Proving that $\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}$ After numerical analysis it seems that

$$
\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}
$$

Could someone prove the validity of such identity?

Using the formula for a geometric series,
$$
\begin{align}
\sum_{k=1}^\infty\frac1{x^{2k}}
&=\frac1{x^2-1}\\
&=\frac12\left(\frac1{x-1}-\frac1{x+1}\right)\tag{1}
\end{align}
$$
Differentiating $(1)$ twice,
$$
\sum_{k=1}^\infty\frac{2k(2k+1)}{x^{2k+2}}
=\frac1{(x-1)^3}-\frac1{(x+1)^3}\tag{2}
$$
Changing the order of summation and applying $(2)$,
$$
\begin{align}
1-\sum_{k=1}^\infty\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}
&=1-\sum_{k=1}^\infty\sum_{j=1}^\infty\frac{2k(2k+1)}{(4j)^{2k+2}}\\
&=1-\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{2k(2k+1)}{(4j)^{2k+2}}\\
&=1-\sum_{j=1}^\infty\left(\frac1{(4j-1)^3}-\frac1{(4j+1)^3}\right)\\
&=1-\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)^3}\\
&=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\tag{3}
\end{align}
$$
The sum in $(3)$ can be generalized as
$$
\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\tag{4}
$$
and is known as the Dirichlet beta function. As shown below, $\beta(3)=\dfrac{\pi^3}{32}$.

We can develop a recurrence for $\beta(2k+1)$. First, the generating function for $\beta(2k+1)$ is
$$
\begin{align}
f(x) &= \sum_{k=0}^\infty \beta(2k+1) x^{2k+1}\\
     &= \sum_{n=0}^\infty \sum_{k=0}^\infty (-1)^n\left(\frac{x}{2n+1}\right)^{2k+1}\\
     &= \sum_{n=0}^\infty (-1)^n\frac{\frac{x}{2n+1}}{1-\left(\frac{x}{2n+1}\right)^2}\\
     &= \frac{x}{2} \sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+x}+\frac{1}{2n+1-x}\right)\\
     &= \frac{x}{2} \sum_{n=-\infty}^{+\infty}(-1)^n \frac{1}{2n+1+x}\\
     &= \frac{x}{4} \sum_{n=-\infty}^{+\infty}(-1)^n \frac{1}{n+\tfrac{1+x}{2}}\\
     &= \frac{x}{4} \pi \csc\left(\pi\tfrac{1+x}{2}\right)\\[3pt]
     &= \frac{\pi}{4} x \sec\left(\frac{\pi}{2}x\right)\tag{5}
\end{align}
$$
where we use $(7)$ from this answer to get
$$
\begin{align}
\sum_{n=-\infty}^{+\infty}\frac{(-1)^n}{n+z}
&=\sum_{n=-\infty}^{+\infty}\frac2{2n+z}-\sum_{n=-\infty}^{+\infty}\frac1{n+z}\\[3pt]
&=\pi\cot(\pi z/2)-\pi\cot(\pi z)\\[9pt]
&=\pi\csc(\pi z)\tag{6}
\end{align}
$$
We can use equation $(5)$ to develop a recurrence relation:
$$
\begin{align}
\frac{\pi}{4} x
&= \cos\left(\frac{\pi}{2} x\right) f(x)\\
&= \sum_{n=0}^\infty\sum_{k=0}^n (-1)^k \frac{(\frac{\pi}{2} x)^{2k}}{(2k)!}\;\beta(2n-2k+1)x^{2n-2k+1}\\
&= \sum_{n=0}^\infty\sum_{k=0}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)x^{2n+1}\tag{7}
\end{align}
$$
For $n=0$, we can use the arctan series to get
$$
\beta(1) = \frac{\pi}{4}\tag{8}
$$
and for $n\gt0$, $(7)$ gives
$$
\beta(2n+1) = -\sum_{k=1}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)\tag{9}
$$
Recursion $(9)$ yields
$$
\begin{align}
\beta(1)&=\frac{\pi}{4}\\
\beta(3)&=\frac{\pi^3}{32}\\
\beta(5)&=\frac{5\pi^5}{1536}\\
\beta(7)&=\frac{61\pi^7}{184320}\\
\beta(9)&=\frac{277\pi^9}{8257536}\\
\beta(11)&=\frac{50521\pi^{11}}{14863564800}\\
\beta(13)&=\frac{540553\pi^{13}}{1569592442880}\\
\beta(15)&=\frac{199360981\pi^{15}}{5713316492083200}\\
\beta(17)&=\frac{3878302429\pi^{17}}{1096956766479974400}\\
\beta(19)&=\frac{2404879675441\pi^{19}}{6713375410857443328000}
\end{align}
$$

First, Mathematica confirms this. For the proof, I would assume that expanding out the zeta and then changing the order of summation will do the trick (you will get sums of the sort
$\sum_{k=1}^\infty 2k(2k+1)\frac{1}{{(4 i)}^{2k+2}},$ which are easy to evaluate (by integrating the sum of the geometric series a couple of times):

$$\frac{2 \left(48 i^2+1\right)}{\left(16 i^2-1\right)^3}$$


Probability that the sum of two discrete random variables is at most 2 Given 2 Random variables and the following sample spaces: $$X = \{0,1,2,3\}$$ and $$Y=\{0,1,2,3\}$$

How could I find the probability that: $$P(X+Y \le 2)$$

My attempt is:$$P(X+Y \le2)=P(X=0)+P(Y=1)*P(X=0)+P(Y=2)*P(X=1)+P(Y=0)*P(X=1)+P(Y=1)*P(X=2)+P(Y=0)$$

EDIT: Wow I came back and had all these responses.... I apologize. But $X$ and $Y$ are the random variables that represent the amount of times a head was flipped from 3 coin tosses. Coin A has 50/50 chance of getting heads or tails. Coin B has 1/4 chance of getting heads.

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

  "Assuming $\ds{X}$ and $\ds{Y}$ are uniformly distributed over their ranges" as $\texttt{@Fimpellizieri}$ said:

The sum is over $\ds{m,n\ \in \braces{0,1,2,3}}$ with the constrain
$\ds{m + n = s}$ for a given $\ds{s \in \braces{0,1,2}}$. The final result is found by adding contributions over $\ds{s \in \braces{0,1,2}}$:
\begin{align}
&\mrm{P}\pars{X + Y \leq 2} = \sum_{s = 0}^{2}\braces{%
{1 \over 4}\sum_{m = 0}^{3}
{1 \over 4}\sum_{n = 0}^{3}\bracks{m + n = s}} =
{1 \over 16}\sum_{s = 0}^{2}\sum_{m = 0}^{3}\sum_{n = 0}^{3}\bracks{n = s - m}
\\[5mm] = &\
{1 \over 16}\sum_{s = 0}^{2}\sum_{m = 0}^{3}\bracks{0 \leq s - m \leq 3} =
{1 \over 16}\sum_{s = 0}^{2}\sum_{m = 0}^{s}1 =
{1 \over 16}\sum_{s = 0}^{2}\pars{s + 1} = {6 \over 16} =
\bbox[5px,border:0.1em groove navy]{3 \over 8}
\end{align}

$P(X+Y\leq2)=$

$1-P(X+Y\geq3)=$

$1-\frac{4}{7}=$

$\frac{3}{7}$.

there are four winning cases  : $3,4,5,6$

between the following $7$ possible sums:

$0,1,2,3,4,5,6$.


Probability of IP address with all digits occuring exactly once I just realized that my IP address have all digits from $1$ to $9$ occurring exactly once, e.g. $91.745.86.23$ (though that is not my real address). Just for fun, could anyone help me calculate the probability of that happening, assuming that all IP addresses have the same probability? I could probably brute force it with a Python script, but I'm interested in how one would do it analytically.

It might be obvious, but remember that each field in an IP address range from $0$ to $255$.

The question can be reformulated to: If we 'concatenate' four numbers between 0 and 255, what's the chance the resulting number contains each of the digits 1 to 9 exactly once.

To answer this question, we need to count all possible permutations of the numbers 1 to 9. In order to get nine digits, the IP address has to either contain three two-digit numbers and one three-digit number, or two three-digit numbers, a two-digit number and a one-digit number. All three-digit numbers have to start with 1 or 2.

Case 1

There is one three-digit number. If it starts with 1, we can distribute the numbers 2 to 9 over the eight other positions, yielding $4\cdot8!$ options (times 4, since the three-digit number can be on any of the four positions). Counting the options starting with 2 is a bit harder. There are 24 three-digit numbers in $\{0,\ldots,255\}$ that can occur, starting with 2. This then leaves 6 digits that can be distributed in any possible way over the remaining 6 spots. As a result, there are $4\cdot24\cdot6!$ of these options.

Case 2

Since there are two three-digit numbers involved, we know one of these has to start with 1 and one of these has to start with 2. Omitting all three-digit numbers starting with 2 and containg 1 from the 24 mentioned in case 1, 14 remain (the lowest being 234). In each case, 4 out of 9 digits are known and the other 5 can be filled in any way we like. We now need to determine the number of orders for these blocks (the three-digit number starting with 1, the three-digit number starting with 2, the two-digit number and the 1-digit number), which is simply $4!$. In total this gives us $4!\cdot5!\cdot15$ possibilities.

As a final result, we get a chance of $$\frac{4\cdot8!+4\cdot24\cdot6!+4!\cdot5!\cdot14}{256^4}=\frac{270720}{256^4}.$$

Fortunately this agrees with the brute-force answer in the comments above.


If the price of coffee is raised by 25%, by what percentage a householder must reduce his consumption of coffee so as not increase his expenditure? If the price of coffee is raised by 25%, by what percentage a householder must reduce his consumption of coffee so as not increase his expenditure?

Price of the coffee is $x $  Expenditure is $y$   Consumption will be $xy$
Increased price of coffee is $1.25x$   Expenditure is y  consumption will be $1.25xy$

Now what to do next?

Hint:

If $p$ is the old price and $q$ is the old consumption, the new consumption $x$ must be such that:
$$
1.25p \cdot x=p\cdot q
$$

so: $x=...$

the percentage variation of consumption is:
$$
\Delta( q)\% =\frac{x-q}{q} \cdot 100=\left(\frac{x}{q}-1  \right)\cdot 100
$$

with ( from the previous equation)

$$x=\frac{q}{1.25}$$

can you do this?


Is it true that $log(i) = \frac\pi2i$ ? If so, are both of these legitimate proofs? They seem too beautiful not to be... Sorry if this is a naive question. I have not yet taken any upper level math courses involving complex numbers. However, in preparation for those courses, together with utilizing the knowledge that mathematicians such as Euler would just play around with infinite series, led me to this result. This seems too beautiful not to be true. If it is true, could someone explain the general framework from which this is derived?

Statement: $$log(i) = \frac\pi2i$$
Proof One: By Euler's Identity $e^{ix} = cos{x} + isin{x}$

So that $$e^{\frac\pi2i} = cos\frac\pi2 + isin\frac\pi2 = i$$
But $$e^{log(i)} = i$$
Thus $$e^{log(i)} = e^{\frac\pi2i}$$
So that $$log(i) = \frac\pi2i$$
Proof Two: Using Power Series
$$log(i) = log(\frac{2i}2) = log\left(\frac{(1+i)(1+i)}{(1+i)(1-i)}\right) = log\left(\frac{1+i}{1-i}\right) = log(1+i) - log(1-i)$$
But $$ log(1+i) = i - \frac{i^2}2 + \frac{i^3}3 - \frac{i^4}4 + ...$$
And $$-log(1-i) = i + \frac{i^2}2 + \frac{i^3}3 + \frac{i^4}4 + ...$$
Thus $$log(1+i) - log(1-i) = 2\{i + \frac{i^3}3 + \frac{i^5}5 + \frac{i^7}7 + ...\} = 2i\{1 - \frac13 + \frac15 - \frac17 + ...\} = 2i\frac\pi4 = \frac\pi2i$$
Therefore $$log(i) = \frac\pi2i$$

Added June 4, 2016: I found a delightful identity for logarithms of complex quantites in general as opposed to the case $a+bi = i$. Since the proof is essentially a generalization of my proof that $log(i) = \frac{\pi}{2}i$ as well as being aesthetically pleasing I thought it was worth posting.

