TITLE: Sum of multi-index factorials
QUESTION [4 upvotes]: Fix $d\in\mathbb{N}\setminus\{0\}$. For $j\in\mathbb{N}\setminus\{0\}$, let
\begin{align*}
[j] = \Big\{\alpha\in \mathbb{N}^d: \sum^d_{i=1}\alpha_i=j\Big\}.
\end{align*}
For $\alpha\in[j]$, define the multi-index factorial $\alpha! =\prod_{i=1}^d (\alpha_i !)$. What is a good bound for
\begin{align*}
\sum_{\alpha \in [j]} (\alpha !)
\end{align*}
in terms of $j$, when $d$ is fixed and $j$ is large?

REPLY [2 votes]: Let us also show for each natural $d$
$$S_{d,j}\sim j!d \tag{1}$$ (as $j\to\infty$),
where $S_{d,j}$ is the sum in question.
The key here is the recursion
$$S_{d,j}=\sum_{b=0}^j b!S_{d-1,j-b} \tag{2}$$
for $d=2,3,\dots$ and $j=0,1,\dots$, with the initial conditions $S_{d,0}=1$ and $S_{1,j}=j!$. The latter initial condition obviously implies (1) for $d=1$.
Proceed by induction on $d$. Take any $d=2,3,\dots$ and $j=0,1,\dots$. Then, by (2) and induction, for any fixed natural $B$
$$S_{d,j}=\sum_{b=0}^j (j-b)!S_{d-1,b} \\
=j!S_{d-1,0}+(j-1)!S_{d-1,1}+S_{d-1,j}+S_{d-1,j-1} \\
+\sum_{b=2}^B (j-b)!S_{d-1,b} +\sum_{b=B+1}^{j-2} (j-b)!S_{d-1,b} \\
=j!+O((j-1)!)+(d-1+o(1))j!+O((j-1)!) \\
+O((j-2)!) +\sum_{b=B+1}^{j-2} (j-b)!S_{d-1,b} \\
=(d+o(1))j!+\sum_{b=B+1}^{j-2} (j-b)!S_{d-1,b}.
$$
Let now $B\ge1$ be large enough so that $S_{d-1,b}\le b!d$ for all $b>B$; such $B$ exists by induction. Then, noting that $(j-b)!b!$ is log-convex in $b\in\{0,\dots,B\}$, we see that for $j\ge3$
$$\sum_{b=B+1}^{j-2} (j-b)!S_{d-1,b}\le \sum_{b=B+1}^{j-2} (j-b)!b!\,d \\
\le \sum_{b=2}^{j-2} (j-b)!b!\,d\le(j-2-1)(j-2)!2!\,d=o(j!).$$
Now (1) follows by the multiline display.


TITLE: Examine function extreme values
QUESTION [2 upvotes]: I am studying for multivariable calculus exam and in homework we always had specific task regarding extreme values: find absolute minima, find local maxima, etc.
In real exam questions are more like "Examine function extreme values".
What are the steps that should be done for complete examination of function $f(x,y)$ extreme values?

REPLY [6 votes]: I will assume that $f$ is "nice". This isn't comprehensive; but will allow you to find, in most cases:
1) The local extrema of $f$ in the interior of its domain.
2) The global extrema of $f$ if $f$ is defined on a "nice" region $R$ with boundary $B$.

To find local extrema in the interior of the domain  of $f$:
First, find the points where the gradient of $f$ is the zero vector or does not exist. These are the candidates (and only candidates) for local extreme values in the interior of the domain of $f$.
Given one these points, the so-called "second partials  test" might be used to determine if the point gives  a local extreme value. I say "might" since, in some cases, this test gives no useful info. If this turns out to be the case, then you need to examine what happens near the critical point in question by looking at $f$ itself (this may be hard).

To find global extrema where $f$ is defined on a "nice" region $R$ with boundary $B$:
You will need to examine the function on the boundary. Here, you will need to find a function of one variable, $g$, that gives the same value as $f$ at any point on the boundary. Usually, you find a parameterization of $B$ , and then substitute into $f$ to obtain $g$. Then find the global extrema of $g$.
You then compare these values with the values of $f$ that are candidates for local extrema in the interior.
Evaluate $f$ at each point in the interior where the gradient is the zero vector or does not exist. Compare these values with the global extreme values found on the boundary and select the largest and smallest of these values.

Candidate points ${\bf x}=(x,y)$ for local extrema:
A function of two variables $f$ can assume an extreme  value only at points where
$
  \nabla f({\bf x})=\bf 0
$ or  does not exist. Here, $\nabla f$ is the gradient of $f$:
$$
 \nabla f({\bf x})=f_x({\bf x})\,{\bf i}+f_y({\bf x})\,{\bf j}.
$$
We call any point where  $
  \nabla f({\bf x})=\bf 0
$ or  does not exist a critical point  of $f$.

Second partials test:
Suppose $f$ has continuous second partial derivatives.
For a critical point $ {\bf x}=(x,y)$ of $f$ in the interior of the domain of $f$,
set $$
  D=f_{xx}({\bf x})f_{yy}({\bf x}) - f_{xy}^2({\bf x}).
$$
Then
$$
\eqalignno{
D\gt0 \hbox{  and  } f_{xx}({\bf x})\gt0 \quad & \Longrightarrow \quad f({\bf x})\ \hbox{is a local min. value.} \cr
D\gt0 \hbox{ and  }f_{xx}({\bf x})\lt 0 \quad & \Longrightarrow\quad f({\bf x})\ \hbox{is a local max. value.}\cr
D\lt 0 \quad & \Longrightarrow\quad f({\bf x})\ \hbox{is a saddle point.}\cr
}
$$

Finding Global Extrema
1) Find the critical points of $f$ contained in the interior of its domain. Evaluate $f$ at each of these points.
2) Find the global maximum and minimum value of $f$ on the boundary of the domain of $f$.
If  the domain of $f$ is all of $\Bbb R^2$, skip this step.
3) Compare the values found in 1) and 2) and select the largest and smallest of these to obtain, respectively, the global maximun and minimum value of $f$.


text:
{\bf Problem.} Given that $x$ and $y$ are nonzero real numbers such that $x+\frac{1}{y}=10$ and $y+\frac{1}{x}=\frac{5}{12},$ find all possible values for $x.$

(Enter your answer as a comma-separated list.)
{\bf Level.} Level 3
{\bf Type.} Intermediate Algebra
{\bf Solution.} Multiplying the first equation by $y$ and the second equation by $x,$ we get \[\begin{aligned} xy+1 &= 10y, \\ xy + 1 &= \tfrac{5}{12} x. \end{aligned}\]Then $10y = \tfrac{5}{12}x,$ so $y = \tfrac{1}{10} \cdot \tfrac{5}{12} x = \tfrac{1}{24}x.$ Substituting into the first equation, we get \[x + \frac{1}{\frac{1}{24}x} = 10,\]or $x + \frac{24}{x} = 10,$ which rearranges to the quadratic $x^2 - 10x + 24 = 0.$ This quadratic factors as $(x-4)(x-6) = 0,$ so the possible values for $x$ are $\boxed{4, 6}.$ (These give corresponding $y$-values $y = \tfrac16, \tfrac14,$ respectively, which, we can check, are valid solutions to the original system of equations.)

set_name: MATH