Prove that $ \sum\limits_{i=1}^ \infty \frac{(-1)^{i+1}}{\sqrt{i}}$ converges, using Abel's summation formula Let $(a_n)$ and $(b_n)$ be two sequences. Let $A_n = \sum_{k=1}^n a_k$ and $A_0 = 0$. Prove that
$$\sum_{n=p}^q a_n b_n =  \sum_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$$
for each $q\geq p \geq 1$
Also, using this result prove that the series
$$ \sum_{i=1}^ \infty \frac{(-1)^{i+1}}{\sqrt{i}}$$
converges.

For the second, converge part, the series continues like $1,\frac{-1}{\sqrt2},\frac{1}{\sqrt3},\frac{-1}{\sqrt4}$. So it converges to 0. But I do not know how to prove first part also using the first part to show this series converges.

Take $a_n=(-1)^{n+1}$ and $b_n=\frac 1 {\sqrt n}$. Check that $A_n$ is $1$ or $0$ for all $n$. Next note that $b_n-b_{n+1}=\frac {\sqrt {n+1}-\sqrt {n}}  {\sqrt {n+1}\sqrt {n}}$ which can be written as $\frac 1 {\sqrt {n+1}\sqrt {n}(\sqrt {n+1}+\sqrt {n})} \leq \frac 1 {n^{3/2}}$. Use the fact that $\sum \frac 1 {n^{3/2}}<\infty$ to complete the argument.

FYI, Determination of the value of the sum.

$S=\sum\limits_{i=1}^ \infty \frac{(-1)^{i+1}}{\sqrt{i}}$

Separate the sum accordance with even and odd terms. If i even then $i=2m$ and if i odd then $i=2m-1$

$S=\sum\limits_{m=1}^ \infty\big( \frac{1}{\sqrt{2m-1}}-\frac{1}{\sqrt{2m}}\big)=\frac{1}{\sqrt{2}}\sum\limits_{m=0}^ \infty\big( \frac{1}{\sqrt{m+\frac{1}{2}}}-\frac{1}{\sqrt{m+1}}\big)=\frac{1}{\sqrt{2}}\sum\limits_{m=0}^ \infty \frac{1}{\sqrt{m+\frac{1}{2}}}-\frac{1}{\sqrt{2}}\sum\limits_{m=1}^ \infty\frac{1}{\sqrt{m}}$

From the definition of the and Hurwitz zeta functions we get:

$S=\frac{1}{\sqrt{2}}\zeta(\frac{1}{2},\frac{1}{2})-\frac{1}{\sqrt{2}}\zeta(\frac{1}{2},1)$, as $\zeta(\frac{1}{2},1)=\zeta(\frac{1}{2})$

and using the equality: $\zeta(s,\frac{1}{2})=(2^{-s}-1)\zeta(s)$ (where $\zeta(s)$ is the Riemann zeta function),
we have the following result:

$S=(1-\sqrt{2})\zeta(\frac{1}{2})\approx 0.60489...$


Why is the error of $O(h^2)$ when using Taylor expansion and centered approximation for the first derivative We know that the approximation of the first derivative by centered approximation is given by
$$ f'(x) =
\frac{f(t+h) - f(x-h)}{2h} + O(h^2)$$
The quality of the above approximation is determined by the error. For smaller values of $h>0$ the error should be as small as possible. To determine this error we use the Taylor expansion

\begin{align*}
y(t+h) &= y(t) + y'(t)h + \frac{y''(t)}{2}h^2+\frac{y'''(t)}{3!}h^3+ O(h^4)\\
y(t-h) &= y(t) - y'(t)h + \frac{y''(t)}{2}h^2-\frac{y'''(t)}{3!}h^3+ O(h^4)\\
y(t+h) - y(t-h) &= 2y'(t)h + 2\frac{y'''(t)}{3!}h^3 + O(h^4)\\
\frac{y(t+h)-y(t-h)}{2h}-y'(t) &= \frac{y'''(t)}{3!}h^2 + O(h^3) = O(h^2)
\end{align*}

an thus we conclude
$$
y'(t) = \frac{y(t+h)-y(t-h)}{2h}+O(h^2)
$$

Here is my question

We could have arbitrarily decided to take the Taylor expansion with the polynomial of fourth and fifth order namely writing the first line as follows:
$$
y(t+h) = y(t) + y'(t)h + \frac{y''(t)}{2}h^2+\frac{y'''(t)}{3!}h^3+\frac{y''''(t)}{4!}h^4+\frac{y'''''(t)}{5!}h^5+ O(h^6)
$$
Thus the last line would have been:
$$
\frac{y(t+h)-y(t-h)}{2h}-y'(t) = \frac{y'''(t)}{3!}h^2 + \frac{y'''''(t)}{5!}h^4 + O(h^5) = O(h^4)
$$
and conclude that the error is of order $= O(h^4)$

Why is this not true?

Because for smaller value of $h$ which is less than $1$, $h^2$ is greater than $h^5$. Hence, if the error of an algorithm is $O(h^2)$ is in $O(h^5)$ for small values of $h$ and not vice versa.


Continuity of composite functions with point discontinuities My colleague and I are stuck on the answer provided by our textbook for the following question.

  Consider the functions $\begin{equation*}f(x) = \begin{cases} x+1 & x \neq 0\\ 0   & x = 0\\ \end{cases} \end{equation*}$ and $\begin{equation*}g(x) = \begin{cases} x & x \neq 1\\ 0   & x = 1\\ \end{cases} \end{equation*}$.

  Show that $f$ is discontinuous at $x=0$ and $g$ is discontinuous at $x=1$ but $f \circ g$ is continuous at $x=0$ and $g \circ f$ is continuous at $x=1$.

We are stuck on:

  $f \circ g$ is continuous at $x=0$

The answer simply provides:

  $\lim_{x \to 0}f(g(x))=\lim_{x \to 0}f(0)=0=f(g(0))$

We can see that $\lim_{x \to 0}g(x)=0$ and the definition of $f$ gives us $f(0)=0$.

The important bit (and reason for posting this question) …

In every reference I can find, the following rule is only said to be true if $f$ is continuous at $\lim_{x \to c} g(x)$.

  $$\lim_{x \to c}f \circ g(x) = f(\lim_{x \to c} g(x))$$

But in our question $c=0$ and $\lim_{x \to 0} g(x)=0$ and $f$ is not continuous at $0$, as seen in the definition of $f$ and stated in the question.

The limit for f(g(x)) is not 0 ... it is 1.  By constructing the composite function manually, it is clear that f(g(x)) = x+1 everywhere except x=0 and x=1.  One can then easily see that the limit must be 1.  However, since f is not continuous at 0, the composite rule, as it is written in the form given, cannot be applied.


Clarifications on proof of Sperner's Theorem Here I am considering a set $X$ with $|X| = n$ and it's power set $ \mathcal P(X)$.

Denote by $X^{(i)}$ the set of elements of $\mathcal P(X)$ with $i$ elements, and consider $\mathcal P(X)$ as a poset ordered by $A < B \Leftrightarrow A \subset B$.

Then in my lecture notes, we have Sperner's Theorem saying the largest size of an anti-chain in $\mathcal P(X)$ is $\binom{n}{\lfloor n/2 \rfloor}$, witnessed by $X^{(\lfloor n/2 \rfloor)}$.

Then, to prove this: on realising that for any chain $C$ and anti-chain $A$ of $\mathcal P(X)$ we have that $|C \cap A| \leq 1$, my notes say that: "it is sufficient to partition $\mathcal P(X)$ into $\binom{n}{\lfloor n/2 \rfloor}$
chains".

I understand that on partitioning $\mathcal P(X)$ into this many chains, then we have that an anti-chain can have at most one non-trivial intersection with each of them. Further, we may notice (with appropriately chosen chains) that $X^{\lfloor n/2 \rfloor}$ is indeed a witness to the case of an antichain of size $\binom{n}{\lfloor n/2 \rfloor}$. Together this tells us the maximal anti-chain size is at least $\binom{n}{\lfloor n/2 \rfloor}.$

Why then do we now know there can't be an anti chain that is longer? From what I can tell this proof has given us the wrong bound (lower instead of upper) on the size of the largest possible anti-chain. I expect that I have misunderstood something crucial here, or missed an assumed result, and I would be very grateful if you could point it out to me.

I apologise if I haven't been clear enough here. I have understood what I said here but likely only because I have my lecture notes in front of me. If there is something worth clarifying please let me know.

In case it is important: the method by which we construct the $\binom{n}{\lfloor n/2 \rfloor}$ chains is by using complete matchings from either $X^{(i-1)}$ to $X^{(i)}$ or from $X^{(i)}$ to $X^{(i-1)}$ (order depending on the size of $i$) by viewing $(X^{(i)},X^{(i-1)})$ as a biregular, bipartite graph.

You have chains $C_1,\ldots,C_m$ where $m=\binom{n}{[n/2]}$. Each element
of $\mathcal P(X)$ is in exactly one $C_j$. So each element of an antichain
$\cal A$ has at most one element in each $C_j$, but each element of $\cal A$ is in some $C_j$, so there are at most $m$ elements in $\cal A$.

I prefer the proof from the LYM inequality....

On the one hand, you have $\cal{P}$ partitioned into $m= {{n} \choose {\lceil \frac{n}{2}\rceil}}$ chains via the proof in the later part of your post @user, which gives the upper bound of $m$ on the size of the largest antichain.