Statement: $$log\left(a+bi\right) = \frac{1}{2}log\left(a^{2}+b^{2}\right) + itan^{-1}\frac{b}{a}$$

Proof: $$log\left(a+bi\right) = log\left({\sqrt{a^{2}+b^{2}}}{\frac{a+bi} {\sqrt{a^{2}+b^{2}}}}\right) = \frac{1}{2}log\left(a^{2}+b^{2}\right) + log\left({\frac{a+bi}{\sqrt{a^{2}+b^{2}}}}\right)$$

Now, $$log\left({\frac{a+bi}{\sqrt{a^{2}+b^{2}}}}\right) = log\left({\frac{1+i\frac{b}{a}}{\sqrt{1+\frac{b^{2}}{a^2}}}}\right) = log\left({1+i\frac{b}{a}}\right) - \frac{1}{2}log\left(1+\frac{b^{2}}{a^2}\right)$$

But, $$log\left(1+i\frac{b}{a}\right) = \left(i\frac{b}{a}\right)-\frac{1}{2}{\left(i\frac{b}{a}\right)}^{2} + \frac{1}{3}{\left(i\frac{b}{a}\right)}^{3} - \frac{1}{4}{\left(i\frac{b}{a}\right)}^{4} +... =
i\left( \left(\frac{b}{a}\right)-\frac{1}{3}{\left(\frac{b}{a}\right)}^{3} + \frac{1}{5}{\left(\frac{b}{a}\right)}^{5} - \frac{1}{7}{\left(\frac{b}{a}\right)}^{7} +... \right) +
\\ \frac{1}{2}\left(\frac{b}{a}\right)-\frac{1}{4}{\left(\frac{b}{a}\right)}^{4} + \frac{1}{6}{\left(\frac{b}{a}\right)}^{6} - \frac{1}{8}{\left(\frac{b}{a}\right)}^{8} +...$$
And,$$ \frac{1}{2}log\left(1+\frac{b^2}{a^2}\right) = \frac{1}{2}\left(\left({\frac{b^{2}}{a^{2}}}\right) - \frac{1}{2}\left({\frac{b^{2}}{a^{2}}}\right)^{2} + \frac{1}{3}\left({\frac{b^{2}}{a^{2}}}\right)^{3} - \frac{1}{4}\left({\frac{b^{2}}{a^{2}}}\right)^{4} +...\right) = \\ \frac{1}{2}\left(\frac{b}{a}\right)^{2}-\frac{1}{4}{\left(\frac{b}{a}\right)}^{4} + \frac{1}{6}{\left(\frac{b}{a}\right)}^{6} - \frac{1}{8}{\left(\frac{b}{a}\right)}^{8} +...$$
So that, $$log\left({1+i\frac{b}{a}}\right) - \frac{1}{2}log\left(1+\frac{b^{2}}{a^2}\right) = i\left( \left(\frac{b}{a}\right)-\frac{1}{3}{\left(\frac{b}{a}\right)}^{3} + \frac{1}{5}{\left(\frac{b}{a}\right)}^{5} - \frac{1}{7}{\left(\frac{b}{a}\right)}^{7} +... \right) = itan^{-1}{\frac{b}{a}}$$
Therefore, $$log(a+bi) = \frac{1}{2}log(a^{2}+b^{2}) + itan^{-1}\frac{b}{a}$$
Example: Let a=0 and b=1 and we have
$$log(i) = itan^{-1}\infty = \frac{\pi}{2}i + 2\pi mi$$

It depends on how you define "logarithm" in the first place. But the simplest and most straightforward approach would be to define

  Definition. $\log x$ means the number $y$ such that $e^y=x$.

With this definition the first two lines of your first proof is a complete, direct proof that $\log i = \frac12\pi i$.

The only problem is that with this reasoning $\log i$ can also be shown to be $\frac52\pi i$ and $\frac 92\pi i$ and $\frac{13}2\pi i\ldots$ because the exponential function of those numbers are also $i$.

This is not just a minor deficiency of the definition, though -- the complex logarithm is inherently multivalued in the precise sense that it is not possible to define a single function that is continuous on all of $\mathbb C\setminus\{0\}$ such that $e^{f(z)}=z$ everywhere.

The usual fudging one does in order to overcome this is to speak of the "principal logarithm", notated $\operatorname{Log}$ with a capital L, and define

  $\operatorname{Log} z$ means the number $y$ such that $e^y=x$ and the imaginary part of $y$ is in $(-\pi,\pi]$.

This is, unfortunately, discontinuous on the negative real axis, but we can't really do better.

Your second proof is an ingenious manipulation, but of a kind that will easily lead to nonsense if you're not careful. Basically you're taking the power series for $\log(1+z)$ which will only work for $|z|\le 1, z\ne -1$ and then just assuming that this extends to a function on all of $\mathbb C\setminus 0$ that satisfies the usual logarithm rules everywhere. But this is not the case -- with the same reasoning we could say
$$ 1 = \Bigl(\frac{1+i}{\sqrt2}\Bigr)^8 $$
so
$$ 0 = \log 1 = 8\log\Bigl(\frac{1+i}{\sqrt2}\Bigr) $$
which requires the logarithm on the right to be $0$ -- but actually the series will yield $\frac14\pi i$.

This is the multi-valuedness of the logarithm that strikes again -- it is not possible for any nontrivial differentiable function on $\mathbb C\setminus\{0\}$ to satisfy $f(ab)=f(a)+f(b)$ everywhere.

From Euler's identity $$e^{i\pi/2} = cos(\pi/2) + i sin(\pi/2) = i
$$

Take the natural log of both sides.

The left $\ln(e^{i\pi/2}) = i\pi/2$ and the right is simply $\ln i$.

Since both are equal,  $i\pi/2 = \ln i.$

So, it is not arbitrary log but ln. Otherwise, your identity is correct.


Is it possible to prove the independence of the parallel postulate without non-euclidean geometry? Euclid's fifth postulate:

  If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.

Every example (that I've seen) of proving that the parallel postulate in Euclid's five postulates is independent of the others  seems to rely on the fact that one can invent a new kind of geometry where the first four postulates hold without the existence of the parallel postulate. Hyperbolic and elliptic  geometry are often used as examples. Is this the only known method or approach to prove the parallel postulate's independence?  Is discovering these curved geometries as counterexamples the only way?

I know two ways of proving that something is not provable:
  - Using the semantic approach, one builds a counter model, i.e. in which all other axioms hold but the parallel postulate fails (this is the usual method for the parallel postulate).
  - Using the syntactic method, you can show that something is not provable more directly.

Together with Michael Beeson and Pierre Boutry, we have given a syntactic proof of the independence of the parallel postulate in the context of Tarski's axioms :
Herbrand's theorem and non-Euclidean geometry, Bulletin of Symbolic Logic, Association for Symbolic Logic, 2015, 21 (2), pp.12

The general idea of the proof is that the parallel postulate (at least some versions of this postulate) allows to construct points which are arbitrarily far from the given points (take two lines which are very close to be parallel), whereas other axioms allows only to double the maximum distance between the points constructed so far.

To explain the difference between a syntactic independence proof and a semantic one, I give an elementary example in the following talk:
 http://dpt-info.u-strasbg.fr/~narboux/slides/Herbrand-Euclid-vulgarization.pdf

Are we agreed on what the other four postulates are? As I recall, the first three do not postulate truths at all, but rather things a geometer is supposed to be able to do: 1) draw a straight line between two points, 2) extend a line indefinitely, 3) draw a circle with any given radius. This is the "ruler and compass" basis of Euclidean geometry. Besides the fifth postulate, only the fourth purports to state a truth: all right angles are equal. If Aristotle was right, that you need at least two premises to draw a conclusion, then I don't see how you could deduce the fifth postulate, a statement of fact, from just one other statement of fact, the fourth.


Do convergence in distribution along with uniform integrability imply convergence in mean? There are at least 2 places in Wikipedia saying that $X_n$ converges to $X$ in mean in and only if $X_n$ converges to $X$ in probability and $X_n$ is uniformly integrable. See the following link for example:
http://en.wikipedia.org/wiki/Convergence_of_random_variables#Properties_4

However, I found a reference in Billingsley's book saying that we only need convergence in distribution along with uniform integrability to imply the convergence in mean. Is Wikipedia wrong, or am I missing something here?

Thanks.

I don't have Billingsley's book here, but I guess what he says is that uniform integrability and convergence in distribution imply convergence of the means (and not in mean), i.e. $E[X_n]$ converges to $E[X]$.

One simple (though not truely elementary) way to see it is to use Skorokhod's theorem stating that convergence in distribution is equivalent to almost-sure convergence of copies of the random variables in some abstract probability state. Then we just apply the usual Vitali theorem about uniform integrability for these copies. It gives convergence in mean of the copies of $X_n$ to the copy of $X$, and therefore convergence of the mean of (the copy of) $X_n$ to that of (the copy of) $X$. I put parentheses since the mean only depends on the law and hence is the same for $X_n$ or a copy of it. This is the result.

Convergence of mean (or any other feature of the distribution for that matter) does not imply weak convergence of the random variables.

Take the example of an infinite sequence of iid random variables of any non degenerate distribution and see what is going wrong.


Self study text for Elementary Number theory What is a very helpful book for elementary number theory? I would like to teach myself and I need help finding a book that would be able to do this.

George Andrew's number theory is fun and good.

Ivan Niven's book is complete and covers continuous fractions really well

Ireland and Rosen's modern number theory is a master piece, but requires some notions of modern algebra

Hardy' s and Wright' s is also a master piece!


Why does $2+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}=(1+1)^n$ $${{n}\choose{0}}+ {{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}+{{n}\choose{n}}=(1+1)^n$$

I don't see why this is true, because (if I'm not mistaken)

\begin{align}&{{n}\choose{0}}+ {{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}+{{n}\choose{n}}\\&=1+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}+1\\
&=2+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}
\end{align}

So my question is why

$$2+{{n}\choose{1}}+{{n}\choose{2}}+\ldots+{{n}\choose{n-1}}=(1+1)^n\;\;\;\;\text{(1)}$$

If anyone requires any context, I am reviewing some set theory, and I came across the proof that the power set of a set $S$ with $n$ elements  had $2^{n}$ elements. Just to reiterate, I'm not looking for a proof of the cardinality of the power set. I just want to know, algebraically, why $(1)$ is true.

Thanks.

There is an easy way to prove the formula.

Suppose you want to buy a burger and you have a choice of 6 different condiments for it - mustard, mayonnaise, lettuce, tomatoes, pickles, and cheese. How many ways can you choose a combination of these condiments for your burger?

Of course, you can choose either 6 different condiments, 5 different condiments, 4 different condiments, etc. So the obvious way to solve the problem is:

$\begin{align}{{6}\choose{6}} + {{6}\choose{5}} + {{6}\choose{4}} + {{6}\choose{3}} + {{6}\choose{2}} + {{6}\choose{1}} = \boxed{64}\end{align}$

But there is a better way. Imagine 6 spaces for 6 condiments:

_____ _____ _____ _____ _____ _____

For every space, there are $2$ possible outcomes: Yes or No, meaning the condiment was chosen or the condiment was not chosen. With $2$ possible outcomes for each space, there are $2^6 = \boxed{64}$ possible ways.

We know that both ways have foolproof logic and will both give identical answers no matter how many condiments there are. So this means we have proven:

$\begin{align}\sum_{k = 0}^{n} \binom{n}{k} = 2^n\end{align}$

Well it just follows from the general formula:

$(a+b)^n=\sum\limits_{k=0}^n \binom {n} {k}\,a^k\,b^{n-k}$,

which you can prove by induction see  here .


Determining Radius of Convergence of Power Series Find the radius of convergence for the following power series:

My workings:
$$\lim_{n\rightarrow ∞}|\frac{(n+1)! (x-1)^{n+1}}{2^{n+1}(n+1)^{n+1}}\centerdot \frac{2^nn^n}{n!(x-1)^n}|$$

$$=\lim_{n\rightarrow ∞}|\frac{1}{(1+1/n)^n}\centerdot \frac{x-1}{2}|$$

$$=\lim_{n\rightarrow ∞}|\frac{x-1}{2e}|$$

$|x-1|<2e$

$|x|<2e+1$; Therefore, $R=2e+1$?

However, my answer is incorrect, as the answer given is $R=2e$

Could any of you tell me where did I go wrong?

