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## Hot answers tagged proof-writing

17

$3^{2m+1}-3=3(3^m-1)(3^m+1)$, and both of the factors $3^m \pm 1$ are even, so their product is divisible by 4.

12

Define $f(q)=(1+q)(1+q^2)(1+q⁴)...(1+q^{{2}^{n}})$, We calculate \begin{align*} (1-q)f(q)&= (1-q)(1+q)(1+q^2)(1+q⁴)...(1+q^{{2}^{n}}) \\ &=(1-q^2)(1+q^2)(1+q^{2^2})...(1+q^{{2}^{n}})\\ &=(1-q^{2^2})(1+q^{2^2})...(1+q^{{2}^{n}})\\ &\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \vdots \\ &=1-q^{2^{n+1}} \end{align*} Now ...

10

Hint: $1+2q+3q^2+\ldots+nq^{n-1}= (q+q^2+\ldots+q^{n})'$

9

You can prove this using induction. Our trivial base case is $n=2$. Now, assume that the statement is true for $n-1$ We can split a $2^n\times 2^n$ board into four $2^{n-1} \times 2^{n-1}$ boards, as shown above. W.L.O.G. the square that we choose to take out is in the upper right quadrant. Since a $2^{n-1} \times 2^{n-1}$ board can be filled with Ls ...

8

The maximal power of $2$ which divides $n!$ is $$v_2(n!)=\lfloor \frac n2 \rfloor+\lfloor \frac n4 \rfloor +\cdots=\sum_{i=1}^{\infty} \lfloor\frac n{2^i}\rfloor$$ We bound this from above by dropping the floor function to see that $$v_2(n!)<n\sum_{i=1}^{\infty} \frac 1{2^i}=n$$

7

Base Case: For $n = 1$, we have: $$1 \times 5 = 5 = \frac{1}{6}(1)(1 + 1)(2(1) + 13)$$ which works. Inductive Hypothesis: Assume that the claim holds for $n' = n - 1$, where $n \geq 2$. It remains to show that the claim holds for $n' = n$. Indeed, observe that: \begin{align*} &1 \times 5 +\cdots + n(n + 4) & \\ &= [1 \times 5 + \cdots + (n - ...

