# Tag Info

13

Hint: Use the inequality $$\left(1-\frac{1}{x}\right)^x>\frac{1}{x-1}$$

11

Forget that $n$ is an integer. For $x > 0$ we have $$\lim_{y\to\infty} \left(1+\frac{x}{y}\right)^y = \lim_{y\to\infty} \left(1+\frac{1}{\frac{y}{x}}\right)^{x\cdot \frac{y}{x}}$$ If $y \to \infty$ then $z:= \frac{y}{x} \to \infty$ as well so change of variables gives that this is equal to $$\lim_{z\to\infty} \left(1+\frac1z\right)^{xz}$$ which is the ...

10

We have $$\frac{n}{m} = \frac{49^{51}}{50^{50}} = 49 \cdot \left(\frac{49}{50}\right)^{50} = 49 \cdot \left(1-\frac{1}{50}\right)^{50} \approx \frac{49}{e} > 1$$ Even if you dont know that for large $n$ $$\left(1-\frac{1}{n}\right)^{n} \approx \frac{1}{e}$$ as long as you're able to tell that $$\left(1-\frac{1}{50}\right)^{50} > \frac{1}{49}$$ you're ...

7

I'd write $m=n-1$. Then the statement reduces to $$\left(1+\frac1m\right)^m<\frac{m^2}{m+1}.$$ It's well-known that $(1+1/m)^m$ increases to $e$, but more naively, $$\left(1+\frac1m\right)^m=1+1+\frac1{m^2}{m\choose 2} +\frac1{m^3}{m\choose 3}+\cdots<1+1+\frac12+\frac16+\cdots<3.$$ But $$\frac{m^2}{m+1}>\frac{m^2-1}{m+1}=m-1>3$$ if $n>4$.

5

Use Bernoulli's inequality with $x=-\frac{1}{50}$ and $r=48$: $$\frac{49^{51}}{50^{50}}=\frac{49^3}{50^2}\cdot\frac{49^{48}}{50^{48}}=\frac{49^3}{50^2}\left(1-\frac{1}{50}\right)^{48}\ge\frac{49^3}{50^2}\left(1-\frac{48}{50}\right)=\frac{2\cdot49^3}{50^3}>1$$

5

A natural number $n$ has $b$ digits iff $10^{b-1}\le n<10^b$. So for this to hold for $a^b$, we clearly need $a<10$, or: $a\le 9$. On the other hand, we need $(\frac{10}{a})^{b-1}\le a$. For given $a$, this can be solved by taking logarithms. We obtain $b\le 21$ for $a=9$ and $b\le 10$ for $a\le8$. So, as perhaps expected, we use $a=9$ and find $$9^{21}... 4 For p > 1 and  \Re(x) > 0 let$$f(x) = \sum_{k=0}^\infty e^{-x p^k}$$for \Re(s) > 0$$F(s)=\int_0^\infty f(x) x^{s-1}dx = \sum_{k=0}^\infty (p^k)^{-s}\Gamma(s)= \frac{\Gamma(s) }{1-p^{-s}}$$and with the residue theorem$$f(x) = \frac{1}{2i\pi} \int_{c-i\infty}^{c+i\infty} F(s) x^{-s}ds = \sum Res(F(s) x^{-s}) \\=\sum_{m =1}^\infty \frac{\...

4

You're just trying to evaluate $10^{765.66} = 10^{0.66} \cdot 10^{765}$. So, in your notation, $x$ is $10^{0.66} \approx 4.57$. The result is $e^{1763.02} \approx 4.57 \cdot 10^{765}$.

3

Here is a proof without induction. $(1-1/n)^n$ is strictly increasing for $n > 1$ and approaches $1/e$ as $n \to \infty$ whereas $1/(n-1)$ is strictly decreasing and approaches $0$ as $n \to \infty$. Hence there is an $N_0$ such that for all $n > N_0$, $$\Big(1 - \frac{1}{n}\Big)^n > \frac{1}{n-1}$$ By little computation, we find that ...