Your calculation is fine, up to the very last step. That is, once, you have $\left | x-1 \right |\leq 2e$, you're done because the series is centered at $x_{0}=1$, so $R=2e$

Actually, you got the answer right. The disk is just not centered at the origin, since the series is centered at 1.


Can you raise a number to an irrational exponent? The way that I was taught it in 8th grade algebra, a number raised to a fractional exponent, i.e. $a^\frac x y$ is equivalent to the denominatorth root of the number raised to the numerator, i.e. $\sqrt[y]{a^x}$. So what happens when you raise a number to an irrational number? Obviously it is not so simple to break it down like above. Does an irrational exponent still have a well formed meaning?

The only example that comes to mind is Euler's identity, but that seems likes a pretty exceptional case. What about in general?

What do we mean when we write $a^b$, say for $a>0$?  The question is a very good one. The answer, unfortunately, is fairly complicated, and the full details
are quite lengthy.

We have a clear understanding of what we mean by $a^2$, or $a^5$.  And from fairly early on, we learn to define $a^n$, where $n$ is negative, as $\frac{1}{a^{-n}}$.

After a while, we develop an understanding of what we mean by something like $a^{3/4}$.  For (we are led to believe) there is a unique positive number $s$ such that $s^4=a$, and then we can define $a^{3/4}$ to be $s^3$.  This idea can be used to define $a^{p/q}$, where $p$ and $q$ are integers.

After a while, we can show, more or less rigorously,  that the laws of exponents that worked for integer powers also work for expressions of the form $x^{p/q}$, where $p$ and $q$ are integers.

However, what do we mean, for example, by $3^{\sqrt{2}}$?  Certainly it is not $3$ multiplied by itself $\sqrt{2}$ times!

There are several ways to resolve the question.  One way is to note that $\sqrt{2}=1.41421356\dots$ and consider the sequence $3^{1.4}$, $3^{1.41}$, $3^{1.414}$, $3^{1.4142}$, and so on.  All these powers make sense, because the exponents $1.4$, $1.41$, $1.414$, and so on, can be expressed as fractions.  But, intuitively, these numbers are getting closer and closer to something, and we define $3^{\sqrt{2}}$ to be that something.  We can do a partial informal verification of the "getting closer and closer" part in this case, by using a calculator.

More formally, let $b$ be a real number, and let $b_1, b_2, b_3, \dots$ be an infinite sequence of rational numbers such that the sequence $(b_n)$ has limit $b$.  It can be shown that the sequence $(a^{b_n})$ has a limit, which is independent of the particular sequence $(b_n)$ that we have chosen, as long as the sequence has $b$ as a limit.

Then we can define $a^b$ as the limit of the sequence $(a^{b_n})$.  With quite a lot of effort, we can then show that the familiar laws of exponentiation hold.

The above approach, though intuitively very natural, is unwieldy.  So in practice, we usually take another approach.

The standard way is to first define the function $\ln x$. Then we define the exponential function $\exp(x)$, also known as $e^x$, as the inverse function of $\ln x$. Or else, depending on taste, we first define the function $\exp(x)$, and then its inverse $\ln x$.  There is a fair variety of (provably equivalent) definitions.

For example, we could define $\ln x$ by
$$\ln x=\int_1^x \frac{1}{t}dt.$$
It is not terribly difficult to show that $\ln$ as defined above satisfies the usual basic "laws of logarithms," and that it is an increasing function, so has an inverse, that we call $\exp$.

Finally, after this background work, we define $a^b$ (for $a>0$) by
$$a^b=\exp(b\ln a).$$
We can then easily verify that in the cases where we already "know" what $a^b$ should be, namely rational $b$, the above definition agrees with our intuition, and that the usual "laws of exponents" hold for this more general notion of power.

Warning: In the entire post, it is assumed that $a$ is a positive real number, and that all exponents are real numbers.  Complex exponentials are a lot more--complex.

The above answers are fine provided that you are only concerned with real numbers.  However, just as $x^{a/b}$ has $b$ solutions in the complex plane, $x$ to the power of an irrational will have an infinite number of solutions.


Check whether the integrand is continuous when evaluating improper integrals In order to evaluate improper integrals, I need to know whether the integrand is continuous between the limits of the integral. For the lower and upper limits, I believe you find out if it's continuous at the points if the limit of the integrand as x tends to either the upper or lower limit exists, but how would I find out if the integrand is continuous over the whole range and not just at the upper and lower limits?

E.g for this integrand how would I find out if it's continuous over the whole range of the limits given?

$$\int_0^\infty e^{-ax}\,\frac{\sin x}{x}\,dx$$

Any help would be much appreciated.

Some functions are, by definition, continuous, and others are not. For example:

polynomials are continuous on their domain
rational functions have a discontinuity when the denominator is equal to $0$
exponential functions are continuous on their domain
split functions often have a discontinuity, when the function jumps or has a removable discontinuity
trigonometric functions such as $\sin$ and $\cos$ are continuous but $
\csc$, $\sec$, $\tan$ and $\cot$ are not continuous. In the latter case, the functions are still integrable because the integral becomes improper yet can still be evaluated.

In your example, $\lim_{x \to 0}\frac{\sin x}{x}$ exists, so the function is not discontinuous at $0$.


Variant of the Contraction Mapping Theorem Let (C, ||.||) be a closed subset of a Banach space, and let f: C -> C be a mapping such that

||f(x) - f(y)|| < ||x - y|| for all x, y in C. Must there exist a fixed point in C that f maps to itself?

Hint $C=[2, \infty) \subset \mathbb R$ and $$f(x)=x+\frac{1}{x}$$


Why does n choose k always yield a whole number? I'm just wondering why the n choose k formula

$n \choose k$ = $\frac{n!}{k!(n-k)!}$

always gets you a whole number. I know that obviously there can only be a whole number of "ways" to choose k items in n items, but I'm more interested on how the factorials divide.

Note that $k<n$. So when you write out what $n!$ contains, you will see $k!$ and $(n-k)!$ are included.


Does Real Eigenvalues mean it is an hermitian Matrix Let us say I know that a given $N\times N$ matrix has all its eigenvalues as real, does it mean, it is hermitian. How do I prove (or disprove) that?

Hint: think about the following matrix:

$$
\left(\begin{array}{cc}
1 & 1\\
0 & 1\end{array}\right)
$$

ABSOLUTELY NOT. It is easy to construct cases with real eigenvalues, even complex coefficients, and not Hermitian.


Curvature of Ellipse We all know that the curvature of a circle is defined by the equation
$$k=\frac{1}{r}$$

What about ellipses?

In terms of major axis $a$, minor axis $b$, $x$ and $y$, what is the curvature of an ellipse?

Thanks lots!

The second answer is the maximum principal curvature, as explained on pages 71-76 of "Vector and Tensor Analysis" by Harry Lass, McGraw-Hill (1950).  The denominator is G^{3/2}, where G is the third term in the first fundamental form for the Ellipse.  The minimal principal curvature is c / ( a \sqrt{G}).


What happens to the $d$ part of the plane equation $x+y+z=d$ when preforming a vector projection onto the plane? So i am working with vector projection and given a vector $v=(3,4,0)$ and a plane $x+y+z=1$ i am supposed to find the projection of $v$ onto the plane. The logical step is to just use the normal vector of the plane $n$ and do a projection $proj_n v=(v*n)/(n*n)*n$. But when i draw up the vectors and the plane, the projection is not onto the plane because of the $D$ part of the plane equation where $D=1$.

Is this the right way to solve these kinds of problems or if not what do i need to do solve it?

The easiest way to find the projection of a vector $v$ to a plane with normal $n$ is to think of  the line $g(t)=v+t\cdot n$ and compute the intersection with your given plane.

As you are given the plane in coordinate form, we can directly see the normal vector to be $n=(1,1,1)$ and we know $v=(3,4,0)$.

Putting it all in the coordinate form equation, we get
$$
3+t+4+t+0+t=1\\
\Leftrightarrow 7+3t =1\\
\Leftrightarrow t = -2
$$

Thus, the projection onto the plane is
$$
g(-2)=(3,4,0)-2\cdot(1,1,1)=(1,2,-1)
$$

A short check shows that the coordinate form fulfills
$$ 1+2-1=1,$$ so this is really inside the plane :)

$\textrm{proj}_n v = \frac{v\cdot n}{n \cdot n} n$ is the component of the vector normal to the plane. So the component of the vector parallel to the plane is $v - \textrm{proj}_n v$, or, in this case, $w=\left( \frac{2}{3},\frac{5}{3},-\frac{7}{3}\right)$.

The value of $d$ doesn't come into it regardless, because we only care about the direction of the vector and the plane. But you can check your work by taking a random point $P$ in the plane, e.g., $(1,1,-1)$, and verifying that $P + w$ is also in the plane.

You can also verify that $w \cdot \textrm{proj}_n v = 0$ and that $w + \textrm{proj}_n v = v$.


The product of digits equal to the sum of digits How to find the number(or numbers ) that has $4$ digits, the product of these digits equal to the sum of these digits ?

First of all, let's observe that all of the digits of such a number cannot be the same. You can just manually check that numbers $1111$, $2222$ and so on don't suit us. It is also clear that all of the digits should be non-zero.

Now suppose that we have such a number. Let $a,\,b,\,c,\,d$ be its digits written in non-ascending order: $a \geqslant b \geqslant c \geqslant d$. Then we have
$$
abcd = a + b + c + d.
$$

From this we have an inequality:
$$
a\cdot bcd < 4a.
$$
This inequality is strict, because at least one of $b, c, d$ is strictly smaller than a. So we have:
$$
bcd < 4,
$$
which is the same as saying
$$
bcd \leqslant 3.
$$
This only leaves us with 3 possible combinations for $(b, c, d)$: $(1, 1, 1)$, $(2, 1, 1)$ and $(3, 1, 1)$.

If $b=c=d=1$, then $a\cdot 1 \cdot 1 \cdot 1 = a + 1 + 1 + 1$, which can't be true.

If $b=2$ and $c=d=1$, then $a \cdot 2 \cdot 1 \cdot 1 = a + 2 + 1 + 1$, which means that $a=4$. This gives us one possible solution: $a=4, b=2, c=d=1$.

If $b=3$ and $c=d=1$, then $a \cdot 3 \cdot 1 \cdot 1 = a + 3 + 1 + 1$, which is impossible.

So, the only solution is $a=4$, $b=2$, $c=d=1$. There are $12$ numbers with such digits.

Must be various combinations of $1,1,2,4$.
I don't think there are any other combinations, I looked at all small numbers...


Number of ways pitchers can be selected for a baseball game . In my Probability  &  Statistics class, my professor put a question on the quiz and $95 \% $ of the class got the question wrong (putting the same answer as me). I understand why my professor said our answer was wrong, but I think the wording of question allows people to interpret the problem in different ways.

"For a baseball game, the managers plans to use one starter and two relievers. If he has $3$ starting pitchers and $6$ relievers available, in how many ways can he select pitchers for the game$?$"

Let me know what you think and then I'll tell you what my professor said and what most students put. Thank you!

I'm guessing that the issue is whether it's an ordered or unordered selection of the relief pitchers.  In a case such as this, it would be good to ask for a clarification, if that is possible.

Depends on whether the order of the relievers matters or not.  If having Smith as the first relief pitcher and Jones as the second is different from having Jones first and Smith second, then there are 3*6*5 = 90 ways to pick the pitchers.  If the order doesn't matter, then the answer is half this, or 45 different ways.

I don't know enough about baseball to know whether this distinction (between "first relief" and "second relief") is actually one that's made in practice.


How can we find the joint probability density function here? Let

$(\Omega,\mathcal A,\operatorname P)$ be a probability space
$(E,\mathcal E,\mu)$ be a measure space
$X:\Omega\to E$ be $(\mathcal A,\mathcal E)$-measurable with $$\operatorname P[X\in B]=\int_Bp\:{\rm d}\mu\;\;\;\text{for all }B\in\mathcal E$$ for some $\mathcal E$-measurable $p:E\to(0,\infty)$
$f:E\to\mathbb R$ be $\mathcal E$-measurable and $$W:=\frac{f(X)}{p(X)}$$
$\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$

  How can we show that $$\operatorname P[(W,X)\in C]=\int_C\delta\left(w-\frac{f(x)}{p(x)}\right)p(x)\:{\rm d}(\lambda\otimes\mu)(w,x)$$ for all $C\in\mathcal B(\mathbb R)\otimes\mathcal E$?

