Suppose we want to prove $n=n+1$ for all (positive) integers $n$. We omit the base case. The induction hypothesis is $k=k+1$ for some $k\in \mathbb N$. Adding $1$ to both sides gives $k+1=k+1+1$, or $(... View answer Accepted answer 16 votes We know that $$y=\sqrt{k\cdot\sqrt{k\cdot\sqrt{\dots}}}$$ We can square the equation and divide by$k$: $$\frac{y^2}k=\sqrt{k\cdot\sqrt{\dots}}$$ But the right hand side is just$y$again, so we ... View answer 10 votes The number of zeros at the and of$n!$is just the number of factors of$5$in$n!$(since the number of factors of$2$in$n!$is always larger than that). So, we need all$n$such that the numbers$...

In general, when $f$ and $g$ differ in (at most) a finite amount of points (in the interval $[a,b]$), we always have $$\int _a^bf(x)\,dx=\int_a^b g(x)\,dx$$ This follows from the definition using ...

The trace of a matrix $M$ is $0$ if and only if the sum of the elements on the (main) diagonal of $M$ is $0$. Since the dimension of all $n\times n$ matrices is $n^2$ and the dimension of its image $\... View answer Accepted answer 8 votes You can do this with the quadratic-arithmetic mean: (this is possible, because$a^2\geq 0$.) $$\sqrt{\frac{a^2+b^2}2}\geq\frac{a+b}2\\ a^2+b^2\geq \frac{(a+b)^2}2=8$$ Now, you only have to proof$\...

You can rewrite the sum to $$\frac {1}{x}\sum_{n=1}^\infty \frac{1}{(\frac n x)^2+1}$$ Because $f(n)=\frac1{n^2+1}$ is strictly decreasing for positive (reals) $n$, we know that \begin{align} \frac ...

For polynomials, the Taylor series around $x=0$ is just the polynomial itself. If you want the series around $x=a$, the general formula (for a function $f$ around $x=a$) $$f(a)+f'(a)(x-a)+\frac{f''(a)... View answer Accepted answer 6 votes Using the substitution a=\frac xy, b=\frac yz and c=\frac zx, we want to prove that$$ \sum_{cyc}\frac {xy}{z^2}\geq \frac xy+\frac yz+\frac zx $$where we are taking cyclic sums. Using AM-GM, ... View answer Accepted answer 6 votes We are given that ∠BCD=θ and ∠BAD=2θ. Also, BD is the angular bisector of ∠ABC. We draw the circle \Gamma with center A and radius |AD|. G is the intersection of \Gamma and AB such ... View answer Accepted answer 6 votes Actually, they do:$$ s=u^2\\ \sqrt s=u $$Differentiating:$$ ds=2u\,du\\ \frac{ds}{2\sqrt s}=du $$Now, we can substitute s=u^2 in the second equation:$$ du=\frac{ds}{2\sqrt u}=\frac{ds}{2u} $$... View answer Accepted answer 6 votes you know that \cos x^2+\sin x^2=1. Because of this, the polynomial can be written as$$ 2(1-\sin^2 x)+\sin x-1=-2\sin ^2 x+\sin x +1=(2\sin x +1)(-\sin x+1) $$EDIT When you do not want to use any ... View answer 5 votes This can be solved using the strategy used in this post: It comes down to calculating$$ \binom{5+3-1}{3}=\binom 73=\frac{7\cdot6\cdot5}{1\cdot2\cdot3}=35 $$What we are doing is the following: We ... View answer 5 votes HINT: Use the fact that$$ r (n+1-r) \geq n $$(quadratic polynomial in r with equality at r=1 and r=n) when 1\leq r\leq n. Then, take the product over r:$$ \left(n! \right)^2=\prod_{r=1}^n ...

Since we only need and evaluate $T(n)$ when $n$ is a power of $2$, say that $a(k)=T(2^k)$. The recursion becomes (I will use $n$ again from now on): $$a(n)=n2^n+\sum_{i=0}^{n-1}a(i)$$ Define $s(n)=\... View answer Accepted answer 5 votes You can use Geogebra, a free program that can be used to created all kinds of geometric images, including this one. First create a circle with center$C$and point on the circle$P$. Then make an ... View answer Accepted answer 5 votes Since$\binom{n}{k}=\frac{n!}{k!(n-k)!}$, we know that any prime$p$that divides$\binom nk$also divides$n!$(since dividing by integers can't add factors). Because$n!$contains only factors$\...

Suppose you want to take an even number of books out of $2n$ books with you on your holiday. The left hand side sums the number of possibilities over all possible amounts $2k$ to take with you. On the ...

We are going to show that both sides count the same number of possibilities and therefore have to be equal for the equation: $$\binom nr 2^r3^{n-r}=\sum_{k=r}^n \binom nk\binom kr2^k$$ Suppose there ...

First, pick any starting point $A$. (Suppose it is at the bottom of the circle.) This can be done in $(2n+1)$ ways. Number the points starting at $A=0$. Let the second point $B$ be at position $i\leq ... View answer Accepted answer 4 votes This is not true in general. Take the following graph$G$for example: We can see that the maximal independent set$M$consists of vertices$4$,$5$and$6$. (If one of the center vertices is in$M$, ... View answer 4 votes Suppose we have$n+2$books in a row numbered from$1$to$n+2$and we want to choose$m+2$of them. Obviously, we can do that in$\binom {n+2}{m+2}$ways, i.e. the right hand side of your equality. ... View answer 4 votes We know that$2^k\pmod n$can only take a finite number of different values ($n-1$to be precise, because$0$is not possible). Since$k\in \mathbb N$and$\mathbb N$is infinite, there must be some ... View answer Accepted answer 4 votes Since Javes has more than$40$books, he has at least$41$books. Thus, Aslam has at least$5\cdot 41=205$books. Since his number of books is a multiple of$4$and$5$, it is a multiple of$20$. Thus,... View answer Accepted answer 4 votes A$k$-degenerate graph needs only to have (at least) one vertex of degree at most$kin every subgraph, while the maximum degree is the largest degree of all vertices, which is completely unrelated. ... View answer 4 votes You can use that \begin{align} a^{k+1}-1&=a(a^k-1)+(a-1)\\ &=a(a-1)(a^k+\cdots+a+1)+(a-1)\\ &=(a-1)(a^{k+1}+\cdots+a^2+a)+(a-1)\\ &=(a-1)(a^{k+1}+\cdots+a+1) \end{align} assuming you ... View answer Accepted answer 4 votes Suppose thatP$and$Q$are both longest paths (and thus have equal length$l$). Assume that they have no vertices in common. There are now two vertex-disjoint paths. Because the graph is connected, ... View answer Accepted answer 4 votes We know that for any$x$(close to$0$): $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - O(x^6)\\ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - O(x^7)$$ We can find$\cos\sin x$by substituting$x\...

The distribution $X=\sum X$ is the Erlang distribution with parameters $\lambda$ and $n$. This distribution has the form $$f(x;n,\lambda)=\frac{\lambda^n x^{n-1} e^{-\lambda x}}{\Gamma(x)}$$ (...
Write $B=(b_1,\dots,b_n)$ and $X=(x_1,\dots,x_n)$ with $b_i$ and $x_i$ the columns of $B$ and $X$ respectively. I assume you are able to solve $Ax=b$. Now, solve the $n$ linear equations $Ax_1=b_1$ to ...