3

Your error mainly occurs in assuming that $y$ can be equal to $4$. It is known (see here) that the infinite 'power-tower' only converges to values within the range $(e^{-1},e)$ for real $x$ (where $x$ is chosen such that the 'power-tower' converges) which does not include $4$.

3

Minimal number of multiplications in worst case $n=4000$: $N_{\min}=\lceil\log_2 4000!\rceil=42\,100$. “Bruteforce” approach, using fast multiplication for power 2, then 3, then 4 etc gives $N_{bf}=\sum_{k=2}^{4000} \lceil\log_2k\rceil=43\,905$. So it's less than 5% inoptimal. I wouldn't be bothered with writing a sophisticated algorithm.

3

You want to find the value of $n$ for which $100n^2 \approx 2^n$. Taking logs, we get $2\log n + 7 \approx n$, so $n \approx 7 + 2\log 7 \approx 13$. Of course, this is just an approximation.

2

Proof sketch: First show that $a=9$ is the best possible value for $a$ (you don't have to show that it is better than any other value, just that no other value can possibly be better). Then find the maximal $b$ that works for $a=9$ through brute force. Alternately: If $a^b$ has $b$ digits, what can be said about $\frac{a^b}{10^b}=\left(\frac{a}{10}\right)^b$...

2

This is just an intuition test of why the properties should be defined like this: The fundamental property of the exponential functions is the following: $b^x\cdot b^y=b^{x+y}$. The other properties are derived in such a way that the mentioned property is satisfied. For example: since we want the equation to be fulfilled $$b^0\cdot b^x=b^{0+x}=b^x$$ we ...

2

Induction step: $$\left(2^{2^{n+1}}-1\right)\left(1+2^{2^{n+1}}\right)=2^{2^{n+1}}+\left(2^{2^{n+1}}\right)^2-1-2^{2^{n+1}}=2^{2^{n+1}\cdot2}-1=2^{2^{n+2}}-1$$ Applied rule:$$\left(a^b\right)^c=a^{bc}$$

2

$$(2^{2^{k+1}})^2 = 2^{2^{k+1}} \cdot 2^{2^{k+1}} = 2^{2^{k+1}+2^{k+1}} = 2^{2 \cdot 2^{k+1}} = 2^{2^{k+2}} \text{.}$$

2

You made a mistake just before the last step("by property of exponent"). Note that $$2*(2^j) = 2^{(j+1)}$$. And $$(a^b)^c = a^{bc}$$ not $a^b^c$

2

You are trying to find the maximum $b$ that satisfies $10^{b-1}<9^b<10^b$ (it is easy to show $a=9$ is best possible) and as you said $b=21$ satisfy that. Also if a "$b$" doesn't satisfy the inequality, then $b+1$ also doesn't. (easy to show that) And $b=22$ doesn't satisfy the inequality. Thus $b=21$ is the maximum number.