Clearly, it's sufficient to prove the claim for all $C\in\mathcal B(\mathbb R)\times\mathcal E$.

Ok, have not done measure theory for a while, and realized where my confusion comes from. I think I am not familiar with the Lebesgue-type integral, where integrands are generalized functions such as Dirac $\delta$. Namely, in MT I'd say a $\delta$ is a kernel $\delta(\mathrm dy|x)$ defined by $\delta(A|x) = 1(x\in A)$. Hence, your integral should be written more formally as
$$
  \mathsf P[(W,X) \in C] = \int\limits_C p(x) \delta(\mathrm dw| g(x))\mu(\mathrm dx)
$$
where $g = f/p$. When you consider a measurable rectangle $C = A\times B$, you get
$$
\begin{align}
\int\limits_{A\times B} p(x) \delta(\mathrm dw| g(x))\mu(\mathrm dx) &= \int\limits_B\left(\int\limits_A \delta(\mathrm dw| g(x))\right)p(x)\mu(\mathrm dx)
\\
&= \int\limits_B1(g(x)\in A)p(x)\mu(\mathrm dx) = \int_{B\cap \{x\in E:g(x)\in A\}} p(x)\mu(\mathrm dx)
\\
&= \mathsf P[X \in B, g(X) \in A] = \mathsf P[X\in B, W\in A] = \mathsf P[(W,X)\in C]
\end{align}
$$


Equilibria and stability/Proving heteroclinic orbits Consider the equation,
  $$
x'' + ax^2x' - x + x^3 = 0,\ x ∈ \Bbb R
$$
  where $0 ≤ a$ is a constant.

  It asked me to find all equilibria and determine their stability/instability.

When $a = 0$, I found the conserved quantity $$E(x,y) = 2y^2 - 2x^2 + x^4$$ with $y$ being equal to $y = x'$. However when $0 ≤ a$, I don't know how to find equilibria and their stability.

Also, it asked me to prove there is no periodic orbits or homoclinic orbits and instead prove there exists a heteroclinic orbits. How to prove them?

Thanks in advance for help.

With $v=x'+\frac a3 x^3$ you get the first order system
$$
x' = v-\frac a3 x^3\\
v' = x-x^3
$$
This gives you the stationary points at $x_k=-1,0,1$ and $v_k=\frac13x_k^3$. For the stability the first thing to try is to compute the eigenvalues of the Jacobian, if that is not sufficient other methods need to be employed.
$$
J=\pmatrix{-ax_k^2&1\\1-3x_k^2&0}
$$
with characteristic equation $$0=λ(λ+ax_k^2)-(1-3x_k^2)=(λ+\tfrac12ax_k^2)^2-(1-3x_k^2+\tfrac14a^2x_k^4)$$ which allows to compute the eigenvalues and examine their real parts.

As observed, $E(x,y)=2y^2−2x^2+x^4$ is a conserved quantity of the equation without the friction term. Computing the time derivative of $E(x,x')$ for the equation with friction results in
$$
\frac{d}{dt}E(x,x')=4x'(x''-x+x^3)=-4ax^2x'^2
$$
This means that any solution curve constantly intersects the level curves of $E$ in direction of its minimal values. As $E(x,y)=2y^2+(x^2-1)^2-1$, these minima can be found at $x'=0$, $x=\pm 1$.

A consequence is that there can be no homoclinic or periodic orbits, as no solution curve can return to the starting level curve of $E$. Second all solution curves outgoing from $(0,0)$ remain inside the region $E<0$, which is contained in $[-\sqrt2,\sqrt2]\times [-\sqrt{\frac12},\sqrt{\frac12}]$, so can only go to one of the stationary points at the minima of $E$ Which means they are heteroclinic  orbits..

Also, it asked me to prove there is no periodic orbits or homoclinic orbits and instead prove there exists a heteroclinic orbits. How to prove them?

Can you expand on this in more detail please?


Suppose $x_n$ is a decreasing sequence of positive reals with $\sum x_n$ converges, must $(n\log n)x_n \to 0$ We are able to show, using the Cauchy criterion (using sum from $n$ to $2n$) that $nx_n \to 0$ Explicitly this is $0<nx_{2n}<\displaystyle \sum_{i=n}^{2n}x_n$ and the result follows from squeeze rule. My question is can we do better?

Intuitively, it feels like we should since $\frac{n\log n}{n^k} \to 0$ for $k>1$. Unfortunately, I am struggling to prove this, Any hints would be much appreciated.

One idea I had was to say that if $\displaystyle \sum_{i=N}^{n} x_i > \sum_{i=N}^{n} \frac{1}{n^k}$ for $k>1$ and for all sufficiently large $n$ then we could somehow compare it to the harmonic series. Could we not say the tail of the series could be made as large as we want this way and so divergence would be inevitable?

Let $N_k = 2^{k^2}$ and define $(x_n)$ by $x_1 = 1$ and $x_n = \frac{1}{N_k \log N_k}$ if $N_{k-1} < n \leq N_k$ and $k \geq 1$. Then $(x_n)$ is positive and (weakly) decreasing. Also,

$$ \sum_{n=1}^{\infty} x_n
= 1 + \sum_{k=1}^{\infty} \frac{N_k - N_{k-1}}{N_k \log N_k}
\leq 1 + \sum_{k=1}^{\infty} \frac{1}{k^2\log 2}
< \infty. $$

On the other hand, $(n \log n) x_n = 1$ for infinitely many $n$, and so, it does not converge to $0$.

Hint: show that the series
$$ \sum_{n=2}^\infty \frac{1}{n (\ln n)(\ln \ln n)}$$
is divergent. From this it follows that $x_n$ must decrease faster, if $\sum_n x_n$ is to be convergent.


If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer I am trying to solve:

  If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer.

If I write $n=2k+1$ and $m=2l+1$ I get stuck at
$$\frac{1}{8}(16k^2 l^2 +4(k+l)^2 +8kl(k+l)+4kl+2(k+l))$$

All odd numbers $a$ satisfy:

$$a^2 \equiv 1 \pmod8$$

So $8 | (nm)^2 - 1$, since $nm$ is an odd number. To see that the above holds, it suffices to check that $1^2 \equiv 3^2\equiv 1\pmod8$, since all odd numbers are either congruent to $1$, $3$, $-3$ or $-1$ modulo 8.

Okay, I'm not the best at writing proofs because I'm only a senior in high school and haven't ever written one except about calculus theorems that my teacher didn't explain, so excuse any terrible structure.

Axiom: Two integers multiplied by each other produces a third integer.
An integer that has a factor of 8 is divisible by 8.

Let n and m be odd integers defined by the following:
n = 2k + 1 and
m = 2l + 1,
where k and l are integers. Therefore, the equation follows:
((nm)^2 - 1) = (((2k + 1)(2l + 1))^2 - 1) = (2k + 1)^2 * (2l + 1)^2 - 1 =
(4k^2 + 4k + 1) * (4l^2 + 4l + 1) - 1.
Note that thus far, I have excluded the 1/8. And this is becoming a pain to type like that and it will become even more confusing, but the proof will continue as follows: Multiply the factors of the two trinomial expressions. Simplify. Each term will have a factor of 8 except + 1 and - 1, which cancel each other. When multiplied by 1/8, this leave factors of 2, k, k^2, l, and l^2. Each of these is an integer (and k and l need not be odd), therefore, the result must also be an integer.


Linear Algebra (Matrixes with powers) Let
$$
M \colon= \left[ \begin{matrix}
2 & -1 \\
2 & 5
\end{matrix} \right]
$$

Find formulas for the entries of $M^n$, where $n$ is a positive integer.

$$ M^n = \text{?} $$

I think simple diagonilization makes sense here for  $M$ has two different eigenvalues, $\{3,4\}$. Two corresponding eigenvectors are $[1,-1]^\top$ and $[1,-2]^\top$. Setting
\begin{align}
P=\begin{pmatrix}
1 & 1\\
-1 & -2
\end{pmatrix}
\end{align}
can be expressed $M$ as $M= P\Delta P^{-1}$ where $\Delta=\operatorname{diag}(3,4)$, that is
\begin{align}
\begin{pmatrix}
2 & -1\\
2 & 5
\end{pmatrix} = \begin{pmatrix}
1 & 1\\
-1 & -2
\end{pmatrix}\begin{pmatrix}
3 & 0\\
0 & 4
\end{pmatrix}\begin{pmatrix}
2 & 1\\
-1 & -1
\end{pmatrix}
\end{align}

Thus
\begin{align}
M^n = \begin{pmatrix}
1 & 1\\
-1 & -2
\end{pmatrix}\begin{pmatrix}
3^n & 0\\
0 & 4^n
\end{pmatrix}\begin{pmatrix}
2 & 1\\
-1 & -1
\end{pmatrix} = \begin{pmatrix}
2\cdot3^n - 4^n & 3^n - 4^n\\
2(4^n-3^n) & -3^n - 2\cdot4^n
\end{pmatrix}
\end{align}

This is of course the same as what was mentioned in one of the comments above (egorovik)

Look for the Eigenvalues and eigenvectors:

$$\det(M-\lambda I)=0$$

Then the eigenvectors:

$$M\mathbf{x}=\lambda\mathbf{x}$$

Put the eigenvectors as columns in a $2\times 2$ matrix $P$, and compute $P^{\textrm{T}}MP$. Note that $P^{\textrm{T}}P=PP^{\textrm{T}}=I$


General question about 'vieta jumping' Suppose I want to prove that a variable posesses a certain property (e.g. is a square). For example if I wanted to prove that $x$ in $\frac{x^2+y^2+1}{xy} = k$ has the property of being a square (It is obviously false, but suppose it is true ). Is it then necessary to fix $x$ and look at the pairs $(y,k)$ (which satisfy the equation), or look at the pairs $(x,y)$ and fix k? The latter is the only version I have seen so far.

( source to the vieta jumping method : www.yimin-ge.com/doc/VietaJumping.pdf  )

In this case the problem is very simple you just need to use Vieta Jumping and find the smallest solution. In other problems that uses vieta's jumping you normally fix one varible an then you find the possible k. Here you have an example http://www.artofproblemsolving.com/community/c6h339649


What can we say about output of Gram–Schmidt process Given $\{x_1, \dots, x_{n-1}\}$ linearly independent vectors and $x_n \in \operatorname{span}\{x_1, \dots, x_{n-1}\}$ and let $\{\hat{x_1}, \dots, \hat{x_{n-1}}, \hat{{x_n}}\}$ be the output of the Gram–Schmidt process.

I have to answer the question: what can we say about $\hat{x_n}$. Of course $\hat{x_n}$ is orthogonal to $\hat{x_1}, \dots \hat{x_{n-1}}$ and $\hat{x_n} \in \operatorname{span}\{x_1, \dots, x_{n-1}\}$ but can we say something more non-trivial?

Since we know that $\DeclareMathOperator{\span}{span}\span(x_i, i=1..n-1) = \span(x_i, i=1..n)$, Gram-Schmidt will produce $\hat x_i$ such that
$$\span(\hat x_i, i=1..n-1) = \span(\hat x_i, i=1..n) \ni \hat x_n$$
and $\langle \hat x_i, \hat x_j \rangle = \delta_{ij}$ (Kronecker delta). Thus we know that there is some representation
$$\hat x_n = \sum_{i=1}^{n-1} \alpha_i \hat x_i$$
and since
$$0 = \langle \hat x_i, \hat x_n \rangle = \sum_{j=1}^{n-1} \alpha_j \langle \hat x_i, \hat x_j \rangle = \alpha_i$$
for all $i=1..n-1$, we have
$$\hat x_n = \sum_{i=1}^{n-1} 0 \cdot \hat x_i = 0$$


Radius of convergence of a complex power series? Let's say I have a series which converges in a radius $R$ about point $a$. Let's say I expand the same about point $b$. How does the radius of convergence change?

My progress so far :- I obtain it to be $R-|a-b|$. However I clearly have points outside this radius where the series converges, such as $b$. What's wrong in this argument?

As Andre said in a comment, all you get is an inequality, because you have in general no idea how the function behaves at or outside the radius of convergence. For example, both $( z \mapsto \frac{1}{z-3} )$ and $( z \mapsto \frac{1}{z^2-9} )$ have a power series around $0$ with radius of convergence $3$, but the first has a power series around $-3$ with radius of convergence $6$ while the second does not even have a power series around $-3$. Also, both have a power series around $-1$ but different radius of convergence.

