}{F} = \displaystyle\int -ω dω\)

Performing the integration, we obtain:

\(\ln|F(ω)| = -\dfrac{ω^2}{2} + C\)

Here, \(C\) represents the constant of integration. To isolate \(F(ω)\), we take the exponential of both sides:

\(F(ω) = e^{-\frac{ω^2}{2} + C}\)

This expression can be rewritten by separating the exponential terms:

\(F(ω) = e^C \cdot e^{-\frac{ω^2}{2}}\)

Let's denote \(A = e^C\), which is a new constant. Thus, the general solution for \(F(ω)\) is:

\(F(ω) = A e^{-\frac{ω^2}{2}}\)

Applying the Initial Condition F(0) = 1

The problem states that \(F(0) = 1\). We will use this initial condition to determine the specific value of the constant \(A\).

Substitute \(\omega = 0\) into the derived expression for \(F(ω)\):

\(F(0) = A e^{-\frac{0^2}{2}}\)

\(F(0) = A e^0\)

\(F(0) = A \cdot 1\)

\(F(0) = A\)

Since we know \(F(0) = 1\), it follows that \(A = 1\).

Therefore, the unique Fourier transform function is:

\(F(ω) = e^{-\frac{ω^2}{2}}\)

This is a Gaussian function in the frequency domain, a common form in signal processing.

Determining f(0) using Inverse Fourier Transform

To find the value of \(f(0)\), we need to apply the inverse Fourier transform formula. The inverse Fourier transform is defined as:

\(f(t) = \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^\infty F(ω) e^{jω t} dω\)

Now, substitute the derived \(F(ω) = e^{-\frac{ω^2}{2}}\) into this formula:

\(f(t) = \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^\infty e^{-\frac{ω^2}{2}} e^{jω t} dω\)

To find \(f(0)\), we set \(t = 0\):

\(f(0) = \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^\infty e^{-\frac{ω^2}{2}} e^{jω \cdot 0} dω\)

\(f(0) = \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^\infty e^{-\frac{ω^2}{2}} \cdot 1 dω\)

\(f(0) = \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^\infty e^{-\frac{ω^2}{2}} dω\)

Evaluating the Gaussian Integral

The integral we obtained is a standard form of the Gaussian integral, \(\displaystyle\int_{-\infty}^\infty e^{-ax^2} dx\), which evaluates to \(\sqrt{\dfrac{\pi}{a}}\).

In our integral, \(\displaystyle\int_{-\infty}^\infty e^{-\frac{ω^2}{2}} dω\), the variable is \(\omega\) and the coefficient \(a\) for \(-\omega^2\) is \(\dfrac{1}{2}\).

Therefore, the integral evaluates to:

\(\displaystyle\int_{-\infty}^\infty e^{-\frac{ω^2}{2}} dω = \sqrt{\dfrac{\pi}{1/2}} = \sqrt{2\pi}\)

Calculating the Value of f(0)

Now, substitute the value of the integral back into the expression for \(f(0)\):

\(f(0) = \dfrac{1}{2\pi} \cdot \sqrt{2\pi}\)

\(f(0) = \dfrac{\sqrt{2\pi}}{2\pi}\)

We can simplify this expression. Recall that \(2\pi\) can be written as \((\sqrt{2\pi})^2\).

\(f(0) = \dfrac{\sqrt{2\pi}}{(\sqrt{2\pi})^2}\)

\(f(0) = \dfrac{1}{\sqrt{2\pi}}\)

Comparing f(0) with 1

To determine the relationship between \(f(0)\) and 1, we need to approximate the value of \(f(0)\).

We know that \(\pi \approx 3.14159\).

So, \(2\pi \approx 2 \times 3.14159 = 6.28318\).

Then, \(\sqrt{2\pi} \approx \sqrt{6.28318}\).

Since \(\sqrt{4} = 2\) and \(\sqrt{9} = 3\), it is evident that \(\sqrt{6.28318}\) is a value between 2 and 3. More precisely, \(\sqrt{6.28318} \approx 2.5066\).

Therefore, \(f(0) = \dfrac{1}{\sqrt{2\pi}} \approx \dfrac{1}{2.5066}\).

Since the denominator \(2.5066\) is greater than 1, the fraction \(\dfrac{1}{2.5066}\) must be less than 1.

Thus, \(f(0) < 1\).

The information that \(f(t)\) is an even function is consistent with our derived \(F(ω)\) being a real and even function, as the Fourier transform of a real and even function is always real and even.

Fourier Transform Solution Steps and Conclusion

The solution involved several key steps:

1. We solved the given first-order differential equation for the Fourier transform \(F(ω)\), obtaining \(F(ω) = A e^{-\frac{ω^2}{2}}\).
2. We used the initial condition \(F(0) = 1\) to find the constant \(A=1\), leading to \(F(ω) = e^{-\frac{ω^2}{2}}\).
3. We applied the inverse Fourier transform formula at \(t=0\) to express \(f(0)\) as an integral: \(f(0) = \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^\infty e^{-\frac{ω^2}{2}} dω\).
4. We evaluated the standard Gaussian integral, finding \(\displaystyle\int_{-\infty}^\infty e^{-\frac{ω^2}{2}} dω = \sqrt{2\pi}\).
5. Finally, we calculated \(f(0) = \dfrac{1}{2\pi} \cdot \sqrt{2\pi} = \dfrac{1}{\sqrt{2\pi}}\).

Since \(\sqrt{2\pi} \approx 2.5066\), \(f(0) \approx \dfrac{1}{2.5066} \approx 0.3989\). This value is clearly less than 1.

Therefore, the correct relationship is \(f(0) < 1\).