The original question was: Plait talks about the “physics of solid bodies” and why, specifically, if you were on the inside of the Sun falling in, the matter “behind” you- closer to the surface- doesn’t affect your acceleration at all, and all that matters is the matter “in front” of you- closer to the core. Why is that?
Physicist: The short, uninteresting answer is that the gravity from any layer above you cancels itself out. If you take any sample layer above you, and you happen to be closer to one side, then you’ll find that the side you’re closer to has more pull on you, but there’s less of it. Conversely, the far side has less pull, but there’s more of it. For a sphere (but not a ring) these forces cancel exactly. So as you fall in you can ignore all the layers above you.
Pick a layer. Anything inside will experience exactly the same amount of pull in every direction, and so, no pull at all.
Answer gravy: One of the greatest tools in the physicist’s tool kit is “Gaussian Surfaces“. They let you shortcut really difficult math problems using pictures and a little reasoning. Even better, you come across smarter than perhaps you deserve, which is a big plus.
A Gaussian surface is nothing more than an invisible bubble that you draw in space. The “inverse square law” of gravity can actually be rewritten as “the total amount of gravity pointing into the bubble is proportional to the amount of matter inside the bubble”. The arrangement of matter (both inside and outside the bubble) certainly changes how gravity points into (or out) of the bubble, but the total amount of gravity pointing through depends only on the amount of matter inside.
(upper left) when the mass is symmetrical and in the middle, then the gravity is exactly the same everywhere on the surface. (upper right) if the matter is off to the side, then gravity will be stronger, or weaker, or point in different directions at different points on the surface, but the total stays the same. (bottom) matter outside of the Gaussian surface can affect how gravity pokes through, but it can't affect the total.
Now say that your bubble is an exact fit around a sphere of matter. Everything is perfectly symmetric, so there’s no reason for gravity to be any stronger or weaker anywhere and, given the amount of mass inside the sphere, you can figure out how strong the gravity is. Now say you add more matter, but uniformly, on top of your original sphere.
In both situations the total amount of matter inside the bubble is the same, and everything is nice and symmetrical, so the gravity along the surface of the bubble is the same.
The matter inside the sphere has remained the same, so the pull at the surface of that sphere remains the same. As a result, so long as the matter above is at least fairly symmetrical (which is the case for any planet or star you can think of), you can ignore the layers above the surface of the bubble.
Specifically, as you fall farther and farther into the Sun (or Earth, or whatever else is round) you can figure out how much gravity you’re feeling by using a Gaussian surface, for which you only need the matter below you. The layers above will exert no pull on you, and you will exert no net pull on them (for every action/force there is an equal and opposite reaction/force).
This part has nothing to do with the question: You can use Gaussian surfaces to prove some surprising things. Specifically: Dyson spheres work, and black holes have no more gravity than the stars they came from.
From the last argument (above) you know that the layers above you have no net gravitational effect on you. But what if you fall a little way into a planet, and suddenly find that the inside of it is completely hollow? One you’re inside all the layers are layers above you. So there’s no gravity at all inside of a large hollow sphere (at least, none caused by the sphere). If you built a really huge sphere around a star you’d have a “Dyson’s sphere”. The sphere doesn’t pull the star, and the star doesn’t pull the sphere. It’s stable no matter where the star is inside the ball. So long as no one shoves anything, everything will just float neutrally right where it is.
(left) the set up for a Dyson Sphere. The perfectly spherical shell has no gravitational effect on anything inside the sphere, including the star. (center) Miles Dyson, inventor of the Dyson sphere, and future inventor of Skynet. (right) an artist's interpretation of a Dyson sphere.
Now, put a Gaussian surface around a star. There’s a certain amount of matter in the star, and that tells you how much gravity is pointing through the surface. If the star shrinks, who cares? Same amount of mass = same amount of gravity.
The gravity through the outer Gaussian surface stays the same, since both contain the same amount of matter. The gravity through the inner Gaussian surface increases dramatically after the star collapses, because it contains all of the star's mass, instead of just a small part of it.
But if you draw a small Gaussian surface around the core of the star you’ll find that the gravity along the surface is small, because there is (relatively) little mass inside of it. If for some reason you found yourself in the center of the Sun, you’d be floating in zero G’s. Point of fact; you’d also be on fire.
