91影视

• A differential equation that uses the wave-like characteristics of particles in a field to describe matter in terms of quantum mechanics. The probability density of a particle in space and time is relevant to the response.
• The time-dependent Schr枚dinger equation is represented as

_{$\mathit{i}\sout{\mathbf{h}}\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}\left|蠄\left(t\right)>=\hat{H}\right|\mathit{蠄}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{>}$}

For $N=1$, balls can be put any of d baskets

So, there are d ways.

For $N=2$

1) We can put two balls in any of d baskets ways$鈬�dways$

2) We can put one ball in one of baskets, and the other in one of the remaining $d-1$

Because balls are identical, we must divide by$2鈬�\frac{d(d-1)}{2}$

Total number:

$d+\frac{d(d-1)}{2}=\frac{d(d+1)}{2}$ways

1) We can put three balls in one ofbaskets$鈬�d$.

2) We can put one in one basket, and two in one of remaining $d-1鈬�d(d-1)$

3) We can put one ball in one of baskets, second in one of the remaining $d-1$, and third in one of remaining $d-2$.

Because balls are identical, we must divide by $3!鈬�\frac{d(d-1)(d+2)}{3!}$

Total number:

$d+d(d-1)+\frac{d(d-1)(d-2)}{6}=d^2+\frac{d(d^2-3d+2)}{6}=\frac{d(d+1)(d+2)}{6}$

For $N=4$

1) We can put four balls in one of d baskets $鈬�d$.

2) We can put one in one basket, and three in one of remaining$d-1鈬�d(d-1)$

3) We can put two balls in one basket, and two in one of the remaining$d-1鈬�\frac{d(d-1)}{2}$

4) We can put two balls in one basket, one in one of the remaining$d-1$, and last one in one of the remaining$d-2鈬�\frac{d(d-1)(d-2)}{2}$

5) All four balls are in different baskets.$鈬�\frac{d(d-1)(d-2)(d-3)}{4!}$

Total number of balls

$$\begin{gathered} d+\frac{d(d-1)}{2}+\frac{d(d-1)(d-2)}{6}+\frac{d(d-1)(d-2)(d-3)}{24}=\frac{d(d+1)(d+2)(d+3)}{24} \\ =d^2+\frac{d{(d-1)}^2}{2}+\frac{d(d-1)(d-2)(d-3)}{24} \\ =\frac{d}{24}\left(24d+12{(d-1)}^2+(d-1)(d-2)(d-3)\right) \\ =\frac{d}{24}\left(12d^2+12+d^3-5d^2+6d-d^2+5d-6\right) \\ =\frac{d}{24}(d^3+6d^2+11d+6) \\ =\frac{d(d+1)(d+2)(d+3)}{24} \end{gathered}$$ So, for number of balls in d_thebasket is

localid="1658227901414" $$\begin{gathered} f(N,d)=\frac{d(d+1)(d+2)+..............(d+N-1)}{N!} \\ f(N,d)=\frac{(d+N-1)!}{N!(d-1)!} \\ f(N,d)=\frac{(d+N)!}{(n+1)!(d-1)!} \end{gathered}$$

By using induction

$\mathrm{N}=0鈬�\mathrm{f}(0,\mathrm{d})=1$

We assume that following expression is valid:$\mathrm{f}(\mathrm{N},\mathrm{d})=\frac{(\mathrm{d}+\mathrm{N}-1)!}{\mathrm{N}!(\mathrm{d}-1)!}$.

We check for$N+1$:

$$\begin{gathered} \left(\begin{array}{c}\mathrm{d}+\mathrm{N}\\ \mathrm{N}+1\end{array}\right)=\left(\begin{array}{c}\mathrm{d}+\mathrm{N}-1\\ \mathrm{N}\end{array}\right)+\left(\begin{array}{c}\mathrm{d}+\mathrm{N}-1\\ \mathrm{N}+1\end{array}\right) \\ =\frac{(\mathrm{d}+\mathrm{N}-1)!}{\mathrm{N}!(\mathrm{d}-1)!}+\frac{(\mathrm{d}+\mathrm{N}-1)!}{(\mathrm{N}+1)!(\mathrm{d}-2)} \\ =\frac{(\mathrm{d}+\mathrm{N}-1)!}{\mathrm{N}!(\mathrm{d}-1)(\mathrm{d}-2)!}+\frac{(\mathrm{d}+\mathrm{N}-1)!}{(\mathrm{N}+1)\mathrm{N}!(\mathrm{d}-2)!} \\ =\frac{(\mathrm{d}+\mathrm{N}-1)!}{\mathrm{N}!(\mathrm{d}-2)!}\left(\frac{1}{\mathrm{d}-1}+\frac{1}{\mathrm{N}+1}\right) \\ =\frac{(\mathrm{d}+\mathrm{N}-1)!}{\mathrm{N}!(\mathrm{d}-2)!}\frac{\mathrm{N}+\mathrm{d}}{(\mathrm{d}+1)(\mathrm{N}+1)} \\ =\frac{(\mathrm{d}+\mathrm{N})!}{(\mathrm{N}+1)!(\mathrm{d}-1)!} \end{gathered}$$

Therefore the derived expression is$(N,d)=\frac{(d+N-1)!}{N!(d-1)!}$ .