# Tag Info

6

It seems that Euler used the letter $e$ to represent the number $2.71828...$ in one of his earliest works, a manuscript entitled "Meditation upon Experiments made recently on the firing of Cannon," dated 1727 (published in 1862). The first "published" $e$ by Euler seems to be in : Leonhard Euler, Mechanica, sive Motus scientia analytice exposita (St. ...

6

\begin{align*} \frac{d}{dt} (1 - e^{-t/\tau}) &= -\frac{d}{dt}e^{-t/\tau} \\ &= - (-1/\tau) e^{-t/\tau} \\ &= \frac{e^{-t/\tau}}{\tau} \\ &= \frac{1 - (1 - e^{-t/\tau})}{\tau} \\ &= \frac{1 - v}{\tau} \end{align*}

3

Hint: It is not so difficult to show that if a $2\times 2$ matrix $A$ has null trace than the exponential is: $$e^A=I \cos \theta +A \dfrac{\sin \theta}{\theta}$$ where $\theta= \sqrt{\det(A)}$. This comes from the definition of exponential as a series: $$e^A=\sum_{n=0}^{\infty}\frac{A^n}{n!}$$ and from the fact that, for such a matrix, we have ...

3

You first have to find a basis in which the matrix has the form $B= D+N$, where $D$ is a diagonal matrix, $N$ is nilpotents and $DN=ND$. Then, since $D$ and $N$ commute, $$\mathrm e^{B}=\mathrm e^{D}\mathrm e^{N}$$ The exponential of a diagonal matrix $D$ is the diagonal matrix with the exponentials of the diagonal elements of $D$ on the diagonal. The ...

3

You integrate with respect to $t$ so $\omega$ is considered as a constant here.

2

$$(1) \quad a \cdot x^{4/5}-b \cdot x=c$$ If you've never heard of U-substitution. It's time to learn, it's very useful. Choose $u=x^{1/5}$ and substitute into $(1)$. $$(2) \quad a \cdot (u^5)^{4/5}-b \cdot u^5=c$$ $$\Rightarrow u^4 \cdot (a-b \cdot u)=c$$ This could be solved, but it'd be very tedious and depend on the constants. I suggest newton's ...

2

Convert to $$v \ln (1-v) = \ln (2).$$ For $0 < v < 1$, the log will be negative, but $v > 0$, so there's no solution in that interval. For $1 - v < 0$ (i.e., $v > 1$) it's not clear what any power of $v$ except for even integer powers actually means. So the only hope is $1 - v > 0$, i.e., $v < 1$, and then by the first case, this ...

2

Since you marked it real analysis: if $v>0$, then $(1-v)<1$ and so $(1-v)^x<1$ for any $x>0$ (in particular, for $x=v$). If $v\leq0$, then $1-v\geq1$, and $(1-v)^{-x}\leq1$ for any $x\geq0$ (in particular, for $-x=v$). Hence, no solutions exist.

2

Write ${1-e^{-x}\over x}$ as $\int_0^1 e^{-tx}\,dt$, reverse the order of integration in the resulting double integral, then integrate.

1

Step 1: Solving for $x$ First, I apply the Laplace transform to $\frac{dx}{dt}$. This gives me $$x'_L(s)=-x(0)+sx_L(s),$$ where $x'_L(s)$ is the Laplace transform of $\frac{dx}{dt}$ and $x_L(s)$ is the Laplace transform of $x(t)$. I'm going to make the assumption that $t-t_0\geq0$ -- i.e., $t\geq t_0$. Since, by definition, $t$ cannot be less than ...

1

I know that online sources such as Wikipedia and Wolfram just state without any proof or extended discussions that the matrix exponential is well-defined and converges for any square matrix. Every matrix has an element of maximal size. $($Obviously, if anything can cause divergence, it's that one$).~$ Let its absolute value be $M.~$ So let us construct ...

1

\begin{align*} \int \frac{1}{\tau_1 - \tau_2}(e^{-t/\tau_1} - e^{-t/\tau_2}) dt &= \frac{1}{\tau_1 - \tau_2}\int e^{-t/\tau_1} dt - \frac{1}{\tau_1 - \tau_2}\int e^{-t/\tau_2} dt \\ &= \frac{-\tau_1}{\tau_1 - \tau_2} e^{-t/\tau_1} - \frac{-\tau_2}{\tau_1 - \tau_2} e^{-t/\tau_2} + C \end{align*} Since $$\int e^{-t/\tau_1} dt = -\tau_1 ... 1 \exp(v \log(1-v)) = 2, v \log(1-v) = \log(2) you have to study the function f:x\rightarrow x\log(1-x) for x<1. 1 Considering your data$$\left( \begin{array}{cc} 300 & 0.9 \\ 600 & 2.1 \\ 900 & 3.8 \\ 1200 & 6.6 \\ 1500 & 9.7 \\ 3000 & 37.5 \end{array} \right)$$as you noticed from a plot, they are nonlinear but look either to be quadratic or exponential. Assuming that time is 0 for 0 words, we can fit the model$$t=a w + b w^2$$... 1$$\lim_{n\to \infty} \left(2e^{it/\sqrt{n}} - e^{2it/\sqrt{n}}\right)^n=\lim_{n\to \infty} ...

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