It then follows
Use induction to prove
We give you a quiz; let 
be positive numbers and let 
be also positive.
It then follows
^{\frac{1}{p}} \leq (a_{1}^q + \cdots + a_{n}^q )^{\frac{1}{q}}. "(a_{1}^p + \cdots + a_{n}^p )^{\frac{1}{p}} \leq (a_{1}^q + \cdots + a_{n}^q )^{\frac{1}{q}}.")
It then follows
Levy flight
A random walk whose steps follow the law of stable distribution is called a Levy flight. I paste some hyperlinks of movies as usual. Please enjoy!
Synthetic Legal Precedent Structures: Levy Flight
Levy Flight Animation
Synthetic Legal Precedent Structures: Levy Flight
Levy Flight Animation
A formula in the combination theory
I was asked to show the following formula in the last part time job:
!} \sum _{j = 0}^{n - k} \frac{(-1)^j}{j!} = 1, \textrm{ for } l = 0, \ldots, n. "\sum_{k = l}^n \frac{1}{(k - l)!} \sum _{j = 0}^{n - k} \frac{(-1)^j}{j!} = 1, \textrm{ for } l = 0, \ldots, n.")
It took me one and a half hours to show this formula X(
}} \\ = & \displaystyle{\sum_{0 \leq k \leq n - l} \frac{1}{k!} \sum_{0 \leq j \leq n - k - l} \frac{(-1)^j}{j!} } \\ = & \displaystyle{\sum_{0 \leq k \leq N} \sum_{0 \leq j \leq N - k} \frac{(-1)^j}{k! j!}} \\ = & \displaystyle{\sum_{0 \leq m \leq N} \sum_{0 \leq j \leq N - m} \frac{(-1)^j}{(m - j)! j!}} \\ = & \displaystyle{\sum_{0 \leq m \leq N} \frac{1}{m!} \sum_{0 \leq j \leq N - m} \frac{m!}{(m - j)! j!}} (-1)^j \\ = & \displaystyle{1 + \sum_{1 \leq m \leq N} \frac{1}{m!} \{ 1 + (-1) \}^m } \\ = & \displaystyle{1} \\ \end{array} "\begin{array}{cl} & \displaystyle{\textrm{(L.H.S.)}} \\ = & \displaystyle{\sum_{0 \leq k \leq n - l} \frac{1}{k!} \sum_{0 \leq j \leq n - k - l} \frac{(-1)^j}{j!} } \\ = & \displaystyle{\sum_{0 \leq k \leq N} \sum_{0 \leq j \leq N - k} \frac{(-1)^j}{k! j!}} \\ = & \displaystyle{\sum_{0 \leq m \leq N} \sum_{0 \leq j \leq N - m} \frac{(-1)^j}{(m - j)! j!}} \\ = & \displaystyle{\sum_{0 \leq m \leq N} \frac{1}{m!} \sum_{0 \leq j \leq N - m} \frac{m!}{(m - j)! j!}} (-1)^j \\ = & \displaystyle{1 + \sum_{1 \leq m \leq N} \frac{1}{m!} \{ 1 + (-1) \}^m } \\ = & \displaystyle{1} \\ \end{array}")
We have replaced k - l by k in the first line and have changed how to take sums by the formula k + j = m in the fourth line. Finally, this formula enable us to compute
^j}{j!} = \begin{cases} 2 & \textrm{ if } n \geq 2 \\ 1 & \textrm{ if } n = 1. \\ \end{cases} "\sum_{k = 0}^n \frac{k^2}{k!} \sum _{j = 0}^{n - k} \frac{(-1)^j}{j!} = \begin{cases} 2 & \textrm{ if } n \geq 2 \\ 1 & \textrm{ if } n = 1. \\ \end{cases}")
It took me one and a half hours to show this formula X(
We have replaced k - l by k in the first line and have changed how to take sums by the formula k + j = m in the fourth line. Finally, this formula enable us to compute
Ito's formula
I have got two kinds of Ito's formula. One is the time-independent version of semimartingales and the other is the time-dependent version of Ito processes. Both of two sorts are recognized as analogues to second order Taylor expansion with integrals in calculus. However, it is very interesting only to subordinate any polynomial to the function. Because I have many martingales with stochastic integrals.