On the other hand, you have been given an explicit antichain with precisely $m$ elements--namely all subsets of $[n]$ with precisely $\lceil \frac{n}{2} \rceil$ elements, this gives the lower bound of $m$ on the size of the largest antichain.

If both the upper bound and the lower bound on the size of the largest antichain is $m$, then it follows that the size of the largest antichain must indeed be $m$.


Moving average model, MA(q) implementation Assuming that I have a Time Series 'X'
which is a 1000 days data of a stock.

can someone share a code sample (in any language)
even pseudo code for the Moving average model order q?

I dont know about your purpose but if you just want to calibrate the model, use R.
i.p
require(smooth)
require(MComp)
sma(data,q)$forecast

Again an R Code:

 weights<-rep(1/q,q) # or any desired weights  ma<-filter(data,weights,sides=1)

 plot(data)  lines(ma)

and you can also take a look at the autocorrelation function  acf(ma,na.action=na.pass)


Find L in $T=2\pi\sqrt{\frac{L}{g}}$ I am given the formula for the period of a pendulum: $T=2\pi\sqrt{\frac{L}{g}}$ along with inputs $T=1$ and $g=32$.

I am to find L, the textbook provides the answer as 10 (to the nearest whole unit). I arrive at 0.8.

I'm not sure I approached this the right way initially. How can I arrive at 10?

$1=2\pi\sqrt{\frac{L}{4\sqrt{2}}}$

$1^2=4\pi^2\frac{L}{32}$  # square both sides to get rid of radical

$\frac{1}{1}=\frac{4\pi^2L}{32}$  # writing the left side a a fraction just makes me see clearer when working with fraction equations

$32=4\pi^2 L$  # multiple out denominator

$32=39.5L$  # $4*(3.14159^22)$ ~ 39.5

$L = 32/39.5=0.8$

How can I arrive at 10?

Here's a screen shot from the textbook in case I've missed anything. It's the second question starting "If the gravity is 32..."

You got an answer of $0.8$. But since the question gave gravity in feet, your final answer is in feet! Since 1 foot is 12 inches, your final answer would be $$0.8\times 12=9.6\,\text{inches}$$


3 coins interpretation of the boy and girl paradox Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

I think I have come up with a new interpretation for this problem. If children where coins, boys were heads and girls were tails, we would have $3$ coins: $2$ of them can be both heads and tails, and the other one is loaded so it can only be heads. We start throwing one of the $2$ normal coins. If it is heads, we throw the other normal coin. If it is tails, we throw the loaded coin. This way, the chance of both coins being heads ends up being ${1 \over 2}\cdot {1 \over 2}$, which is ${1 \over 4}$. This is neither ${1 \over 3}$ nor ${1 \over 2}$, which are the classic answers to this problem.

Is my interpretation wrong?

This would be easier with two coins, both fair.

Version (a): toss both coins.  If both are tails, start again.  Otherwise  record what happened.  About $\frac{1}{3}$ of records will show two heads.

Version (b): toss the first coin.  If it is tails, start again.  Otherwise toss the second coin and record what happened.  About $\frac{1}{2}$ of records will show two heads.

Your version is closer to "There is a family who wants at least one son.  If their first child is a boy they will have another child, but if it is a girl then they intend to use adoption or gender-selective abortion to make sure their second is a boy." That indeed would give a probability of $\frac14$ for the probability of two boys, but it is not either of the standard forms of the question.

In your method you are calculating based on the chance that if the first throw of normal coin is a head, however in the original problem one head is always guaranteed


What is the probability that the gambler is ruined$?$ A gambler has one rupee in his pocket.He tosses an unbiased normal coin unless either he is ruined or unless the coin has been tossed for a maximum of five times.If for each head he wins a rupee and for each tail he looses a rupee,then what is the probability that the gambler is ruined$?$

I made the sample space $S=\left\{H,T,HH,HT,HHH,HHT,HTH,HTT,HHHH,HHTH,HHHT,HHTT,HTHH,HTHT,HHHHH,HHHHT,HHTHH,HHHTH,HHHTT,HHTHT,HHTTH,HHTTT,HTHHH,HTHHT,HTHTH,HTHTT\right\}$

Now the cases when he is ruined are $T,HHTTT,HTT,HTHTT$
So i calculated probability that the gambler is ruined=$\frac{4}{26}$.But the book answer is $\frac{22}{32}$

What is wrong in my solution.What is the right way to solve this problem?

Some of your outcomes will never happen.  He can't get a H and then quit, nor would he quit after HTH, etc.  And your samples are not all equally likely to happen.  HHHTH for example will occur only 1 in 32 times, whereas HTT will occur 1 in 8 times, etc.

consider the ways he can be ruined and calculate those probabilities.

T  has a 1/2 probability.
HHTTT has a 1/32 prob,HTT has a 1/8 and HTHTT has 1 1/32.  The probability is
1/2 + 1/32 + 1/8 + 1/32 is 22/32.

The correct answer is $\frac {11}{16}$...the ways to be ruined are

  $$T,HTT,HHTTT,HTHTT$$

their respective probabilities are

  $$\frac 1 2, \frac 1 8, \frac {1}{32}, \frac {1}{32}$$

and the sum of these probabilities gives answer $\frac {11}{16}$.


Coefficient of Determination for High Order Polynomials Because the Coefficient of Determination makes use of the mean of y, is the Coefficient of Determination meaningless for higher order polynomials (higher order than linear polynomials)? I'm assuming this because I think it's easy to visualise how the variation from the mean gives a representation for how much of the variance is not described by the best fit line for linear graphs, but not for, say, fifth or sixth order polynomials.

Coefficient of Determination makes perfect sense for the linear regression of $y$ on $x_1$, $x_2$, ... In particular, it makes sense for the linear regression of $y$ on $x$, $x^2$, ...  -  a polynomial is a LINEAR combination of powers of $x$!


Can manholes be made in other shapes than circles, that prevent the cover from being able to fall down its own hole? Circular manholes are great because the cover can not fall down the hole. If the hole were square, the heavy metal cover could fall down the hole and kill some man working down there.
Circular manhole:
Can manholes be made in other shapes than circles, that prevent the cover from being able to fall down its own hole?
Semi rigid math formulation:
Let us say that we have an infinite matematical 2D plane in 3D space. In this plane is a hole of some shape. Furthermore we have a flat rigid 2D figure positioned on one side of the plane. This figure has the same shape as the hole in the plane, but infinitesimal larger.
Is it possible to find a shape, where there is no path twisting and turning the figure that brings the figure through the hole?
Here is one such shape (only the black is the the shape):
But if one put the restriction on the shape, that it needs to be without holes (topological equivalent to a circle in 2D), then I can not answer the question!?
Edit:
Because of the huge amount of comments and answers not about math, I fell the need to specify that:
I am not interested in designing manholes. I am interested in the math inspired by the manhole problem.

Any manhole cover bounded by a curve of constant width will not fall through.  The circle is the simplest such curve.

I hope manhole of equilateral Triangle Shape will also work.


Interesting Differentiation Technique @HansEngler Left the following response to this question regarding "bad math" that works,

  Here's another classical freshman calculus example:

  Find $\frac{d}{dx}x^x$.

  Alice says "this is like  $\frac{d}{dx}x^n = nx^{n-1}$, so the answer is $x x^{x-1} = x^x$."
  Bob says "no, this is like  $\frac{d}{dx}a^x = \log a \cdot a^x$, so the answer is $\log x \cdot x^x$."
  Charlie says "if you're not sure, just add the two terms, so you'll get partial credit".

  The answer  $\frac{d}{dx}x^x = (1 + \log x)x^x $ turns out to be correct.

In this comment, @joriki asserts that this is not "bad math" but rather a legitimate technique,

  You get the derivative of any expression with respect to $x$ as the sums of all the derivatives with respect to the individual instances of $x$ while holding other instances constant.

I had never previously seen such a technique so naturally I tested it on a few examples, including $\frac{d}{dx} \left( x^{ \sin x}\right)$, etc. and it provided the correct result. The following three questions arose,

$ \ \ $  1. What is the proof of its is validity?
$ \ \ $  2. Are there any examples where this technique outshines standard methods?

Look at $y = f(u(x),v(x))$:
$$\frac{dy}{dx} = \frac{\partial f}{\partial u}\cdot\frac{du}{dx}+\frac{\partial f}{\partial v}\cdot\frac{dv}{dx}$$
Now, note that $x^{\sin x}$ can be written as:
$$y=x^{\sin x} = f(x,\sin x), \ f(u,v) = u^v$$
So that:
$$\frac{dy}{dx} = \frac{\partial f}{\partial u}\cdot\frac{du}{dx}+\frac{\partial f}{\partial v}\cdot\frac{dv}{dx}=
vu^{v-1} \cdot\frac{du}{dx} + \log u \cdot u^v \cdot\frac{dv}{dx}$$
In fact, this holds for any number of terms (since so does the chain rule), where each time we only look at the derivative by $x$ of only one of the terms, multiply those by the "internal" derivative, and sum them all up.

