# Tag Info

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How does one actually multiply something by itself half a time ? Zen Buddhism has a similar question: What is the sound of one hand clapping ? When my father told me, in passing, one day, that $\sqrt x=x^{1/2}$ I had pretty much the same reaction. But then I started thinking to myself: What is the fundamental property of an n-th root, $\sqrt[\Large^n]x$ ...

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You appear to understand what a square root is just fine; $\sqrt{x}$ is any number who's square happens to equal $x$. If you're asking “how do I compute that?”, well there's several possible algorithms for that. As @TrevorWilson points out, Wikipedia has an entire page on the subject. One method is to pick a number, square it, and see if the ...

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"how does one actually multiply something by itself half a time?" You answer that indirectly, from the definition of the square. The square of the square root of a number is that same number, or $$\left(\sqrt x\right)^2=x.$$ It turns out that this definition is consistent with a fractional exponent. For integer exponents, the following rule holds: ...

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You are correct about your interpretation of the $e$ stuff. Indeed, this value is $2.2250738585072014 \cdot 10^{-308}$ Which is, to be sure, a number. It's just that, in order to properly store this number, the calculator will need a fair bit of memory to store it accurately, and you've essentially attempted to exhaust it.

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$1^i$ is defined as $e^{i \ln 1}$ $\ln 1$, though, it's not our real logarithm, but it's the complex one; the point is that is a multi-valued function; to be more precise, $$\ln z = \ln_\mathbb R |z| + (\arg z + 2k\pi)i$$ Where with $\ln_\mathbb R$ I mean the usual real logarithm. So $\ln 1 = \ln_\mathbb R 1 + (2k\pi)i = 2k\pi i$ Hence $$1^i = e^{i \ln ... 10$$\lim_{n\to\infty} \frac{3^n}{2^n+3^n}=\lim_{n\to \infty}\frac{1}{(\frac{2}{3})^n+1} =1.$$Since \frac{2}{3}<1 , so (\frac{2}{3})^n\to 0 as n\to \infty. 10 Observe that 68=17 \times 4 and 51 = 17 \times 3, therefore 4^{68}=(4^4)^{17} and 5^{51} = (5^3)^{17}. Compute the base, 4^4= 256 is greater than 5^3 = 125. Then observe that 3 \times 23 = 69, so 4^{68} = (4^3)^{23}/4 = 64^{23}/4, which is much much much greater than 12^{23}. Therefore 12^{23} is the smallest. Quantitatively, 5^{51} ... 7 First note that$$12^{23}=(3\cdot4)^{23}=3^{23}\cdot4^{23}<4^{23}\cdot4^{23}=4^{46}\;,$$so 12^{23} is definitely less than 4^{68}. Next, 51 and 68 are both multiples of 17, so let’s see whether that fact leads to anything nice. 4^{68}=(4^4)^{17}=256^{17}, and 5^{51}=(5^3)^{17}=125^{17}, so clearly 5^{51}<4^{68}. It only remains to ... 7 Asking what a square root "is", to a mathematician's ear, means asking how it is defined: if I may read into what you're saying, that's not what you're trying to figure out here. But knowing how some other things are defined will help. So, let's start with the real numbers: what you know as streams of decimal digits are generally defined in some way or ... 7 Let n be an integer that is not the square of any integer. Two versions of the question have already been answered, namely What is the definition of \sqrt{n}, i.e. what does it mean for a given number to be a square root (or the positive square root) of n? How can we approximate \sqrt{n}, i.e. how can we find rational numbers (or decimals) ... 6 Note that 8=2^3. Compare it to 2^n and conclude that n=3. 5 Let x=\prod_i {p_i}^{m_i}, where m_i\ge 1 and p_i prime, then y=\prod_i {p_i}^{n_i} for n_i\ge 0. Plugging this in the equation have 2007 m_i = n_i x = n_i \prod_j {p_j}^{m_j}. Suppose that p_j\nmid 2007, then {p_j}^{m_j} \mid m_j which implies 2^{m_j} \le m_j, contradiction, hence p_j \mid 2007. Also {p_i}^{m_i}\mid 2007 m_i. Note ... 