# Tag Info

8

Feynman tells the story in one of his books of anecdotes. http://www.ee.ryerson.ca/~elf/abacus/feynman.html $12$ is a very good first approximation and the linear term of the series expansion suffices to get high precision. $$\sqrt[3]{1728 + d} = 12\sqrt[3]{1+x} = 12 + 4x + O(x^2)$$ where $d = 1.03$ and $x = \frac{d}{1728}$ is, in Feynman's words, ...

5

Well, since $\frac 1x \log a = 0$, $a$ must be equal to $1$ and $x$ can be absolutely any real number whatsoever!

3

Try proving that for every natural number $m\geq 1$, $n!$ is eventually larger than $m^n$ (in the sense that $n! > m^n$ for all $n>N$, for some $N$). This implies that $\limsup_n (n!)^{-1/n} < 1/m$ for all $m$.

3

Rewrite as $$\frac18\cdot\frac1{\Bigl(1-\dfrac x2\Bigr)^3}\quad\text{and}\quad\frac12\cdot\frac1{\Bigl(1-\dfrac x{32}\Bigr)^{1/5}},$$ and use the binomial formula.

3

Applying Parseval's identity (see e.g. Proof of Parseval's identity) to $$f'(z) = \sum_{n=0}^\infty (n+1)a_{n+1} z^n$$ gives $$\int_{0}^{2 \pi} \lvert f'(re^{i\theta})\rvert^2 \, d\theta = 2 \pi \sum_{n=0}^\infty (n+1)^2 \lvert a_{n+1} \rvert^2 r^{2n}$$ for $0 \le r < 1$. If $f$ is injective on the unit disk $D$ then $$\text{area}\, f(D) = ... 2 M-test works if you're working on a bounded interval - just need that \vert \arctan x \vert \le \vert x \vert and that \sum_i \frac{1}{i^2} is convergent. A tidy proof of this for x>0 comes from the following:$$0\le\frac{1}{1+t^2}\le1\implies0\le\int_0^x\frac{1}{1+t^2}dt\le\int_0^x1dt\implies0\le\arctan x\le x$$and an entirely analogous proof ... 2 For starters,$$ \frac{d}{dx}\arctan\frac{x}{k^2} = \frac{\frac{1}{k^2}}{1+\left(\frac{x}{k^2}\right)^2} =\frac{1}{k^2+\frac{x^2}{k^2}}$$and for x\to 0$$ \sum_{k\geq 1}\frac{1}{k^2}=\frac{\pi^2}{6}.$$Moreover, by the AM-GM inequality$$ \frac{1}{k^2+\frac{x^2}{k^2}}=\frac{1}{\frac{k^2}{2}+\frac{k^2}{2}+\frac{x^2}{k^2}}\leq \frac{2}{k^2+|x|\sqrt{8}} $$... 2 Denote \sum a_n x^n the given series then$$\left\vert\frac{a_{n+1}}{a_n}\right\vert=\left(1+\frac1n\right)^{-n}\xrightarrow{n\to\infty}\frac1e$$so by the ratio test, the radius of convergence is e. 2 There are at least n/2 numbers in \{1,\ldots,n\} that are greater than or equal to \frac{n}2, so$$n! \ge \left(\frac{n}2\right)^{n/2} \implies (n!)^{1/n} \ge \sqrt{\frac{n}2} \to \infty.$$2 Denote \sum a_n x^n this series. We have$$\frac{a_{n+1}}{a_n}=\left(1+\frac1n\right)^n\frac1{n+1}\xrightarrow{n\to\infty} e\times0=0$$so by the ratio test the radius of convergence is R=\frac10=+\infty. 2 If we try the ratio test,$$ \frac{\frac{(n+1)^{n+1}x^{n+1}}{((n+1)!)^2}}{\frac{n^nx^n}{(n!)^2}} =\frac{(n+1)^{n+1}x}{n^n(n+1)^2}=\left(1+\frac1n\right)^n\frac{x}{n+1}\to0. $$As n\to\infty. So the series converges for all x. 2 One may recall that, as x \to 0, by the Taylor series expansion, we have$$ \frac1{1+x}=1-x+O(x^2). $$We assume g\neq0 and  u \to 0. Then we have$$ \begin{align} \frac{g}{gu+\cdots+u^g}&=\frac1{u+u\left(1/g+\cdots+u^{g}/g\right)} \\\\&=\frac1{u}\frac1{1+u\left(1/g+\cdots+u^{g-1}/g\right)} ...

2

If you look at the left-hadn side of the equals sign, you have \begin{align} \frac{g}{gu + \cdots + u^g} - \frac{1}{u} &= \frac{1}{u}\frac{g}{g + {g \choose 2} u + \cdots + u^{g-1}} - \frac{1}{u} \\ &= \frac{1}{u}\frac{1}{1 + \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}} - \frac{1}{u} \end{align} If you then expand the first term ...

2

Expanding using the binomial theorem and applying upper negation gives\begin{align} A=\left(\frac 1{1+t}\right)^a=\sum_{r=0}^\infty\binom{-a}r t^r =\sum_{r=0}^\infty \underbrace{(-1)^r\binom{a+r-1}{r}t^r}_{f(r)}\\ B=\left(\frac 1{1+t}\right)^b=\sum_{r=0}^\infty\binom{-b}r t^r= \sum_{r=0}^\infty \underbrace{(-1)^r\binom{b+r-1}{r}t^r}_{g(r)}\end{align} ...