I beg to differ with the suggested differential to the x to the power of x. The result would be best expressed using ln x or the log to base e of x. Ie

d/dx(x ^x) = x ^x (ln x + 1)

When using log(believed to be in base 10), the result will be expressed as

d/dx(x ^x) = x^x ln10(log x + log e)

Problems of this form could all be solved by taking the log or ln of both sides and implicitly differentiating with respect  to x


Let G = (V, E) be a simple undirected graph with n = |V | ≥ 1 vertices... Let G = (V, E) be a simple undirected graph with n = |V | ≥ 1 vertices. A subset U ⊆ V of the vertices is called a vc-set if for every edge {i, j} ∈ E either i ∈ U or j ∈ U (or both). Let $U^∗$ be a vc-set for G that has the smallest size possible (meaning there does not exist any other vc-set $U^{'}$
such that (|$U^{'}$| < |$U^∗$|). A subset W ⊆ V of the vertices of G is called an in-set if for all i, j ∈ W,
{i, j} $\notin$ E. Let $W^∗$ be a largest possible in-set in G (meaning there is no other in-set, $W^{'}$, such that |$W^∗$| < |$W^{'}$|). Prove that |$W^*$| = n − |$U^∗$
|.
I tried to use the basics of graph theory but was of no use as I got stuck and back to the starting point. If you could help me I would be really grateful thanks a ton!

We use the fact: the complement of vc-set is an in-set and vice versa.

V\W* is vc-set then |V\W*| $\geq$ |U*| which means |V\W*| $=$ n - |W*| $\geq$ |U*|.

V\U* is in-set then |V\U*| $\leq$ |W*| which means  |V\U*| $=$ n - |U*| $\leq$ |W*|.

Putting both inequalities together yields n $=$ |U*| + |W*|

You can show, that for every x,y which belong to V\U* holds that {x,y} is not an edge. Otherwise x or y is in U* (contradiction). So |W*| is greater than or equal to |V\U*|.
Also there exists no other W' with more vertices than V\U*. Because V\W' would be vertex cover smaller than U (contradiction).

There for n = |U*| + |V\U*| = |U*| + |W*|.

Sorry for my broken english.


Applications of equidistribution What are applications of equidistributed sequences ?

I'm looking for examples of problems/questions in fields (not necessarily mathematical) where equidistribution is an ad-hoc tool towards a solution.

Before I list some applications of equidistribution, I need to maek clear
what equidistribution is and is not.  Especially what it is, because many
people have a rather narrow view of it.

First of all, equidistribution is neither a sub-area of number theory or
analysis nor a technique.  It is a principle and as such it permeates every
field of mathematics to some extent (as we will see, this includes some
hardcore algebra).

Secondly, equidistribution is not a game played only on $[0,1]$ or the real
numbers.  Kuipers and Niederreiter's classical text considers
equidistribution in general topological spaces and groups.

Thirdly, sequences are not the only thing that can be equidistributed.  Any
system involving a parameter is amenable to the notion, whether the index set
is the integers, the reals, a real or complex vector space or a suitable
index category.  All you need is a directed object, a notion of averaging on
the target and some completeness.

Fourthly, when experts speak of equidistribution they do not necessarily mean
a limiting distribution that gives a fair share of the sequence to every
corner of the space, although that is what the name implies and it is the
most important limiting distribution.  What the experts mean is that the
sequence does have a limiting distribution.  In fact, equidistribution
theory should be better named 'limiting distribution theory'.

Finally, finding a limiting distribution is not the end of the investigation,
it is a little below the midpoint.  A limiting distribution is not good
enough for practical reasons, you want to know what happens before we reach
the limit; this is where the crucial notion of discrepancy comes in, and that
can get pretty complicated if your host space has rich geometry: then you
have to define lots of discrepancies adapted to various shapes in your space,
and because discrepancy is not a limiting statement it is very delicate with
respect to the shape.  I am mentioning this because discrepancy estimates is
one of the most applicable aspects of equidistribution theory.

Now on to the applications.

As a principle, since equidistribution is a limit of counting processes,
naturally it helps simplify difficult counting problems by passing to the
limit.  For example:

Counting number of closed geodesics of some bounded length on a
hyperbolic surface, or more generally totally geodesic tori in symmetric
spaces.  See
http://www.math.harvard.edu/~ctm/papers/home/text/papers/mixing/mixing.pdf
for an approachable introduction.
Counting prime numbers; the prime number theorem is an equidistribution
theorem, and the Riemann Hypothesis an optimal discrepancy bound.  By
bounding this discrepancy optimally, you guarantee that there are no notable
gaps in the distribution of prime numbers that could be exploited to optimize
brute force factorization (for cracking public key cryptosystems) by avoiding
'gappy' regions of the search space.
Proving density of a sequence in a space.  This is a paradoxical but very
common technique: you have a sequence in a space that you want to prove is
dense.  Prove the stronger statement that it is equidistributed and you are done.
Counting and more generally the distribution of rational points on
algebraic groups, their homogeneous spaces and algebraic varieties.  Algebraic varieties over a field $k$ can have many
points in an algebraic closure, but how do we know about the existence of
points in $k$ or finite extensions?  For a large class of varieties this
question is one of equidistribution (by the paradoxical scheme
equidistribution implies density).  See
Yuri Tchinkel's amazing exposition here:
http://www.cims.nyu.edu/~tschinke/papers/yuri/08cmi/cmi4.pdf
An amazing instance of the previous theme is found in conjunction with
Weil's conjectures.  In proving the Weil conjectures, Deligne formulated his
equidistribution theorem linking in a fascinating way projective varieties and
affine (compact real) groups; this formed part of the inspiration for Katz
and Sarnak's proof that the statistics of zeros of zeta functions of
algebraic varieties come from random matrix (Haar) ensembles on appropriate
compact groups or symmetric spaces of compact type.  The entire proof is
found in Katz & Sarnak, Random Matrices, Frobenius Eigenvalues, and Monodromy.
Manjul Bhargava's groundbreaking work on the number of low-degree field
extensions of bounded discriminant is based on a counting problem that
involves equidistribution techniques.  This is one case (of the many) where the object
being equidistributed is an algebraic structure!  For a flavor, see
the paper http://arxiv.org/pdf/1309.2025 or dive directly into the sequence
of Bhargava's Annals papers (the original based on his thesis is not too
hard to read due to his clear writing; it gets more and more involved with
each successive paper though).  On the very first page of
http://annals.math.princeton.edu/wp-content/uploads/annals-v162-n2-p10.pdf
you have an equidistribution statement :)
For a more applied flavor, numerical integration and the entire field of
approximation theory hinge on finding discrete sequences in a space that are
distributed well enough to replace integration by weighted summation on the
sequence.  The basic results that makes this possible is the Koksma-Hlawka
inequality which states, roughly, that the difference between the integral
and a summation over a sample is bounded by the discrepancy of the sample
times the variation of the integrand.  This has been extended to a vast
number of settings and is one of the most useful mathematical statements outside of mathematics.
In fact, the best place to look for applications of equidistribution
outside mathematics is everywhere.  That is because equidistribution goes
hand in hand with the ubiquitous sampling problem.  Sampling by independent
trials is not a good idea, because independence guarantees that you will
have long strings of bias, and you do not know where they will be.  What you
want is low discrepancy samples and that is the purview of equidistribution.
Physicists care about distribution of eiqenfunctions of Hamiltonians on manifolds as the
primal example of quantum chaos.  The mathematical statement is called
quantum ergodicity and is essentially an equidistribution statement, see the
wiki article http://en.wikipedia.org/wiki/Quantum_ergodicity and Sarnak's
book Arithmetic Quantum Chaos for a splendid introduction (not only to  the
arithmetic theory, it also has a good exposition on the general physical
problem!).  The book is freely available online!

There are so many more applications outside of mathematics that this post
would never need to end.  However, I need to stop somewhere so I will let you
follow the links above and branch out.  The starting point for the theory is
Kuipers and Niederreiter Uniform Distribution of Sequences, but as I have
indicated above, there is hardly a mathematical field that does not exploit
the principle of equidistribution.  For those that don't, well, what are you waiting for?

EDIT: Oh dear, I forgot to mention the overwhelming number of applications of equidistribution in Computer Science.  Because I am too tired for a big edit, if you are a computer scientist and want to see such applications, I wholeheartedly recommend Chazelle's book The Discrepancy Method.  WARNING: some of the proofs there are WRONG and some of the exercises undoable.  However, the prose and topic selection makes for an excellent introduction to discrepancy theory in CS and related fields!

This is not too broad, but it reminds me of a solution to a specific problem:

Show that the sequence $tan(n) ; n=1,2,3,...$ diverges.

We (can) show this by using equidistribution of $\pi$ ( and therefore $\pi/2) mod1$, meaning the decimal parts of $\{\pi, 2\pi,..., n\pi,...\}$ are dense in the unit interval. It follows that the sequence $$a_n =n $$ will be indefinitely close to (but never actually equal to, by irrationality of $\pi$) to $(4k \pm 1)\pi/2$ , so that $tan(n)$ will blow up
for infinitely-many n, meaning beyond any specific/fixed value of n. So $tan(n)$ diverges. EDIT: What  I mean here is that the decimal parts of $\{\pi,2pi,..., n\pi,...\}$ are dense on the unit interval. I stand by it despite the downvote.