All you know is that if $b$ is within the original convergence disk around $a$, then we can guarantee that the power series around $b$ certainly will converge within a disk that fits in $a$, such as the disk of radius $R - |a-b|$.

Use |z-a|≧|z-b|-|a-b| to get the result as R+|a-b|


Numerical approximation of Levy Flight I'm trying to produce a computer simulation of a Levy Flight in 2-dimensions; an approximation would be ok. Please excuse the simplistic level of this question: my maths is very rusty.

My proposed method of plotting the Levy Flight is as follows: Start at the origin, and plot a connected series of points, each being in a random direction $\theta$ and a distance f from the previous point, where the size of f conforms to a power law distribution. Wikipedia suggests this should be of the form $y=x^{-a}$ where $1<a<3$.

My problem is that I can't see how to generate values of f using a random number generator: I understand that I can use for example x = rand(1) which will with equal probability (allowing for the limitations of numerical accuracy and pseudo-randomness) take any value between 0 and 1. But here's where my limited math skills start to let me down: to transform this into a suitable value of f, I think I would need to use a 'cumulative distribution function' cdf(x), which maps my equally probable values of x onto some corresponding values of f. I also suspect that cdf(x) would basically be related the integral of $y=x^{-a}$, however I can't see exactly how to do this:

Do I need to somehow scale $y=x^{-a}$ so the area under it sums to 1 (in order to be a valid probability density function)? I can't see how this is possible, unless I approximate using discrete values of $x$ and/or choosing a limited interval for values of $x$.
Alternatively, could I do an approximation to this by the following method:(1) select a nominal start value of f, say 1. (2) if rand(1)>t (where t is an arbitrary threshold value) then set f=f*k (say k=1.2) and repeat step 2, otherwise stop and use the current value of f.

If someone could offer some pointers, I would be very grateful. Thank you.

EDIT:

Thank you to Chris for the prompt answer which works perfectly! So for $\alpha = 3$ and $x_{\mathrm{min}} = 1$, when I choose 100,000 random values of u in [0,1] I get this distribution of values of $F^{-1}(u) = x_{\mathrm{min}} (1-u)^{-1/\alpha}$:

... which is just what I'm after (there's a little blip at the end for truncated values of $x>8$).

In fact, since u is evenly distributed I can get exactly the same distribution using $F^{-1}(u) = u^{-1/\alpha}$ which is shorter.

Here's a typical flight plot (for $\alpha=1.5$, 1000 steps):

EDIT: Matlab code for the above example:

alpha=1.5;
s=1000;
x=zeros(1,s);
y=zeros(1,s);

for n=2:s;
    theta=rand*2*pi;
    f=rand^(-1/alpha);
    x(n)=x(n-1)+f*cos(theta);
    y(n)=y(n-1)+f*sin(theta);
end;

figure('color', 'white');
axis equal off;
line(x,y);

To transform a uniformly distributed random variable into another distribution, you need to use the inverse cumulative distribution function. That is, if $F$ is a cumulative distribution function corresponding to the probability density $f$, and $u$ is a uniform random variable in [0,1] then

$$x = F^{-1}(u)$$

is distributed according to $F$. The cumulative distribution function for a pure power law distribution is

$$F(x) = 1 - \left( \frac{x}{x_{\mathrm{min}}} \right)^{-\alpha}$$

where $x_{\mathrm{min}}$ is the minimum value that your random variable can take, and therefore the inverse distribution function is

$$F^{-1}(u) = x_{\mathrm{min}} (1-u)^{-1/\alpha}$$

If you don't want to introduce an artificial minimum, you can consider two-sided power law distributions. These are not pure power laws (you have to fudge the distribution around zero so that the density is not infinite) but they are asymptotically power laws in the tails.

One such distribution is the Student's t distribution with $\nu$ degrees of freedom, which behaves like a power law with exponent $\nu$ as its values tend to $\pm\infty$.


Solve the equation ${\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}} = 9x-3$ I tried factoring the expression inside the square root, but that does not seem to help. Squaring the equation makes it even more terrible.

Can anyone provide a hint about what should be done?

Notice that

$$\left({\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}}\right) \left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) = 9x-3.$$
Thus,

$$9x-3 =  \left(9x-3\right)\left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) \implies $$

$$9x-3 = 0 \implies x=1/3$$ or

$${\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}} = 1.$$

But the minimal value of $x^2-x+1$ is $3/4$, which implies that
$$ 2{\sqrt {x^2-x+1}} \geq \sqrt{3} > 1.$$

Therefore $x =1/3$ is the only real solution.

Move $2\sqrt {x^2 - x + 1}$ to the rhs, then square both sides.


Propositional calculus - I can't get why the answer for this test question is what it is Consider the following premises.

If A = B then B = C.
B != C.
If C > D then D < E.
F != G and A = B.
A = B or C > D.

Alternatives:

a) F != G

b) F != G and D < E

c) A = B

d) B = C or D < E

e) D < E

The answer sheet says the correct alternative is B.

I am confused, because (1) and (2) seems to imply that A != B. In which case (4) would be contradictory. That's why I am wondering if the question itself is wrong.

I suspect that there is a typo on your source's part, namely being that (4) should be a disjunction.

Making this correction, we have

(6) $A\neq B$. Take the contrapositive of (1) and applying modus ponens with (2).

(7) $C>D$. Use disjunctive syllogism on (5) with (6).

(8) $D<E$. Apply modus ponens on (3) with (7).

(9) $F\neq G$. Apply disjunctive syllogism on (modified 4) with (6).

(10) b) Conjunction between (8) and (9).

Thus b) would then indeed be a correct answer.

(1) and (2) do not imply that A != B. Indeed is an if, so that IF  A = B, then we have that A = B = C. However in 2, it is not stated that A=B, we only know that B != C,which implies that A != B.

To prove the equality we see that:
 from 4) F!=G and A=B, which implies 5) A=B or C > D.
Hence from 3) we get the answer b!

Notice that the passage from 5 to 3 works as 'or' works when one of the propositions or both are true, so we can take  C > D to be true.


Can unexpected anamalous probabilities lead to expectations If you were told I flip a coin 10 times and 9 were heads.

Would it be most logical to assume the other was tails if they're was no reason for it to be heads more often.

I read an article saying more serial killers were born in November and if there is no reason then shouldnt it even out over time?

There is no correlation between a past event and future if it is a truly random event, therefore, the probability will remain $0.5$ (for a two-sided coin).


Rewriting an equation to include a condition I have a simple equation: a = b(c/d)

but it has a condition: if b > 1, then b = 1

How do I write this condition into the equation?

Your question is really not a physics question, but a mathematical one. However, the way to add in the condition is the following:

$$
a = \begin{cases}\left(\frac{c}{d}\right) & {\rm if}\,b>1 \\ b\left(\frac{c}{d}\right) & {\rm otherwise}\end{cases}
$$

The long brace is the cases environment in latex: $\begin{cases} ...stuff...\end{cases}$


$\mathbb{C}\mathbb{P}^1$ is homeomorphic to $S^2$ I just read on wikipedia that the the complex projective line is homeomorphic to the riemann sphere. How do I prove this? But, before that I have an extremely silly doubt that has been eating me. In the complex projective line, antipodal points are identified, but in the Riemann sphere they are clearly distinct. Then how can the two be homeomorphic.

I found out the following map from the complex projective line to $S^2$. But how do I prove this is a homeomorphism? Finding the inverse of this functions seems complicated, and then how do I prove that the inverse map of a open subset is a open subset, and vice versa?

$f:\mathbb{C}\mathbb{P}^1 \to S^2 = [z:w] \to \frac{(2Re(w\bar{z}), 2Im(w\bar{z}),|w|^2-|z|^2)}{(|w|^2+|z|^2)} $

Sorry I'm late to the party; I would comment on Berci's post but I'm not allowed to comment just yet.

While I'm sure you can actually prove that's a homeomorphism by computing inverses and showing that everything's continuous/smooth, I think most people would balk at actually carrying out such computations, and would simply accept that it can be done without actually doing it out.

But if you want, there is another way to see this fact using CW-complexes: on pages 6-7 of Hatcher's book he describes the standard CW-structure for $\mathbb{CP}^n$.
In this structure you get one $2k$-cell for each $k$ between $0$ and $n$, and no other cells.

This helps here, because it means that the CW-structure of $\mathbb{CP}^1$ is just a $2$-cell glued onto a $0$-cell ($=$ a point).
But there is only one way to do that, and that's to map all of the boundary of the $2$-cell to the point, which gives you the $2$-sphere.

This is actually a very interesting/useful formalism used in physics, usually mentioned as the Bloch sphere.
I will only provide a sketch. Indeed, $\mathbb{C}\mathbb{P}^1$ can be regarded as the set of rank-1 projections on $\mathbb{C}^2$, which can be denoted as $|\hat{n}\rangle \langle \hat{n}|$ (physicist notation) where $|\hat{n}\rangle$ is the unit vector in $\mathbb{C}^2$ being projected on. Then by the Bloch sphere isomorphism, one can show that
$$
|\hat{n}\rangle \langle\hat{n}| =\frac{1}{2}(I+\hat{n}\cdot\sigma)
$$
For some unique unit vector $\hat{n}\in \mathbb{S}^2$, where $\sigma=(\sigma_1,\sigma_2,\sigma_3)$ is the vector of Pauli matrices. The usual proof of this statement basically follows from the fact that
$$
|0\rangle \langle 0| = \frac{1}{2} (I+\sigma_3)
$$
Where $\sigma_3 |0\rangle = |0\rangle$, and that if $U\in SU(2)$ corresponds to a rotation $R\in SO(3)$ which take $e_3$ to $\hat{n}$, then
$$
U|0\rangle \langle 0|U^* =\frac{1}{2}(I+\hat{n}\cdot\sigma)
$$
One can also easily check that this is indeed the isomorphism the OP mentioned, i.e., if $\hat{n}$ is the unit vector in $\mathbb{S}^2$ specified by the spherical coordinates $(\theta,\phi)$, then the state $|\hat{n}\rangle$ is uniquely defined (up to a global phase) as
$$
|\hat{n}\rangle =(\cos{(\theta/2)}e^{-i\phi/2},\sin{(\theta/2)}e^{+i\phi/2})
$$


A zero sum subset of a sum-full set I had seen this problem a long time back and wasn't able to solve it. For some reason I was reminded of it and thought it might be interesting to the visitors here.

Apparently, this problem is from a mathematics magazine of some university in the United States (sorry, no idea about either).

So the problem is:

Suppose $S \subset \mathbb{Z}$ (set of integers) such that

1) $|S| = 15$

2) $\forall ~s \in S, \exists ~a,b \in S$ such that $s = a+b$

Show that for every such $S$, there is a non-empty subset $T$ of $S$ such that the sum of elements of $T$ is zero and $|T| \leq 7$.

Update (Sep 13)

Here is an approach which seems promising and others might be able to take it ahead perhaps.

If you look at the set as a vector $s$, then there is a matrix $A$ with the main diagonal being all $1$, each row containing exactly one $1$ and one $-1$ (or a single $2$) in the non-diagonal position such that $As = 0$.

The problem becomes equivalent to proving that for any such matrix $A$ the row space of $A$ contains a vector with all zeroes except for a $1$ and $-1$ or a vector with all zeroes except $\leq 7$ ones.

This implies that the numbers in the set $S$ themselves don't matter and we can perhaps replace them with elements from a different field (like say reals, or complex numbers).

A weaker statement, where we allow elements in $T$ to be repeated, can be proved as below:

Since we can look at the set $\{-s | s\in S \}$ we may assume there are at most $7$ positive numbers in $S$. Let each positive number be a vertex, from each vertex $s$ we draw an arrow to any vertex $a$ such that $s=a+b$. Since if $s>0$, one of $a,b$ must be positive, there is at least one arrow from any vertex. So there must be a cycle $s_1,\cdots,s_n=s_1$ with $n\leq 8$. We can let $T$ consists of $s_i-s_{i+1}$, $1\leq i\leq n-1$.

In computer science, the subset sum problem is an important decision problem in complexity theory and cryptography. There are several equivalent formulations of the problem. One of them is: given a set (or multiset) of integers, is there a non-empty subset whose sum is zero? For example, given the set {\displaystyle {-7,-3,-2,5,8}} {\displaystyle {-7,-3,-2,5,8}}, the answer is yes because the subset {\displaystyle {-3,-2,5}} {\displaystyle {-3,-2,5}} sums to zero. The problem is NP-complete, meaning roughly that while it is easy to confirm whether a proposed solution is valid, it may inherently be prohibitively difficult to determine in the first place whether any solution exists.

