Riemann surface
matlab

Problem 1.2
(a)
\begin{aligned} & \left\langle x^2\right\rangle=\int_0^h x^2 \frac{1}{2 \sqrt{h x}} d x=\left.\frac{1}{2 \sqrt{h}}\left(\frac{2}{5} x^{5 / 2}\right)\right|0 ^h=\frac{h^2}{5} \ & \sigma^2=\left\langle x^2\right\rangle-\langle x\rangle^2=\frac{h^2}{5}-\left(\frac{h}{3}\right)^2=\frac{4}{45} h^2 \Rightarrow \sigma=\frac{2 h}{3 \sqrt{5}}=0.2981 h . \end{aligned} (b) \begin{aligned} & P=1-\int{x_{-}}^{x_{+}} \frac{1}{2 \sqrt{h x}} d x=1-\left.\frac{1}{2 \sqrt{h}}(2 \sqrt{x})\right|{x{-}} ^{x_{+}}=1-\frac{1}{\sqrt{h}}\left(\sqrt{x_{+}}-\sqrt{x_{-}}\right) . \ & x_{+} \equiv\langle x\rangle+\sigma=0.3333 h+0.2981 h=0.6315 h ; \quad x_{-} \equiv\langle x\rangle-\sigma=0.3333 h-0.2981 h=0.0352 h . \ & P=1-\sqrt{0.6315}+\sqrt{0.0352}=0.393 . \end{aligned}

Problem 1.3
(a)
\begin{aligned} & 1=\int_{-\infty}^{\infty} A e^{-\lambda(x-a)^2} d x . \quad \text { Let } u \equiv x-a, d u=d x, u:-\infty \rightarrow \infty . \ & 1=A \int_{-\infty}^{\infty} e^{-\lambda u^2} d u=A \sqrt{\frac{\pi}{\lambda}} \Rightarrow A=\sqrt{\frac{\lambda}{\pi}} . \end{aligned}
(b)
\begin{aligned} \langle x\rangle & =A \int_{-\infty}^{\infty} x e^{-\lambda(x-a)^2} d x=A \int_{-\infty}^{\infty}(u+a) e^{-\lambda u^2} d u \ & =A\left[\int_{-\infty}^{\infty} u e^{-\lambda u^2} d u+a \int_{-\infty}^{\infty} e^{-\lambda u^2} d u\right]=A\left(0+a \sqrt{\frac{\pi}{\lambda}}\right)=a . \ \left\langle x^2\right\rangle & =A \int_{-\infty}^{\infty} x^2 e^{-\lambda(x-a)^2} d x \ & =A\left{\int_{-\infty}^{\infty} u^2 e^{-\lambda u^2} d u+2 a \int_{-\infty}^{\infty} u e^{-\lambda u^2} d u+a^2 \int_{-\infty}^{\infty} e^{-\lambda u^2} d u\right} \ & =A\left[\frac{1}{2 \lambda} \sqrt{\frac{\pi}{\lambda}}+0+a^2 \sqrt{\frac{\pi}{\lambda}}\right]=a^2+\frac{1}{2 \lambda} . \ \sigma^2 & =\left\langle x^2\right\rangle-\langle x\rangle^2=a^2+\frac{1}{2 \lambda}-a^2=\frac{1}{2 \lambda} ; \quad \sigma=\frac{1}{\sqrt{2 \lambda}} . \end{aligned}

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