A question regarding finding the minimal polynomial associated with a field extension . Say we have the field extension $\Bbb Q(w,\sqrt[3]{5})$ over $\Bbb Q$, where w is the primitive cubed root of unity.
I know that the minimum polynomial of $\sqrt[3]{5}$ is $x^3-5$.
I want to figure out the degree of the extension;

I know we can use the tower law $|\Bbb Q(w, \sqrt[3]{5});\Bbb Q|=|\Bbb Q(w, \sqrt[3]{5});\Bbb Q(\sqrt[3]{5})||\Bbb Q(\sqrt[3]{5});\Bbb Q|=x.3$ as $x^3-5$ has degree 3

But how does one find the minimum polynomial of the extension  of degree x

Clearly $w \not \in \Bbb Q(\sqrt[3]{5})$ so the degree is at least $2$. Now notice that $x^2+x+1$ is the minimum polynomial.


The question is about the homeomorphic image of right closed interval; I know the answer but I am a little confused. Let $f:\mathbb{R}\to \mathbb{R}$ be  a continuous map. Which one of the following can not be the image of $(0,1]$ under $f$?

$\{0\}$
$(0,1)$
$[0,1)$
$[0,1]$

(I am really confused about the mappings of intervals, please suggest the links. ). Also I know that the 2nd option is correct but I am not sure about how. Please explain.

I only treat $(0,1)$, the other cases are not difficult. Assume that a continuous map $f : \mathbb R \to  \mathbb R$ has the property $f((0,1]) = (0,1)$. Then $f([0,1]) = (0,1) \cup \{f(0)\}$. On the left hand side we have a compact subset of $\mathbb R$, but the set on the right hand side cannot be compact because $[0,1]$ is the smallest compact set containing $(0,1)$. This shows that no such $f$ exists.

Note that there exist continuous maps $g : (0,1] \to  \mathbb R$ such that $g((0,1]) = (0,1)$. As an example take
$$g(x) = \frac{1}{2} +  \frac{1-x}{2}\sin  \frac{1}{x} .$$
As $x \to 0$, $g(x)$ oscillates increasingly rapidly around the line $y =  \frac{1}{2}$ and comes arbitrarily close to the lines $y = 0$ and $y = 1$. However, there is no limit $\lim_{x\to 0} g(x)$.

$(0,1)$ is the image under $f(x)=\frac{1}{\pi}\arctan(x)+\frac{1}{2}$, e.g.

$\{0\}$ is the image under the obvious constant map.

$[0,1)$ is the image under $f(x)=\frac{x^2}{1+x^2}$.

$[0,1]$ under $$f(x)=\begin{cases} 0 & x \le 0\\
                                   x & x \in [0,1]\\
                                   1 & x \ge 1\\
                     \end{cases}$$

which is continuous too.

That is for $\Bbb R$.

We can find $-x + 1$ to get $[0,1)$ and a similar casewise function to get $[0,1]$ and $\{0\}$ is easy as before. Only $(0,1)$ as image seems tricky..

For $(0,1]$ onto $(0,1)$ consider what point $f(1)$ can be. Suppose it's $0 < p < 1$, then $f[(0,1)]$ must be connected so $p$ must be the image of at least one other $x \in (0,1)$ under $f$, otherwise $f[(0,1)] = (0,1)\setminus \{p\}$ would be disconnected. Then argue that $f^{-1}[\{p\}]$ cannot be finite (using components again). But this set must have a limit point in $[0,1]$ ...

Admittedly, it's a tricky argument, but off hand I see no better one. Of course it's multiple choice and I showed how to get the other 3 as image, so elimination also works. Maybe that was the idea of the question.


Evaluating limits with variable exponent and an exponent function with fraction I face a problem when trying to evaluate the following limits.

$$\lim\limits_{x\rightarrow 0} \left(\frac 14\left(\dfrac{e^2}{x} +3\right)\right)^{4x}.$$
Please help. Thank you.

Assume the limit exists and has value $L$.  Then

$$\begin{align}\log{L} &= \lim_{x \to 0} 4 x \log{\left ( \frac{e^2/x + 3}{4}\right)}\\ &=\lim_{x \to 0} 4 x\log{\frac{e^2}{4 x}} \\ &= \lim_{x \to 0} 4 x (2 - \log{4 x}) \\ &= 0 \end{align}$$

because $\lim_{y \to 0} y \log{y} = 0$.  Then $L=1$.

If $x$ is not an exponent, this is how you do it: let $h=1/x$. The problem now becomes $limit$ $as$ $h$ $approaches$ $infinity$ $of$ $(h*e^2+3)^4/(4^4*h)$. This gives the indeterminate form of $infinity/infinity$, so let's apply L'hopitals rule. $Limit$ $as$ $h$ $approaches$ $infinity$ $of$ $[4e(h*e+3)^3]/4^4$. This should be clear now that the limit approaches $infinity$.


Can one define the range of csc(x) with one prædicate in {y∈ℝ| ... }? I'm unfamiliar with set-builder notation and want to know whether the above can be concisely written with disjunction or union without definining two sets, so I only have to define the domain once. The only advice I've gotten so far is that doing it like that is not possible.

The set is a a union of two disfunction sets which can be rearranged as as one set consisting of two parts, principal and(intersection with) minus_[repeating ]complement


Difference between two inequality symbols I have come across the symbol $ \leqq$ in a paper I am reading. Is there any difference between this symbol and the symbol $ \leq $?

No, they are the same. You should consider them the same symbol in different 'fonts' rather than two different symbols.

Y ≦ X

Means Y is than or equal to every value of X for each correspending component in vector inequality.

Difference is that ≤ is used in calculus and ≦ in vector analysis.

Also see http://en.wikipedia.org/wiki/List_of_mathematical_symbols


Partial fractions and using values not in domain I'm studying partial fraction decomposition of rational expression. In this video the guy decompose this rational expression:

$$ \frac{3x-8}{x^2-4x-5}$$

this becomes:

$$\frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1} $$

$$[(x-5)(x+1)]\times \frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1}\times[(x-5)(x+1)]$$

$$3x-8 = A(x+1) + B(x-5)$$

then he substitutes $x$ with $-1$ this is when I feel something is wrong.

because $x=-1$ is not in the domain of the original function and all the statement below the original function hold true only if they comply within the domain of the original function isn't it? so at the time this guy substitute $x=-1$ the statement become false?

I'm newbie in math, please explain to me with easy understanding and step by step.

When you get to the stage $3x-8 = A(x+1) + B(x-5)$, you've forgotten about the partial fractions and all you're trying to do is determine a polynomial identity. i.e: what $A$ and $B$ will make $A(x+1) + B(x-5)$ identical to $3x-8$ for all values of $x$, this includes $x=-1$ and  $5$.

Once you've got those values, then you divide both sides of the polynomial identity by $(x+1)(x-5)$ with the added restriction that $x\neq -1, 5$. So this gives you $$\frac{3x-8}{(x+1)(x-5)} = \frac{A(x-5) + B(x+1)}{(x+1)(x-5)} = \frac{A}{x-5} + \frac{B}{x+1}$$

which holds for all $x \neq -1, 5$. (which is what let us do the cancelling bit on the RHS)

Some explanation: the motivation for using $x=-1$ and $5$ is that is efficient in the sense that it makes one of the terms in the identity $0$, speeding up the process of finding the values of the coefficients. If you wanted to, you could use any other two values such as $x=3$ and $4$ to get two simultaneous equations in $A$ and $B$, allowing you to determine their values, just not as efficiently.

Observe that the equation

\begin{eqnarray}
\frac{3x-8}{(x+1)(x-5)}=\frac{A}{x-5}+\frac{B}{x+1}
\end{eqnarray}

is valid only for $x\in \mathbb{R}\backslash \{-1,5\}$, as you have rightly pointed out. However, the equation

\begin{eqnarray}
3x-8=A(x+1)+B(x-5)
\end{eqnarray}

is valid for all $x\in \mathbb{R}$.


Help solving $5^{-4} \left(\frac{2x+5}{8}\right)^2 = 49$ I am trying to solve a simple formula but I get lost in the orders of operation.

The formula is:

$5^{ -4}{ \left( \frac { 2x+5 }{ 8 }  \right)  }^{ 2 } =49$

The easy parts:
  $5^{-4} = 0.0016$
  $\sqrt { 49 }  = 7$

So that gives me:

$-0.0016(2x+5/8) = 7$

Am I now allowed to to divide both sides of the equal sign with X to get:

$-0.0016(2+5/8) = 7/x$

?

And if that is correct, then the order of operation is as follows, right?