Now when the Star collapses, all of the matter is drawn into a tiny region. Both spheres (see diagram on the right) now contain all of the star’s matter, and thus the same total amount of gravity pokes through them. The only difference is that the inner sphere is smaller, so the gravity has to be more intense to get the same total as the outer sphere.
Black holes do have very intense gravity, but only in the region where the star used to be.
for the position of a particle, then you can find the the momentum wave function, 
, by taking the 
. Now, you can define the uncertainty as the standard deviation of the probability function, which is a really good way to go about it.
, and the square of its Fourier transform, 
, is always greater than some constant. In what follows you’ll find some useful stuff such as 
. A physicist would recognize this as 
. The difference comes about because the Fourier transform that takes you from the position wave function to the momentum wave function involves an 
, and 
. (For the physicists out there who were wondering what happened to their precious h’s)
is the 
for 
which makes the seemingly infinite sums always have only a finite number of terms. In fact, for N and K fixed, the answer is a polynomial with respect to the variable p.
” reads “not heads equals tails”![\begin{array}{ll}&P(E_{j+1})\\i)&=P(E_{j+1}\cap E_j\cap A_j)+P(E_{j+1}\cap E_j\cap \overline{A_j})+P(E_{j+1}\cap \overline{E_j}\cap A_j)+P(E_{j+1}\cap \overline{E_j}\cap \overline{A_j})\\ii)&=\left[P(E_{j+1}\cap E_j\cap A_j)+P(E_{j+1}\cap E_j\cap \overline{A_j})\right]+P(E_{j+1}\cap \overline{E_j}\cap A_j)+P(E_{j+1}\cap \overline{E_j}\cap \overline{A_j})\\iii)&=\left[P(E_{j+1}\cap E_j)\right]+P(E_{j+1}\cap \overline{E_j}\cap A_j)+P(E_{j+1}\cap \overline{E_j}\cap \overline{A_j})\\iv)&=P(E_j)+P(E_{j+1}\cap \overline{E_j}\cap A_j)+P(E_{j+1}\cap \overline{E_j}\cap \overline{A_j})\\v)&=P(E_j)+P(E_{j+1}\cap \overline{E_j}\cap A_j)+0\\vi)&=P(E_j)+P(E_{j+1}|\overline{E_j}\cap A_j)P(\overline{E_j}\cap A_j)\\vii)&=P(E_j)+pP(\overline{E_j}\cap A_j)\\viii)&=P(E_j)+pP(\overline{E_j}| A_j)P(A_j)\\ix)&=P(E_j)+pP(\overline{E_j}| A_j)qp^{K-1}\\x)&=P(E_j)+qp^KP(\overline{E_{j-k}})\\xi)&=P(E_j)+qp^K\left[1-P(E_{j-k})\right]\\xii)&=P(E_j)+qp^K-qp^KP(E_{j-k})\end{array}](http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7D%26P%28E_%7Bj%2B1%7D%29%5C%5Ci%29%26%3DP%28E_%7Bj%2B1%7D%5Ccap%20E_j%5Ccap%20A_j%29%2BP%28E_%7Bj%2B1%7D%5Ccap%20E_j%5Ccap%20%5Coverline%7BA_j%7D%29%2BP%28E_%7Bj%2B1%7D%5Ccap%20%5Coverline%7BE_j%7D%5Ccap%20A_j%29%2BP%28E_%7Bj%2B1%7D%5Ccap%20