Chapter 7: Q36E (page 281)

Prove that if the function $e^{i \sqrt{D} φ}$ is to meet itself smoothly when $φ$ changes by $2 π$ , $\sqrt{D}$ must be an integer.

Short Answer

Expert verified

For the function $e^{i \sqrt{D} φ}$ to have a period of $2 π$ , $\sqrt{D}$ must be an integer.

Step by step solution

A concept:

The functions which repeat itself after regular interval of time is called a Cyclic function and that interval in which it is repeating itself is called the period of that cyclic function.

As you know that, by Euler’s equation,

$e^{ix} = \cos (x) + i \sin (x)$

Value of the function at φ=0 and φ=2π :

Let, the function be

$F (φ) = e^{i \sqrt{D} φ} = \cos (\sqrt{D} φ) + isin (\sqrt{D} φ)$

Where, $D = - \frac{1}{Φ} \frac{\partial^{2} Φ}{\partial φ^{2}}$ and $φ$ is the Azimuthal Angle.

Now at $φ = 0$ :

$F (0) = e^{i \sqrt{D} 0} = \cos (\sqrt{D} 0) + isin (\sqrt{D} 0) = 1$

Again, if the function meets itself at $φ = 2 π$ ,

$F (2 π) = F (0) = 1$

Value of :

If, $F (2 π) = 1$

$e^{i \sqrt{D} 2 π} = 1 \cos (\sqrt{D} 2 π) + isin (\sqrt{D} 2 π) = 1$ ….. (1)

In equation (1), if the real part is 1 the imaginary part should be zero and for that to hold, must be an integer $\sqrt{D}$

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Prove that if the function $e^{i \sqrt{D} φ}$ is to meet itself smoothly when $φ$ changes by $2 π$ , $\sqrt{D}$ must be an integer.

Short Answer

Step by step solution

A concept:

Value of the function at φ=0 and φ=2π :

Value of :

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Prove that if the functioneiDφis to meet itself smoothly whenφchanges by 2π, D must be an integer.

Short Answer

Step by step solution

A concept:

Value of the function at φ=0 and φ=2π :

Value of :

One App. One Place for Learning.

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Prove that if the function $e^{i \sqrt{D} φ}$ is to meet itself smoothly when $φ$ changes by $2 π$ , $\sqrt{D}$ must be an integer.