TITLE: Limit of trig function raised to trig function
QUESTION [1 upvotes]: I need help on evaluating this limit:
$\lim_{x\to0}(\frac{\sin{x}}{x})^{\frac{1}{1-\cos{x}}}$. So far I have tried setting the limit to L and taking the ln of both sides. So $\ln{L}=\lim_{x\to0}(\frac{1}{1-\cos{x}}\ln{(\frac{\sin{x}}{x})})$. I then tried using L'Hopital on the indeterminate 0/0 to no luck. Any hints would be appreciated.

REPLY [0 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\,\mrm{sinc}:\mathbb{R} \to \mathbb{R}}$ is an even  function such that $\ds{\,\mrm{sinc}'}$ is an odd function, $\ds{\,\mrm{sinc}''}$ is an $\ds{\color{#f00}{even}}$ function, $\ds{\,\mrm{sinc}'''}$ is an odd function and so on.

  $\ds{\,\mrm{sinc}\pars{x} =
\left\{\begin{array}{lcl}
\ds{\sin\pars{x} \over x} & \mbox{if} & \ds{x \not = 0}
\\
\ds{1} & \mbox{if} & \ds{x = 0}
\end{array}\right.}$

\begin{align}
\lim_{x \to 0}\bracks{\sin\pars{x} \over x}^{1/\bracks{1-\cos\pars{x}}} & =
\exp\pars{\lim_{x \to 0}{\ln\pars{\mrm{sinc}\pars{x}} \over 1 - \cos\pars{x}}} =
\exp\pars{\lim_{x \to 0}{{\mrm{sinc}'\pars{x}} \over \sin\pars{x}\,\mrm{sinc}\pars{x}}}
\\[5mm] & =
\exp\pars{\lim_{x \to 0}{{\mrm{sinc}'\pars{x}} \over \sin\pars{x}}} =
\exp\pars{\lim_{x \to 0}{{\mrm{sinc}''\pars{x}} \over \cos\pars{x}}}
\\[5mm] & =
\exp\pars{\lim_{x \to 0}{\mrm{sinc}''\pars{x}}} =
\exp\pars{\lim_{x \to 0}\bracks{-\,{1 \over 3} + {x^{2} \over 10} + \,\mrm{O}\pars{x^{4}}}}
\\[5mm] & =
\bbx{\expo{-1/3}} \approx 0.7165
\end{align}


TITLE: Prove that the Jordan measure of set remains the same under $\phi$
QUESTION [0 upvotes]: The problem is as follows (from a multivariable calculus course, so preferably a solution without deep measure-theory):
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuously differentiable function. We define $$\phi:\mathbb{R}^2 \rightarrow \mathbb{R}^2, \phi(x, y)=(x+f(x+y), y - f(x+y)).$$
Prove that the Jordan measure (area) of any Jordan measurable set $E$ is equal to $\phi(E)$.
My experience with the Jordan measure is only integral-related, so I'm unfamiliar with any of its properties beyond the fact that it can be used to define areas and volumes. Also, I seem to remember that it can be defined analogously to the Lebesgue measure as follows:
$$\mu(E)=\inf\{\sum_{k=1}^{n}\mu(B_{k}) | B_{k} \space \textrm{is a finite sequence of open balls such that} \space E \subseteq \bigcup\limits_{k=1}^{n} B_{k}\},$$ but I'm neither sure that this is useful, nor 100% sure if it's true, for that matter. Any hint would be appreciated.

REPLY [0 votes]: This is a preliminary answer and I'm grateful for any (even the most minute) criticisms, nitpickings and comments on my solution, as I barely know what Jordan measures are (although I believe I got the gist of it).
If $E$ is Jordan measurable, then the integral $$I=\iint_E \,dx\,dy$$  exists. By substituting $(x,y) \mapsto (x+f(x+y),y-f(x+y))$, we get $$I=\iint_{\phi(E)}J(\phi)dxdy,$$ where $J(\phi)$ (extremely informaly) denotes the Jacobian of the substitution, which is equal to: $$\begin{vmatrix}
1+f'(x+y) & f'(x+y)\\
-f'(x+y) & 1-f'(x+y)
\end{vmatrix}=1,$$ which completes the proof.
Are there any auxiliary statements that need to be proven or at least noted before I can claim that this substitution will work?


