continuity - Problem of continuous real valued function

Which ofthe following statements are true?

a.If $f:\mathbb R\to\mathbb R$ is injective and continuous, then it is strictly monotonic.

b.If $f\in C[0,2]$ is such that $f(0)=f(2)$,then there exists $x_1,x_2$ in [0,2] such that $x_1-x_2=1$ and $f(x_1)=f(x_2).$

c.Let $f$ and $g$ be continuous real valued function on $\mathbb R$ such that for all $x\in \mathbb R$ wehave $f(g(x))=g(f(x.))$ If there exists $x_0\in \mathbb R $ such that $f(f(x_0))=g(g(x_0))$ then there exists $x_1\in\mathbb R$ such that $f(x_1)=g(x_1).$

geometry - Numbers of circles around a circle

"When you draw a circle in a plane of radius $1$ you can perfectly surround it with $6$ other circles of the same radius."

BUT when you draw a circle in a plane of radius $1$ and try to perfectly surround the central circle with $7$ circles you have to change the radius of the surround circles.

How can I find the radius of the surround circles if I want to use more that $6$ circles?

ex :

$7$ circles of radius $0.4$

$8$ circles of radius $0.2$

Complex power series and radius of convergence

Let $c$ be a non-zero complex number, and consider the power series
\begin{equation}
S(z)=\frac{z-c}{c}-\frac{(z-c)^2}{2c^2}+\frac{(z-c)^3}{3c^3}-\ldots.
\end{equation}

By using the Ratio Test, or otherwise, show that the series has radius of convergence $|c|$. By differentiating term by term, show that $S'(z)= \frac{1}{z}$.

I've never done power series in complex analysis, so this is what I've attempted so far:

\begin{equation}
S(z)=\frac{z-c}{c}-\frac{(z-c)^2}{2c^2}+\frac{(z-c)^3}{3c^3}-\ldots\\
=\sum_{n=0}^{\infty} \frac{(-1)^n (z-c)^n}{nc}.
\end{equation}
Let $x_{n}=\frac{(-1)^n}{nc}$. Using the Ratio test, we have:
\begin{equation}

lim_{n \rightarrow \infty} \lvert \frac{x_{n+1}(z-c)^{n+1}}{x_{n}(z-c)^n} \rvert = \lvert z-c \rvert lim_{n \rightarrow \infty} \lvert - \frac{n}{n+1} \rvert = -\lvert z-c \rvert <1.
\end{equation}

I'm pretty sure this is wrong somewhere, but I have no idea how to continue to show that the radius of convergence is $|c|$.

Answer

Notice (hint):

First of all:

• When $n$ 'starts' with $0$ than we've got a problem, because we get (dividing by $0$):

$$\frac{(-1)^0(z-c)^0}{0c}$$

• Use the ratio test, to proof that this series converges, when $|c-z|<1$.


• So:

$$\sum_{n=1}^{\infty}\frac{(-1)^n(z-c)^n}{cn}=-\frac{\ln(1+z-c)}{c}\space\space\space\space\space\space\text{when}\space|c-z|<1$$

real analysis - Is there a bijection between $[0, infty)$ and $(0,1)$

I tried $\tan(x)$, and $\log(x)$, but seems it does not work, so I wonder is there a bijection or not?

geometry - Viviani's theorem and ellipse

With reference to this post,

A conjecture related to Viviani's theorem,

it was suggested (thanks to user Aretino) that, in the top picture (where the triangle is equilateral and the ellipse pass through two vertex of the triangle and tangent there to two sides), $P$ belongs to the arc of ellipse if $\gamma^2=2\alpha\beta$. Here, $\alpha,\beta,\gamma$ are the segments whose lengths are the distances of $P$ from each side.

My first question is:

How can I prove this?

The second question is related to the second picture:

Does $P'$ belong to the arc of circle if and only if $\gamma'^2=2\alpha'\beta'$?

Notice that, in the first picture the segments $\alpha,\beta,\gamma$ sum up to the altitude of the triangle, whereas, in the second one, the segments $\alpha',\beta',\gamma'$ sum up to the side of the triangle.

The most general issue is:

Is there any relationship between the two scenarios?

Answer

For the first part please refer to my answer to the original question. I will repeat here the argument for the second case, but there must be an error in your question, because the locus of points such that $\gamma'^2=2\alpha'\beta'$ is the same ellipse as in the answer to the first part. This is obvious, because
$\alpha'=(2/\sqrt3)\alpha$, $\beta'=(2/\sqrt3)\beta$,
$\gamma'=(2/\sqrt3)\gamma$.