The problem can be equivalently formulated as: given the integers or natural numbers {\displaystyle w_{1},\ldots ,w_{n}} w_{1},\ldots ,w_{n} does any subset of them sum to precisely {\displaystyle W} W?[1] Subset sum can also be thought of as a special case of the knapsack problem.[2] One interesting special case of subset sum is the partition problem, in which s is half of the sum of all elements in the set.

The complexity of the subset sum problem can be viewed as depending on two parameters, N, the number of decision variables, and P, the precision of the problem (stated as the number of binary place values that it takes to state the problem). (Note: here the letters N and P mean something different from what they mean in the NP class of problems.)

The complexity of the best known algorithms is exponential in the smaller of the two parameters N and P. Thus, the problem is most difficult if N and P are of the same order. It only becomes easy if either N or P becomes very small.

If N (the number of variables) is small, then an exhaustive search for the solution is practical. If P (the number of place values) is a small fixed number, then there are dynamic programming algorithms that can solve it exactly.

Efficient algorithms for both small N and small P cases are given below.

There are several ways to solve subset sum in time exponential in {\displaystyle n} n. The most naïve algorithm would be to cycle through all subsets of {\displaystyle n} n numbers and, for every one of them, check if the subset sums to the right number. The running time is of order {\displaystyle O(2^{n}n)} O(2^{n}n), since there are {\displaystyle 2^{n}} 2^{n} subsets and, to check each subset, we need to sum at most {\displaystyle n} n elements.

A better exponential time algorithm is known which runs in time O(2N/2). The algorithm splits arbitrarily the N elements into two sets of N/2 each. For each of these two sets, it stores a list of the sums of all 2N/2 possible subsets of its elements. Each of these two lists is then sorted. Using a standard comparison sorting algorithm for this step would take time O(2N/2N). However, given a sorted list of sums for k elements, the list can be expanded to two sorted lists with the introduction of a (k + 1)st element, and these two sorted lists can be merged in time O(2k). Thus, each list can be generated in sorted form in time O(2N/2). Given the two sorted lists, the algorithm can check if an element of the first array and an element of the second array sum up to s in time O(2N/2). To do that, the algorithm passes through the first array in decreasing order (starting at the largest element) and the second array in increasing order (starting at the smallest element). Whenever the sum of the current element in the first array and the current element in the second array is more than s, the algorithm moves to the next element in the first array. If it is less than s, the algorithm moves to the next element in the second array. If two elements with sum s are found, it stops. Horowitz and Sahni first published this algorithm in a technical report in 1974.[3]

The problem can be solved in pseudo-polynomial time using dynamic programming. Suppose the sequence is

x1, ..., xN
sorted in the increasing order and we wish to determine if there is a nonempty subset which sums to zero. Define the boolean-valued function Q(i, s) to be the value (true or false) of

"there is a nonempty subset of x1, ..., xi which sums to s".
Thus, the solution to the problem "Given a set of integers, is there a non-empty subset whose sum is zero?" is the value of Q(N, 0).

Let A be the sum of the negative values and B the sum of the positive values. Clearly, Q(i, s) = false, if s < A or s > B. So these values do not need to be stored or computed.

Create an array to hold the values Q(i, s) for 1 ≤ i ≤ N and A ≤ s ≤ B.

The array can now be filled in using a simple recursion. Initially, for A ≤ s ≤ B, set

Q(1, s) := (x1 == s)
where == is a boolean function that returns true if x1 is equal to s, false otherwise.

Then, for i = 2, …, N, set

Q(i, s) := Q(i − 1, s) or (xi == s) or Q(i − 1, s − xi),  for A ≤ s ≤ B.
For each assignment, the values of Q on the right side are already known, either because they were stored in the table for the previous value of i or because Q(i − 1,s − xi) = false if s − xi < A or s − xi > B. Therefore, the total number of arithmetic operations is O(N(B − A)). For example, if all the values are O(Nk) for some k, then the time required is O(Nk+2).

This algorithm is easily modified to return the subset with sum 0 if there is one.

The dynamic programming solution has runtime of {\displaystyle O(sN)} O(sN) where {\displaystyle s} s is the sum we want to find in set of {\displaystyle N} N numbers. This solution does not count as polynomial time in complexity theory because B − A is not polynomial in the size of the problem, which is the number of bits used to represent it. This algorithm is polynomial in the values of A and B, which are exponential in their numbers of bits.

For the case that each xi is positive and bounded by a fixed constant C, Pisinger found a linear time algorithm having time complexity O(NC) (note that this is for the version of the problem where the target sum is not necessarily zero, otherwise the problem would be trivial).[4] In 2015, Koiliaris and Xu found the {\displaystyle {\tilde {O}}(s{\sqrt {N}})} {\tilde {O}}(s{\sqrt {N}}) algorithm for the subset sum problem where {\displaystyle s} s is the sum we need to find.[5]


Natural deduction: predicate logic proof (Prenex form) I'm pretty familiar with proofs in propositional logic, but not so much with predicate logic.

I'm trying to prove the following (which can be used during construction of prenex normal form).

If $x$ is not free in $\psi$, then:
$\forall x\phi\rightarrow \psi $
if and only if
$\exists x(\phi \rightarrow \psi)$

I personally use tree proofs, but any notation is fine.
I don't really know how to start this one to be honest.

Here's a proof sketch for the "if" direction, where the proof more or less writes itself. Obviously you have to use $\exists$-Elim, since you have an $\exists$ premiss which you need to make use of. So you already know the proof has to look like this (Fitch style --re-arrange for a tree!)

$\exists x(\phi(x) \to \psi)\\
\quad\quad |\quad \phi(a) \to \psi\\
\quad\quad |\quad \vdots\\
\quad\quad |\quad \forall x\phi(x) \to \psi\\
\forall x\varphi(x) \to \psi$

Then how to fill in the dots is pretty obvious since you'll have to use Conditional Proof -- i.e. assume the antecedent of the conditional you want to prove and aim for the consequent. Easy!

$\exists x(\phi(x) \to \psi)\\
\quad\quad |\quad \phi(a) \to \psi\\
\quad\quad |\quad \quad |\quad \forall x\phi(x)\\
\quad\quad |\quad \quad |\quad \phi(a)\\
\quad\quad |\quad \quad |\quad \psi\\
\quad\quad |\quad \forall x\phi(x) \to \psi\\
\forall x\varphi(x) \to \psi$

The other direction is trickier. But try using reductio (always a good plan if nothing else looks an obvious move) so the overall shape of the proof will look like this:

$\forall x\varphi(x) \to \psi\\
\quad\quad |\quad \neg\exists x(\phi(x) \to \psi)\\
\quad\quad |\quad \vdots\\
\quad\quad |\quad \bot\\
\neg\neg\exists x(\phi(x) \to \psi)\\
\exists x(\phi(x) \to \psi)$

You can complete the proof like this (think through why this is now an obvious way, and what justifies each step).

$\forall x\varphi(x) \to \psi\\
\quad\quad |\quad \neg\exists x(\phi(x) \to \psi)\\
\quad\quad |\quad \quad |\quad \phi(a) \to \psi\\
\quad\quad |\quad \quad |\quad \exists x(\phi(x) \to \psi)\\
\quad\quad |\quad \quad |\quad \bot\\
\quad\quad |\quad \neg(\phi(a) \to \psi)\\
\quad\quad |\quad \vdots\\
\quad\quad |\quad \phi(a)\\
\quad\quad |\quad \neg\psi\\
\quad\quad |\quad \forall x\phi(x)\\
\quad\quad |\quad \psi\\
\quad\quad |\quad \bot\\
\neg\neg\exists x(\phi(x) \to \psi)\\
\exists x(\phi(x) \to \psi)$

Fill in the dots by propositional calculus reasoning. $\forall$ intro is justified from $\phi(a)$ since $a$ appears in no undischarged assumption on which $\phi(a)$ depends.

Here's an informal deduction.   Express it in your own formal system.

Take the premise: $\forall x~\phi ~\to~\psi$
Derive the contraposition: $\lnot\psi\to\lnot\forall x~\phi$
Assuming $\lnot\psi$, derive that there is some $x$ where $\lnot\phi$.
Deduce then that, there is some $x$ where $\lnot\psi\to\lnot\phi$, which is to say $\phi\to\psi$
Therefore $\exists x~(\phi\to\psi)$ is derived from the premise

Take the premise: $\exists x~(\phi\to\psi)$
Thence there is some witness $x$ where $\phi\to\psi$ holds
Assuming $\forall x~\phi$, derive that for the witness, $\phi$ holds, and apply modus ponens.
Thus deduce that $\forall x~\phi ~\to~\psi$ is derivable from the premise.


Adjoint operator of $L^\infty$ Lets denote with $(\Omega,\Sigma,\mu)$ a $\sigma$-finite measurble space with a linear, continuous operator $$T : L^\infty \to L^\infty.$$ Does this always imply the existence of a linear, continuous operator $$T^*: L^1 \to L^1 $$
that solves the integral equation
$$ \int_\Omega Tg (x) f(x) \mu(dx) = \int_\Omega g(x) T^*f (x) \mu(dx)  $$
or in short notation
$$ <Tg,f>= <g,T^*f>$$
for all $f \in L ^1, \, g \in L^\infty$, where $<,>$ denotes the integral of an $L^\infty$ function times an $L^1$ function.

Here is a brief example.  This is the type of thing you will learn in a course on Banach spaces.  Or a course on functional analysis with a good chapter on Banach spaces.   The general fact is that space $L^1$ is not reflexive.  I have adapted this to the particular case of $L^1, L^\infty$.

Let's use measure space $[0,1]$ with Lebesgue measure $\lambda$.  For $g \in L^\infty$ and $f \in L^1$, define the  pairing $\langle g, f\rangle = \int fg\,d\lambda$.  This pairing identifies $L^\infty$ with the isometric dual space of $L^1$.  Topological words used will refer to the norm toplogies of the spaces.

Let the sequence $r_n \in L^\infty$ be the functons $1, \sin(n \pi x), \cos(n \pi x)$ in some order.  This sequence separates points of $L^1$.  That is, if $f \in L_1$ and $\langle r_n,f\rangle = 0$ for all $n$, then $f=0$.

Now the space $L^\infty$ is non-separable, so the closed span $N$ of $\{r_1, r_2, r_3, \cdots\}$ is not the whole space.  By the Hahn-Banach theorem, there is a continuous linear functional $\xi$ on $L^\infty$ such that $\xi$ vanishes on $N$ but $\xi$ is not identically zero.  Define an operator $T \colon L^\infty \to L^\infty$ by $T(g) = \xi(g)\mathbb 1$.  Here I wrote $\mathbb 1$ for the constant function $1$.  Thus $T(g)$ is a certain scalar $\xi(g)$ times the constant $1$.

Now suppose there is $T^* \colon L^1 \to L^1$ with
$$
\langle g, T^*(f)\rangle = \langle T(g), f\rangle \qquad\text{for all $f \in L^1, g \in L^\infty$}
$$
Write $\mathbf 1$ for the constant $1$ function in $L^1$.  What is $T^*(\mathbf 1)$ ?  (We claim it has contradictory properties.)  Write $F = T^*(\mathbf 1)$.

For all $g$ in $L^\infty$,
$$
\langle g, F\rangle =\langle g, T^*(\mathbf 1)\rangle = \langle T(g), \mathbf 1\rangle = \langle \xi(g)\mathbb 1, \mathbf 1 \rangle = \xi(g) .
$$
Now for all $n$, we have $\langle r_n, F\rangle = \xi(r_n) = 0$.  Since the $r_n$ separate points of $L^1$, we have $F = 0$.  But then for all $g \in L^\infty$,
$$
\xi (g) = \langle g, F\rangle = 0 ,
$$
so $\xi$ is the zero linear functional.  This contradicts the choice of $\xi$.

See also: http://en.wikipedia.org/wiki/Hermitian_adjoint

The operation $\langle\cdot\rangle$ denotes the scalar product in $L^2(\Omega,\Sigma,\mu)$.


