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## Hot answers tagged exponentiation

5

$(n+1)!=2\cdot 3\cdot \dots\cdot(n+1)$ here a product of $n$ numbers all are at least 2 so the result follows...

5

The comments have already pointed out where you have misunderstood this question. However, as an answer to your question: \begin{align*}\frac{\mathrm{d}}{\mathrm{d}x}\Bigg(\frac{5}{9}x^\frac{2}{3} \Bigg)&= \frac{5}{9}\frac{\mathrm{d}}{\mathrm{d}x}(x^\frac{2}{3})\\ &= \frac{5}{9}\Bigg(\frac{2}{3}\Bigg)x^{\frac{2}{3}-1}\\ &= ...

5

You have actually (not) proved that if $0=0^0$, then $0^0\ne 1$ which needs no argument at all. Moreover, your argument is wrong: you use $\ln(t)$, which is not defined for $t=0$. So you can draw no conclusion about $t=0$. Under fairly standard conventions, the function $x\mapsto x^x$ is only defined for $x>0$; more generally, one sets $$... 4 z = 0 is always a solution. Others, for m, d \ne 0, are$$ z = -d^{-1} \left(1/m + W(-e^{-1/m}/m) \right) $$where W is a branch of the Lambert W function. EDIT: I suspect you're interested in real solutions where m is real. For m < 0, -\exp(-1/m)/m) > 0 so the principal branch of Lambert W can be used, but it will give z = 0. For ... 4 HINT:$$\frac{(n+1)!}{2^n}=\frac{2}{2}\frac{3}{2}\frac{4}{2}\cdots \frac{n-1}{2}\frac{n}{2}\frac{n+1}{2}$$4 One may recall that, as x \to \infty,$$ \lim_{x \to \infty} \frac{\ln^b x }{x^a}=0,\quad a>0, \tag1 $$whatever the value of b is. Then using (1) with x:=n, a=-\alpha>0, we obtain$$ \lim_{n \to \infty} \frac{n^\alpha}{\ln^\beta n }=\lim_{n \to \infty} \frac{\ln^{-\beta} n}{n^{-\alpha} }=0,\tag2 $$as announced. 4 It is an indeterminate form and as such cannot be assigned any value. It is better expressed as \lim_\limits{{x\to \infty}\\,\\{y\to 0}}x^y. As commented by Did, it is true that x^y has no limit when x\to \infty and y\to 0. And \infty^0 has no definite meaning in mathematics. It is basically some kind of a meaningless statement where the ... 3 Hint:$$x^2=\left(\frac{1}{2}\right)^x\implies x=\sqrt{\left(\frac{1}{2}\right)^x}\implies x=\sqrt{\left(\frac{1}{2}\right)^\sqrt{\left(\frac{1}{2}\right)^\sqrt{\left(\frac{1}{2}\right)^\sqrt{\left(\frac{1}{2}\right)^{........}}}}}$$3 You can use Lambert W, the inverse to x\mapsto xe^x:$$\begin{align} x^2 &=\frac{1}{2^x}\\ x^22^x &=1\\ x^2e^{x\ln(2)} &=1\\ xe^{x\ln(2)/2} &=\pm1\\ x\ln(2)/2 e^{x\ln(2)/2} &=\pm\ln(2)/2\\ W\left( x\ln(2)/2 e^{x\ln(2)/2}\right) &=W\left(\pm\ln(2)/2\right)\\ x\ln(2)/2 &=W\left(\pm\ln(2)/2\right)\\ x ...

