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I believe there is a sort of convention going around that no sentence should start with a mathematical symbol, even in a case like the example you provide. But as far as I know it's a typographical convention and not a semantic or grammatical one: if you start a sentence like $G$ is a simple group... then the typesetting can lead to some confusion ...

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For $x\in(0,\frac\pi 2)$ we have $$\cos x+\sin x-1=\cos^2\left(\frac x2\right)-\sin^2\left(\frac x2\right)+2\sin\left(\frac x2\right)\cos\left(\frac x2\right)-1\\=-2\sin^2\left(\frac x2\right)+2\sin\left(\frac x2\right)\cos\left(\frac x2\right)=2\sin\left(\frac x2\right)\left(\underbrace{\cos\left(\frac x2\right)-\sin\left(\frac ... 2 In my experience, there is no plan of making a proof bidirectional. Really, you just try to prove P\implies Q and see what happens. When you have your proof that P implies Q, you take a good look at it and try to reverse every step of it. The usual case is that you can reverse some of the steps, but not all of them, but sometimes you get lucky and can ... 2 When you get an expression of the form f(n)<\varepsilon, what you'd like to do is rewrite the inequality in terms of \varepsilon, that is, you'd like to write g(\varepsilon )<n, for some g. Then you take n a natural number that satisfies this inequality and you're done. Often times f(n) is not simple enough (sometimes even impossible) to ... 2 The usual induction approach, but you need something special at n=2. It is clearly true for n=1. Now prove n=2 from the triangle inequality. Then if it is true for k numbers, prove it is true for k+1 numbers by grouping the first k into one group and combine them with the last one using your proof of n=2 1 The event on the LHS in english can be described as "The event that either A but not B or B but not A occurs". This event is the union of two disjoint sets A \cap B^c and A^c \cap B. They are disjoint because$$(A\cap B^c) \cap(A^c \cap B)=(A\cap A^c)\cap (B\cap B^c)=\emptyset\cap\emptyset=\emptyset$$The probability of the union of two disjoint sets ... 1 As said in the above comments, the cited statements are meaningful as "advices" about proof-strategy ... nothing is deterministic in this topic. What is sound is the observation : if you are trying to prove P \equiv Q, it is wrong to start your write-up of the proof with the unjustified statement P \equiv Q... He says the (obvious but useful) ... 1 An example of what Velleman refers to might be the following: Claim \lambda  is an eigenvalue of the matrix A if and only if it solves the characteristic equation det(A-\lambda I)=0. Proof If \lambda is an eigenvalue of A then there is a x\neq0 such that Ax=\lambda x, implying that (A-\lambda I)x=0 some x\neq 0, implying that ... 1 Here's one way to look at this: Suppose \nabla is the unique covariant derivative operator associated with the metric tensor g on the surface S, i.e. it is the unique symmetric connection on S with \nabla g = 0. It is well know that the geodesics \gamma(t) of g satisfy \nabla_{\mathbf w}\mathbf w = 0, \tag{1} where \mathbf w is the ... 1 We are given a-b \equiv 0 \pmod {m_1 m_2}. This means that (a-b) = km_1 m_2, for a constant k We want to prove that (a - b) = j m_1, for some other constant j. It should be clear from here. 1 HINT 1 Plot the graphs of$$ y (x) = x \\ y(x) = e^{-x} $$See where/whether they intersect. HINT 2 As already commented; try to find two real points/values of x such that the sign for  x - e^{-x} changes. 1 For 0<x<\pi/2, we have 0<\sin x <1; thus$$\sin x>\sin^2 x,\text{ for }0<x<\pi/2.$$Similarly,$$\cos x>\cos^2 x,\text{ for }0<x<\pi/2.$$Thus, for 0<x<\pi/2,$$ \sin x+\cos x >\sin^2 x+\cos^2 x=1. $$You could also see why the result holds by considering a right triangle whose hypotenuse has unit length ... 1 This is my humble suggestion: Cantor's Diagonalisation Method I should probably provide a justification why it is controversial?? But I think this post sort of takes care of it.. 1 It is a good idea to draw the function first: From this we guess that the function is differentiable except at x=-1 and x=0. Your formula for f above is correct, and from this we see that for x \in \mathbb{R} \setminus \{-1,0\}, the function f is differentiable. For x=-1, we see that {f(-1+h)-f(-1) \over h} = 0 for h \in (0,1) and ... 1 I don't think there is a reasonable general answer to this very genuine phenomenon: that's just the way the (mathematical) world is. In a related area, most of modern cryptography (and thus of modern economics: think banking) relies on the fact that it is very easy to multiply two huge prime numbers p, q but extremely difficult, given just their product ... 1 This is a comment and not an answer to the post Just for your curiosity, the antiderivative is given without any restriction by$$\frac{x^{a+1} \, _2F_1\left(1,\frac{a+1}{b};\frac{a+1}{b}+1;-x^b\right)}{a+1}$$and the integral between 0 and \infty is given by$$\frac{\pi \csc \left(\frac{\pi (a+1)}{b}\right)}{b}  if $\Re(a-b)<-1\land ... 1 Another way: Count the number of paths of length$2n$on the integers taking$0$to itself. Since there need to be$n$lefts and$n$rights, the total number is${\displaystyle {2n \choose n}}$. The number of paths where the first$n$moves contains$k$rights and the second$n$moves contains$n-k$rights is${\displaystyle {n \choose k}{n \choose n - k} = ...

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