A circle is indeed the locus of $P$ when $\gamma'^2=\alpha'\beta'$.
In this case it is easier to find the answer using cartesian orthogonal coordinates, with $x$ axis along side $AB$ of the triangle (see diagram below).
For simplicity I also set $AB=AC=1$.
Coordinates $(x,y)$ of point $P$ have a simple relation with $\alpha'$ and $\beta'$:
$$
y={\sqrt3\over2}\alpha',\quad x=\beta'+{1\over2}\alpha'

\quad\text{and}\quad
\gamma'=1-\alpha'-\beta'\quad
\text{by the given assumption.}
$$
Plugging these into $\gamma'^2=\alpha'\beta'$ gives a simple equation for $P$:
$$
(x-1)^2+\bigg(y-{1\over\sqrt3}\bigg)^2={1\over3}.
$$
This is the equation of a circle with center $O=(1,1/\sqrt3)$ and radius $1/\sqrt3$.

summation - Sum with irrational powers and binomial coefficients

What is the value of:
$$\sum_{k=1}^{n} {\left(\left(-1\right)^{n-k}{n+1 \choose k+1}\left(\frac{\sqrt5}{1+\sqrt5}\left(3+\sqrt5\right)^k-\frac{\sqrt5}{1-\sqrt5}\left(3-\sqrt5\right)^k-\frac{5}{2}\right)\right)}?$$
I am stuck because of the binomial coefficient there, because without it the sum would just be a bunch of geometric series.

Answer

$\sum_{k=1}^{n} {\left(\left(-1\right)^{n-k}{n+1 \choose k+1}\left(\frac{\sqrt5}{1+\sqrt5}\left(3+\sqrt5\right)^k-\frac{\sqrt5}{1-\sqrt5}\left(3-\sqrt5\right)^k-\frac{5}{2}\right)\right)}?
$

Here's a start.

I'm feeling too tired right now
to do more.

$\begin{array}\\
\sum_{k=1}^{n} {n+1 \choose k+1} x^k
&=\sum_{k=2}^{n+1} {n+1 \choose k} x^{k-1}\\
&=\frac1{x}\sum_{k=2}^{n+1} {n+1 \choose k} x^{k}\\
&=\frac1{x}\left(-1-(n+1)x+\sum_{k=0}^{n+1} {n+1 \choose k} x^{k}\right)\\
&=\frac1{x}\left((1+x)^{n+1}-1-(n+1)x\right)\\
\end{array}

$

Note that
$(3-\sqrt{5})(3+\sqrt{5})
=4
$.

If $x = -(3+\sqrt{5})$,

$\begin{array}\\

\sum_{k=1}^{n} {n+1 \choose k+1}(-1)^k (3+\sqrt{5})^k
&=\frac1{-(3+\sqrt{5})}\left((1-(3+\sqrt{5}))^{n+1}-1+(n+1)(3+\sqrt{5})\right)\\
&=\frac{-(3-\sqrt{5})}{4}\left((-1)^{n+1}(2+\sqrt{5})^{n+1}-1+(n+1)(3+\sqrt{5})\right)\\
\end{array}
$

That's all.

number theory - N and M are positive integers having same digits but in different order, N + M = 10^10, then prove that N is divisible by 10.

"N and M are positive integers having same digits but in different order, $$N + M = 10^{10}$$ then prove that N is divisible by 10."

I have tried solving the question but to no avail. Please help.

Answer

Suppose that the last two digits $p$ and $q$ of $M$ and $N$ are not zero. It means that $p+q=10$.

Case 1: $p\ne q$

Let us first consider the case where $p\ne q$. I will use concrete values for $p,q$ but you can replace them with any other two. Suppose $p=6$, $q=4$.

The numbers $M,N$ are:

$$.........6 \\
.........4$$

Now, number 4 has to appear somewhere in the first number:

$$..4......6 \\
.........4$$

Under that number we must have digit 5 in the second number:

$$..4......6 \\
..5......4$$

Now you have 5 in the second number, it has to appear somewhere in the first:

$$..4.5....6 \\
..5......4$$

The matching number under it has to be 4:

$$..4.5....6 \\
..5.4....4$$

But now you have two 4s in the second number so you have to add one more to the first number and again add one 5 below it:

$$..4.5..4.6 \\
..5.4..5.4$$

But now you have the wrong count of 5s... and you are clearly in the infinite loop that you cannot exit.

Case 2: $p=q=5$

We have ten digit numbers both ending in 5:

$$.........5\\
.........5
$$

We have to fill 9 remaining positions in each number.

Missing digits come in pairs with their sum equal to 9 (digits cannot be equal, obviously). For example:

$$.3.......5\\
.6.......5
$$

But to keep the same digits in both numbers you have to put 6 into the first number and 3 into the last number:

$$.3..6....5\\
.6..3....5

$$

We have 7 positions left. Pick any two digits and you will have only five empty positions left:

$$.3..62.7.5\\
.6..37.2.5
$$

Eventually, you will end up with two incomplete numbers with all the same digits, just in a different order, and one empty position in each one. Whatever you choose to put there will break the "symmetry", your numbers won't have the same collection of digits.

Based on these two cases, we have a conclusion: the only way to construct the requested numbers is to start with $p=q=0$.