# Tag Info

11

Short answer: It depends on your convention and how you define exponents. Long answer: There are a number of ways of defining exponents. Usually these definitions coincide, but this is not so for $0^0$: some definitions yield $0^0=1$ and some don't apply when both numbers are zero (leaving $0^0$ undefined). For example, given nonnegative whole numbers $m$ ...

10

Note: This post contains only numerics; see the answers by robjohn and Jack D'Aurizio for actual math. Numerical computations show that $$k(n) = \lfloor n\log 2-0.039721\rfloor \quad \text{ for } \ 2\le n<944029$$ which is kind of nice. However, this empirical formula breaks down for $n=944029$. What is worse, it turns out there is no $c$ such that ...

10

The first thought that comes into my head is to write $$e - \sum_{k=0}^n \frac{1}{k!} = \sum_{k=n+1}^\infty \frac{1}{k!},$$ so that the given sum is equivalent to a double sum: \begin{align*} \sum_{n=1}^\infty \sum_{k=n+1}^\infty \frac{1}{k!} &= \sum_{k=2}^\infty \sum_{n=2}^k \frac{1}{k!} \\ &= \sum_{k=2}^\infty \frac{1}{(k-2)!k} \\ &= ... 7 A very important inequality about the exponential is\tag1e^x\ge 1+x\qquad \text{for all }x\in\mathbb R$$(with equality iff x=0). Hence if k\ge 1 we can let x=-\frac1k and obtain$$\tag2 1-\frac 1k\le e^{-\frac 1k}. $$Since 1-\frac1k\ge 0, we can take dth power on both sides (d\ge1) to obtain$$\tag3 \left(1-\frac 1k\right)^d\le e^{-\frac ...

5

For $x > 1$, $$\frac{e^x}{\ln x} > \frac{1 + x + \frac{x^2}{2!}}{x} > \frac{x}{2} \to \infty.$$ The relevant inequalities can be proved by elementary means; see for example here and here.

5

Borrowing @Jason's notation for: $$S_n(x)=\sum_{i=1}^{n-1}\left(\frac{i}{n}\right)^x$$ we have that: $$S_n(x)=-1+(1+1/n)^x\cdot S_{n+1}(x)\leq -1+e^{x/n}\cdot S_{n+1}(x),\tag{1}$$ so in order to have $S_n(x_n)<1$, it is sufficient that $S_{n+1}(x_n)<2\cdot e^{-x_n/n}$. Now I state, for later proof: $$... 5$$ \sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right) =\sum_{n=1}^\infty \int_0^1 \exp(u) \frac{(1-u)^{n}}{n!} du $$as everything is positive:$$ \sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right)= \int_0^1 \exp(u) \sum_{n=1}^\infty \frac{(1-u)^{n}}{n!} du \\= \int_0^1 \exp(u)(\exp(1-u) - 1) du = e - \int^1_0 \exp u du = 1 $$4 Using the inequality$$ e^{-x/(1-x/k)}\le\left(1-\frac xk\right)^k\le e^{-x}\tag{1} $$and setting x=\frac{ik}{n} we get$$ \begin{align} \frac1{n^k}\sum_{i=0}^{n-1}i^k &=\frac1{n^k}\sum_{i=1}^n(n-i)^k\\ &=\sum_{i=1}^n\left(1-\frac in\right)^k\\ &\le\sum_{i=1}^ne^{-ik/n}\\ &\le\frac1{e^{k/n}-1}\tag{2} \end{align} $$Thus, k=\log(2)n is an ... 4 I was having some luck by defining$$ S_n(x) = \sum_{i=1}^{n-1} \left(\frac{i}{n}\right)^x. $$In the very least, you get a reasonable justification for the \log{2} coefficient. With minimal algebra, one sees$$ S_n(x) = \left(\frac{n-1}{n}\right)^x \left[1+S_{n-1}(x)\right]. $$Define x_n such that S_n(x_n)=1. We write x_{n}=x_{n-1}+\Delta_n. So, ... 3 If \sec^{-1}(e^x)=u,0\le u\le\pi based on the definition of Principal value But, \displaystyle e^x>0 for finite real x \displaystyle\implies 0\le u<\frac\pi2  \implies \cot u\ge0 Now as \displaystyle\cot^2u=\frac1{\tan^2u}=\frac1{\sec^2u-1} \displaystyle\implies\cot u=+\frac1{\sqrt{\sec^2u-1}} 3 Hint: look at the Wikipedia article on the Cauchy product. In general, if \sum_{n=0}^\infty a_n and \sum_{n=0}^\infty b_n both converge, and at least one converges absolutely, we have$$\left(\sum_{n=0}^\infty a_n\right)\left(\sum_{n=0}^\infty b_n\right)=\sum_{n=0}^\infty\left(\sum_{k=0}^n a_{n-k}b_k\right).$$Apply this to the product of series you are ... 3 Consider the following sequences: 0,0^\frac{1}{2},0^\frac{1}{3},0^\frac{1}{4},\cdots (\frac{1}{2})^0,(\frac{1}{3})^0,(\frac{1}{4})^0,\cdots The first evaluates to a sequence of 0s that would imply a limit of 0^0\implies0 as each term is zero. The second evaluates to a sequence of 1s that would imply a limit of 0^0\implies1 as each term is one. ... 3 By convexity (or by any of a number of other approaches, for example because \log(1+x)\leqslant x for every x\gt-1), the graph of the exponential is above its tangent at 0, that is, \color{red}{\text{For every real number}\ x,}$$\color{red}{1+x\leqslant\mathrm e^x.}$$Apply this to x/n and raise to the power n, this yields the first ... 3 The idea behind saying "if you multiply nothing you get 1" is called the empty product, which comes up often in discrete math and abstract algebra. A similar argument, for example, motivates the definition that 0! should be 1. This is one justification among several that 0^0 should be defined as 1. Alternatively, for the case of a positive n: ... 2 Well since a,b\in\mathbb{R}^+ (positive reals), you know that a= \exp(x) and b=\exp(y) for some x,y\in\mathbb{R}. Then use the fact that \log is the inverse function to \exp in conjunction with properties of exponentials. To see the second one, note that n\log(a) = \underbrace{\log(a)+\cdots+\log(a)}_{\text{n times}}. I think the last one ... 2 This is a system of the form$$ x'=A(t)x, $$where x\in\mathbb R^n, A(t)\in C(\mathbb R;\mathbb R^{n\times n}) and most important$$ A(s)A(t)=A(t)A(s). \tag{1} $$Satisfaction of (1) implies that the solution of$$ x'=A(t)x,\,\,\,x(0)=\xi_0, $$is equal to$$ x(t)=\exp \Big(\int_{0}^t A(s)\,ds\Big)\,\xi_0. $$In our case$$ ...