$-0.0016((2/8)+5) = 7/x$

Am I messing it up completely? :)

So, walking through this step-by-step, in case someone finds this question in the future and needs a more detailed answer, we have

$$5^{-4}\left(\frac{2x+5}{8}\right)^2 = 49.$$

Now, our order of operations tells us to take care of "parenthesis" first. However the expression inside the parenthesis is about as simple as it can get, so we'll leave that alone, for now, and instead take of the exponents. As you calculated, $5^{-4} = 0.0016$, so we have

$$0.0016\left(\frac{2x +5}{8}\right)^2 = 49.$$

We still have an exponent that we would like to get rid of. We can do so by taking the square root of both side, as you did in your attempt, but making sure to square root everything on both sides. This gives us:

$$\sqrt{0.0016 \left(\frac{2x+5}{8}\right)^2} = \sqrt{49}$$
$$\sqrt{0.0016}\left(\frac{2x + 5}{8}\right) = \sqrt{49}.$$

Now, we'll divide each side of our equation by $\sqrt{0.0016}$, and we have

$$\frac{2x + 5}{8} = \frac{\sqrt{49}}{\sqrt{0.0016}} = \sqrt{\frac{49}{0.0016}},$$

The right hand side of this equation is either $175$, or $-175$. So, we have

$$\frac{2x+5}{8} = 175 \ \ \ \text{ or } \ \ \ \frac{2x+5}{8} = -175,$$

and these expressions give us

$$2x + 5 = 1400 \ \ \ \text{ or } \ \ \ 2x + 5 = -1400$$

$$2x = 1395 \ \ \ \text{ or } \ \ \ 2x = -1405$$

$$x =697.5 \ \ \ \text{ or } \ \ \ x = -702.5$$

I would have opted for immediately taking the square root of both sides, if just to simplify the numbers!  $\sqrt{5^{-4}\left(\frac{2x+ 5}{8}\right)^2}= \sqrt{49}$ or $5^{-2}\left(\frac{2x+ 5}{8}\right)= 7$.  Multiplying on both sides by $5^2= 25$, $\frac{2x+ 5}{8}= 175$.  Multiply on both sides by 8, $2x+ 5= 1400$.  I let you do the rest!


How long will it take Marie to saw another board into 3 pieces? So this is supposed to be really simple, and it's taken from the following picture:

Text-only:

  It took Marie $10$ minutes to saw a board into $2$ pieces. If she works just as fast, how long will it take for her to saw another board into
  $3$ pieces?

I don't understand what's wrong with this question.  I think the student answered the question wrong, yet my friend insists the student got the question right.

I feel like I'm missing something critical here. What am I getting wrong here?

Haha! The student probably has a more reasonable interpretation of the question.

Of course, cutting one thing into two pieces requires only one cut! Cutting something into three pieces requires two cuts!

  -------------------------------  0 cuts/1 piece/0 minutes
  ---------------|---------------  1 cut/2 pieces/10 minutes
  ---------|-----------|---------  2 cuts/3 pieces/20 minutes

This is a variation of the "fence post" problem: how many posts do you need to build a 100 foot long fence with 10 foot sections between the posts?

Answer:  11  You have to draw the problem to get it...See below, and count the posts!

|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|
0-----10----20----30----40----50----60----70----80----90---100

Something is wrong, certainly. Several answers have debated the question of whether it was the student or the teacher who was wrong. I see another possibility: the question is wrong.

The student's been taught, in a maths lesson, about proportions or "rule of three" or something similar. So it's natural for the student to assume that the question's purpose is to test the student's ability to use that technique. And also to assume that the question is sound, and that the student isn't responsible for looking for errors in it and dealing with them (if any). So even if the student spots the out-by-one error, they might assume that that is merely an error in the way the question was presented, and then go on to answer the proportions question that they think was intended.

"15 minutes" and "20 minutes" are both reasonable responses to a question which was unsound in what the circumstances seem to be. The correct thing would've been to present the underlying proportions question in a different way.


How find this limit $\lim_{x\to 0}\frac{1}{x^4}\left(\frac{1}{x}\left(\frac{1}{\tanh{x}}-\frac{1}{\tan{x}}\right)-\frac{2}{3}\right)=？$ Find this following limit
$$\displaystyle \lim_{x\to 0}\dfrac{1}{x^4}\left(\dfrac{1}{x}\left(\dfrac{1}{\tanh{x}}-\dfrac{1}{\tan{x}}\right)-\dfrac{2}{3}\right)=？$$

My try: since $$\tanh{x}=\dfrac{e^{x}-e^{-x}}{e^x+e^{-x}}$$
then
$$\dfrac{1}{\tanh{x}}-\dfrac{1}{\tan{x}}=\dfrac{e^{x}-e^{-x}}{e^x+e^{-x}}-\dfrac{1}{\tan{x}}$$

then I can't,Thank you

$$
\coth x=\frac{1}{x}+\frac{x}{3}-\frac{x^3}{45}+\frac{2x^5}{945}-\frac{x^7}{4725}+... \\
\cot x =\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2x^5}{945}-\frac{x^7}{4725}-... \\[0.9cm]
\text{Given limit is}\\ \lim_{x \to 0} \frac{1}{x^4}\left(\frac{1}{x}\left(\frac{2x}{3}+\frac{4x^5}{945} + ...\right)-\frac{2}{3}\right) \\
= \lim_{x \to 0} \frac{1}{x^4}\left(\frac{4x^4}{945} + ...\right) \\
=\frac{4}{945}
$$

Don't worry if you can't remember these power series. You can derive them in a jiffy using Bernoulli numbers. But that's for another day.

Take the power series expansion of tanh to fifth order, invert it.  Then take power series expanion of tan to fifth order, invert it.  Take the difference.  The rest is clear sailing. Note that the expansions will only have 3 nonzero terms since the functions are odd.


Derivative of e^x with respect to y I recently came across a question that asked for the derivative of $e^x$ with respect to $y$. I answered $\frac{d}{dy}e^x$ but the answer was $e^x\frac{dx}{dy}$. How is that the answer? I am confused.

If $x$ is a function of $y$ then the given answer follows by the chain rule:

$$\frac {\text d}{\text dy} \left(e^x\right) = e^x \cdot \frac {\text d}{\text dy}(x)$$

I feel the question needs to be clearer. It needs to be mentioned what exactly is $y$. Like one of the comment says, $y$ may not be a dependent variable. In that case, $\frac{\partial{e^{x}}}{\partial{y}}$ equates to zero. However, in case $x$ is a function of $y$ then, $\frac{de^{x}}{dy}$ can be written as $$\frac{de^{x}}{dy}×\frac{dx}{dx}$$ $$=\frac{de^{x}}{dx}×\frac{dx}{dy}$$ This gives the answer as, $$e^{x}×\frac{dx}{dy}$$


How many distinct ways can the number be written as product of $3$ factors? How many distinct ways can the number $126$ be written as a product of $3$ positive integer  factors?
I found that the prime factors are $126=2\times3\times3\times7$. But how to get number of different ways out of this?

We're trying to count factorizations $126=abc$ with $a\le b\le c$.  Note that, except for $1\cdot1\cdot126$ and $3\cdot3\cdot14$, the rest have $a\lt b\lt c$, since $3$ is the only repeated prime factor of $126$.  Let $m$ denote the number with $a\lt b\lt c$, so the number we want to find is $m+2$.

If we remove the condition $a\le b\le c$, the number of factorizations $126=abc$ is $6m+3\cdot2=6m+6$.  But it's also $3\cdot6\cdot3=54$, since the $2$ and $7$ can be assigned to any of the three factors and the pair of $3$'s can be assigned in $6$ different ways.  Setting $6m+6=54$, we find $m=8$, so the answer we're after is $8+2=10$.

As you said, prime factors of 126 are 126=2*3*3*7

Then in order to make it as a product of 3 positive integer factors, you need to pick two out of four from these prime numbers and multiply them to get an integer, leaving the other two as prime. But since number 3 repeat twice, you need to divide by 2!

So there are $\frac{4C2}{2!} = 3$ ways


Proof that vector product satisfies right-hand rule Let $\mathbf{a}=\left(a_{1},a_{2},a_{3}\right)$ and $\mathbf{b}=\left(b_{1},b_{2},b_{3}\right)$ be vectors in $\mathbb{R}^3$. Then the only two distinct unit vectors that are perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ are those that point in the directions of:

$$\mathbf{u}=\left(\begin{array}{c}
a_{2}b_{3}-a_{3}b_{2}\\
a_{3}b_{1}-a_{1}b_{3}\\
a_{1}b_{2}-a_{2}b_{1}
\end{array}\right)$$

and

$$\mathbf{v}=\left(\begin{array}{c}
a_{3}b_{2}-a_{2}b_{3}\\
a_{1}b_{3}-a_{3}b_{1}\\
a_{2}b_{1}-a_{1}b_{2}
\end{array}\right)$$

where of course $\mathbf{u}=-\mathbf{v}$.

How do we prove that $\mathbf{u}$ satisfies the right-hand rule, while $\mathbf{v}$ satisfies the left-hand rule?

(A similar question has been asked several times before: e.g. 1, 2, 3, 4, 5. I have looked at all the answers and can't find a single one that I can understand and that I'm satisfied explains it. Hence, I'm posting this question again, worded slightly differently.

In particular, I've deliberately omitted any reference to the fact that by convention, we choose $\mathbf{u}$ and not $\mathbf{v}$ as the vector product $\mathbf{a}\times\mathbf{b}$. This fact seems to be the most popular "answer" but does not actually answer the above question.)

For a definite answer, you need to define what a "right hand" is, and that's difficult.

However, note that we have a choice between two distinct cross products for every choice of non-collinear input vectors. For the particular choice that $\mathbf a$ is the first standard base vector $\mathbf e_x$, and $\mathbf b$ is the second standard base vector $\mathbf e_y$,, it turns out that $\mathbf u=\mathbf e_z$ and $\mathbf v=-\mathbf e_z$, its negative. So if we agree that $\mathbf e_x$, $\mathbf e_y$, $\mathbf e_z$ in that order form a right hand configuration, we must choose $\mathbf u$ instead of $\mathbf v$ at least in that situation. For all other cases, one can show that it is possible to continuously transform any given non-collinear $\mathbf a$ and $\mathbf b$ to $\mathbf e_x$ and $\mathbf e_y$ (without them being collinear at any intermediate point of time). In such a transformation, $\mathbf u$ and $\mathbf v$ are also continuous functions of the time and never coincide. So it akes perfect sense to make the consistent convention of picking $\mathbf u$ as the vector product and call this "right hand".

