Easy way to remember Taylor Series for log (1+x)? Assuming $|x| lt;1$, if one can easily remember that $$ \\dfrac{1}{1-x}=\\sum_{n=0}^{\\infty}x^{n} $$ then one can derive the following \\begin{eqnarray*} \\mbox{log
Simplify K! (K+1)! - Mathematics Stack Exchange Pretend K=3 That means (K+1)= 4 This means you'd be dividing 3*2*1 by 4*3*2*1 Consider how you'd cancel out multiples by dividing them Like how (2 (5+x)) 2 would just equal 5+x Following that idea we'd pretty much be able to cancel out every number in the numerator, so long as its also in the denominator This would end up canceling every number except for 4 which equals (K+1) Essentially
Diffeomorphism of $\\mathbb{C}P^1$ and $S^2$. CP1 C P 1 $\mathbb{C}{P}^{1}$ can be canonically identified with the Alexandrov compactification of C C $\mathbb{C}$, which is the Riemann sphere, a natural identification of that with S2 S 2 ${S}^{2}$ is given by stereographic projection
Normalize values to sum 1 but keeping their weights 40 10 I need to format these numbers so that they form the sum 1, but they should keep their "weight" In this specific case: 40 would become 0 80 10 would become 0 2 But if I have more values (like 40, 10, 25, 5 for example), I am really lost because I don't know the formula
Binomial expansion of $(1-x)^n$ - Mathematics Stack Exchange I'm not sure how appropriate it is to answer questions this old, but compared to the methods above, I feel the easiest way to see the answer to this question is to take a = -x And substitute that into the binomial expansion: (1+a)^n This yields exactly the ordinary expansion Then, by substituting -x for a, we see that the solution is simply the ordinary binomial expansion with alternating
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