2

Hint. Clearly $$\frac{1}{5}\sum_{j=1}^5\mathrm{e}^{\omega^j x}=\sum_{n=0}^\infty\frac{x^{5n}}{(5n)!}$$ where $\omega=\mathrm{e}^{2\pi i/5}$, since $$\sum_{j=1}^5 \omega^{jn}=\left\{\begin{array}{ccc} 5&\text{if}& 5\mid n, \\ 0&\text{if} &5\not\mid n. \end{array}\right.$$ But $$\omega=\cos (2\pi/5)+i\sin (2\pi/5), \,\,\omega^2=\cos ... 2 Let's examine this WolframAlpha output: From there, it's easy to see geometrically what the solution set looks like. (It sounds like you knew this much.) We also see the solution set expressed in terms of a mysterious function W, which is a special function called the Lambert W function. You can click the approximate form button to find that ... 2 For the first question, the problem is basically to solve the equation$$f(x)=2^{x/8}- x=0$$The solutions are obtained using Lambert function and the solution are$$x_1=-\frac{8 W\left(-\frac{\log (2)}{8}\right)}{\log (2)}x_2=-\frac{8 W_{-1}\left(-\frac{\log (2)}{8}\right)}{\log (2)}$$which are respectively equal to 1.10 and 43.56. So, the ... 2 Setting$$ a_n=\frac{n^n}{n!}, $$we have$$ \frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\frac{(n+1)^n\cdot n!}{n!\cdot n^n}=\frac{(n+1)^n}{n^n}=\left(1+\frac{1}{n}\right)^n \quad \forall n. $$Since$$ \lim_n\frac{a_{n+1}}{a_n}=\lim_n\left(1+\frac{1}{n}\right)^n=e>2, $$there is an N \in \mathbb{N} such that$$ ...

2

Your best bet is to just store the values in an array, and when you want a value, access the array element at the index, i.e. a[1] for level 1, a[2] for level 2... a[199] for level 199. There's probably no nice formula for this problem. Here's why. Launching off from Peter's answer, let's assume you want to recreate the table he linked with a closed form ...