1

You're looking at $$\sum_{n=0}^{\infty}\left(\frac{n x}{(n!)^{2/n}}\right)^n.$$ If the expression in parentheses ever drops below $1$ and stays there, then the series converges geometrically. If you know that $(n!)^{1/n}\sim n/e$, then you can see that this will eventually happen for any $x$, since the expression in parentheses is asymptotic to ...

1

If $a =1$, any $x\neq 0$ will do. Otherwise, you would need $1/x=0$ which of course never happens (in the usual $\Bbb{R}$)

1

You could say $\lim_{x\to\infty}a^{1/x}=1$, but has been already noted there is no specific value for $x$ that results in $a^{1/x}$ equaling $1$.

1

Note that $$\sqrt[n]{n!}=e^{\frac1n \log(n!)} \tag 1$$ Next, we can write \begin{align} \frac1n \log(n!)&=\frac1n \sum_{k=1}^n\log(k)\\\\ &=\log(n)+\frac1n \sum_{k=1}^n\log(k/n) \tag 2 \end{align} Finally, note that $$-1=\int_0^1 \log(x)\,dx\le \frac1n \sum_{k=1}^n\log(k/n)\le 0 \tag 3$$ Putting together $(1)-(3)$, we find that $$\frac ... 1 For n>1 we have (n!)^{-1/n}= (\prod_{j=1}^n(1/j))^{1/n}< (\sum_{j=1}^n (1/j))/n= (1/n)O(\ln n)=o(1). You can also use Stirling's Formula : n!=(1+d_n)(n/e)^n\sqrt {2 \pi n} where |d_n|<1/6 n for n\geq 1. For this Q, nothing as precise as this is needed: When m\geq 2 we have \ln m>\int_{m-1}^m \ln x \;dx.\; So \ln ... 1 I upvoted the other answer as it comes from the other book of anecdotes, but one way to calculate this without Feynman's experience might be the following. You can start with the linear approximation, and then you have to calculate (\dfrac{1.03}{3})/12^2. \dfrac{1.03}{3}\approx .34333, but that doesn't lend itself to division by 12^2, so you can ... 1 Since an infinite geometric series converges iff \;|r|<1\; , with \;r= the series fixed ratio, the first given series already tells you that the first three options are false. For the second series you can use the \;n-th root test for the absolute value: ... 1 Let x=e^t. We are interested in the behaviour of \frac{t}{e^t}as t\to\infty. Use the fact that for positive t we have e^t\gt 1+t+\frac{t^2}{2!}\gt \frac{t^2}{2}. 1 Let's start with \displaystyle\lim_{x\to\infty}\frac{\ln(x)}{x} = L. Then we can take \displaystyle\lim_{x\to\infty}e^\frac{\ln(x)}{x} = e^L which is the same as saying \displaystyle\lim_{x\to\infty}(e^{\ln(x)})^\frac{1}{x} = e^L. By properties of exponential and logarithm functions, we get \displaystyle\lim_{x\to\infty}x^\frac{1}{x} = e^L. Hopefully ... 1 Hint: f is continuous for |x|<R - conclude that c_0=f(0)=0. It is also differentiable there, and f'(x)=\sum_{n=0}^\infty (n+1)c_{n+1}x^n. - Use Rolle to find a sequence of zeroes of f' converging to 0. 1 Supposing that not all the c_k are 0, let n_0=\min{\{n\geq 0, c_n\neq 0\}} Then f(x)=\sum_{k=n_0}^\infty c_kx^k=x^{n_0}(c_{n_0}+\sum_{k=n_0+1}^\infty c_kx^{k-n_0}) The power series g(x)=\sum_{k=n_0+1}^\infty c_kx^{k-n_0} has the same radius as f and g(0)=0. g being continuous at x=0, there is a neighborhood V of 0 such that ... 1$$\int_{z-1}^{+\infty} x^{-2} dx\int_{0}^{1} x^{-z} dx \int_{-\infty}^{+\infty} x^2\sqrt{\frac{1-z}{2\pi}}e^{\displaystyle-\frac{x^2(1-z)}{2}}dx $$1 There exists a known function, Jacobi theta function, defined as$$ \vartheta(z; \tau):=\sum_{n=-\infty}^\infty w^{n^2}\eta^n \tag1 $$where w=:e^{\pi/\tau} and where \eta:=e^{2i\pi z} . You may have a look at many interesting properties of \vartheta(\cdot;\cdot). In particular, the Jacobi triple product tells us that for complex numbers w ... 1 It's a second order differential equation, so there are 2 arbitrary constants in the general solution. As you could see, there was a constraint on c_2, but there is none on c_0 and c_1, and there is no mutual constraint between the latter two constants. It seems most convenient to separate things out into 2 series, one where all the exponents are ... 1 We can see that each solution is completed characterized by two numbers c_0 and c_1. Also, the terms that are influeced by c_0 do not overlap with the terms that are influenced by the terms influenced by c_1. In particular observe the relationship:$$y(c_0,c_1)=c_0y(1,0)+c_1y(0,1)

1

As a complement to the previous answers, observe that you can give initial conditions to select a unique solutions. The choice of initial data corresponds with the choice of the constanta $c_0$ and $c_1$: $c_0=1$ and $c_1=0$ corresponds to $y(0)=1$ and $y'(0)=0$. $c_0=0$ and $c_1=1$ corresponds to $y(0)=0$ and $y'(0)=1$.

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