%5Coverline%7BE_j%7D%5Ccap%20%5Coverline%7BA_j%7D%29%5C%5Cii%29%26%3D%5Cleft%5BP%28E_%7Bj%2B1%7D%5Ccap%20E_j%5Ccap%20A_j%29%2BP%28E_%7Bj%2B1%7D%5Ccap%20E_j%5Ccap%20%5Coverline%7BA_j%7D%29%5Cright%5D%2BP%28E_%7Bj%2B1%7D%5Ccap%20%5Coverline%7BE_j%7D%5Ccap%20A_j%29%2BP%28E_%7Bj%2B1%7D%5Ccap%20%5Coverline%7BE_j%7D%5Ccap%20%5Coverline%7BA_j%7D%29%5C%5Ciii%29%26%3D%5Cleft%5BP%28E_%7Bj%2B1%7D%5Ccap%20E_j%29%5Cright%5D%2BP%28E_%7Bj%2B1%7D%5Ccap%20%5Coverline%7BE_j%7D%5Ccap%20A_j%29%2BP%28E_%7Bj%2B1%7D%5Ccap%20%5Coverline%7BE_j%7D%5Ccap%20%5Coverline%7BA_j%7D%29%5C%5Civ%29%26%3DP%28E_j%29%2BP%28E_%7Bj%2B1%7D%5Ccap%20%5Coverline%7BE_j%7D%5Ccap%20A_j%29%2BP%28E_%7Bj%2B1%7D%5Ccap%20%5Coverline%7BE_j%7D%5Ccap%20%5Coverline%7BA_j%7D%29%5C%5Cv%29%26%3DP%28E_j%29%2BP%28E_%7Bj%2B1%7D%5Ccap%20%5Coverline%7BE_j%7D%5Ccap%20A_j%29%2B0%5C%5Cvi%29%26%3DP%28E_j%29%2BP%28E_%7Bj%2B1%7D%7C%5Coverline%7BE_j%7D%5Ccap%20A_j%29P%28%5Coverline%7BE_j%7D%5Ccap%20A_j%29%5C%5Cvii%29%26%3DP%28E_j%29%2BpP%28%5Coverline%7BE_j%7D%5Ccap%20A_j%29%5C%5Cviii%29%26%3DP%28E_j%29%2BpP%28%5Coverline%7BE_j%7D%7C%20A_j%29P%28A_j%29%5C%5Cix%29%26%3DP%28E_j%29%2BpP%28%5Coverline%7BE_j%7D%7C%20A_j%29qp%5E%7BK-1%7D%5C%5Cx%29%26%3DP%28E_j%29%2Bqp%5EKP%28%5Coverline%7BE_%7Bj-k%7D%7D%29%5C%5Cxi%29%26%3DP%28E_j%29%2Bqp%5EK%5Cleft%5B1-P%28E_%7Bj-k%7D%29%5Cright%5D%5C%5Cxii%29%26%3DP%28E_j%29%2Bqp%5EK-qp%5EKP%28E_%7Bj-k%7D%29%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\begin{array}{ll}&P(E_{j+1})\\i)&=P(E_{j+1}\cap E_j\cap A_j)+P(E_{j+1}\cap E_j\cap \overline{A_j})+P(E_{j+1}\cap \overline{E_j}\cap A_j)+P(E_{j+1}\cap \overline{E_j}\cap \overline{A_j})\\ii)&=\left[P(E_{j+1}\cap E_j\cap A_j)+P(E_{j+1}\cap E_j\cap \overline{A_j})\right]+P(E_{j+1}\cap \overline{E_j}\cap A_j)+P(E_{j+1}\cap \overline{E_j}\cap \overline{A_j})\\iii)&=\left[P(E_{j+1}\cap E_j)\right]+P(E_{j+1}\cap \overline{E_j}\cap A_j)+P(E_{j+1}\cap \overline{E_j}\cap \overline{A_j})\\iv)&=P(E_j)+P(E_{j+1}\cap \overline{E_j}\cap A_j)+P(E_{j+1}\cap \overline{E_j}\cap \overline{A_j})\\v)&=P(E_j)+P(E_{j+1}\cap \overline{E_j}\cap A_j)+0\\vi)&=P(E_j)+P(E_{j+1}|\overline{E_j}\cap A_j)P(\overline{E_j}\cap A_j)\\vii)&=P(E_j)+pP(\overline{E_j}\cap A_j)\\viii)&=P(E_j)+pP(\overline{E_j}| A_j)P(A_j)\\ix)&=P(E_j)+pP(\overline{E_j}| A_j)qp^{K-1}\\x)&=P(E_j)+qp^KP(\overline{E_{j-k}})\\xi)&=P(E_j)+qp^K\left[1-P(E_{j-k})\right]\\xii)&=P(E_j)+qp^K-qp^KP(E_{j-k})\end{array}")
. If you have a run of K heads in the first j trials, of course you’ll have a run in the first j+1 trials. v) The zero comes from the fact that if the first j terms 
and 
).