5 The given equation can be rewritten as$$2^{2x^2-4x+2}+2^{x^2-2x+4}=2^7$$Now take 2^{(x-1)^2}=z which gives you the equation$$z^2+8z=2^7\implies z=8,-16$$Taking only the positive solution gives you, (assuming x to be real)$$(x-1)^2=3\implies x=1\pm \sqrt{3}$$4 We have 2^{2-\ln(x)} = \dfrac4{2^{\ln(x)}} and 2^{2+\ln(x)} = 4\cdot2^{\ln(x)}. Hence, setting 2^{\ln(x)} = a, we obtain$$\dfrac4a + 4a = 8 \implies a^2 + 1 =2a \implies a =1 \implies 2^{\ln(x)} = 1 \implies x = 1$$4 The best way to solve problems like this is using logarithms. If you have an equation$$10 ^x = y$$where y is any positive number, and you wish to find x, then the value of x will be (by definition) \log y (this is what the \log button on your calculator is for). For example, to solve$$10^x = 110$$we calculate$$x=\log 110= 2.0413...$$There ... 4 Expanding on what has already been said, I want to emphasize that a^0=1 is a convention. There is no computation to be made, because there is no obvious meaning to "multiply with itself zero times". The reason why the convention is reasonable is what has been mentioned by Thorben's answer and Gregory's comment: the relation a^{m+n}=a^ma^n is so nice ... 3 Since n^k=n^{k+0}=n^k \cdot n^0. This short computation suggests hardly that n^0 should be 1. 3 Looking at the source paper (Michaillat and Saez, "Aggregate Demand, Idle Time, and Unemployment" (2014)) that you linked, I think the function that you'd want to invert is actually$$ g(x) = \left(1 + \frac{c}{(1+x^b)^{-1/b} - c}\right)^{\epsilon - 1} (1 + x^{-b})^{-1/b} $$(where I guess you set \epsilon = 2 for simplicity) so that g(x) = a. Anyways, ... 3 There is (at least) one option - however depending on whether a solution for the problem of "fractional iteration of logarithm" is available (aka: "Tetration". Note that in the tetration-forum there is an exchange on various approaches to this problem). Assume for the moment that a version of that fractional iteration is available, and let us ... 3 1^i is a multi-valued quantity$$e^{i\theta}=\cos\theta+i\sin\theta\implies e^{2n\pi i}=1\implies1^i=\color{RED}{e^{-2n\pi }}$$for any integer n. 3$$\large e^{5i\pi/6}=e^{5i\pi/6-i\pi+i\pi}=e^{-i\pi/6+i\pi}=e^{-i\pi/6}e^{i\pi}=-e^{-i\pi/6}$$3 We do have fractional roots but, I write it like -$$x^{\frac{a}{b}}=C \to x=C^{\frac{b}{a}}$$But anyways there are a lot of ways you can represent math symbols and I can't find anything wrong in your representation too. 3 (a^b)^c=a^{bc} is not valid when b has imaginary part otherwise one would have$$e^{-4\pi^2}=(e^{2i\pi})^{2i\pi}=1^{2i\pi}=1$$And this is absurd 3 This is indeed a number, it is extremely near by 0. Your number is approximately$$2 \cdot 10^{-308} = 0.\underbrace{00\ldots 0}_{307 \text{ zeros}}2 \; ,$$which is slightly bigger than zero. I don't know what you want to program, but one often wants to test, if such a result is zero. You should never do something like this: res = 2.2E-10; if (res == ... 3 It's come to my attention that I don't actually understand what a square root really is (the operation). I think your intuition is telling you, correctly, that "square rooting" really isn't an operation—except for under some very special circumstances. As you probably know, when n is a whole number, an nth root of x is defined as a number whose ... 3 Lazy people should use be careful to avoid having to perform a division in each step of Newton's method. They should consider the equation:$$\frac{1}{x^2}- \frac{1}{y} = 0$$This then yields the recursion:$$x_{n+1} = \frac{x_n}{2}\left(3-\frac{x_n^2}{y}\right)$$Here, you only have division by fixed numbers in each step. 3 You're right, in the sense that if a \neq 0 and b > 0 are real numbers, then the graph y = a \cdot b^{x} does not cross the x-axis. It's enough to show b^{x} \neq 0 for all x. The reason comes down to the law of exponents$$ b^{x + x'} = b^{x} \cdot b^{x'}\quad\text{for all $x$ and $x'$.}  If $b^{x_{0}} = 0$ for some $x_{0}$, then writing ...

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This is normally called a "Bit field."

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You should study the definitions of logarithms and exponential functions to understand not only the square root but expressions like $a^x$ when $x$ is rational. Spivak's Calculus book has an amazing chapter about that. After you learn some basic definitions and useful theorems finally you will realize that if $a>0$ then for all $x$, you can write ...

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It uses three facts: that $\sqrt{a}=a^{1/2}$ and $(a^b)^c=a^{bc}$ and $(ab)^c=a^cb^c$. $320 = 32 \times 10 = 2\times2\times2\times2\times2\times2\times5=2^6\times5$. So $(2^65)^{1/2}=2^35^{1/2}$.

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