# 数学代写|微积分代写Calculus代写|MATH1111

## 数学代写|微积分代写Calculus代写|Euler’s Gamma Function

Since $e^{-t} t^x$ vanishes at $t=0$ and $t=\infty$ for $x>0$ fixed, and the integrands are positive, by the fundamental theorem, we obtain
\begin{aligned} \Gamma(x+1) &=\int_0^{\infty} e^{-t} t^x d t \ &=-\left.e^{-t} t^x\right|_0 ^{\infty}+x \int_0^{\infty} e^{-t} t^{x-1} d t \ &=0+x \Gamma(x)=x \Gamma(x) . \end{aligned}
Note that this identity is true whether or not $\Gamma(x)$ is finite. We derive $\Gamma(n)=$ $(n-1)$ ! by induction. The statement is true for $n=1$ since $\Gamma(1)=1$ from above. Assuming the statement is true for $n, \Gamma(n+1)=n \Gamma(n)=n(n-1) !=n$ !. Hence, the statement is true for all $n \geq 1$. Now, we show that $\Gamma(x)$ is finite for all $x>0$. Since the integral $\int_0^1 e^{-t} t^{x-1} d t \leq \int_0^1 t^{x-1} d t=1 / x$ is finite for $x>0$, it is enough to verify integrability of $e^{-t} t^{x-1}$ over $(1, \infty)$. Over this interval, $e^{-t} t^{x-1}$ increases with $x$, hence, $\int_1^{\infty} e^{-t} t^{x-1} d t \leq \int_1^{\infty} e^{-t} t^{n-1} d t \leq \Gamma(n)$ for any natural $n \geq x$. But we already know that $\Gamma(n)=(n-1) !<\infty$, hence, the result.

Because of this result, we define $x !=\Gamma(x+1)$ for $x>-1$. For example, in Exercise 5.4.1, we obtain $(1 / 2) !=\sqrt{\pi} / 2$.

We already know (linearity $\S 4.4$ ) that the integral of a finite sum of continuous functions is the sum of their integrals. To obtain linearity for infinite sums, we first derive the following.

Theorem 5.1.2 (Monotone Convergence Theorem (For Integrals)). Let $0 \leq f_1 \leq f_2 \leq f_3 \leq \ldots$ be an increasing sequence of nonnegative functions ${ }^1$, all defined on an interval $(a, b)$. If
$$\lim {n>\infty} f_n(x)=f(x), \quad a{n / \infty} \int_a^b f_n(x) d x=\int_a^b \lim _{n / \infty} f_n(x) d x=\int_a^b f(x) d x$$

## 数学代写|微积分代写Calculus代写|Gauss’ Arithmetic-Geometric Mean

Given $a>b>0$, their arithmetic mean is given by
$$a^{\prime}=\frac{a+b}{2}$$
and their geometric mean by
$$b^{\prime}=\sqrt{a b}$$
Since
$$a^{\prime}-b^{\prime}=\frac{a+b}{2}-\sqrt{a b}=\frac{1}{2}(\sqrt{a}-\sqrt{b})^2>0,$$
these equations transform the pair $(a, b), a>b>0$, into a pair $\left(a^{\prime}, b^{\prime}\right)$, $a^{\prime}>b^{\prime}>0$. Gauss discovered that iterating this transformation leads to a limit with striking properties.

To begin, since $a$ is the larger of $a$ and $b$ and $a^{\prime}$ is their arithmetic mean, $a^{\prime}b$. Thus, $bb>0$, this gives a strictly decreasing sequence $\left(a_n\right)$ and a strictly increasing sequence $\left(b_n\right)$ with all the $a$ ‘s greater than all the $b$ ‘s. Thus, both sequences converge (Figure 5.3) to finite positive limits $a_, b^$ with $a_* \geq b^>0$. Fig. 5.3. The AGM iteration. Letting $n \nearrow \infty$ in (5.3.2), we see that $a_=\left(a_+b^\right) / 2$ which yields $a_=b^$. Thus, both sequences converge to a common limit, the arithmetic-geometric mean (AGM) of $(a, b)$, which we denote
$$M(a, b)=\lim {n \nearrow \infty} a_n=\lim {n \nearrow \infty} b_n .$$
If $\left(a_n^{\prime}\right),\left(b_n^{\prime}\right)$ are the sequences associated with $a^{\prime}=t a$ and $b^{\prime}=t b$, then, from (5.3.2) and (5.3.3), $a_n^{\prime}=t a_n$, and $b_n^{\prime}=t b_n, n \geq 1, t>0$. This implies that $M$ is homogeneous in $(a, b)$,
$$M(t a, t b)=t \cdot M(a, b), \quad t>0$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|欧拉函数

\begin{aligned} \Gamma(x+1) &=\int_0^{\infty} e^{-t} t^x d t \ &=-\left.e^{-t} t^x\right|_0 ^{\infty}+x \int_0^{\infty} e^{-t} t^{x-1} d t \ &=0+x \Gamma(x)=x \Gamma(x) . \end{aligned}

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