2

Yes it can (we are assuming that $p \not \mid a$). We know that: $$(a^{\frac{p-1}{2}})^2 \equiv a^{p-1} \equiv 1 \pmod{p}$$ And since for any number $x$ such that $x^2 \equiv 1 \pmod{p}$ we have $p \mid x^2-1=(x-1)(x+1)$ and therefore $p \mid x-1$ or $p \mid x+1$ and therefore $x \equiv \pm 1 \pmod{p}$ we can conclude that: $$a^{\frac{p-1}{2}} \equiv \pm 1 \... 2 The claim x^x<(x-1)^{x+1} is equivalent to$$ x\ln x<(x+1)\ln(x-1). $$Consider f(x)=(x+1)\ln(x-1)-x\ln x and prove that f'(x)=\ln\left(1-\frac1x\right)+\frac2{x-1}\to 0 as x\to +\infty, f''(x)=-\frac{x+1}{x(x-1)^2}<0 for x>1. Then f'(x)>0 and f increases. From f(5)>0 it follows that f(x)>0 for all x\ge 5. 2 In this answer, it is shown that \left(1+\frac1{n-1}\right)^n is decreasing. That means that its reciprocal \left(1-\frac1n\right)^n is increasing. Thus, for n\ge2, we have$$ \left(1-\frac1n\right)^n\ge\frac14\tag1 $$Therefore,$$ \begin{align} \frac{49^{51}}{50^{50}} &=49\left(1-\frac1{50}\right)^{50}\\ &\ge\frac{49}4\tag2 \end{align} $$... 2 Apply binomial theorem resp. the binomial series, then$$ \left(1+\frac1n\right)^{nx}=1+x+\sum_{k=2}^{\infty}\frac{x(x-\frac1n)...(x-\frac{k-1}n)}{k!} $$and$$ \left(1+\frac xn\right)^{n}=1+x+\sum_{k=2}^{\infty}\frac{(1-\frac1n)...(1-\frac{k-1}n)}{k!}x^k $$Both expansions converge in their coefficients to the same limit \frac{x^k}{k!}, the complicated ... 2 In general$$\lim_{x\to \infty} \left( 1+\frac{a}{x} \right) ^{bx}=e^{ab}$$and they are only based on the definition of the number e$$e:=\lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^n $$Here is a simple proof:$$\begin{align} \lim_{x\to \infty} \left( 1+\frac{a}{x} \right) ^{bx} &= \lim_{x\to \infty} \left[ \left( 1+\frac{a}{x} \right)^{x/a} \...

2

Consider that for $x>0$, $n$ and $\frac nx$ both tend to infinity so that $$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=\lim_{n\to\infty}\left(1+\dfrac xn\right)^{n/x}=e.$$ Then by continuity, you can raise to the $x^{th}$ power inside the limit, $$\lim_{n\to\infty}\left(1+\dfrac1n\right)^{nx}=\lim_{n\to\infty}\left(1+\dfrac xn\right)^n=e^x.$$ For ...

2

Let $d = \gcd(i,p-1)$. Note that $(g^x)^i \equiv (g^y)^i \mod p$ iff $g^{ix - iy} \equiv 1 \mod p$, and since the multiplicative order of $g \mod p$ is $p-1$, this is the case iff $i(x-y)$ is a multiple of $p-1$, and thus iff $x-y$ is divisible by $(p-1)/d$. Thus the images of $g, g^2, \ldots, g^{(p-1)/d}$ are all distinct, and by adding multiples of $(p-... 1 Symbolab used the rule$(a^m)^n=a^{m×n}$, which does not always work when$a$is negative and$m$and$n$are not integers; see this answer. 1 Briggs outlines an approach in his paper Abundant Numbers and the Riemann Hypothesis. Another method would be to multiply the primes found in the integer sequence A073751. 1 Given some$\delta > 0,$the correct exponent (to build a Colossally Abundant Number by prime factorization) for some prime$p$is $$\left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor \; - \; 1.$$ This is Theorem 10 on page 455 of Alaoglu and Erdos (1944). For a fixed$\delta,$the exponents either stay the same ... 1 I would suggest trying it backwards: given some$\epsilon >0$, it is easy to find an$N$such that$\frac{\sigma(n)}{n^{1+\epsilon}} < 3/2^{1+\epsilon}$for all$n \geq N$(because$\sigma(n) \leq n\ln{n}+n$). Then you just optimize$\frac{\sigma(n)}{n^{1+\epsilon}}$over$2 \leq n \leq N$. 1 Knowing the following fairly common limit is very useful: $$\lim_{x \rightarrow \infty}\left(1+\frac{1}{x} \right)^x=e \approx 3$$ $$50^{50}<49^{51}$$ $$\left(\frac{50}{49}\right)^{50}<49$$ $$\left(1+\frac{1}{49}\right)^{50}<49$$ $$\left(1+\frac{1}{49}\right)^{50} \approx e \approx 3$$ $$3<49$$ Therefore we conclude that$50^{50}<49^{51}\$

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