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6

More generally, if we have two lines $A_1x+B_1y+C_1=0$ and $A_2x+B_2y+C_2=0,$ then they are perpendicular if and only if $$A_1A_2=-B_1B_2.$$ Now, if neither line is vertical or horizontal (that is, the $A_i$s and $B_i$s are non-zero), then we can divide by $B_1B_2$ and obtain the result that the product of their slopes is $-1.$ But if, say, $B_1=0$ (so that ...

4

You can use the formula (valid for $0<x<\pi$) $$\cos\Bigl(\frac{x}{2}\Bigr)=\sqrt{\frac{1+\cos x}{2}}$$ repeatedly. So, for the first terms, $$\cos(\pi/4)=\frac{\sqrt{2}}{2},\quad \text{and}\quad \cos(\pi/8)=\sqrt{\frac{1+\cos(\pi/4)}{2}}=\sqrt{\frac{2+\sqrt{2}}{4}}=\frac{\sqrt{2+\sqrt{2}}}{2}.$$ Do you recognize the first two factors? Do you see ...

3

In general, one can write a product of sums as a sum of a products: $$\left(\sum_{i \in I} x_i\right)\left(\sum_{j \in J} y_j \right) = \sum_{i \in I, j \in J} x_i y_j.$$ One cannot, however, in general reverse this process, that is, write a sum of products, $$\phantom{(\ast) \qquad} \sum_{k \in K} x_k y_k, \qquad (\ast)$$ as a product of sums. (In this ...

3

You can observe that $$\frac1{i(i+1)}=\frac1i-\frac1{i+1}\;,$$ so the series telescopes: $$\sum_{i=1}^n\frac1{i(i+1)}=\sum_{i=1}^n\left(\frac1i-\frac1{i+1}\right)=\frac11-\frac1{n+1}=\frac{n}{n+1}\;.$$ For the second problem, ...