Binomial coefficients
A student came and asked me to compute the following finite series in part time job yesterday: let n be an arbitrary natural number, then show
} & \displaystyle{ \binom{n}{0} + \binom{n + 1}{1} + \cdots + \binom{n + k}{k} = \binom{n + k + 1}{k} } ; \\ \textrm{(ii)} & \displaystyle{ \binom{k}{k} + \binom{k + 1}{k} + \cdots + \binom{n}{k} = \binom{n + 1}{k + 1}, \textrm{ for } n \geq k. } \\ \end{array} "\begin{array}{cl} \textrm{(i)} & \displaystyle{ \binom{n}{0} + \binom{n + 1}{1} + \cdots + \binom{n + k}{k} = \binom{n + k + 1}{k} } ; \\ \textrm{(ii)} & \displaystyle{ \binom{k}{k} + \binom{k + 1}{k} + \cdots + \binom{n}{k} = \binom{n + 1}{k + 1}, \textrm{ for } n \geq k. } \\ \end{array}")
It is easy to show both (i) and (ii) by induction. However, some smart people know that a binomial polynomial expansion yields the results. I refer to only (i) and leave (ii) the readers as an exercise.
Note that
^{n + l} = \cdots + \binom{n + l}{l} x^n + \cdots, \mathrm{ for } 0 \leq l \leq k. } "\displaystyle{ (x + 1)^{n + l} = \cdots + \binom{n + l}{l} x^n + \cdots, \mathrm{ for } 0 \leq l \leq k. }")
Summing over,
^{n + l} = \cdots + \sum_{0 \leq l \leq k} \binom{n + l}{l} x^n + \cdots. } "\displaystyle{ \sum_{0 \leq l \leq k} (x + 1)^{n + l} = \cdots + \sum_{0 \leq l \leq k} \binom{n + l}{l} x^n + \cdots. }")
The left hand side in the above equality is
^{n} \{ 1 - (x + 1)^{k + 1} \} }{1 - (x + 1)} = \frac{(x + 1)^{n + k + 1} - (x + 1)^n}{x}. } "\displaystyle{ \frac{(x + 1)^{n} \{ 1 - (x + 1)^{k + 1} \} }{1 - (x + 1)} = \frac{(x + 1)^{n + k + 1} - (x + 1)^n}{x}. }")
The coefficient of x to the pover of n in the right hand side is equal to
It is easy to show both (i) and (ii) by induction. However, some smart people know that a binomial polynomial expansion yields the results. I refer to only (i) and leave (ii) the readers as an exercise.
Note that
Summing over,
The left hand side in the above equality is
The coefficient of x to the pover of n in the right hand side is equal to
My favorite formula
I had a mathematical chat with some friends in the last month. One of them said, "My favorite formula is partial fraction decomposition":
} = \frac{x - (x - 1)}{x(x - 1)} = \frac{1}{x - 1} - \frac{1}{x}, "\frac{1}{x(x - 1)} = \frac{x - (x - 1)}{x(x - 1)} = \frac{1}{x - 1} - \frac{1}{x},")
and the other one said, "I like the integration by parts formula":
 g^{'}(x) \textrm{d} x = f(x) g(x) - \int f^{'}(x) g(x) \textrm{d} x. "\int f(x) g^{'}(x) \textrm{d} x = f(x) g(x) - \int f^{'}(x) g(x) \textrm{d} x.")
