# Tag Info

8

Starting equation is equivalent to this one: $$\left(\frac{2011}{2014}\right)^x + \left(\frac{2012}{2014}\right)^x + \left(\frac{2013}{2014}\right)^x =1.$$ Denote $a=\frac{2011}{2014}, b=\frac{2012}{2014}, c=\frac{2013}{2014}$. Function $f(x)=a^x+b^x+c^x-1$ is continuous and decreasing (since $\ln a$, $\ln b$,$\ln c$ are negative): $$f'(x) = a^x\ln a ... 7 The logarithm of the expression under the limit can be rewritten as$$\sum_{k=0}^{2^{2^n-1}}\ln\left(1+\frac{1}{2^{2^n}+2k}\right)=\sum_{k=0}^{2^{2^n-1}}\frac{1}{2^{2^n}+2k}+O\left(2^{-2^n}\right).$$Denoting N=2^{2^n-1}, it is easy to see that the limit of the logarithm can be computed as the limit of a Riemann sum: ... 6$$2^n+2^n=2 \cdot 2^n=2^{n+1}$$6 We distribute just as we would distribute$$(a + b)(c + d) = a(c+d) + b(c + d) = ac + ad + bc + bd.(2x^3 - x)(\color{blue}{\sqrt x + 2x^{-1}}) = 2x^3(\color{blue}{\sqrt x + 2x^{-1}}) - x(\color{blue}{\sqrt x + 2x^{-1}})\\ =2x^3\sqrt x + 4x^3x^{-1} - x\sqrt x - 2xx^{-1}\\ = 2x^{6/2}x^{1/2} + \frac {4x^3}{x} - x^{2/2}x^{1/2} - \frac {2x}{x}\\ = 2x^{6/2 ...

6

You can factor $x^4+1 = (x^2 + \sqrt{2} x + 1) (x^2 -\sqrt{2}x + 1)$. This allows you to write the integrand as $\frac{a_1 x + b_1}{x^2 + \sqrt{2}x + 1} + \frac{a_2 x + b_2}{x^2- \sqrt{2}x + 1}$. You would then rewrite the denominator in the form of $(x-u)^2 + v$ and rewrite the numerator as $a_i (x-u) + w$, from which you can do a change of variable to ...

6

Hint: \eqalign{e^{4x}+3e^{2x}-28=0&\implies (e^{2x})^2+3(e^{2x})-28=0\\&\implies e^{2x}=\ldots\\&\implies x\:\:\:=\ldots} And I think what "$e$ to the $x$ squared" means is $(e^{x})^2$, in this case it is equal to $e^{2x}$ since one have $(a^{m})^n=a^{mn}.$

5

A rotation matrix $R$ is orthogonally diagonalizable with eigenvalues of absolute value one, i.e., $$R=U^* D U,$$ where $D=\mathrm{diag}(d_1,\ldots,d_n)$, with $\lvert d_j\rvert=1$, for all $j=1,\ldots,n$, and $U^*U=I$. Clearly, as $\lvert d_j\rvert=1$, there exists a $\vartheta_j\in\mathbb R$, such that $$d_j=\mathrm{e}^{i\vartheta_j}, \quad ... 5 The answer can't be zero solutions. Let f(x)=2011^x+2012^x+2013^x-2014^x Obviously f(x) is continuous (sum of continuous functions), and f(0)=2. But what about the behavior as x\to\infty? 4 By Laguerre's extension of Descartes' rule of sign, The number of real roots of Dirichlet polynomials of the form$$\alpha_1 \beta_1^x + \alpha_2 \beta_2^x + \cdots + \alpha_n \beta_n^x \quad\text{ subject to }\quad \begin{cases}\alpha_1, \ldots, \alpha_n \ne 0\\\beta_1 > \beta_2 > \cdots \beta_n > 0\end{cases} $$is no more than number of ... 4$$\log \left(1+\tfrac{1}{2^{2^n}}\right)\left(1+\tfrac{1}{2^{2^n}+2}\right)\cdots\left(1+\tfrac{1}{2^{ 2^n+1}}\right) = \log \left(1+\tfrac{1}{2^{2^n}}\right)+ \log \left(1+\tfrac{1}{2^{2^n}+2}\right)+\cdots + \log\left(1+\tfrac{1}{2^{ 2^n+1}}\right)$$Expanding on Alex' idea let t = 2^{2^n} which is getting very large. Notice we get only even numbers: ... 4$$5^x=2 \cdot 3^x \Rightarrow \left (\frac{5}{3} \right )^x=2 \Rightarrow \log_{\frac{5}{3}} \left (\frac{5}{3} \right )^x= \log_{\frac{5}{3}} 2 \Rightarrow x= \log_{\frac{5}{3}} 2 $$4 If we try with$$ \lim_{x\to 0} \frac{\log(a^x+b^x)-\log 2}{x} $$and apply l'Hôpital's theorem, we get$$ \lim_{x\to 0}\frac{a^x\log a+b^x\log b}{a^x+b^x}=\frac{\log a+\log b}{2}= \log\sqrt{ab}. $$It's just the derivative of x\mapsto (a^x+b^x)/2 at 0, of course. However,$$ \lim_{x\to0}\frac{\log(1+x)}{x}=1 $$so that$$ \lim_{x\to 0} ...