![\begin{array}{ll}&S(N,K)\\i)&=S(N-1,K)+qp^K-qp^KS(N-K-1,K)\\ii)&=\left[S(N-2,K)+qp^K-qp^KS(N-K-2,K)\right]+qp^K-qp^KS(N-K-1,K)\\iii)&=S(N-2,K)+2qp^K-qp^K\left[ S(N-K-1,K)+S(N-K-2,K)\right]\\iv)&=S(N-3,K)+3qp^K-qp^K\left[ S(N-K-1,K)+S(N-K-2,K)+S(N-K-3,K)\right]\\v)&=S(N-(N-K),K)+(N-K)qp^K-qp^K\left[ S(N-K-1,K)+\cdots+S(N-K-(N-K),K)\right]\\vi)&=S(K,K)+(N-K)qp^K-qp^K\left[ S(N-K-1,K)+\cdots+S(0,K)\right]\\vii)&=p^K+(N-K)qp^K-qp^K\sum_{r=0}^{N-K-1} S(r,K)\\viii)&=p^k+(N-K)qp^K-qp^K\sum_{r=K}^{N-K-1} S(r,K)\end{array}](http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7D%26S%28N%2CK%29%5C%5Ci%29%26%3DS%28N-1%2CK%29%2Bqp%5EK-qp%5EKS%28N-K-1%2CK%29%5C%5Cii%29%26%3D%5Cleft%5BS%28N-2%2CK%29%2Bqp%5EK-qp%5EKS%28N-K-2%2CK%29%5Cright%5D%2Bqp%5EK-qp%5EKS%28N-K-1%2CK%29%5C%5Ciii%29%26%3DS%28N-2%2CK%29%2B2qp%5EK-qp%5EK%5Cleft%5B%20S%28N-K-1%2CK%29%2BS%28N-K-2%2CK%29%5Cright%5D%5C%5Civ%29%26%3DS%28N-3%2CK%29%2B3qp%5EK-qp%5EK%5Cleft%5B%20S%28N-K-1%2CK%29%2BS%28N-K-2%2CK%29%2BS%28N-K-3%2CK%29%5Cright%5D%5C%5Cv%29%26%3DS%28N-%28N-K%29%2CK%29%2B%28N-K%29qp%5EK-qp%5EK%5Cleft%5B%20S%28N-K-1%2CK%29%2B%5Ccdots%2BS%28N-K-%28N-K%29%2CK%29%5Cright%5D%5C%5Cvi%29%26%3DS%28K%2CK%29%2B%28N-K%29qp%5EK-qp%5EK%5Cleft%5B%20S%28N-K-1%2CK%29%2B%5Ccdots%2BS%280%2CK%29%5Cright%5D%5C%5Cvii%29%26%3Dp%5EK%2B%28N-K%29qp%5EK-qp%5EK%5Csum_%7Br%3D0%7D%5E%7BN-K-1%7D%20S%28r%2CK%29%5C%5Cviii%29%26%3Dp%5Ek%2B%28N-K%29qp%5EK-qp%5EK%5Csum_%7Br%3DK%7D%5E%7BN-K-1%7D%20S%28r%2CK%29%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\begin{array}{ll}&S(N,K)\\i)&=S(N-1,K)+qp^K-qp^KS(N-K-1,K)\\ii)&=\left[S(N-2,K)+qp^K-qp^KS(N-K-2,K)\right]+qp^K-qp^KS(N-K-1,K)\\iii)&=S(N-2,K)+2qp^K-qp^K\left[ S(N-K-1,K)+S(N-K-2,K)\right]\\iv)&=S(N-3,K)+3qp^K-qp^K\left[ S(N-K-1,K)+S(N-K-2,K)+S(N-K-3,K)\right]\\v)&=S(N-(N-K),K)+(N-K)qp^K-qp^K\left[ S(N-K-1,K)+\cdots+S(N-K-(N-K),K)\right]\\vi)&=S(K,K)+(N-K)qp^K-qp^K\left[ S(N-K-1,K)+\cdots+S(0,K)\right]\\vii)&=p^K+(N-K)qp^K-qp^K\sum_{r=0}^{N-K-1} S(r,K)\\viii)&=p^k+(N-K)qp^K-qp^K\sum_{r=K}^{N-K-1} S(r,K)\end{array}")
![\begin{array}{ll}i)&S(N,K)=p^K+(N-K)qp^K-qp^K\sum_{r=K}^{N-K-1} S(r,K)\\ii)&=p^K+(N-K)qp^K-qp^K\sum_{r=K}^{N-K-1} \left[p^k+(r-K)qp^K-qp^K\sum_{\ell=K}^{r-K-1} S(\ell,K)\right]\\iii)&=p^K+(N-K)qp^K-qp^K\sum_{r=K}^{N-K-1}p^k-qp^K\sum_{r=K}^{N-K-1}(r-K)qp^K+\left(qp^K\right)^2\sum_{r=K}^{N-K-1}\sum_{\ell=K}^{r-K-1} S(\ell,K)\\iv)&=p^K+(N-K)qp^K-qp^K\sum_{r=1}^{N-2K}p^k-\left(qp^K\right)^2\sum_{r=0}^{N-2K-1}r+\left(qp^K\right)^2\sum_{r=2K+1}^{N-K-1}\sum_{\ell=K}^{r-K-1} S(\ell,K)\\v)&=p^K+(N-K)qp^K-p^k(N-2K)qp^K-\left(qp^K\right)^2\frac{(N-2K)(N-2K-1)}{2}+\left(qp^K\right)^2\sum_{r=2K+1}^{N-K-1}\sum_{\ell=K}^{r-K-1} S(\ell,K)\\vi)&=p^K+{N-K\choose 