These two formula have a common point: they follow from an idea
Indeed, the first formula obviously follows from this idea. The second formula also follows, since we have the differentiation of product-type functions from the definition of differentiation as follows.
 g(x) \}^{'} } \\ \notag =& \displaystyle{\lim_{h \to 0} \frac{f(x + h) g(x + h) - f(x) g(x)}{h}} \\ \notag =& \displaystyle{\lim_{h \to 0} \frac{f(x + h) g(x + h) - f(x + h) g(x) + f(x + h) g(x) - f(x) g(x)}{h}} \\ \notag =& \displaystyle{f (x) \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} + \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} g (x)} \\ \notag =& \displaystyle{f (x) g^{'} (x) + f^{'} (x) g (x)} \\ \notag \end{array} "\begin{array}{cl} & \displaystyle{ \{ f(x) g(x) \}^{'} } \\ \notag =& \displaystyle{\lim_{h \to 0} \frac{f(x + h) g(x + h) - f(x) g(x)}{h}} \\ \notag =& \displaystyle{\lim_{h \to 0} \frac{f(x + h) g(x + h) - f(x + h) g(x) + f(x + h) g(x) - f(x) g(x)}{h}} \\ \notag =& \displaystyle{f (x) \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} + \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} g (x)} \\ \notag =& \displaystyle{f (x) g^{'} (x) + f^{'} (x) g (x)} \\ \notag \end{array}")
My favorite mathematical formula is this simplest one:
^2 = a^2 - 2ab + b^2. "(a - b)^2 = a^2 - 2ab + b^2.")
Once we understand the true meaning of this formula, we immediately obtain many formulas including the relation between arithmetic means and geometric means, Schwarz's inequality and the upper bound of symmetric forms.
and the other one said, "I like the integration by parts formula":
These two formula have a common point: they follow from an idea
Indeed, the first formula obviously follows from this idea. The second formula also follows, since we have the differentiation of product-type functions from the definition of differentiation as follows.
My favorite mathematical formula is this simplest one:
Once we understand the true meaning of this formula, we immediately obtain many formulas including the relation between arithmetic means and geometric means, Schwarz's inequality and the upper bound of symmetric forms.
Binomial coefficient and imaginary number
In the last part time job, I find the problem that we compute the following sum of binominal coefficients.
^n \binom{n}{n} "\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots + (-1)^n \binom{n}{n}")
As you know, we can compute by using the binominal theorem. The answer is 0. The last result implies the following formula:
Here, 2j is the greatest number which is not more than n. Then, we can also compute:
Here, jk is the greatest integer which is not more than n. The computation of this is a littel tricky. But we can compute by the following binominal expansions.
^n, (1 + \omega )^n, \cdots, (1 + \omega^{k - 1} )^n "(1 + 1)^n, (1 + \omega )^n, \cdots, (1 + \omega^{k - 1} )^n")
Here,
. "\omega := \exp\left ( \frac{2 \pi }{k} i \right ).")
Indeed, since
l} = 0 \textrm{ for every } l \geq 1 \textrm{ such as } k \nmid l , "1 + \omega^l + \omega^{2l} + \cdots + \omega^{(k - 1)l} = 0 \textrm{ for every } l \geq 1 \textrm{ such as } k \nmid l ,")
summing each expansion, we have the result:
^n + (1 + \omega )^n + \cdots + (1 + \omega^{k - 1} )^n}{k}. "S_{n, k} = \frac{(1 + 1)^n + (1 + \omega )^n + \cdots + (1 + \omega^{k - 1} )^n}{k}.")
As you know, we can compute by using the binominal theorem. The answer is 0. The last result implies the following formula:
Here, 2j is the greatest number which is not more than n. Then, we can also compute:
Here, jk is the greatest integer which is not more than n. The computation of this is a littel tricky. But we can compute by the following binominal expansions.
Here,
Indeed, since
summing each expansion, we have the result:
Some useful exercises
You can easily find some exercise made by LaTeX. For example, if you would like to see some exercises of the theory of stochastic processes, search by
Try it later!
exercise stochastic process filetype:pdf
Try it later!
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