$a,b,c,d$ are real numbers such that.... Suppost that $a,b,c,d$ are real numbers such that $$a^2+b^2=1$$ $$c^2+d^2=1$$ $$ac+bd=0$$

I've to show that $$a^2+c^2=1$$  $$b^2+d^2=1$$ $$ab+cd=0$$

Basically,I've no any idea or tactics to tackle this problem. Any methods? Thanks in advance.

EDITED.

The given hint in the books is $S:=(a^2+c^2-1)^2+(b^2+d^2-1)^2=(ac+bd)^2$

Just for fun, here is a more advanced solution. Let

$g = \left( \begin{array}{cc} a & b \\
           c & d \end{array} \right)$

The equations in the hypothesis are equivalent to $gg^T=I_2$. The equations in the conclusion are equivalent to $g^Tg = I_2$. To get from the hypothesis to the conclusion, simply notice that $gg^T=I_2$ implies that $g^T = g^{-1}$.

Remark: in a more advanced language, the orthogonal group $O(n)$ is closed under transpose. This exercise is the $n=2$ special case.

It is boring. From $ac+bd=0$, $a=\frac{-bd}{c}$, put in $a^2+b^2=1$ the value of $a$, you have $\frac{b^2}{c^2}(d^2+c^2)=1$ now from $d^2+c^2=1$ you have $b^2=c^2$. similar arguments show that $a^2=d^2$. and together with $a=\frac{-bd}{c}$ you have all your results.

Put some effort to play with equations. Consider them all non-zero. If one is zero then obvious.

good luck


Regular polygons and Pythagoras Let $L_n:$ the side length of a regular $n$-polygon inscribed in a  unit
fixed circle.
We have an interesting relationship:
$L_6^2+L_6^2=L_4^2$
$L_6^2+L_4^2=L_3^2$
$L_{10}^2+L_6^2=L_5^2$

There are more solutions: $L_m^2+L_n^2=L_p^2$ ?

This will not answer the question as phrase, but:

The last one you list is Proposition 10 of Book XIII of Euclid's Elements.  It was used by Ptolemy in the second century for constructing a trigonometric table.
The circle need not be of unit radius.  You could have just said "inscribed in a circle".


Convergence in $L^p(U)$ of smooth functions acted on by a $C^1$ function with $|U|<\infty$ Assume $U$ is bdd and $G:\mathbb{R}\rightarrow \mathbb{R}$ is $C^1$ with $G'$ bdd. If we have smooth $f_n \rightarrow f$ in $L^p(U)$, do we necessarily have that $Gf_n \rightarrow Gf$ in $L^p(U)$?

I feel if $U$ was connected I could just apply the MVT, but as this is not necessarily the case I'm at a loss.

HINT. Establish the following inequality:
$$\lvert G(y)-G(z)\rvert^p\le C\lvert y-z\rvert^p,$$
for a constant $C>0$ that is independent on $y$ and $z$.

Forget $C^1$ and smoothness. Just let $G$ be any bounded measurable function on $U$ and assume $f_n \to f$ in $L^p(U).$ Then $\|Gf_n-Gf\|_p \le \|G\|_\infty\|f_n - f\|_p \to 0.$


Prove $B(p,q) + B(p+1,q) + B(p+2,q) + ... = B(p,q-1) $ where $B$ is beta function and $q > 1$ Prove $B(p,q) + B(p+1,q) + B(p+2,q) + ... = B(p,q-1) $ where $B$ is beta function and $q > 1$.

I tried with some basic formulas for beta function, mathematical induction but just cannot get some good idea and solve this. If anyone have some idea, please help. I will be grateful.

By definition, starting from the LHS of your problem,
$$
\sum_{n=p}^\infty\beta(n,q) = \sum_{n=p}^\infty\int_0^1t^{n-1}(1-t)^{q-1}dt
$$
Since $t$ is always in $[0,1]$ things converge nicely (I think uniformly) and we are free to interchange integral and sum:
$$
\sum_{n=p}^\infty\beta(n,q) = \int_0^1\sum_{n=p}^\infty t^{n-1}(1-t)^{q-1}dt
$$
Again, for this range of $t$, the sums converge:
$$
\sum_{n=p}^\infty t^{n-1}(1-t)^{q-1} = t^{p-1}\sum_{m=0}^\infty t^{m}(1-t)^{q-1}
= t^{p-1}\frac1{1-t}(1-t)^{q-1}=  t^{p-1}(1-t)^{q-2}
$$
Finally, placing this back into the integral, we have
$$
\sum_{n=p}^\infty\beta(n,q) =  \int_0^1t^{p-1}(1-t)^{q-2}dt =\beta(p,q-1)
$$


Finding $P(X<2Y)$ given joint pdf $f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2 + y^2}}$ for $x,y\in\mathbb R$ The joint pdf of $X$ and $Y$ is $$f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2 + y^2}}\,; \quad x,y\in \mathbb{R}$$ Find $P(X<2Y)$.

I have tried this:

$$\int_{-\infty}^{\infty}\int_{-\infty}^{2y} f(x, y)\, dx\,dy$$

@StubbornAtom suggested polar transformations.
So here it goes,
Let $X = r \cos\theta, \, Y = r \sin\theta$.
Then The integral gets transformed as $\int \int re^{-r}d\theta$.
Could some one help find the new limits.

The joint distribution of $(X,Y)$ is rotationally symmetric, since the density depends only on the distance from the origin. This means that the conditional distribution of $(X,Y)$ given $X^2+Y^2$ is uniformly distributed on the circle of radius $r=\sqrt{X^2+Y^2}$. Writing $X=r\cos\Theta$ and $Y=r\sin\Theta$ with $\Theta\in [0,2\pi)$ uniformly random, we have
$$
\mathbb P(X<2Y\mid X^2+Y^2)=\mathbb P(\cos\Theta<2\sin \Theta)$$$$=\mathbb P\Bigl(\tan\Theta>\frac{1}{2}, \cos\theta > 0\Bigr)+\mathbb P\Bigl(\tan\Theta<\frac{1}{2}, \cos\theta < 0\Bigr),$$
which can be seen to equal
$$\mathbb P\Bigl(\tan^{-1}\frac{1}{2}<\Theta<\pi+\tan^{-1}\frac{1}{2}\Bigr)=\frac12.
$$
Since the conditional probability is $\tfrac12$ independent of $X^2+Y^2$, it follows that the unconditional probability is also $\tfrac12$.

So this is my version that I have understood from @pre-kidney's answer please note that this is just an elaboration of his answer.

$f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2+y^2}} \ ;\  -\infty<\ x \ <\infty\ ; -\infty < y < \infty$

$X=rcosθ,Y=rsinθ$

$f(r, \theta) = f(\theta).f(r) = (\frac {1}{2\pi} ).(re^{-r})\ \ ;\ \ 0 < r < \infty\ ;\ 0 <  \theta < 2\pi$

Now, $P(X < 2Y) = P(rcos\theta < 2rsin\theta) = P(tan\theta > \frac {1}{2}) = P(\theta > tan^-\frac {1}{2}) = \frac {1}{2\pi} \int_{tan^-\frac {1}{2}}^{2 \pi}d\theta = \frac {1}{2\pi} [\theta]_{tan^-\frac {1}{2}}^{\pi + tan^-\frac {1}{2}} = \frac {1}{2\pi}[\pi + tan^-\frac {1}{2} - -tan^-\frac {1}{2}]) = \frac {1}{2\pi} . \pi = 0.5$


Finding one sided limit using L'Hospital's rule $\lim_{x\to 0^+}\left(x^\sqrt{x}\right)=?$

How do I turn it into a fraction? Is L'Hospital's rule even applicable?

Hint: Consider $\log$, so $\sqrt{x}\log x=\dfrac{\log x}{\dfrac{1}{\sqrt{x}}}$.

given limit = $e^{\sqrt{x}\ln x} $

we know $\ln x < (x-1) < x $

$0 \leq\sqrt{x}\ln x <  \sqrt{x} x $

this limit converges to 0 as $x\to 0$ by squeeze theorem

i.e., given limit converges to $e^0= 1$ as $x \to 0$


Equation of circle touching a parabola Suppose we have a parabola $y^2=4x$  . Now, how to write equation of circle touching parabola at $(4,4)$ and passing thru focus?
I know that for this parabola focus will lie at $(1,0)$ so we may assume general equation of circle and satisfy the points in it . hence
$$x^2 +y^2 + 2gx + 2fy +c =0$$ should be equation of general circle but on satisfying given points we get 2 equations and 3 variables . I think I am missing something, what to do?

the tangent to the parabola at $(4, 4)$ has slope $1/2$ so the radius has slope $-2.$ let the center of the circle touching the parabola $y^2 = 4x$ at $(4,4)$ be
$x = 4 + t, y = 4 - 2t$. now equating the radius $$5t^2 = (3+t)^2 + (4-2t)^2$$
you can find $t$ which will give you the center and the radius of the circle.

If what you mean by "touching" is "intersect", then this is the solution. Your circle is defined by its center $(x_c,y_c)$ and its radius $R$:
$$(x-x_c)^2+(y-y_c)^2=R^2$$
Now, you want all three points $(1,0)$, $(4,4)$ and $(4,-4)$ to be on the circle, so you hav to solve the system:
\begin{cases}(1-x_c)^2+(0-y_c)^2=R^2\\(4-x_c)^2+(4-y_c)^2=R^2\\(4-x_c)^2+(-4-y_c)^2=R^2\end{cases}
which gives
$$x_c=\dfrac{31}{6},\ \ y_c=0,\ \ R=\dfrac{25}{6}$$
or
$$\left(x-\dfrac{31}{6}\right)^2+y_c^2=\dfrac{625}{36}$$


Show that not exists any polynomial function such that $f(x) = \log (1+x)$. Does anyone have any idea on that problem?

  Let $f : \mathbb{R} \to \mathbb{R}$ be a polynomial function. Show that not exists any $f$ such that $f(x) = \log (1+x)$.

It's easy to show that $a_0 = 0$ and $a_1 = 1$. But after i don't have any idea. Any point? Thanks!

Maybe there are many ways to show this; but what I have in mind is this:
If $f$ is a polynomial of degree $k$, then its $(k+1)$-th derivative is zero.
But this is not the case for $\log(1+x)$; the $(k+1)$-th derivative of $\log(1+x)$ does not vanish.

If $\log(1+x)$ is polynomial, $f$ is a transcendental polynom function which doesn't exist.


If $C$ is symmetric and $\operatorname{rank}C=1$, then there is a $u\in\mathbb R^n$ and a $\alpha\in\mathbb R$ such that $C=\alpha uu^T$ Let $C\in\mathbb R^{n\times n}$. I want to show, that

If $0\ne C=uv^T$ for some $u,v\in\mathbb R^n$, then $$\operatorname{rank}C=1$$
If $C$ is symmetric and $\operatorname{rank}C=1$, then there is a $u\in\mathbb R^n$ and a $\alpha\in\mathbb R$ such that $$C=\alpha uu^T$$

(1.) is easy, since $$C=(v_1u,\ldots,v_nu)\;,$$ $\operatorname{rank}C$ is at most $1$. Since $C\ne 0$, the statement follows.

However, I don't know what I need to do for (2.). Clearly, $$\operatorname{rank}(\alpha uu^T)=1\;,$$ by $(1.)$. So, maybe we should assume that $$C=\alpha uu^T+A\;,$$ for some $A\in\mathbb R^n$, and show, that this leads to a contradiction to $\operatorname{rank}C=1$. Can we do this? The inequality $$\operatorname{rank}(X+Y)\le\operatorname{rank}X+\operatorname{rank}Y$$ doesn't seem to help.

Since $C$ has rank $1$, the column space of $C$ has dimension $1$, which means every column of $C$ is a multiple of some vector $u$. Hence
$$C=[v_1 \vec{u}, \cdots , v_n\vec{u}]=\vec{u}\vec{v}^T$$
for some vector $\vec{v}=(v_1, \dots, v_n)^T$.

Now if $C$ is symmetric, then $C=C^T=\vec{v}\vec{u}^T$. This means every column of $C$ is also a multiple of $\vec{v}$. Hence $\vec{u}$ is a multiple of $\vec{v}$, i.e.,

$$\vec{u}=k\vec{v}$$

You can see the result follows straightforwardly.

considering $C= uu^T + A$ doen't solve the problem because there might be a matrix $\lambda$ so that $\alpha uu^T + A = \beta\lambda \lambda^T$ so $C$ again can be written as  $\beta\lambda \lambda^T$.
It seems that you have already know what matrix $u$ is!


