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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\\ ...


3

$$e^A=I+A+\frac{1}{2}A^2+\cdots$$ $$=UU^{-1}+UDU^{-1}+\frac{1}{2}UDU^{-1}UDU^{-1}+\cdots$$ $$=UU^{-1}+UDU^{-1}+\frac{1}{2}UD^2U^{-1}+\cdots$$ $$=U\bigg(I+D+\frac{1}{2}D^2+\cdots\bigg)U^{-1}$$ $$Ue^DU^{-1}$$


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 $k$th 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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