91影视

• A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field.
• Its answer is related to a particle's probability density in space and time.

The probability of crossing is

$P=P\left(y\right)P(x)$

Where y and x are some parameters that we will use.

Suppose that the distance between the end of the needle that have a hole and the the closest line to it to bey , so that yhave the interval between 0 and 1 (i.e., l ), and the projection along some direction x is in the interval between -l and l (i.e., $-l鈮�x鈮�l$).

The condition of crossing a line that is above is $x+y鈮�l$(i.e., $x鈮�l-y$).

Where the condition of crossing line that is below is $x+y鈮�0$(i.e., $x鈮�-y$).

So, for a given value if y the probability of crossing (by make usage of problem 1.12 is

role="math" localid="1658552457032" $$\begin{gathered} P\left(x\right)={鈭�}_{-l}^{-y}p\left(x\right)dx+{鈭�}_{l-y}^{l}p\left(x\right)dx \\ ={鈭�}_{-l}^{-y}\frac{1}{蟺\sqrt{l^2-x^2}}dx+{鈭�}_{l-y}^l\frac{1}{蟺\sqrt{l^2-x^2}}dx \\ =\frac{1}{蟺}\left[{\sin }^{-1}{\overline{)\left(\frac{x}{l}\right)}}_{-l}^{-y}+{\sin }^{-1}{\overline{)\left(\frac{x}{l}\right)}}_{l-y}^l\right] \\ =\frac{1}{蟺}\left[-{\sin }^{-1}\left(\frac{y}{l}\right)+2{\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(\frac{l-y}{l}\right)\right] \\ P\left(x\right)=\frac{1}{蟺}\left[蟺-{\sin }^{-1}\left(\frac{y}{l}\right)-{\sin }^{-1}\left(\frac{l-y}{l}\right)\right]...\left(1\right) \end{gathered}$$

Now, using the normalizing condition we can find $蚁\left(y\right)$, where it is equal to $1/l$ because all the values of y are equally likely.

Thus eq. (1) become,

localid="1658553112340" $$\begin{gathered} P_{\mathrm{crossing}}=\frac{1}{蟺}{鈭�}_0^l\left[蟺-{\sin }^{-1}\left(\frac{y}{l}\right)-{\sin }^{-1}\left(\frac{l-y}{l}\right)\right]dy \\ =\frac{1}{蟺}{鈭�}_0^l\left[蟺-{\sin }^{-1}\left(\frac{y}{l}\right)\right]dy \end{gathered}$$

To integrate the second term of the integrand we use integration by parts, so we get

localid="1658552944578" $$\begin{gathered} P_{cros\sin g}=\frac{1}{蟺l}{\left[蟺l-2\left\{y{\sin }^{-1}\left(\frac{y}{l}\right)+l\sqrt{1-\frac{y^2}{l^2}}\right\}\right]}_0^l \\ =1-1+\frac{2}{蟺} \\ =\frac{2}{蟺} \end{gathered}$$

Therefore, the probability that the needle will cross a line is $2/蟺$which is equal to (approximately) $0.63662$.