Is Keno a fair game? This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with probability, which perhaps yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

In the game of Keno, from the numbers $1-80$, a player chooses three, and makes a $\$1$ bet.
Then twenty numbers are drawn.
If all three of his numbers are among the twenty, he is paid $\$42$ (gain of $\$41$).
If two of his numbers are among the twenty, he is paid $\$1$ (break even).
If fewer than two of his numbers are among the twenty, he loses.
What are his chances ?
How can this be made a fair game ?

My thoughts:
A player's entry can be any of $C(80,3)$ combinations $= 82160$
From the the $20$ numbers there are $C(20,3)$ winning combinations $= 1140$
$\frac{1140}{82160} = .0138753 =$ Probability of having three winning numbers.
But now I am stuck.

P(win) = P(choose 3 correct #s) = ${20\choose 3}/{80\choose 3} = \frac{57}{4108}$

P(break even) = P(choose 2 correct and 1 wrong #) = ${20\choose 2}\cdot{60\choose 1}/{80\choose 3} = \frac{570}{4108} $

P(lose) = $1 - \frac{57}{4108} - \frac{570}{4108} = \frac{3481}{4108}$

Let x dollars be the net winnings if all 3 #s are guessed correctly.

For a fair game, net winnings = 0 = $\frac{57x}{4108} - \frac{3481}{4108}$, which yields

x = $\frac{3481}{57}, \approx 61.07 dollars $ , (paid $62.07)

This is what I think could be correct...

$P$ (win) = The probability that all $3$ of his numbers are among the $20$ chosen is as you said   = $20c3/80c3 = 0.01388$

$P$ (break even)= (not sure if this is correct) The probability that 2 of his numbers are successful = ($20c2 \times 20c1 \times3c2)/80c3$ $=$ $0.1388$

The way I calculated this was because there are 20c2 ways of choosing 2 correct combinations, and then 20c1 ways of choosing a wrong number, but there are 3c2 ways of this whole event happening anyway (correct correct wrong, correct wrong correct, wrong correct correct)

$P$ (lose) = 1- ($0.01388+0.1388$) $= 0.8474$

So, if he does say $100$ bets, each betting $\$1$, on average $84.74$ of the bets will be lost, thus $\$84.74$ is lost.

On average $13.88$ bets will result in no net outcome.

On average, $1.388$ bets will result in a win of $41 dollars.

So, net outcome = $41 \times 1.388 - 84.74 = - \$27.83$ dollars.

So this game isn't fair.

In order to make this game fair, we solve this equation:

$1.388x-84.74=0$

$x=61.05 $

Thus for the game to be fair, one must win $\$61.05$ (be paid  $\$62.05$) if all $3$ of his numbers come in the chosen $20$.


conditional mutual information; by how much are two heads better than one? I have two observers (Y and Z) making a subjective evaluation. There is a relatively high rate of disagreement between the two, but over a "population" of observations it appears that opinions somewhat, but not entirely, balance each other out in ability to correctly identify  label (which is disease or no disease, and we will call label = X). So:
$I(X;Y) > 0$ and $I(X;Z)> 0$, but $I(X;Y) \ne I(X;Z)$

By chain rule of mutual information (2.5.3 in Cover) it appears that:
I(X; Y,Z) = I(X; Z) + I(X; Y | Z) (first-- is this correct?)
Assuming both Y and Z are not noise, nor completely redundant, then the conditioning variable Z improves upon information Y has with X. The question is if improvement by Y as the conditioning variable leads to the same result.
Does

$$I(X; Y) + I(X; Z | Y)   =    I(X; Z) + I(X; Y | Z)$$

Would there be a simple proof if the above is true? Or am I off base?
The reason for the question is that discordance in subjective observations is typically considered a bad thing (i.e. a low kappa statistic). It may be that if both observations have relevance then discordance can be a good thing-- a low kappa statistic may be a promising number. How much will one observer offer another and vice versa.

Thank-you leonbloy for your help. Based on your comment I found theorem 2.8.1 in Cover and Thomas 2nd edition page 34. From my medical background what this symmetry means is that two observers may be capable, but one observer (Y) might be considered "smarter", or of greater stature. But, the lesser (Z ="junior" chap) can be helpful (informative). This is radical in some circles, but supports some data I have. An example is possible value of "double reads" in the interpretation of mammograms for breast cancer (X= cancer). This will prompt a follow-on question. Again thanks.

The first equation $$I(X; Y,Z) = I(X; Y) + I(X;  Z \mid Y) \tag{1}$$

is indeed correct, it's the chain rule for mutual information and is easily derivable from the chain rule for entropies. By simmetry, it's also true:

$$I(X; Y,Z) = I(X; Z) + I(X; Y \mid Z) \tag{2}$$

Equating the right sides:

$$ I(X; Y) + I(X;  Z \mid Y) =  I(X; Z) + I(X; Y \mid Z) \tag{3}$$

so, yes, your equation is true. I'm not sure what's your doubt.

By definition:
$$\begin{align}\mathsf I(X; Y, Z) =~& \sum_{z\in\{0,1\}}\sum_{y\in\{0,1\}}\sum_{x\in\{0,1\}} \mathsf p_{X,Y,Z}(x,y,z)\log\dfrac{\mathsf p_{X,Y,Z}(x,y,z)}{\mathsf p_X(x)\,\mathsf p_Y(y)\,\mathsf p_Z(z)}
\\[1ex] =~& \sum_{y\in\{0,1\}}\sum_{z\in\{0,1\}}\sum_{x\in\{0,1\}} \mathsf p_{X,Y,Z}(x,y,z)\log\dfrac{\mathsf p_{X,Y,Z}(x,y,z)}{\mathsf p_X(x)\,\mathsf p_Y(y)\,\mathsf p_Z(z)}
\\[1ex] =~& \mathsf I(X; Z, Y)
\end{align}$$


What are some interesting un-intuitive problems in probability aside from Monty Hall? Does anyone know of some interesting and almost strange problems in probability?

I know that probability is sometimes notorious for being mind-bending and un-intuitive! (Monty Hall is already an obvious one)

Shuffle a deck of cards (standard $52$-card bridge deck) and deal one card. What's the probability that it's an ace? The answer is $\frac1{13},$ nothing unintuitive about that.

Shuffle a deck of cards and deal one card after another until an ace is dealt, then deal one more card. What 's the probability that the last card is an ace? Intuitiveness is a matter of opinion; personally, I find it kind of unintuitive that the probability is still $\frac1{13}.$

(More generally, for $n\ge k\ge2,$ if balls are drawn one by one without replacement from an urn containing $n$ balls of which $k$ are black, then the probability that the first black ball is immediately followed by another black ball is equal to $k/n.$)

An interesting introductory probability problem is one of basic geometry.  Say you have a circle with an inscribed equilateral triangle.  The question is, if you were to randomly draw a chord on this circle, what is the probability that it will be greater than the side length of the triangle?

You will come to 2 (or possibly more) conclusions.  Thus, this problem emphasizes the importance of being precise when stating a question in probability.  Saying "randomly selecting a chord on the circle" is the ambiguous statement that allows one to arrive at several conclusions about the probability.


Symmetric random walk with bounds can anyone help me with this:
We are considering a symmetric random walk that ends if level 3 is reached or level -1 is reached. Start=0

What is the expected number of walks? So I am looking for: $E[{\tau}]$ with $\tau$=the stopping time.

Basic argument: One asks for $t_0=\mathbb E_0(\tau)$ where $t_x=\mathbb E_x(\tau)$ for every $-1\leqslant x\leqslant3$. By the (simple) Markov property after one step, $t_x=1+\frac12(t_{x-1}+t_{x+1})$ for every $0\leqslant x\leqslant2$. By definition, $t_{-1}=t_3=0$. This is an affine system of $5$ equations with $5$ unknowns. Solve it. This yields $t_x=(3-x)(x+1)$ for every $-1\leqslant x\leqslant3$. In particular $t_0=3$.

Less basic argument: For every $n\geqslant0$, let $x_n$ denote the position after $n$ steps, $\mathfrak X_n$ the sigma-algebra generated by $(x_k)_{0\leqslant k\leqslant n}$, and $z_n=(x_n+1)(3-x_n)+n$. Then $z_0=3$, $z_\tau=\tau$, and $(z_n)_{n\geqslant0}$ is a martingale with respect to the filtration $(\mathfrak X_n)_{n\geqslant0}$, hence $\mathbb E(z_0)=\mathbb E(z_\tau)$, that is, $\mathbb E(\tau)=3$.

In full generality, the same argument shows that, for every $(a,b)$, the first hitting time of $\{a,b\}$ starting from $a\leqslant x\leqslant b$ has mean $(b-x)(x-a)$.

I asked a similar question to poster and was directed here, but found the level of complexity of the arguments very difficult to sift through for my level of math experience.

It turns out the answer is incredibly simple - I was shocked.

For anyone who just needs the solution for application purposes, the answer is A*B, where A is the upper bound and B is the lower bound.

Quote from wikipedia:
If a and b are positive integers, then the expected number of steps until a one-dimensional simple random walk starting at 0 first hits b or −a is ab. The probability that this walk will hit b before −a is a/(a+b), which can be derived from the fact that simple random walk is a martingale.


Limit of convergent sequences Suppose that $\{a_n\}$ and $\{b_n\}$ are convergent sequences. Show that if $\lim a_n < \lim b_n$, then there exists a natural number $N$ such that $a_n < b_n$ holds for all $n\geq N$.

I don't know how to solve this problem. Could someone help, thanks in advance.

We have $$\lim_na_n=A<B=\lim_nb_n.$$ By definition

$$\forall \epsilon>0\exists N_1\in\mathbb{N}: n\ge N_1\implies A-\epsilon<a_n<A+\epsilon$$ and

$$\forall \epsilon>0\exists N_2\in\mathbb{N}: n\ge N_2\implies B-\epsilon<b_n<B+\epsilon.$$ In particular, for $\epsilon=\frac{B-A}{2}$ we have that

$$\exists N_1\in\mathbb{N}: n\ge N_1\implies a_n<A+\frac{B-A}{2}=\frac{A+B}{2}$$ and

$$\exists N_2\in\mathbb{N}: n\ge N_2\implies \frac{A+B}{2}=B-\frac{B-A}{2}<b_n.$$ Thus, we have that

$$n\ge \max\{N_1,N_2\}\implies a_n<\frac{A+B}{2}<b_n.$$

Hint: Study the sequence $c_n = b_n - a_n$. What is the sign of $\lim_{n \to \infty}c_n$? What does that say about the sign of $c_n$ for large $n$? What does that say about the relationship between $a_n$ and $b_n$ for correspondingly large $n$?


Find integers $x$ and $y$ such that $8^x-9^y=431$ Find integers $x$ and $y$ such that $8^x-9^y=431$

My working:
By taking mod 9 and 16, I got $x$ odd and $y$ even.
Also $8^x>431\implies x\ge 3$
For $x=3$ I got $y=2$


Stuck with homework (limit exercise) - squeeze theorem Given function $f:\mathbb R→\mathbb R$ for which $|f(x)-2|≤x^2$.
Find the limits:

$\lim_{x\to0}f(x)$.
$\lim_{x\to0}\frac{f(x)-\sqrt{x+4}}{x}$.

I can solve question (a) very easily (squeeze theorem) and the answer is $\lim_{x \rightarrow 0} f(x)= 2$.
But when I start solving question (b), I'm in a dead end. The book says that the answer of question (b) is $\lim_{x \rightarrow 0 }\frac {f(x)-\sqrt{x+4}}{x} = -\frac{1}{4}$.
Perhaps I'm doing an algebra mistake somewhere because I end up always at $-\frac{1}{0}$ which is not possible. Could someone please give me a hint ?
Thanks

You can split your limit into two parts:

Part 1: $\frac{f(x)-2}{x}$ which tends to zero.

Part 2: $\frac{2-\sqrt{x+4}}{x}$ which tends to $-1/4$ by L'Hopital's rule.

a) $lim f(x)$ for $x→0 =2$
because

$0<=|f(x)-2|<=x^2$, by squeeze theorem,

as $x$ goes to $0,0<=|f(x)-2|<=0.
so |f(x)-2|=0$,

hence $lim f(x)$ for $x→0 =2$