3

$f(x)=\log\log x$ is a concave function on $[2,+\infty)$, since its derivative is $f'(x)=\frac{1}{x\log x}$ and its second derivative is $f''(x)=-\frac{1}{x^2}\left(\frac{1}{\log x}+\frac{1}{\log^2 x}\right)$. The Hermite-Hadamard inequality hence gives: $$\sum_{n=2}^{N}\log\log n \leq \int_{3/2}^{N+1/2}\log\log x\,dx = \left.\left(x\log\log ... 2 Using geometric and arithmetic mean and Stirlings formula you get$$ \left(\prod_{k=2}^n\ln(k)\right)^{1/(n-1)}\le\frac{\sum_{k=2}^n\ln(k)}{n-1} =\frac{\ln(n!)}{n-1}\le\frac{\ln(\sqrt{2\pi(n+1)})+n·(\ln(n)-1)}{n-1} $$2 We only have to consider the last 2 digit of the product. Hence we can do multiplication modulo 100 by hand:$$\begin{align}527356 \cdot 12456 &\equiv 56 \cdot 56 \pmod {100}\\&\equiv3136 \pmod {100}\\&\equiv 36 \pmod {100}\end{align}$$so that x = 6 and y = 3. 2 As noted in other answers the slope of a vertical line is not defined, so simply we can not speak of the product of the two slopes of the x and the y axis. If you want you can accord the fact that these two lines are orthogonal with the general rule m m'=-1 using a limit. Take a line y=mx amd a perpendicular line y=m'x=-\dfrac{1}{m}x, than the ... 2 Let \displaystyle P = \left(1+\frac{1}{5}\right)\cdot \left(1+\frac{1}{5^2}\right)\cdot \left(1+\frac{1}{5^4}\right)...........\left(1+\frac{1}{5^{64}}\right) Now Multiply both side by \displaystyle \left(1-\frac{1}{5}\right)\;, We get \displaystyle P\cdot \left(1-\frac{1}{5}\right) = \underbrace{\left(1-\frac{1}{5}\right)\cdot ... 2 Let (x,y) \in (C \times D) \cap (A \times B). Then (x,y) \in C \times D and (x,y) \in A \times B. That is x \in C and y \in D and x \in A and y \in B. Hence, x \in C \cap A and y \in D \cap B. Therefore C \cap A \not = \emptyset and D \cap B \not = \emptyset. Observe that this is true for any sets A,B,C and D as Milo Brandt ... 2 Your product can be easily rewritten as \dfrac B{A^2}, where A=\displaystyle\prod_{n=1}^\infty\bigg(1-\dfrac{x^2}{n^2}\bigg) with x=\dfrac16, and B=\displaystyle\prod_{n=2}^\infty\frac{n^3+1}{n^3-1}, which, believe it or not, is telescopic, and yields a rational number as result. (B has been evaluated on this site several times). So the ... 1 One can say that for x to +infinity on y=\sqrt{x^2-4x} that \sqrt{x^2-4x}=\sqrt{x^2(1-4/x^2)}=x\sqrt{1-4/x^2}=x because the square root term goes to 1. Therefore the slant asymptote is y=x This is an example of the incorrect use of limit laws since the first term, x has a limit that does not exist for x to infinity. There is a slant asymptote for ... 1 Maybe this is the kind of thing you are looking for: Define the sequence x_n=(-1)^n. "Therefore", we have that: 1=\lim x_n^2"="(\lim x_n)^2. "So", \lim x_n=1 or -1. Also, you could use x and \frac{1}{x} if they think that the argument "0.anything=0" is valid. 1 We can use the Euler-Maclaurin Sum Formula to get a fairly accurate value for this product. Using Taylor series gives$$ \begin{align} &\log\left(2-e^{1/n}\right)\\ &\small=-\frac1n-\frac1{n^2}-\frac1{n^3}-\frac{13}{12n^4}-\frac5{4n^5}-\frac{541}{360n^6}-\frac{223}{120n^7}-\frac{47293}{20160n^8}-\frac{36389}{12096n^9} \end{align} $$Applying the ... 1 You can prove something more general:$$\forall x\in\mathbb{R}\backslash\{1\},\,\forall n\in\mathbb{N},\,\prod_{i=0}^n\left(1+x^{2^i}\right)=\dfrac{x^{2^{n+1}}-1}{x-1}$$using some methods like induction, juantheron's answer,... 1 How is 0*\infty=-1? Is it really? No, 0 times \infty is not equal to -1. In fact, the product isn't even defined. It is not a question of this somehow giving a contradiction, it just isn't defined. The rule you are referring to says that: Given two lines with slopes m_1 and m_2 (real numbers) then the lines are perpendicular if and only if ... 1 Given 527356\times 12456 = Now Calculation for last 2 Digit of the Given product, We will calculate 56\times 56 = .....36 So we get x=6 and y=3. So we get x+y= 9 1 The space X=(\mathbb{Q}\times\mathbb{Q})/({\sim}_1\times{\sim}_2) is a quotient of a sequential space and hence sequential, so it suffices to show the space Y=(\mathbb{Q}/{\sim}_1)\times(\mathbb{Q}/{\sim}_2) is not sequential. Fix a sequence (\alpha_n) of positive irrational numbers converging to 0. For each n\in\mathbb{N}, choose a sequence of ... 1 Without really using induction, you can use telescoping:$$\prod_{i=2}^n \left(1-\frac{1}{i^2}\right)=\prod_{i=2}^n\frac{i^2-1}{i^2}=\prod_{i=2}^n\frac{(i-1)(i+1)}{i^2}=\frac{1\cdot 3}{2\cdot 2}\cdot \frac{2\cdot 4}{3\cdot 3}\cdot \frac{3\cdot 5}{4\cdot 4}\cdots \frac{(n-1)(n+1)}{n\cdot n}=\frac{1\cdot \not3}{2\cdot \not 2}\cdot \frac{\not ...

1

Induction Hypothesis: ${\displaystyle \prod_{i=2}^{i=n} (1-\frac{1}{i^2})} = \frac{n+1}{2n}$ Now, we need to prove it for $n+1$ case. ${\displaystyle \prod_{i=2}^{i=n+1} (1-\frac{1}{i^2})} = (1-\frac{1}{(n+1)^2}) *{\displaystyle \prod_{i=2}^{i=n} (1-\frac{1}{i^2})} = (1-\frac{1}{(n+1)^2})*\frac{n+1}{2n} = \frac{n^2 + 2n}{(n+1)(2n)} = \frac{n+2}{2(n+1)}$ ...

1

Using Abel's summation we have $$\sum_{n=2}^{N}\log\left(\log\left(n\right)\right)=\sum_{n=2}^{N}1\cdot\log\left(\log\left(n\right)\right)=\left(N-1\right)\log\left(\log\left(N\right)\right)-\int_{2}^{N}\frac{\left\lfloor t\right\rfloor -1}{t\log\left(t\right)}dt$$ where $\left\lfloor t\right\rfloor$ is the floor function and using $\left\lfloor ... 1 After working at this for a few days, I think I got a good proof. To start, the proof for the case of$N$being odd still holds. Rather than trying to prove the general case, I found a proof for the case of$N$being even. I will start with the result at the end of the question.$\$\sum_{i=1}^N \frac{x_i}{x_m} \prod_{j\ne i}|x_j-x_m| = \sum_{i=1}^N ...

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