# Tag Info

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My chief understanding of the exponential and the logarithm come from Spivak's wonderful book Calculus. He devotes a chapter to the definitions of both. Think of exponentiation as some abstract operation $f_a$ ($a$ is just some index, but you'll see why it's there) that takes a natural number $n$ and spits out a new number $f_a(n)$. You should think of ...

14

$2^\pi$ or $3^\sqrt2$ (or any other irrational power, really). What does this mean? $$a^\pi=a^{3.1415\ldots}=a^{3\ +\ 0.1\ +\ 0.04\ +\ 0.001\ +\ 0.0005\ +\ \cdots}=a^3\cdot a^{0.1}\cdot a^{0.04}\cdot a^{0.001}\cdot a^{0.0005}\cdots$$ $$a^\sqrt2=a^{1.4142\ldots}=a^{1\ +\ 0.4\ +\ 0.01\ +\ 0.004\ +\ 0.0002\ +\ \cdots}=a^1\cdot a^{0.4}\cdot a^{0.01}\cdot ... 10 One reason your intuition might break down here is that exponentiation - even integer exponentiation - is inherently dimensionless. If a square has area of 5\mathrm{cm}^2 then you can talk about its sides having length \sqrt{5\mathrm{cm}^2} = \sqrt{5}\mathrm{cm}, but there are no units for x that give 2^x a sensible set of units — try to ... 10 Note: This post contains only numerics; see the answers by robjohn and Jack D'Aurizio for actual math. Numerical computations show that$$k(n) = \lfloor n\log 2-0.039721\rfloor \quad \text{ for } \ 2\le n<944029$$which is kind of nice. However, this empirical formula breaks down for n=944029. What is worse, it turns out there is no c such that ... 9 We seem to have a real-world understanding of exponentials, for example when we say that a certain quantity is increasing "exponentially". To make the ingredients of this intuition precise, consider a continuous function f:\mathbb{R}\to \mathbb{R} (think of this as describing a quantity f(t) dependent on time t). Then we can define f to be ... 7 To get to the problem here, we start with \sqrt{(-1)^2}. Please note that, before dealing with the square root, that (-1)^2\neq -1^2. (-1)^2=(-1)(-1)=1, while -1^2=-(1^2)=-1. We can rewrite this using rational exponents, so$$\sqrt{(-1)^2}=((-1)^2)^{1/2}$$I believe here, though we can just remedy the situation using our order of operations. ... 6 As I said in the comment, the problem with this series of equalities is that we cannot generally say that (a^{p})^q = a^{pq}. So, in this instance,$$ ((-1)^{2})^{1/2} \neq (-1)^{2/2} $$The equation (a^{p})^q = a^{pq} will hold, however, as long when either p and q are both integers or a is a positive number. 6 Since \infty is not a real number, the expression \infty^0 is not interpreted like 3^0, but rather as the following: We are given two sequences of real numbers, a_n and b_n, such that \lim a_n=0 and \lim b_n=\infty, and now we calculate the limit of the exponentiation, \lim b_n^{a_n}. If a_n=0 for all n, then the limit will be 1, ... 5 To be honest, exponentiation loses a nice real-world interpretation pretty quick. What 3^{\sqrt{2}} means isn't really easy to say, but what it is is more-or-less just \displaystyle\lim_{x\to\sqrt{2}} 3^x, which we can make sense of as we can prove f:\mathbb{R}^+ \mapsto \mathbb{R}^+,\ x\mapsto 3^x is a continuous function. If you were to require some ... 5 The binomial series is "just" the Taylor series of (1+x)^{\alpha} at x=0. Start deriving f(x)=(1+x)^{\alpha} by x and you get \alpha(1+x)^{\alpha-1}, then \alpha(\alpha-1)(1+x)^{\alpha-2} for the first derivative etc. The nth derivative is$$\frac{d^n}{dx^n} (1+x)^{\alpha} = \alpha(\alpha-1)\cdots(\alpha-n+1)(1+x)^{\alpha-n}$$So the Taylor ... 5 Borrowing @Jason's notation for:$$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}$$so in order to have S_n(x_n)<1, it is sufficient that S_{n+1}(x_n)<2\cdot e^{-x_n/n}. Now I state, for later proof:$$ ...

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, ...

