# Tag Info

15

Conventionally $a^{b^c}$ means $a^{(b^c)}$. The other way of parsing it, $(a^b)^c$, yields a result equal to $a^{bc}$. In particular $(2^3)^4 = 2^{3\times 4} = 2^{12} = 4096$.

12

Fix $x\in\mathbf R$. By Bernoulli's inequality (here $n$ has to be large enough, so that $a_n x/n>-1$), $$\Bigl(1+\frac{a_nx}{n}\Bigr)^n\geq 1+a_n x.$$ On the other hand, since $$\Bigl(1+\frac{a}{n}\Bigr)^n\leq e^a$$ for all $a\in\mathbf R$, $$\Bigl(1+\frac{a_nx}{n}\Bigr)^n\leq e^{a_n x}.$$ Now use the sandwich theorem.

10

If one first defines $$e:=\lim\limits_{n\to\infty} \left(1+\frac 1n\right)^n$$ and then goes on defining a real-valued function $$\exp:\mathbb R\rightarrow\mathbb R,~x\mapsto\exp(x):=e^x,$$ one has $e^x=\exp(x)$ by definition but then has to show $\displaystyle e^x=\sum\limits_{k=0}^\infty \frac{x^k}{k!}$. This involves finding the derivative of $\exp$ ...

9

The next step is $$\left(2^x\right)^2-10\cdot2^x+16=0$$ wich is a quadratic equation in $2^x$. Then $2^x=5\pm3$ and $$x=1\text{ or }x=3.$$

9

Hint. One may recall that $$e^{-x}=\frac1{e^x}$$

8

The guiding question goes like this: What happens when you divide $1$ by a really, really large number?

8

$x=\dfrac{\log7}{\log25}$ $y=\dfrac{\log 125}{\log 7}$ $\implies xy = \dfrac{\log 125}{\log 25}= \dfrac{\log 5^3}{\log 5^2} = \boxed {\dfrac{3}{2}}$

7

Look closer at your graph! (I am going to use $k$ and $n$ here.) I get e.g. $$\frac{1}{100000}\left(\sum_{k=1}^{100000}\ln(k)\right)\approx \ln(100000)-0.999933$$ This suggests an asymptotic of $\ln(n)-1$. Indeed: $$\begin{array}{ll} \displaystyle \frac{1}{n}\sum_{k=1}^n\ln(k) & \displaystyle =\ln(n)+\frac{1}{n}\sum_{k=1}^n \big(\ln(k)-\ln(n)\big) \\ ... 7 There are a number of type-setting situations with algebraic expressions that cause problems when you try to enter them into a one-line input calculator. For example$$\frac{2+3}{4+5}$$must be entered into a calculator as$$(2+3)/(4+5)$$The horizontal line in the fraction implies brackets around the expression in the numerator and denominator. Similarly, ... 6 Elaborating on what Jack said, assume we have an nth root instead of a square root:$$y = \sqrt[n]{\log_x{\exp{\sqrt[n]{\log_x{\exp{\sqrt[n]{\log_x{\exp{\cdots}}}}}}}}}$$Then$$y = \sqrt[n]{\log_x{\exp\left(y\right)}}y = \sqrt[n]{y\log_x{e}}y^n = y\log_x{e}y^{n-1} = \log_xe$$Obviously, with n = 2, n-1 = 1, meaning y itself ... 6$$y=\sqrt{\log_x{\exp{\sqrt{\log_x{\exp{\sqrt{\log_x{\exp{\cdots}}}}}}}}}\implies y=\sqrt{\log_x\exp(y)}=\sqrt{y\log_xe}\\ \therefore y=\log_xe$$6 The complex exponential function needs some definition. There are several ways one could define it: Power series As a solution to a differential equation As the unique holomorphic extension of the real exponential function to the whole plane Via the limit definition \lim_{n \to \infty} (1+\frac{z}{n})^n Via Euler's formula ... One can take any one of ... 6 x^{2n}=(x^2)^n=\frac{1}{2^n}. So if n\neq 0, for x\in \mathbb{R} you have x^2=\frac{1}{2} then x=\frac{1}{\sqrt{2}} or x=-\frac{1}{\sqrt{2}}. 5 Usually one defines \exp x=\displaystyle \sum_{n=0}^\infty \dfrac{x^n}{n!}. It is easy to show it converges for all x (even for complex values) and it satisfies the differential equation with initial condition:$$y'=y,\quad y(0)=1.$$Now this differential equation implies the functional ralation:$$\exp(x+y)=\exp x\cdot\exp y$$which in turn implies ... 5 By Taylor series with Lagrange remainder$$ e^x=\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}+\frac{(e^x)^{(2n)}|_t}{2n!}x^{2n}=\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}+\frac{e^t}{2n!}x^{2n} $$where |t|<|x|. And$$ ...