I'm partial to a proof published in The College Mathematics Journal, Vol. 46, No. 3, May 2015, p. 215 - 217, perhaps because I wrote it, although I'm sure that thousands of others came up  with the same proof. You can read and download the complete paper for free at https://www.academia.edu/18549734/An_Even_Simpler_Proof_of_the_Right-Hand_Rule However, you will have to sign up on Academia.edu, so if you don't want to do that here's a proof of the main case:

Imagine you're way up on the positive z-axis, looking down on the x-y plane, and a and b both have initial points at the origin. Let $\textbf{a}_\textbf{p} = (a{_1}, a{_2}, 0 )$ and $\textbf{b}_\textbf{p} = (b{_1}, b{_2}, 0)$ be respectively the projections of a and b onto the x-y plane. Assume that $\textbf{a}_\textbf{p}$ and $\textbf{b}_\textbf{p}$ are not collinear. Look at the smaller of the two angles formed by $\textbf{a}_\textbf{p}$ and $\textbf{b}_\textbf{p}$.

The right-hand rule says that from this perspective, if a is to the right of b then the z-component of a x b will be positive, since your thumb will point up, and if a is to the left of b then the z-component of a x b will be negative, since your thumb will point down. Note that the left-right orientation of $\textbf{a}_\textbf{p}$ and $\textbf{b}_\textbf{p}$ is the same as the left-right orientation of a and b, i.e., if a is to the right of b then $\textbf{a}_\textbf{p}$ is to the right of $\textbf{b}_\textbf{p}$, and if a is to the left of b then $\textbf{a}_\textbf{p}$ is to the left of $\textbf{b}_\textbf{p}$.

Let $\alpha$ and $\beta$ be the measures of the angles which $\textbf{a}_\textbf{p}$ and $\textbf{b}_\textbf{p}$ respectively form with the positive x-axis, where $0 \leq \alpha, \beta < 2\pi$, and as usual, positive angle measurements correspond to counterclockwise rotations. Note that a x b and $\textbf{a}_\textbf{p}$ x $\textbf{b}_\textbf{p}$ have the same z-component,

$$a{_1}b{_2} - a{_2}b{_1} = (\left|\textbf{a}_\textbf{p}\right|\cos\alpha)(\left|\textbf{b}_\textbf{p}\right|\sin\beta) - (\left|\textbf{a}_\textbf{p}\right|\sin\alpha)(\left|\textbf{b}_\textbf{p}\right|\cos\beta )\\ = \left|\textbf{a}_\textbf{p}\right|\left|\textbf{b}_\textbf{p}\right|[(\sin\beta)(\cos\alpha) - (\sin\alpha)(\cos\beta)]\\ = \left |\textbf{a}_\textbf{p}\right|\left|\textbf{b}_\textbf{p}\right|\sin(\beta - \alpha)$$

Suppose a is to the right of b. Then $\textbf{a}_\textbf{p}$ is to the right of $\textbf{b}_\textbf{p}$, so $0 < \beta - \alpha < \pi$ (e.g., if $\textbf{a}_\textbf{p}$ is in Quadrant I and $\textbf{b}_\textbf{p}$ is in Quadrant II) or $-2\pi < \beta - \alpha < -\pi$ (e.g., if $\textbf{a}_\textbf{p}$ is in Quadrant IV and $\textbf{b}_\textbf{p}$ is in Quadrant I). In each of these cases, $\sin(\beta - \alpha) > 0$, so the z-component of a x b is positive, as stated by the right-hand rule. Similarly, if a is to the left of b then $\textbf{a}_\textbf{p}$ is to the left of $\textbf{b}_\textbf{p}$, so $-\pi < \beta - \alpha < 0$ or $\pi < \beta - \alpha < 2\pi$, so $\sin(\beta - \alpha) < 0$, thus the z-component is negative.


Probability of a run of 3 heads when I flip a coin $n$ times I'm wondering if there is a nice solution for this problem. As stated, I flip a coin $n$ times, and want the probability of a run of 3 (or more) heads appearing within it. For example, if I toss a coin 9 times, an example that would include a run of 3 heads is TTHHHTHTH, as is THTHHHHHT, but TTHHTHTTH is not.

Generalisations to runs of $m$ (or more) heads also welcome!

The time $T_{m+1}$ needed to get a run of $m+1$ heads is $$T_{m+1}=T_m+1$$ with probability $p_H=\frac12$ and $$T_{m+1}=T_m+1+T'_{m+1}$$ with probability $p_T=1-p_H=\frac12$, where $T'_{m+1}$ is distributed like $T_{m+1}$ and independent of $T_m$.

Thus, the generating functions $g_m(s)=E(s^{T_m})$ are such that

  $$g_{m+1}(s)=\frac12s(g_m(s)+g_m(s)g_{m+1}(s))\tag{$\ast$}$$

From there, the solution is standard. First, $(\ast)$ can be rewritten as $$g_{m+1}(s)=\frac{sg_m(s)}{2-sg_m(s)}$$ hence $$\frac1{g_{m+1}(s)}-\frac{s}{2-s}=\frac2s\left(\frac1{g_m(s)}-\frac{s}{2-s}\right)$$ that is, using $g_0(s)=1$,$$\frac1{g_m(s)}=\left(\frac2s\right)^m\left(1-\frac{s}{2-s}\right)+\frac{s}{2-s}$$ from which every $g_m(s)$ follows as

  $$g_m(s)=\frac{s^m(2-s)}{2^{m+1}(1-s)+s^{m+1}}$$

For example, for $m=1$, $$g_1(s)=\frac{s^2}{2-s}$$ that is, $T_1$ is a shifted geometric random variable. More generally, recall that each function $g_m$ fully encodes the distribution of $T_m$ since, by definition, $$g_m(s)=\sum_{k=0}^\infty P(T_m=k)s^k$$ thus, expanding $g_m(s)$ as a power series yields every $P(T_m=k)$. If one is interested to observe a run of length $m$ during the $k$ first draws, one would simply compute $$P(T_m\leqslant k)$$

Let $H_m$ be the event that the $m$th toss is heads. Then 3 heads is $$A_m := H_m \cap H_{m+1} \cap H_{m+2}$$

where $m = 1$ to $n-2$. Thus, by inclusion-exclusion, a probability of a run of 3 heads is

$$P(\bigcup_{k=1}^{n-2} \bigcap_{i=0}^2 H_{k+i}) = P(\bigcup_{k=1}^{n-2} A_k)$$

$$ = \sum_{i=1}^{n-2} P(A_i) - \sum_{1 \le i < j \le n-2} P(A_i \cap A_j) + \sum_{1 \le i < j < k \le n-2} P(A_i \cap A_j \cap A_k)$$

Now, $P(A_i \cap A_j \cap A_k) = 0$ unless $i+2=j+1=k$ in which case we have $P(A_i \cap A_j \cap A_k) = P(H_k)$

Similarly, $P(A_i \cap A_j) = 0$ unless $i+1=j$ or $i+2=j$ in which cases we respectively have $P(A_i \cap A_j) = P(H_j \cap H_{j+1})$ and $P(A_i \cap A_j) = P(H_j)$


Confusion between function and multivalued function. "What is a function?" can be  answered as "Single-valued relations are called functions". But how can "What are the multi-valued function?" be answered?
Will someone clarify my doubt why multi-valued functions are not violating the classical definition of function?
EDIT
This is what Wikipedia says on multivalued functions:

In mathematics, a multivalued function from a domain X to a codomain Y is a heterogeneous relation. However, in some contexts such as the complex plane (X = Y = ℂ), authors prefer to mimic function theory as they extend concepts of the ordinary (single-valued) functions.
In this context, an ordinary function is often called a single-valued function to avoid confusion.
The term multivalued function originated in complex analysis, from analytic continuation. It often occurs that one knows the value of a complex analytic function $f(z)$ in some neighbourhood of a point $ z=a$. This is the case for functions defined by the implicit function theorem or by a Taylor series around $ z=a$. In such a situation, one may extend the domain of the single-valued function $f(z)$ along curves in the complex plane starting at  $a$. In doing so, one finds that the value of the extended function at a point $z=b$ depends on the chosen curve from $ a$ to $ b$; since none of the new values is more natural than the others, all of them are incorporated into a multivalued function. For example, let $f(z)=\sqrt {z}$, be the usual square root function on positive real numbers. One may extend its domain to a neighbourhood of $z=1$ in the complex plane, and then further along curves starting at $z=1$, so that the values along a given curve vary continuously from $\sqrt {1}=1$. Extending to negative real numbers, one gets two opposite values of the square root such as $\sqrt {-1}=\pm i$, depending on whether the domain has been extended through the upper or the lower half of the complex plane. This phenomenon is very frequent, occurring for $n$th roots, logarithms and inverse trigonometric functions.
To define a single-valued function from a complex multivalued function, one may distinguish one of the multiple values as the principal value, producing a single-valued function on the whole plane which is discontinuous along certain boundary curves. Alternatively, dealing with the multivalued function allows having something that is everywhere continuous, at the cost of possible value changes when one follows a closed path (monodromy). These problems are resolved in the theory of Riemann surfaces: to consider a multivalued function ${\displaystyle f(z)}$ as an ordinary function without discarding any values, one multiplies the domain into a many-layered covering space, a manifold which is the Riemann surface associated to  $f(z)$.