1}qp^K-p^K{N-2K\choose 1}qp^K-{N-2K\choose 2}\left(qp^K\right)^2+\left(qp^K\right)^2\sum_{r=2K+1}^{N-K-1}\sum_{\ell=K}^{r-K-1} S(\ell,K)\end{array}](http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7Di%29%26S%28N%2CK%29%3Dp%5EK%2B%28N-K%29qp%5EK-qp%5EK%5Csum_%7Br%3DK%7D%5E%7BN-K-1%7D%20S%28r%2CK%29%5C%5Cii%29%26%3Dp%5EK%2B%28N-K%29qp%5EK-qp%5EK%5Csum_%7Br%3DK%7D%5E%7BN-K-1%7D%20%5Cleft%5Bp%5Ek%2B%28r-K%29qp%5EK-qp%5EK%5Csum_%7B%5Cell%3DK%7D%5E%7Br-K-1%7D%20S%28%5Cell%2CK%29%5Cright%5D%5C%5Ciii%29%26%3Dp%5EK%2B%28N-K%29qp%5EK-qp%5EK%5Csum_%7Br%3DK%7D%5E%7BN-K-1%7Dp%5Ek-qp%5EK%5Csum_%7Br%3DK%7D%5E%7BN-K-1%7D%28r-K%29qp%5EK%2B%5Cleft%28qp%5EK%5Cright%29%5E2%5Csum_%7Br%3DK%7D%5E%7BN-K-1%7D%5Csum_%7B%5Cell%3DK%7D%5E%7Br-K-1%7D%20S%28%5Cell%2CK%29%5C%5Civ%29%26%3Dp%5EK%2B%28N-K%29qp%5EK-qp%5EK%5Csum_%7Br%3D1%7D%5E%7BN-2K%7Dp%5Ek-%5Cleft%28qp%5EK%5Cright%29%5E2%5Csum_%7Br%3D0%7D%5E%7BN-2K-1%7Dr%2B%5Cleft%28qp%5EK%5Cright%29%5E2%5Csum_%7Br%3D2K%2B1%7D%5E%7BN-K-1%7D%5Csum_%7B%5Cell%3DK%7D%5E%7Br-K-1%7D%20S%28%5Cell%2CK%29%5C%5Cv%29%26%3Dp%5EK%2B%28N-K%29qp%5EK-p%5Ek%28N-2K%29qp%5EK-%5Cleft%28qp%5EK%5Cright%29%5E2%5Cfrac%7B%28N-2K%29%28N-2K-1%29%7D%7B2%7D%2B%5Cleft%28qp%5EK%5Cright%29%5E2%5Csum_%7Br%3D2K%2B1%7D%5E%7BN-K-1%7D%5Csum_%7B%5Cell%3DK%7D%5E%7Br-K-1%7D%20S%28%5Cell%2CK%29%5C%5Cvi%29%26%3Dp%5EK%2B%7BN-K%5Cchoose%201%7Dqp%5EK-p%5EK%7BN-2K%5Cchoose%201%7Dqp%5EK-%7BN-2K%5Cchoose%202%7D%5Cleft%28qp%5EK%5Cright%29%5E2%2B%5Cleft%28qp%5EK%5Cright%29%5E2%5Csum_%7Br%3D2K%2B1%7D%5E%7BN-K-1%7D%5Csum_%7B%5Cell%3DK%7D%5E%7Br-K-1%7D%20S%28%5Cell%2CK%29%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\begin{array}{ll}i)&S(N,K)=p^K+(N-K)qp^K-qp^K\sum_{r=K}^{N-K-1} S(r,K)\\ii)&=p^K+(N-K)qp^K-qp^K\sum_{r=K}^{N-K-1} \left[p^k+(r-K)qp^K-qp^K\sum_{\ell=K}^{r-K-1} S(\ell,K)\right]\\iii)&=p^K+(N-K)qp^K-qp^K\sum_{r=K}^{N-K-1}p^k-qp^K\sum_{r=K}^{N-K-1}(r-K)qp^K+\left(qp^K\right)^2\sum_{r=K}^{N-K-1}\sum_{\ell=K}^{r-K-1} S(\ell,K)\\iv)&=p^K+(N-K)qp^K-qp^K\sum_{r=1}^{N-2K}p^k-\left(qp^K\right)^2\sum_{r=0}^{N-2K-1}r+\left(qp^K\right)^2\sum_{r=2K+1}^{N-K-1}\sum_{\ell=K}^{r-K-1} S(\ell,K)\\v)&=p^K+(N-K)qp^K-p^k(N-2K)qp^K-\left(qp^K\right)^2\frac{(N-2K)(N-2K-1)}{2}+\left(qp^K\right)^2\sum_{r=2K+1}^{N-K-1}\sum_{\ell=K}^{r-K-1} S(\ell,K)\\vi)&=p^K+{N-K\choose 1}qp^K-p^K{N-2K\choose 1}qp^K-{N-2K\choose 2}\left(qp^K\right)^2+\left(qp^K\right)^2\sum_{r=2K+1}^{N-K-1}\sum_{\ell=K}^{r-K-1} S(\ell,K)\end{array}")
and 