TITLE: Finding the Units Digit to 7 to the 2945
QUESTION [3 upvotes]: How would I go about finding the Units Digit to 7^(2945)?
I know that:
7^0 = 1
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
...
7^9 = 40353607

REPLY [2 votes]: when u are interested only in the unit digit you can multiply step by step like u did in your post but you can forget the other digits.
So you have $$7^2 = 49$$ however you forget what was it exactly and u only remember that $$7^2 = x\cdot 10 + 9 $$
you multiply again
$$7^3=7^2 \cdot 7 = 70x+63 = (7x+6) \cdot 10+3$$
yet again you only remembering that it looked like
$$7^3=x_3 \cdot 10 +3$$
as you can see you dont have to remember $7^9 = 40353607$, $7^9 =x_9 \cdot 10 +7$ will be enougth
so now we can repeat you what u did but make it simpler. We can note that

if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 7
if the $7^n$ ends with $7$ then $7^{n+1}$ ends with 9
if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 3
if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 1

so the sequence of units digits will be $(1,7,9,3,1,7,3,9,\ldots)$
now you want 2945th element of that sequence. The value repeats itself evey 4 steps.$$a_n=a_{n-4}$$
and also
$$a_n=a_{n-20}$$
$$a_n=a_{n-100}$$
so u can say that:
$$a_{2945}=a_{45}=a_5=a_1=7^1=7$$


text:
\begin{definition}[Definition:Equivalent Topological Bases]
Let $X$ be a [[Definition:Set|set]].
Let $\mathbb S_1$ and $\mathbb S_2$ be [[Definition:Subset|subsets]] of $\powerset X$, the [[Definition:Power Set|power set]] of $X$.
Let $\mathbb S_1$ and $\mathbb S_2$ be used as a [[Definition:Synthetic Basis|synthetic basis]] or [[Definition:Synthetic Sub-Basis|synthetic sub-basis]] to [[Definition:Generated Topology|generate topologies for $X$]].
Let $\tau_1$ and $\tau_2$ be the [[Union from Synthetic Basis is Topology|topologies arising from $\mathbb S_1$ and $\mathbb S_2$]] respectively.
Then $\mathbb S_1$ and $\mathbb S_2$ are '''equivalent''' {{iff}} $\tau_1 = \tau_2$.
\end{definition}

config: wiki


TITLE: problem in realizing "piece of cake"
QUESTION [1 upvotes]: I have a problem in realizing of this question. As it is mentioned in the picture, we have a cake which is sliced roughly in half, the largest part being chosen each time and the other pieces discarded. I cannot understands the proportions it considers for a random cut:

And why it just considers 2/3 and 3/5 in computing the E[log C1] ? Where are 1/3 and 2/5?

REPLY [1 votes]: Here's basically what's going on in this problem: At each turn, the cake is replaced by some portion of it, which is either $2/3$ (with probability $3/4$), or $3/5$ (with probability $1/4$).  The other pieces are always discarded.
Suppose that it was always $2/3$ (with probability $1$).  Then each turn, the cake would be $2/3$ of its previous size, which means that after $k$ turns, it would be $(2/3)^k$ of its original size:
$$
\text{Size}_k = \text{Size}_0 \left(\frac23\right)^k
$$
or equivalently,
$$
\frac{\text{Size}_k}{\text{Size}_0} = \left(\frac23\right)^k
$$
However, the actual situation is that $1/4$ of the time, the ratio is $3/5$.  That means that of the $k$ fractional factors on the right-hand side, about a fourth of them are $3/5$ and not $2/3$.  Heuristically,
$$
\frac{\text{Size}_k}{\text{Size}_0} \sim \left(\frac23\right)^{3k/4}\left(\frac35\right)^{k/4}
$$
where I intend the $\sim$ to mean that "varies approximately as".  If you take the log of both sides, you get
$$
\log \frac{\text{Size}_k}{\text{Size}_0} \sim \frac{3k}{4} \log \frac23 + \frac{k}{4} \log \frac35
$$
and then dividing both sides by $k$ gives you the provided solution (provided you replace the informal $\sim$ with a more rigorous limit demonstration).


TITLE: The Domain of this Calculus II Function
QUESTION [0 upvotes]: I am trying to find the domain of this function:

I thought the domain to be (2, infinity) because (x-2) is 0 when x = 2 and ln(x-2) is undefined when x = 2. However the correct answer is (2, 2 + e) U (2 + e, infinity). What causes a problem when x = 2 + e in this problem?
Thanks

REPLY [1 votes]: Note that
$$(1 - \ln (2 + e - 2))^2 = (1 - \ln (e))^2 = (1-1)^2 = 0$$