3

After some calculation you have \begin{align*} 24 &\equiv 5 \bmod 19 \\ 24^2 &\equiv 25 \equiv 6 \bmod 19 \\ 24^4 &\equiv 36 \equiv -2 \bmod 19 \\ 24^8 & \equiv 4 \bmod 19 \\ 24^{16} & \equiv 16 \equiv -3 \bmod 19 \end{align*} Now multiply: $$24^{31} \equiv 5\cdot 6 \cdot (-2) \cdot 4 \cdot -3 \equiv (30) \cdot (24) \equiv 11 \cdot 5 ... 3 With perhaps a little less arithmetic, 2^2=4\equiv23\pmod{19}, and 4\times5=20\equiv1\pmod{19}, so 24\equiv5\equiv4^{-1}\equiv2^{-2}\pmod{19}. By Fermat's little theorem,$$23^{32}=2^{2\times32}=2^{64}\equiv2^{64-7\times18}\equiv2^{-62}\equiv2^{-2\times31}\equiv24^{31}\pmod{19}3  24 \equiv 5 \bmod 19   23 \equiv 4 \bmod 19   5 \cdot 4 \equiv 1 \bmod 19   5^{-31} 4^{32} \equiv 4^{31} 4^{32} \equiv 4^{63} \equiv 4^9 = 2^{18} \equiv 1 \bmod 19  3 Using Euler's formula: -1=\exp(\ln(-1))=\exp(\ln(1)+i\arg(-1))=\exp(0+i(2k+1)\pi) =\exp(i(2k+1)\pi) for k\in \Bbb{Z} Thus: (-1)^{\sqrt{2}}=\exp(i(2k+1)\pi\sqrt{2})=\cos((2k+1)\sqrt{2}\pi)+i\sin((2k+1)\sqrt{2}\pi) 3 With \zeta=e^{2\pi\mathrm i/p} as Gerry suggested, \begin{align} \sum_{l=0}^{\left\lfloor\frac mp\right\rfloor}\binom m{lp} &=\sum_{j=0}^m\binom mj\frac1p\sum_{k=0}^{p-1}\zeta^{jk}\\ &=\frac1p\sum_{k=0}^{p-1}\sum_{j=0}^m\binom mj\zeta^{jk}\\ &=\frac1p\sum_{k=0}^{p-1}\left(1+\xi^k\right)^m\\ ... 3e^A=I+A+\frac{1}{2}A^2+\cdots=UU^{-1}+UDU^{-1}+\frac{1}{2}UDU^{-1}UDU^{-1}+\cdots=UU^{-1}+UDU^{-1}+\frac{1}{2}UD^2U^{-1}+\cdots=U\bigg(I+D+\frac{1}{2}D^2+\cdots\bigg)U^{-1}Ue^DU^{-1}$$3 Notice the following: \frac{10^{n^{2}}}{10^{(n+1)^{2}}}=\frac{10^{n^{2}}}{10^{n^2+2n+1}} \frac{10^{n^{2}}}{10^{n^2+2n+1}}=\frac{10^{n^{2}}}{10^{n^2}10^{2n}10^{1}} \frac{10^{n^{2}}}{10^{n^2}10^{2n}10^{1}}=\frac{1}{10^{2n+1}} So, I hope now, it should be obvious what happens as n gets infinitely large. 3 Choosing a definite argument for \;i\; , say \;\frac\pi2\; , we get$$i^i=e^{i\log i}=e^{i\left(\log|i|+i\frac\pi2\right)}=e^{-\pi/2}$$2 Hint:$$10^{(n+1)^2}=10^{n^2+2n+1}=10^{n^2}10^{2n+1}.$$2$$\lim_{n\to\infty}\frac{10^{n^{2}}}{10^{(n+1)^{2}}}=\lim_{n\to\infty}\frac{10^{n^{2}}}{10^{n^2}\cdot10^{2n+1}}=\lim_{n\to\infty}\frac{1}{10^{2n+1}}=0$$2 Hint: Let y=x^{1/x}. Now take the natural log of both sides.$$\ln(y)=\ln(x^{1/x})=\frac{1}{x}\ln(x).$$Now you can differentiate both sides and solve to find y'. I'll even do the left side for you:$$\frac{d}{dx}\ln(y)=\frac{y'}{y}=\frac{y'}{x^{1/x}}.$$2 Yes, it is true. If you start with exponent k, as in n^k, the kth differences will be k! As a result, if you begin with some polynomial a_k n^k + a_{k-1} n^{k-1} + \cdots all the lower degree terms will have disappeared, and the final will be row k with number a_k k! That way you know what a_k is, if you had enough terms to begin with. If ... 2 Find the prime factorization of your number n and take the greatest common divisor d of the exponents. Then \sqrt[k]n is an integer if and only if k divides d (and we may also want k>1). You want the radix base to be as big as possible, i.e., you are looking for the smallest k. If d>1, that would be \sqrt[p]n where p is the ... 2 Hint :$$108^2-92^2=(108-92)(108+92)$$2 x^{-1/3} = \frac{1}{x^{1/3}}, not \frac{1}{x^3}. 2 See the terms having factor 11 are 11,22,33,..121,...220,..297 so excluding 121,242 we have 25 numbers which give only one 11 and 121,242 gives two 11 so total exponent is 11^{25}.11^4=11^{29} 1 I doubt it has a name, since the proof is so straightforward.$$\sum_{i=0}^{n-1}{b^i} = \frac{b^n-1}{b-1} \le b^n-1 < b^n$$The key part is the equality, which is called a geometric series or the sum of a geometric progression. The inequalities are consequences of your restrictions on b and n. 1 If n is odd, then n \land 2^{i} = 1, so you can "divide both sides" by 2^i. This is not true if n is even: 2^3 \equiv 2^2 \pmod 4, but 2 \not \equiv 1 \pmod 4  1 Hint. You may set X=2^x, then solve, for X>0, the inequation$$ X+\frac8X>6 $$or$$ \frac{X^2-6X+8}X>0. $$Can you take it from here? 1 The above answers are fine provided that you are only concerned with real numbers. However, just as x^{a/b} has b solutions in the complex plane, x to the power of an irrational will have an infinite number of solutions. 1 You can write it as$$e^{\sqrt{2}\ln(i^2)} now $\ln(i^2)=\ln(1)+i\arg(i^2)=0+i(2k+1)\pi$ thats it. You can continue now. Log of complex number is $log(z)=log(|z|)+arg(z)$

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