you can carry out the process a couple more times, and you’ll find that:![\begin{array}{l}S(N,K)=p^K\left[1-{N-2K\choose 1}qp^K+{N-3K\choose 2}\left(qp^K\right)^2\cdots\right]+\left[{N-K\choose 1}qp^K-{N-2K\choose 2}\left(qp^K\right)^2+{N-3K\choose 3}\left(qp^K\right)^3\cdots\right]\\=p^K\sum_{T=0}^\infty {N-(T+1)K\choose T}(-qp^K)^T-\sum_{T=1}^\infty {N-TK\choose T}(-qp^K)^T\end{array}](http://s.wordpress.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bl%7DS%28N%2CK%29%3Dp%5EK%5Cleft%5B1-%7BN-2K%5Cchoose%201%7Dqp%5EK%2B%7BN-3K%5Cchoose%202%7D%5Cleft%28qp%5EK%5Cright%29%5E2%5Ccdots%5Cright%5D%2B%5Cleft%5B%7BN-K%5Cchoose%201%7Dqp%5EK-%7BN-2K%5Cchoose%202%7D%5Cleft%28qp%5EK%5Cright%29%5E2%2B%7BN-3K%5Cchoose%203%7D%5Cleft%28qp%5EK%5Cright%29%5E3%5Ccdots%5Cright%5D%5C%5C%3Dp%5EK%5Csum_%7BT%3D0%7D%5E%5Cinfty%20%7BN-%28T%2B1%29K%5Cchoose%20T%7D%28-qp%5EK%29%5ET-%5Csum_%7BT%3D1%7D%5E%5Cinfty%20%7BN-TK%5Cchoose%20T%7D%28-qp%5EK%29%5ET%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\begin{array}{l}S(N,K)=p^K\left[1-{N-2K\choose 1}qp^K+{N-3K\choose 2}\left(qp^K\right)^2\cdots\right]+\left[{N-K\choose 1}qp^K-{N-2K\choose 2}\left(qp^K\right)^2+{N-3K\choose 3}\left(qp^K\right)^3\cdots\right]\\=p^K\sum_{T=0}^\infty {N-(T+1)K\choose T}(-qp^K)^T-\sum_{T=1}^\infty {N-TK\choose T}(-qp^K)^T\end{array}")
. If this does not happen, then at least one tail must occur within the first K coin flips. Let’s suppose that j is the position of the first tail, and by assumption it satisfies 
. Then, the probability of having K or more heads in a row in the entire set of coins (given that the first tail occurred at 
) is simply the probability of having K or more heads in a row in the set of coins following the jth coin (since there can’t be a streak of K or more heads starting before the jth coin due to j being smaller or equal to K). But this probability of having a streak of K or more after the jth coin is just S(N-j,K). Now, since the probability that our first tail occurs at position j is the chance that we get j-1 heads followed by one tail, so it is 
. That means that the chance that the first tail occurs on coin j AND there is a streak of K or more heads is given by 
. Hence, the probability that the first K coins are all heads, OR coin one is the first tails and the remainder have K or more heads in a row, OR coin two is the first tails and the remainder have K or more heads in a row, OR coin three is the first tails and…, is given by:
and 
(click to see it enlarged):
we have in a particle’s position and the uncertainty 
we have in the particle’s momentum cannot both be very small at the same time. In particular, the product of these uncertainties is greater than a constant (
). This implies a fundamental limit to the knowledge that is possible because we can know 
accurately or 
accurately, but not both.