4

The order of operations is a convention that give us what operation should we do before. In the given examples there are three operations (given in the priority):operation in parenthesis, power,multiplication and subtraction so we have $$-2^2=-4\leftarrow\text{the power has the priority before the subtraction}$$ $$(-2)\cdot(-2)=4\leftarrow\text{the ... 4 Of course it is not a coincidence. Historically, you first define b^n for n a positive integer, and you go from there. Thing is that with this approach you need quite an insight to define exponentiation for all positive real numbers, and to deduce the existence and properties of the exponential function. You then define the logarithm as its inverse, and ... 3 There are infinitely many solutions (x,y) to your problem, all of them must satisfy$$x^2-79y^2=1 \iff x^2 = 1+79y^2 \iff |x| = \sqrt{1+79y^2} \\\iff x = \sqrt{1+79y^2} \quad \text{ or } \quad x = -\sqrt{1+79y^2} .$$And so you can write the set S of all the solutions (x,y) as follow$$S = \left\{(\sqrt{1+79t^2},t): t \in \mathbb{R}\right\} \cup ...

3

The Pell equation $$x^2 - Ny^2 = 1$$ (where $N$ is not a square) always has an infinite number of integer solutions, since there infinite units in the ring $\mathbb{Z}[\sqrt{N}]$. The "least" non-trivial solution can be found by expanding $\sqrt{N}$ into a continued fraction: the Legendre's theorem grants that such an expansion is periodic and the least ...

3

Because $-2^2$ means that you take the number $2$, raise it to the second power ($^2$), and then you consider its additive inverse ($-$). So, $2$ raised to the second power is $4$, whose additive inverse is $-4$. This is because exponentiation has a higher priority and it is the first thing you have to do; hence you compute the power before doing anything ...

3

First use simple fact, that $\displaystyle\lim_{y \to 0}\frac{\ln(1+y)}{y}=1$, so: $$\lim_{x \to 0}\frac{\ln(\frac{a^x+b^x}{2}-1+1)}{y}=\lim_{x \to 0}\frac{\ln(\frac{a^x+b^x}{2}-1+1)}{\frac{a^x+b^x}{2}-1} \cdot \lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{y}=\\=1 \cdot \lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{x}$$ Now $\lim_{x \to ... 3$2^n + 2^n = (1+1)*2^n = (2^1)*(2^n) = 2^{(1+n)}$The properties used here were$a^1=a$and$a^{(b+c)} = (a^b)*(a^c)$.$a^n+a^n = a^{(n+1)}$only if$a = 2$(or$0$, if$n>0$), but$a*a^n = a^{(n+1)}$for all$a$and$n$. 3 It is possible that the book intended to write the expression $$\displaystyle\frac{1}{\frac{1}{\frac{1}{5^{-2}}}}$$ This evaluates to $$\displaystyle\frac{1}{\frac{1}{\frac{1}{5^{-2}}}}=\frac{1}{\frac{1}{5^2}}=5^2=25$$ 3 This kind of equation does not show solutions which could be expressed analytically and only numerical methods, such as Newton, should be used. But before, you could notice that this function varies extremely fast and, in opinion, it should be better to rewrite the equation as $$\log(2011^x+2012^x+2013^x)=\log(2014^x)$$ If there were only one term, it will ... 2$\sum_{s=0}^{r-1} e^{\frac{-2 \pi i s k}{r}}$equals$0$unless$k=0$, in which case it equals$r$. So your expression is equal to $$\frac{1}{r} R_N(x^0) r = R_N(1) = 1$$ 2 The explanation is that the book is in error (assuming the question is accurately stated). $$\Large{\frac{\frac{1}{1}}{\frac{1}{5^{-2}}}}=\Large{\frac{1}{\frac{1}{5^{-2}}}}=\frac{1}{5^{2}}=\frac{1}{25}$$ 2 In the general case we have $$f(x)=g(x)+o(g(x)),\quad \text{when}\; x\to a\in\overline{\Bbb R}\iff f(x)\sim_a g(x)$$ which's equivalent to (the case$a=\infty$isn't very different) $$\forall \epsilon>0,\;\exists \delta>0,\;|x-a|<\delta\implies |f(x)-g(x)| <\epsilon|g(x)|$$ and in the given case we have simply ... 2 The simple answer to the second part of the question is no. For example,$3^n + 3^n \ne 3^{n+1}.$For any positive integer$m$, however, $$\underbrace{m^n + m^n + \ldots + m^n}_{m \text{ terms}} = m(m^n) = m^{n+1}.$$ 2 I think other answers are overcomplicated: just take logarithms to your favourite base to get $$x\log 5=\log 2+x\log 3$$This is a linear equation for$x$, which you should be able to solve. That is equivalent to what others have said, but the first step is easier. 2 No. If$|z|<1$the limit is$0$, if$|z|>1$the limit is$\infty$or does not exist since$|z|^n\to\infty$. If$|z|=1$then$z=e^{i\theta}$for$\theta\in[0,2\pi)$and$z^n=e^{in\theta}$. Unless$\theta=0$, i.e.$z=1$, the sequence will rotate the point on the unit circle by$\theta$counterclockwise at each step, and there is no limit. 2 It is true that$-2^2$is ambiguous (unless you know the convention), because$-(2^2) \ne (-2)^2$. (And, indeed, some calculators or programing languages may do it using their own convention, different from the mathematicians' convention.) The mathematicians' convention is$-2^2 = -(2^2)$. Why? Presumably because we often need to write$-(2^2)$. But if ... 2 Note that $$|x^3 - y^3| = |x^3 + (-y^3)| \leq |x^3| + |(-y^3)| = |x^3| + |y^3|$$ The$\leq$part comes from the triangle inequality ($|x+y| \leq |x| + |y|$) 2 For a continuous random variable$X$, the median is sometimes defined as the value$x$such that$\Pr[X \le x] = \Pr[X > x]= \dfrac{1}{2}$. So, if we assume that$A,B$are i.i.d.$\text{Uniform}[0,1]$, and we let$X = A^B$, then the median$x$satisfies:$\dfrac{1}{2} = \Pr[X \ge x] = \Pr[A^B \ge x] = \Pr[B \le \dfrac{\ln x}{\ln A}] = ...

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