2

$P(z)= a^z = E(A(z))$, where $E(z)=\exp(z)$ and $A(z)=z \log a$. Both $E$ and $A$ are continuous functions and so is $P = E\circ A$, being a composition of continuous functions. This argument assumes you have proved that $\exp$ is continuous, which should come from its power series definition. The function $A$ is simply a scaling.

2

Note that the algebraic identity $$\lambda t+\frac{(t-\tau)^2}{2\sigma^2}=\tau\lambda-\frac12\sigma^2\lambda^2+\frac{(t-\tau+\sigma^2\lambda)^2}{2\sigma^2}$$ and the change of variable $s=t-\tau+\sigma^2\lambda$ yield $$\int_{-\infty}^{\infty}\exp\left(-\lambda t\right)\,\exp\left(-\frac{(t-\tau)^2}{2\sigma^2}\right)\mathrm ... 2 If f is zero free and it is defined in a simply-connected domain, you can define a logarithm of f as$$ r(z)=\int_{z_0}^z \frac{f'(w)\,dw}{f(w)}. $$This integral is univalent, as f'/f is holomorphic in a simply-connected domain. Clearly f(z)=f(z_0)\exp\big(r(z)\big). Then h(z)=w_0\exp\big(r(z)/t\big), where w_0^t=f(z_0). 2 If we set$$ f(z)=\mathrm{e}^{z}-1, $$then$$ \frac{f'(z)}{f(z)}=\frac{\mathrm{e}^{z}}{\mathrm{e}^{z}-1}, $$and$$ \int_\Gamma\frac{3\mathrm{e}^{z}\,dz}{1-\mathrm{e}^{z}}=-3\int_{|z|=4\pi/3}\frac{f'(z)\,dz}{f(z)}  But $\frac{1}{2\pi i}\int_{|z|=4\pi/3}\frac{f'(z)\,dz}{f(z)}$ is the number of roots of $f(z)=\mathrm{e}^z-1=0$ in the interior of $\Gamma$. ...

2

For problem $(ii)$, note that $\lvert e^z\rvert \leqslant 1$ on the closed left half-plane, and $z^5$ takes purely imaginary values on the imaginary axis. That makes it a straighforward application of Rouché's theorem too. For problem $(iii)$, note that $f$ is real (that is, $f(\mathbb{R}) \subset \mathbb{R}$, hence $f(\overline{z}) = \overline{f(z)}$ ...

1

Hint: Currently, you have it so that you get the value you want at $-3$ when $x=0$, and the value you want at $-2$ when $x=1$, and so on. Therefore, you want have an $x$ which is $3$ larger than it should be. How should you change your equation to reflect this? Response to the comments: One of the subtleties in PEMDAS/BEDMAS is that certain things are ...

1

$e^{\cos\theta+i\sin\theta}$ can be explained as follows. i) $e^{i\theta}$ is a rotation about $\theta$ radians in the complex plane. $e^{\cos\theta+i\sin\theta}=\exp({e^{i\theta}})$ is the a rotation in $\mathbb{C}^1$ represented in the space $X$, where $\exp:\mathbb{C}^1\to X$. ii) The range of $\cos\theta$, and $\sin\theta$ is $[-1,1]$. So, as you've ...

1

Injectivity If you know that $\exp' = \exp$, use the fact that $\exp'(x) = \exp > 0 \;\; \forall \, x$. Alternatively, if you know that $\exp (a + b) = \exp a \exp b$, then assume $\exp x = \exp y$. Conclude $\exp(x - y) = 1$. Assume WLOG $x \ge y$, and then use the series expansion of $\exp(x - y)$ to show $x - y = 0$, hence $x = y$. Surjectivity ...

1

We show that $\lim_{x \to \infty} \frac{\sum{1 \leq i \leq n}log(i)}{n log(a)} = \infty$. Indeed, $\sum_{1 \leq i \leq n}log(i) > \sum_{n/2 \leq i \leq n}log(i)$. Note that for all $i \geq n/2$, we have $log(i) \geq log(n/2) = log(n)-1$. Hence, we have $\sum_{n/2 \leq i \leq n}log(i) \geq \frac{n}{2}log(n) - \frac{n}{2}$. Therefore, \$\sum_{1 \leq i \leq ...

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