and 
, and the difference in longitudes is 
, then the true arcangle between them is:
)
, where 
![E[A_0B_{22.5}]=P(same)-P(different)=\cos^2{(22.5)}-(1-\cos^2{(22.5)})=\frac{1}{\sqrt{2}}](http://s.wordpress.com/latex.php?latex=E%5BA_0B_%7B22.5%7D%5D%3DP%28same%29-P%28different%29%3D%5Ccos%5E2%7B%2822.5%29%7D-%281-%5Ccos%5E2%7B%2822.5%29%7D%29%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D&bg=ffffff&fg=000000&s=0 "E[A_0B_{22.5}]=P(same)-P(different)=\cos^2{(22.5)}-(1-\cos^2{(22.5)})=\frac{1}{\sqrt{2}}")
![E[A_0B_{22.5}]+E[A_{45}B_{22.5}]+E[A_{45}B_{67.5}]-E[A_0B_{67.5}]=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{-1}{\sqrt{2}}=2\sqrt{2}](http://s.wordpress.com/latex.php?latex=E%5BA_0B_%7B22.5%7D%5D%2BE%5BA_%7B45%7DB_%7B22.5%7D%5D%2BE%5BA_%7B45%7DB_%7B67.5%7D%5D-E%5BA_0B_%7B67.5%7D%5D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%2B%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%2B%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D-%5Cfrac%7B-1%7D%7B%5Csqrt%7B2%7D%7D%3D2%5Csqrt%7B2%7D&bg=ffffff&fg=000000&s=0 "E[A_0B_{22.5}]+E[A_{45}B_{22.5}]+E[A_{45}B_{67.5}]-E[A_0B_{67.5}]=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{-1}{\sqrt{2}}=2\sqrt{2}")
! But that’s a violation of Bell’s inequality! But the existence of each measurement (whether or not you actually do that measurement) is all you need for Bell’s inequality! So if the inequality is false, then the result of those measurements 
. Do this twice and you get 
. So if you measure in the 0° direction to find A0, then go back and change the angle by 1° and repeat this until you’re measuring at 90°, then:
.
where x and y are real numbers, and i is the square root of -1. We can then think of z as a point on the (x,y) plane with x being the position along the horizontal (i.e. real) axis, and y the position along the vertical (i.e. imaginary) axis. To understand the operation of adding and multiplying complex numbers though, we need to think about them slightly differently.
represents not the point (x,y) but rather, the directed line segment which extends from (0,0) in the complex plane to (x,y) in the complex plane (unless otherwise stated we will assume here that our vectors are emanating from the origin (0,0)). You can think of a vector as simply representing a magnitude (or length) and a direction. The line segment extending from (1,2) to (-4,5) is the same as the one extending from (-4,5) to (1,2), but the vectors represented would be different because they would be pointing in opposite directions. Vectors are often drawn as arrows.
.
is the vector in the complex plane that extends from (0,0) to 
. Note that 
, we can also piggy back z1 at the end of z2 to get the same result. A slightly different view is achieved if we think of our first number 
as being a vector, and our second number 
as being a point. In this case, 
of z satisfies
.
is a complex number since it is the sum of a real and imaginary number. In particular, we have
. In our case though, since 
and 
this gives:
.
. This provides a new way of thinking about a complex number: as a vector specified by its length and angle. What happens when we multiply two such numbers 
and 
together? Well, we have
.
is a new complex number representing a vector of length 
pointed at angle 
. Therefore, we can think of the action of 
and rotate by an angle 
.
.
= 
= 
= 
= 
+ 
+ … + 
+ 