5

This can be proved purely combinatorially using nothing more than the binomial theorem. In fact, I claim that the polynomial $P_n(x)=\left(\sum\limits_{k=0}^{2n-1} \frac{x^k}{k!}\right)\left(\sum\limits_{\ell=0}^{2n-1}(-1)^\ell \frac{x^\ell}{\ell!}\right)-1$ is even, has no constant term, and has no positive coefficients, which is sufficient to show that it ...

4

Set $z=2^x$ to obtain $z^2-10z+16=0$ and solve the quadratic

4

I'm looking at integral (1) and seeing a Laplace transform convolution integral. Note that $$\int_0^t dt' \, f(t') g(t-t') = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, F(s) G(s) e^{s t}$$ where $$F(s) = \int_0^{\infty} dt \, f(t) e^{-s t}$$ $$G(s) = \int_0^{\infty} dt \, g(t) e^{-s t}$$ In this case, we may define $$f_1(t) = t^{-3/2} ... 4 In order to eliminate e-function, it can be done as follows$$B=A\cdot (1-e^{-x})(1-e^{-x})=\frac{B}{A} (1-e^{-x})=\frac{B}{A} e^{-x}=1-\frac{B}{A}e^{-x}=\frac{A-B}{A}$$Now, taking logarithm as follows$$\ln e^{-x}=\ln\left|\frac{A-B}{A}\right|-x\ln e=\ln\left|\frac{A-B}{A}\right|x=-\ln\left|\frac{A-B}{A}\right|$$... 3 To solve an expression like x^{2n}=\frac{1}{2n} for x, you need to use logarithms. As was quite rightly pointed out, if n\neq 0 and x\in\mathbb{R}, then$$(x^2)^n=\frac{1}{2^n}\Rightarrow x^2=\frac{1}{2}\Rightarrow x=\frac{1}{\sqrt{2}}.$$If n=0, then we have a blow-up singularity. Hence we will need to travel in an arbitrarily small circuit ... 3 Notice,$$\lim_{n\to \infty}\left(1+a_n\frac{x}{n}\right)^n$$Using binomial expansion of \left(1+a_n\frac{x}{n}\right)^n & neglecting higher power terms, we get$$=\lim_{n\to \infty}\left(1+na_n\frac{x}{n}\right)=\lim_{n\to \infty}\left(1+a_n x\right)=\lim_{n\to \infty} 1+x\lim_{n\to \infty}a_n = 1+x(0)=1 $$3 In standard mathematics the notation A \mathrel{^\wedge} B is not used; one does A^B instead. In some computer science (programming) applications the infix circumflex operator notation A \mathrel{^\wedge} B is used (or rather the ASCII version A^B). What you point out is that this operator \mathrel{^\wedge} is not associative. So does A ... 3 HINT : Let (e^x)^2=e^{2x}=t. Then, you'll have$$t^2-4t=-3$$3 no just consider the 1 by 1 case. i.e. real numbers. Let X=10 and the series doesn't converge. 3 I assume that the integral of interest I is$$I=\int_0^{\infty}e^{-\left(u^2+\frac{\alpha^2}{16tu^2}\right)}du$$First, we let a^2=\frac{\alpha^2}{16t} and write$$\begin{align} I&=\int_0^{\infty} e^{-a\left(\frac{u^2}{a}+\frac{a}{u^2}\right)}du\\\\ &=e^{-2a}\int_0^{\infty} e^{-a\left(\frac{u}{\sqrt{a}}-\frac{\sqrt{a}}{u}\right)^2}du\tag 1 ...

3

The property that $(a^n)^m=(a^m)^n=a^{mn}$ is only defined for real numbers. But $\sqrt{-1}=i$ is a complex number.

2

I think computation via some kind of Jordan decomposition is pretty much the best you could do for calculation by hand. I suspect that you might make things slightly easier if you used something like Jordan real form, if you're comfortable exponentiating rotation matrices.

2

Looking here, there is an interesting expansion for large arguments $$E_1(x)=\frac{e^{-x}}{x}\sum_{k=0}^\infty \frac {k!}{(-x)^k}$$ For $x=25$, using $n$ terms, we have $$S_1=5.332970444146184 \times 10^{-13}$$ $$S_2=5.350747012293338 \times 10^{-13}$$ $$S_3=5.348613824115680 \times 10^{-13}$$ $$S_4=5.348955134224105 \times 10^{-13}$$ ...

2

Generally, power series representation is not a good choice for larger input, even if the radius of convergence if infinite. In this case, you need about 90 leading terms just simply to achieve the desired order of magnitude (and more terms for the accurate digits). Indeed, Mathematica 10.2 confirms that if we let  S_n = -\gamma - \log 25 - ...

2

Without seeing your code it is difficult. But I guess it is catastrophic cancellation. $E_1(25) \approx 5.3488997553\times 10^{-13}$ and the single terms are much larger, e.g. the for $n=10$ the value is $2628070.75729$ and for $n=24$ the term is $238585088.1445781!$ You will get similar problems is you compute $e^{-x}$ or $\sin(x)$ for $x=25$ using the ...

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