Multi-valued functions are violating the (classical) definition of functions.

efficiently you can say "relation between variable (or 's') are called functions"


What is function $\sin$ without $x$? I'm solving a problem set in ODE:

This is the first time in my life that I see such functions $\sin,\cos,\tan$ without argument $x$.

I would like to ask if $\sin,\cos,\tan$ mean $\sin x,\cos x,\tan x$ or they mean something else.

Thank you so much!

Usually the convention is that if $f:X\to Y$ is a function, then $f(x)\in Y$ is the value at the point $x\in Y$. As you can do arithmetic with function values, you can do arithmetic with functions, applying the operation pointwise. Thus $h=f+g$ is $h(x)=f(x)+g(x)$ for all $x$.

This gives a little complication if you want to write equations like 1) without too much decorations like writing $\forall x\in X$. One possible interpretation would then be that $x$ has a double meaning, it is the variable name and, where appropriate, a function itself, the identity function. So that 1) would be in this convention $xy'+y=2x$, avoiding the function evaluation notation.

In 3-5, the pendulum is moving to the extreme other side regarding the function evaluation notation in that $\sin$ etc. is used as function name in the same way as $f$ above. This is a very unusual notation, but not formally wrong. But it will look very strange in larger terms, $\sin+\cos$ is tolerable, but how to interpret $\sin\cos$?

$\sin$ is an abbreviation for $\sin(x)$, in the same manner that $y'$ is an abbreviation for $y'(x)$ and $y$ for $y(x)$.


How prove exist continous g(x) such f(g(x))=x for $f(x)=\frac{2x^3-3}{3(x^2-1)}$? Let $f(x)=\frac{2x^3-3}{3(x^2-1)}$ for $x\neq {1,-1}$. Prove exist continous g(x) satisfy f(g(x))=x and g(x)>x with every x is belong to R.

$\frac{2g(x)^3-3}{3(g(x)^2-1)}=x$ but how solve? Maybe another way?

Plotting $f$ one sees that its graph consists of three pieces. In particular $f$ is monotonically increasing on the interval $J:=\ ]1,\infty[\ $, and one has
$$\lim_{x\to1+}f(x)=-\infty,\qquad\lim_{x\to\infty} f(x)=\infty\ .$$

Therefore the restriction $f\restriction J$ maps $J$ homeomorphically onto ${\mathbb R}$, and there is a continuous inverse $g:\ {\mathbb R}\to J$. In particular $(f\restriction J)\circ g={\rm id}_{\mathbb R}$, or
$$f\bigl(g(x)\bigr)=x\qquad\forall x\in{\mathbb R}\ .$$
The graph of $f\restriction J$ lies completely to the right of the line $y=x$, which implies $g(y)>y$ for all $y\in{\mathbb R}$.

$f(g(x))=x$ so $g(x)$ is inverse of $f(x)$ so nothing but reflection of f(x) on y=x. proceed , you might have to check at $x=1$ though


Understanding the idea of the proof of "For two ideals $I$ and $J$ prove that prove that $I(R/J)=(I+J)/J$" I was reading this question, but I could not understand the proof of why for two ideals $I$ and $J$ we have $I(R/J)=(I+J)/J$.
Here is the proof given in this link:

Well, if you use the definition, it's obvious: an element of $I\cdot R/J$ is  the ideal of $R/J$ generated by $I$, so it is the set of finite sums of the form:
$$\Bigl\{\sum_k i_kr_k+J\bigm\vert i_k\in I, r_k\in R\Bigr\}.$$
Now, as $I$ is an ideal in $R$, the set of these finite sums u=is nothing but $I$ itself, whence the formula
$$I\cdot R/J=(I+J)/J.$$
Note that, by the second isomorphism theorem,  the latter quotient is isomorphic to $I/I\cap J$.

Here is what I know:
I know how to show that $IM$ is a submodule of $M$ and hence it is an ideal of $M.$
Here is what I do not understand in the solution above:
1- Why $I \cdot R/J$ must be generated by $I$?
2- Why "as $I$ is an ideal in $R$, the set of these finite sums u=is nothing but $I$ itself" and why that leads to that $I(R/J)=(I+J)/J$ are not we moding out by $J$ in the RHS of this formula?
Could anyone explain these points to me please?

Given commutative unital rings $R$ and $S$ such that there exists a ring homomorphism $\varphi : R \to S,$ we define the extension of an ideal $I$ of $R$ to be the ideal $$I^e = \varphi(I) S = {\left \{\sum_{k = 1}^n s_k \varphi(i_k) \,|\, n \geq 0 \text{ is an integer, } i_k \in I, \text{ and } s_k \in S \text{ for all } k \right \}}$$ of $S.$ For an ideal $J$ of $R,$ there is a ring homomorphism $\pi : R \to R/J$ defined by $\pi(r) = r + J.$ Putting this together with the above shows that $I^e = \pi(I)(R/J)$ consists of the elements $$\sum_{k = 1}^n (r_k + J)(i_k + J) = \sum_{k = 1}^n (r_k i_k + J) = {\left(\sum_{k = 1}^n r_k i_k \right)} + J.$$ Often, the map $\pi$ is omitted, and this ideal is simply written $I(R/J).$ From here, it is straightforward to verify that $I(R/J) = (I + J)/J$ by showing the containments $\subseteq$ and $\supseteq.$


Prove that $x^4-x-1$ is irreducible over $\mathbb{Q}$ Prove that $f(x)=x^4-x-1$ is irreducible in $\mathbb{Q}[x]$.

All methods I know failed. I can only exclude that $f$ admits a factorization with a factor of degree 3, because in this case $f$ would have a root in $\mathbb{Q}$, and I can prove that this is not the case. But I can't exclude $f=gh$ with $g,h$ both of degree $2$. I also know that $f$ has two real roots and a pair of conjugate complex roots, but don't know how to use this. I know that if $f$ were reducible over $\mathbb{Q}$ then it would be reducible over $\mathbb{Z}$, but again I don't know how to deduce irreducibility. What can be done in this case to prove that $f$ is irreducible?

In this case  you can just look at $f$ in $\mathbb F_2[x]$. The only irreducible quadratic polynomial is $x^2+x+1$ and it doesn't divide $x^4+x+1=x^2(x^2+1)+(x^2+x+1)$.

You can use the Rational Root Theorem to solve this problem. The possible roots under consideration are $\pm 1$ (because the roots under consideration are $\frac{p}{q}$, (where $p$ is the coefficient of the last term and $q$ is the coefficient of $x^4$ in $f(x)$), and neither of them are roots of $f(x)$.


How do I sum 1/6( e^t + e^2t + .... + e^6t)? I have a question about a sum when calculating moment generating function.
The question is : "Find the moment generating function for each of these two random variables. (i) $X$ = outcome a die toss, $p(x) = P[X= x] = \frac{1}{6}$ for $x = 1,2,3,4,5,6$. I don't understand how the sum in the answer $e^t \cdot \frac{1}{6} + e^{2t} \cdot \frac{1}{6} +...+ e^{6t}\cdot \frac{1}{6} = \frac{1}{6}e^t\frac{e^{6t}-1}{e^t-1}$.
Why doesn't it work to have this as the sum of a finite geometric series with the common ratio being $e^t$?

I think it doesn't work with finite geometric series because $e$ is not rational number, but I am not sure


How many of each ticket were sold in one day? Child tickets - $\$7$

Adult Tickets - $\$10$

Senior Tickets - $\$5$

Day one sold $678$ tickets for $\$5,812$

Day two sold $535$ tickets for $\$4,541$

How many of each ticket were sold on day one and two?

If someone could even just explain how to set it up or what method to use I would appreciate it a lot. Thank you!

Let $x$ be the number of child tickets sold, $y$ adults, and $z$ seniors. Then we know for day one
$$
\$x + \$y + \$z = \$ 5812
$$
and
$$
x + y + z = 678.
$$
What are the dollar amounts associated with $x,y$, and $z$? What does day two look like?

Child tickets - $7

Adult Tickets - $10

Senior Tickets - $5

A is the number of Adult tickets, C is the number of Child tickets, and S is the number of Senior tickets in what follows below.

A1 + C1 + S1 = 678 (Day 1)

A2 + C2 + S2 = 535 (Day 2)

Solution:

Day one: Sold tickets 678 for $5,812

A1 = 420, C1 = 161, S1 = 97 (Day 1)

Day two: Sold tickets 535 for $4,541

A2 = 215, C2 = 38, S2 = 425 (Day 2)

It must be remembered that other answers are also possible.