6

It is obviously true for $2\times 2$. As an advice, when you see $2^n$ try to divide it into two parts consisting of $2^{n-1}$ and use induction. In your case, divide the whole board into $4$ $2^{n-1}\times 2^{n-1}$ boards. Edit: Without loss of generality, suppose we remove a square in the up-right quadrant. Then, on the up-right we have $2^{n-1}\times 2^... 6 Perhaps more than a bit of overkill, but you could get a generating function for all the integrals$J_n = \int_{-3}^3 (x^2-3)^n \; dx: \eqalign{g(t) &= \sum_{n=0}^\infty t^n J_n\cr & = \int_{-3}^3 \sum_{n=0}^\infty t^n (x^2-3)^n\; dx \cr &= \int_{-3}^3 \dfrac{dx}{1 + 3 t - t x^2}\cr &= \dfrac{2}{\sqrt{3t^2+t}} \text{arctanh}\left( \dfrac{... 6 There's actually the stronger result3^{2m +1} - 3 = 3(9^m - 1) = 24\sum_{i=1}^m 9^{m-i}$$5 Let f: \mathbb{R} \rightarrow [-1,1], f(x)=\sin(x). Then let g(x)=x, h(x)=x+2 \pi. f is surjective and f \circ h= f \circ g, but we clearly don't have h=g 5 For \theta in \mathbb{R} define I(\theta) = \int_{-3}^{3}{({x^2-3\theta})^3 dx}. Note that the integrand is a polynomial in \theta of degree 3 and because you are integrating in x, then I(\theta) is a polynomial in \theta of degree less than or equal to 3. Because of this, I(\theta) agrees with its Taylor polynomial of order 3 centered ... 5 A proof by induction has two parts, and you have to prove each of them: (i)\quad A(n) implies A(n+1), for all natural numbers n; (ii)\quad A(1) is true. Together, these two facts imply that A(n) is true for all natural numbers n. For example, to see A(3) is true (given these two facts), we argue as follows: We know A(1) is true by (... 4 Here is the easy way. Travel along the unit circle x units in the positive direction (counterclockwise). You are at angle x radians. The coordinates of the point are (\cos(x), \sin(x)). The distance from the point to the x-axis is \sin(x) for 0 < x < \pi; this is shorter than the distance along the circular arc, which is x. ... 4 You can use an easy corollary of the Mean value theorem: Let f and g be continuous functions on an interval [a,b], differentiable on (a,b). If f(a)\le g(a) and f'(x)<g'(x) on (a,b), then f(x)<g(x) for all x \in (a,b). Note that actually, it is enough to prove it for the interval (0,\frac\pi2). 4 You can take a natural number N_0 larger than the candidate real N. Larger n only makes \frac{1}{n} smaller anyway. For the second one, it suffices to consider neighbourhoods of the form [0,a), even (why?). 4 Your idea sounds good. I don't know where you got stuck, but here's something that should help you for a proof by induction : You can prove that if v_1,\dots,v_n are linearly dependent, then v_1,\dots,v_{n-1} are linearly dependent. Indeed, suppose$$\alpha_1v_1+\dots+\alpha_n v_n=0\tag{A}\label{A}$$with \alpha_i\neq 0. Applying \mathbb{L}, you get ... 4 No, this is not correct: you’ve proved that if \sum_{n\ge 1}a_n^2 converges absolutely, then it converges, which is not at all what you’re supposed to be doing. You have assumed precisely the result that you are supposed to be proving, and you’ve ignored the hypothesis. You should be assuming that \sum_{n\ge 1}a_n converges absolutely and using that ... 4$$\int_{-3}^{3}(x^2-3)^3\,dx = 81 \int_{-1}^{1}(3u^2-1)^3\,du \stackrel{IBP}{=}81\left(16-18\int_{-1}^{1}x^2(3x^2-1)^2\,dx\right)\tag{1}$$Now we may exploit the fact that the last integral is a L^2 norm. Since, in terms of Legendre polynomials:$$ x(3x^2-1) = \frac{4}{5}\,P_1(x) + \frac{6}{5}\,P_3(x) \tag{2}$$we have:$$ \int_{-1}^{1} x^2(3x^2-1)^2\,dx ... 4 There are generally several ways to approach these types of divisibility problems. I am showing two of them. By Mathematical Induction: Puttingm=0$we have$3^{2m+1}-3=0$, which is obviously divisible by$4$. Now let$4\mid 3^{2m+1}-3$for some$m\in N_0$. Let$3^{2m+1}-3=4k$, for some$k\in N_0$. Now$3^{2(m+1)+1}-3=9(3^{2m+1}-3)+24=9\cdot 4k+24=4(9k+6)$,... 4 The identity is false at$x=\pi/6.$Left side$-\sqrt{3}/2,$right side$-3/2.$Added: Suppose we want the left side to come out$-\tan 2x \cdot G,$for some$G$or other. After division of the sides by$\tan 2x$this gives$G=1-\cos 2x.$So as stated the identity would only hold provided$1-\cos 2x=\sin 2x.The latter is not an identity, one could solve ... 4 \begin{align} S&=1+2q+3q^2+\qquad\cdots\qquad \qquad+nq^{n-1}\\ qS&=\qquad q+2q^2+3q^3+\cdots +\quad(n-1)q^{n-1}+nq^n \\ \text{Subtracting,}&\\ (1-q)S&=1+\;\ q \ +\ q^2 +\ q^3+\cdots \qquad \qquad +q^{n-1}-nq^n\\ &=\frac {\;\ 1-q^n}{1-q}-nq^n\\ S&=\frac{1-q^n-nq^n(1-q)}{(1-q)^2}\\ &=\frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2}\qquad\... 4 It is not true that 6 points guarantees qualification. Consider the scenario where three teams achieve 2 wins and 1 loss and the fourth team loses all their games: Team A beats teams B and D, loses to C Team B beats teams C and D, loses to A Team C beats teams A and D, loses to B Team D loses to teams A,B,C. Then teams A, B, and C all have 6 points but ... 4 Suppose you draw a right triangle such that the two legs have lengths a and b respectively. This is clearly possible, since one can construct line segments of any integer length and can construct right angles. Now, the hypotenuse has to have some length c, and this c must satisfya^2+b^2=c^2$$by the Pythagorean theorem. However, there is exactly ... 4 Yes, the triangle , \{ (0, 0), (0, b), (a, 0) \} . 4 The irrationality of \sqrt{2}^{\sqrt{2}} is a trivial consequence of the Gelfond-Schneider theorem. 4 I think it is much clearer to stipulate that U is an open set. Expressing it as an element of the topology does not increase the clarity of what you are saying. 4 This is clear: the components of f are continuous. Now use the universal property of product spaces. 3 By Legendre's theorem$$\nu_2(n!) = \sum_{m\geq 1}\left\lfloor\frac{n}{2^m}\right\rfloor\leq \sum_{m\geq 1}\frac{n}{2^m}=n.$Equality may hold only if$2^k\parallel n$, but in such a case the$(k+1)$-th terms of the central sums fullfill a strict inequality. 3 To show that the set is countable, you need to put all its elements into a list. Look at how you would do this for$Q$itself, and (here's a hint) think about how to do this for$Q \times Q$. 3 You can define an (obvious) injection$\mathcal{B}\to\Bbb Q\times\Bbb Q$so that$\mathcal B$is in bijection with a subset of$\mathbb{Q}\times\Bbb{Q}$, which is countable because$\Bbb Q\$ is countable.

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