4

This isn't a complete answer, but perhaps something that might be part of one. Consider the equation $2^x \equiv x \pmod{100}$. The smallest positive solution is $x=36$. It also turns out that $36^{36} \equiv 36 \pmod{100}$. Thus if we have $2^{{36}^k} \pmod{100}$ it will reduce to $2^{36 * 36 * \dots * 36} \equiv (2^{36})^{36 * \dots * 36} \equiv 36^{36 * ... 3 Before proving your observations, we need to know a few things, I'll present them as lemmas. Before doing so, I'll introduce a notation. Let's denote $$a_b=\underset{b\text{ times}}{\underbrace{a^{a^{a^{\cdots^a}}}}}$$ for$a,b\in\mathbb Z^+$. Lemma 1. We have$2^k\mid 7_k-7_{k-1}$for all$k\geq2$. Proof. By induction. It's true for$k=2$, because ... 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$n$: ... 3 If the ".k" is multiply by$k$then: $$S=1+e^{6x\pi 2}+e^{6x\pi 4}+ e^{6x\pi 6}$$ $$Se^{6x\pi 2}=e^{6x\pi 2}+e^{6x\pi 4}+e^{6x\pi 6}+ e^{6x\pi 8}$$ Substract the two expressions: $$Se^{6x\pi 2}-S=(e^{6x\pi 2}+e^{6x\pi 4}+e^{6x\pi 6}+ e^{6x\pi 8})-(1+e^{6x\pi 2}+e^{6x\pi 4}+ e^{6x\pi 6})$$ The only terms which remains are$1$and$e^{6x\pi 8}$: ... 3 Let us consider$a^b$. Then, the concept of exponentiation can be understood as the value of$e^{b\ln a}$, since both this expression and$a^b$are the same. So, for example, if$b$is an imaginary number$di$, then$a^b$denotes a rotation by$d\ln a$in the complex plane. However, one can also consider arbitrary exponents$A^B$. This can be interpreted as ... 3 Personally I would square root first because the students know how to do this, giving$(8i)^3=-512i$. As with all square root questions there are two possible answers - if you cube first and then take the square root you get$512i$. This suggests that there is a hidden convention at work - indeed there is, and it is a more subtle thing when complex numbers ... 3 First of all$(\sqrt {3x})^2 = 3x$. And furthermore, $$(x + \sqrt 3)^2\neq x^2 + 3$$ Recall, when squaring a binomial: $$(a + b)^2 = a^2 + 2ab + b^2$$ Starting from the beginning: we need for$3x \geq 0 \iff x \geq 0$for the left hand side of the following equation to be defined in the reals: $$\sqrt{3x} = x + \sqrt 3$$ Squaring both sides gives us ... 3 I think that this one is due to Courant: For all$n\in\mathbb{N}$define$a_{n}$such that $$a_{n}=\sqrt[n]{n}-1$$ Note that for every$n\in\mathbb{N}$,$a_{n}\geqslant0$. Now rewrite as: \begin{eqnarray*} n & = & \left(a_{n}+1\right)^{n}\\ & = & \sum_{k=0}^{n}\binom{n}{k}a_{n}^{k}\\ & \geqslant & \binom{n}{2}a_{n}^{2} ... 2 Take any number$r>1$. Eventually$r^n>a$and so eventually$\sqrt[n]a<r$. Now take$r<1$. Eventually$r^n<a$and so eventually$\sqrt[n]a>r$. Calculating$\lim\ r^n$rigorously is a little tricky and depends on how deep you want to go, see this answer for one approach. I did also use the fact that the$n$-th root function is increasing, ... 2$\infty^0=\infty^{^\frac1\infty}=\sqrt[\infty]\infty$. Now,$\sqrt[n]x$is the quantity which, when multiplied n times with itself, gives x. So: which quantity, multiplied an infinite number of times with itself, yield infinity as a result ? Obviously, any real number$>1$has this property, as well as infinity itself. So the result is rightly left ... 2 To draw a parallel to multiplication: If we consider the expression$e\cdot \sqrt5$, I could tell you that this represents the area of a rectangle with side lengths$e$cm and$\sqrt5$cm. Or maybe$e \cdot \pi$is the cost of$\pi$kg of material that costs$edollars per kg. Of course these quantities would not be exact, but the underlying ... 2 We have \begin{align} \frac{(2^{3n+4})(8^{2n})(4^{n+1})}{(2^{n+5})(4^{8+n})} &= \frac{(2^{3n+4})(2^{3})^{2n}(2^2)^{n+1}}{(2^{n+5})(2^2)^{8+n}} \\ &=\frac{2^{3n+4+6n+2n+2}}{2^{n+5+16+2n}} \\ &=\frac{2^{11n+6}}{2^{3n+21}} \\ &=2^{(11n+6)-(3n+21)} \\ &=2^{8n-15} \end{align} Also from the original problem, \begin{align} ... 2 First thing that comes to mind: Use the series representatione^x = \sum_{i=1}^\infty \frac{x^n}{n!}$. Because$21!>2^{64}$, calculating After some more thorough consideration it seems that in order for $$\left\lfloor e^x \right\rfloor = \left\lfloor \sum_{n=0}^{n} \frac{x^n}{n!} \right\rfloor$$ to produce an exact value, we need$n = O(x)$. That ... 2 Hint:$(10-k)^n$is$k^n\pmod{10}$if$n$is even and$10-k^n\pmod{10}$if$n$is odd. Thus$c:k\mapsto k^n\pmod{10}$is such that$c(10-k)$is$c(k)$in the first case (axial symmetry with respect to the line$k=5$) and$10-c(k)$in the second case (central symmetry with respect to the point$(5,5)$). 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 It can be shown that$\displaystyle (1 + x)^{\alpha} = \sum_{n = 0}^{\infty} \binom{\alpha}{n}x^n$when$|x| < 1$. This is not really precalculus though... EDIT: I'll add a proof for the sum: Let$s(x)$be the sum of the series. Then$\begin{align} (1 + x)s'(x) &= (1 + x)\sum_{n = 0}^{\infty} n\binom{\alpha}{n}x^{n-1} \\ &= \sum_{n = ...

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