具體數學--(有限微分法)

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derivative and difference

• derivative $Df(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

• difference $\triangle f(x) = f(x+1)-f(x)$

operator:They operate on functions to give new functions.They are
functions of functions that produce functions.

x to the m falling and x to the m rising

• x to the m falling

  $x^{\underline{m}}=x(x-1)...(x-m+1), m\in Z$

• x to the m rising

  $x^{\overline{m}}=x(x+1)...(x+m-1), m\in Z$

example

• $n!=n(n-1)...1=n^{\underline{n}}=1^{\overline{n}}$

• $\triangle (x^{\underline{m}})=mx^{\underline{m-1}}$

important analog

$p(x+1)=p(x)$ leads to no change when you apply difference operator.That
is why we should use $C\equiv p(x+1)=p(x)$ for indefinite sum

An important fact is that

$\sum_a^bg(x)\delta x = \sum_{k=a}^{b-1}g(k)=\sum_{a\leqslant k<b}g(k), (a,b\in Z, b\geqslant a)$

Now, we will introduce an method which will help us to get closed form
of some sums. The path is

1. Some sums like the following formula.

   $\sum_{a\leqslant k<b}g(k)=\sum_{a}^{b}g(x)\delta x$

2. If we can find an indefinite sum or anti-difference function f such
   that $g(x)=f(x+1)-f(x)$ we can get closed form of the original
   question directly by indefinite sum
   $\mathbf{\sum_{a\leqslant k<b}g(k)=\sum_{a}^{b}g(x)\delta x=f(b)-f(a)}$

We have to complete this method into other defintion regions. such as
$b<a$

$\sum_a^bg(x)\delta x=f(b)-f(a)=-(f(a)-f(b))=-\sum_b^a g(x)\delta x$

It means that

$\sum_a^b+\sum_b^c=\sum_a^c$

the same as

$\int_a^b+\int_b^c=\int_a^c$

Try it

$\mathbf{\sum_{0\leqslant k<n} k^{\underline{m}}=\frac{k^{\underline{m+1}}}{m+1}|_0^n=\frac{n^{\underline{m+1}}}{m+1}}$

1. How to get the closed form of $\sum_{0\leqslant k<n}k$

   • what is the f(x) when $g(x)=x$ ?

     $g(x)=x=x^{\underline{1}} \Rightarrow f(x)=\frac{n^{\underline{2}}}{2}$

   • It means

     $\sum_{0\leqslant k<n}k = \frac{n^{\underline{2}}}{2}|_0^n=n(n-1)/2$

2. Let’s go to our question mentioned at previous section. How to get
   the closed form of $\sum_{k=0}^{n-1}k^2=\sum_{0\leqslant k<n}k^2$

   what is the $g(x)$? It is
   $k^2=k^{\underline{2}}+k^{\underline{1}}\Rightarrow f(x)=\frac{x^{\underline{3}}}{3}+\frac{x^{\underline{2}}}{2}$

   $\sum_{0\leqslant k<n}k^2=\frac{n^{\underline{3}}}{3}+\frac{n^{\underline{2}}}{2}=\frac{1}{3}n(n-\frac{1}{2})(n-1)$

3. $\sum_{0\leqslant k\leqslant n}k^3$

   • $g(k)=k^{\underline{3}}+3k^{\underline{2}}+k^{\underline{1}}$

   • $f(k)=\frac{k^{\underline{4}}}{4}+k^{\underline{3}}\frac{k^{\underline{2}}}{2}$

One more tip.

$(x+y)^{\underline{2}}=x^{\underline{2}}+2x^{\underline{1}}y^{\underline{1}}+y^{\underline{2}}$

ln(x)

We already have

$\left. \begin{aligned} x^{\underline{3}}&=x(x-1)(x-2)\\ x^{\underline{2}}&=x(x-1)\\ x^{\underline{1}}&=x\\ x^{\underline{0}}&=1 \end{aligned} \right.$

What is $x^{\underline{-1}}$?

$x^{\underline{-m}}=\frac{1}{(x+1)(x+2)...(x+m)}, m>0$

Let’s compare

$\left. \begin{aligned} x^{m+n}&=x^mx^n\\ x^{\underline{m+n}}&=x^{\underline{m}}(x-m)^{\underline{n}} \end{aligned} \right.$

Coming back to
$\sum_a^bm^{\underline{m}}\delta x=\frac{m^{\underline{m+1}}}{m+1}|_a^b,m\neq -1$

$\left. \begin{aligned} \int_a^bx^{-1}dx&=lnx|_a^b\\ x^{\underline{-1}}&=\frac{1}{x+1}=f(x+1)-f(x) \end{aligned} \right.$

It is not hard to see that

$f(x)=\frac{1}{1}+\frac{1}{2}+...+\frac{1}{x}$

It means that $H_x$ is the discrete analog of the continuous $lnx$

$\sum_a^b x^{\underline{m}}\delta x= \left\{ \begin{aligned} \frac{x^{\underline{m+1}}}{m+1}|_a^b, m\neq 1\\ H_x|_a^b,m=-1 \end{aligned} \right.$

$e^x$

$D(e^x)=e^x$
$\triangle f(x)=f(x+1)-f(x)=f(x) \Leftrightarrow f(x+1)=2f(x)$

It means that $f(x)=2^x$. $2^x$ is the discrete analog of continuous
$e^x$.

Let’s look at a more general case
$\triangle(c^x)=c^{x+1}-c^x=(c-1)c^x$

$\sum_{a\leqslant k<b}c^k=\sum_a^bc^x\delta x=\frac{c^x}{c-1}|_a^b=\frac{c^b-c^a}{c-1}, c\neq 1$

conclusion

More analog

Unfortunately, there is no nice chain rule for
$\triangle f(g(x))$.However, there is a nice analog rule for multiple
rule.

$D(uv)=uDv+vDu$

$\triangle (uv)=u\triangle v+v(x+1)\triangle u=u\triangle v+E(v)\triangle u$

$E(f(x))=f(x+1)$

$\sum u\triangle v=uv-\sum Ev\triangle u$

Example

1. $\left. \begin{aligned} \sum x2^x \delta x&=x2^x-\sum 2^{x+1}\delta x\\ &=x2^x-2^{x+1}+C \end{aligned} \right.$

   $\left. \begin{aligned} \sum_{k=0}^{n}k2^k&=\sum_{0}^{n+1}x2^x \delta x\\ &=x2^x-2^{x+1}|_0^{n+1}\\ &=(n-1)2^{n+1}+2 \end{aligned} \right.$

2. $\left. \begin{aligned} \sum xH_x\delta x&=\frac{x^2}{2}H_x-\sum \frac{(x+1)^{\underline{2}}}{2}x^{\underline{-1}}\delta x\\ &=\frac{x^{\underline{2}}}{2}H_x-\frac{1}{2}\sum x^{\underline{-1}}\delta x\\ &=\frac{x^{\underline{2}}}{2}H_x-\frac{x^{\underline{2}}}{4}+C \end{aligned} \right.$

   $\sum_{0\leqslant k<n} kH_k=\sum_0^nxH_x\delta x=\frac{n^{\underline{2}}}{2}(H_n-\frac{1}{2})$