Why is $\int \frac{f'(x)}{f(x)}=\ln |f(x)|$ ignored in differential equations? I've noticed that whenever $\int \frac{f'(x)}{f(x)}$ comes up in a differential equation the answer is always given as $\ln f(x)$ rather than $\ln |f(x)|$ as I was taught it should be. Is it because of the arbitrary constant? In other words, since $$\int \frac{f'(x)}{f(x)}=\ln |f(x)|+\ln A$$ for some constant $A$, then the answer is $\ln A|f(x)|$ and because $A$ can be positive or negative it follows that there is no point including the absolute signs? Hence the answer is given as $\ln f(x)+C$ for some constant $C$ rather than $\ln |f(x)|+C$. Is this why?

Review of Integrating Factors

Consider integrating factors in the context of linear equations, say $\dfrac{\mathrm{d}y}{\mathrm{d}t}+p\left(t\right)y=g\left(t\right)$. (I'll often drop the input $t$s from now on.) Our hope is to multiply the equation by some function $\mu$ to turn the left side into the derivative of a product, say $\left(\mu y\right)'$, so that we can integrate both sides and divide by $\mu$ to solve for $y$. Note that by the product rule, we have $\left(\mu y\right)'=\mu'y+\mu y'$. To get $\left(\mu y\right)'=\mu*\left(y'+py\right)$, we should look for $\mu$ that satisfy $\mu'=\mu p$. (That's a separable equation.)

Absolute values in integrating factors mostly don't matter

Note that if $\mu\left(t\right)$ is a valid integrating factor (in the sense that $\mu'\left(t\right)=\mu\left(t\right)p\left(t\right))$, then define $\nu(t)=-\mu(t)$ and note that $\nu'\left(t\right)=-\mu'\left(t\right)=-\mu\left(t\right)p\left(t\right)=\nu\left(t\right)p\left(t\right)$. But $\nu'=\nu p$ is exactly the equation $\nu$ should satisfy to be an integrating factor! So the negative of an integrating factor works just as well.

Now suppose you solve $\mu'=\mu p$ and get a solution like $\mu\left(t\right)=\left|\omega\left(t\right)\right|$, thanks to an integral of $\dfrac{\omega'\left(t\right)}{\omega\left(t\right)}$. On an interval where $\omega\left(t\right)$ is always nonnegative, $\mu\left(t\right)=\omega\left(t\right)$, and you don't need the absolute value. On an interval where $\omega\left(t\right)$ is always nonpositive, $\mu\left(t\right)=-\omega\left(t\right)$, so $\nu\left(t\right)=-\mu\left(t\right)=-\left(-\omega\left(t\right)\right)=\omega\left(t\right)$, and you still don't need to worry about the absolute value, since $\nu$ works fine, too.

Therefore, if you ignore the absolute value, you get valid solutions on these intervals where $\omega$ doesn't change sign.

In practice, absolute values in integrating factors don't matter at all

A limitation of the above discussion is that there's no guarantee the solutions on separate intervals will be able to connect up when $\omega\left(t\right)=0$. However, that's okay, because in practice an issue like that would usually arise because the original differential equation has a problem at that point.

For example, with something like $ty'+3y=te^{t}$, you would rewrite as $y'+\dfrac{3}{t}y=e^{t}$ and initially get an integrating factor of $\left|t\right|^{3}$. By the above argument, for positive $t$ or negative $t$, you can get away with $t^{3}$ and learn that $t^{3}y=\int_{0}^{t}e^{x}\,\mathrm{dx}+C$. But what about at $t=0$? Well, that would make $t^3y$ not tell you about $y$, it would make $\dfrac{3}{t}$ undefined, and would make $ty'+3y=te^{t}$ not tell you anything about $y'$. In fact, the solutions for positive or negative $t$ diverge as $t$ approaches $0$, so you had no hope of finding a solution on an interval like $\left[-1,1\right]$ anyway.

Mathematicians doing ODE usually assume complex number calculations.  In that setting, the assertion
$$
\int\frac{dx}{x} = \log |x| + C
\tag{1}$$
is simply wrong.  $\log |x|$ is not complex differentiable at any point.  In particular, it does not have derivative $1/x$.

But, as you note, you can still write
$$
\int \frac{dx}{x} = \log x + C
\tag{2}$$
even when $x<0$.  The $\log$ and the $C$ may both be non-real complex numbers.  But only beginners would worry about that.

I think formulas of the type (1) were invented by textbook writers within the last 50 years or so.


How would I find this eigenvalue? I'm told to let $A$ be the matrix of the linear transformation $T$ and without writing $A$, find an eigenvalue of $A$ and describe the eigenspace. The first is to let $T$ be the transformation on $\mathbb R^2$ that reflects points across some line through the origin. Linear transformations always confuse me so I'm not really sure how to go about doing this.

I thought that possibly the standard matrix of a reflection across the origin would be our $A$ and then we find the eigenvalue from that. Is that somewhere in the right direction?

Since you are asked to work in $\mathbb{R}^{2}$ so geometric thinking will be of immense help. Note that you are asked to find the eigen-value of the matrix of reflection across some line "passing" through the origin not "about" the origin. You can easily see the difference by drawing some points in $xy$ plane and the reflecting them about the origin and then across some line passing through the origin.
\Now recall that a linear transformation from some vector space to another is completely determined by its action on the basis of the domain vector space. So, in particular you can see the reflections of the standard basis of $\mathbb{R}^{2}$ across the given line and then you will be able to determine the transformation. Write the matrix of that transformation and then eigenvalue(s) can be found easily. For understanding point of view you can start with the $x$ or $y$ axis as the line through origin and try to draw pictures and take the advantage of working in $\mathbb{R}^{2}$.


Doubt about an operation involving vectors Given $\vec{u}=(1;2;-1)$ and $\vec{a}=(1;-3;4)$, calculate $\vec{u}-\frac{3}{\vec{u}} . \vec{a}$.

Well, I think that the operation $\frac{3}{\vec{u}}$ is not defined, is it? In case it's possible, what should I do?

You are correct; there's no concept of dividing by a vector.

It may be a typo in the problem you're solving -- perhaps it should be $ \vec u - \frac3{\vec u\cdot \vec a}\cdot \vec a$ or $\vec u - \frac{3}{|\vec u|}\cdot\vec a $?

Or perhaps it is some idiosyncratic notation that you'll have to ask the person who gave you the problem about.

Perhaps
$\frac1{\vec{u}}
$
is a vector such that
$\frac1{\vec{u}}\times\vec{u}
=I
$.


Slowing down divergence 2 Let $f(x)$ and $g(x)$ be positive nondecreasing functions such that
$
\sum_{n>1} \frac1{f(n)} \text{ and } \sum_{n>1} \frac1{g(n)}
$

diverges.

(Why) must the series $$\sum_{n>1} \frac1{g(n)+f(n)}$$ diverge?

No, $\sum\frac{1}{f(n)+g(n)}$ need not diverge.  We define such a pair $f,g$ inductively.  Let $f(1) = g(1) = 2$.

Suppose $f(n)$ and $g(n)$ have been defined for $1\leq n\leq N$.  Let $L$ be the largest value taken by $f(n)$ or $g(n)$ for $1\leq n\leq N$.  Define $f(N+1) = f(N+2) = \cdots = f(N + L) = L$ and $g(N+i) = 2^iL$ for $1\leq i\leq L$.  Define $g(N+L+1) = \cdots = g(N+L+2^LL) = 2^LL$ and $f(N+L+i) = 2^{L+i}L$ for $i\leq 1\leq 2^LL$.  The extends the domain of definition of $f(n)$ and $g(n)$ to $1\leq n\leq N+L+2^LL$.  Repeat to define $f$ and $g$ on all of $\mathbb{N}$.

Note that $f$ and $g$ are positive and nondecreasing.  In each stage of this process there are $L$ times when $f$ takes the value $L$ and $2^LL$ times when $g$ takes the value $2^LL$.  Therefore $\sum_{n=N+1}^{N+L+2^LL} \frac{1}{f(n)} > 1$ and $\sum_{n=N+1}^{N+L+2^LL} \frac{1}{g(n)} > 1$: at least $1$ is added to the sum of the reciprocals of each series at each step.  Since we repeat this process ad infinitum, adding at least $1$ to each series at each step,  $\sum_{n=1}^\infty \frac{1}{f(n)}$ and $\sum_{n=1}^\infty \frac{1}{g(n)}$ both diverge.

Note that by construction, $\max(f(n+1),g(n+1)) = 2\max(f(n),g(n))$ for all $n$.  Since $f(1)=g(1)=2$, we have $\max(f(n),g(n))=2^n$ for all $n$.   Therefore $\frac{1}{f(n)+g(n)}\leq \frac{1}{\max(f(n),g(n))} = 2^{-n}$ for all $n$, so $\sum_{n=1}^\infty\frac{1}{f(n)+g(n)}$ converges by comparison with the geometric series $2^{-n}$.

This is a follow-up to Gortaur's "answer" (and isn't a real answer either).

If $a_n$ and $c_n$ are as above, then to show that $\sum_nc_n=\infty$, I believe that it is sufficient to show that
if $\;1=i_0< i_1<i_2<\dots\;$ is any infinite sequence of indexes, then $\sum_jd_j=\infty$ where
$$
d_j\;=\;a_{i_k} \;\;\textrm{where}\;\; i_{k-1}\leqslant j<i_k.
$$
This reduces the question to one about a single non-increasing positive sequence. I haven't been able to find a counter-example to this.


Special sums of numbers The natural numbers from 1 to 100 are written next to each other in any order. Now all sums of three adjacent numbers are calculated. How many of these sums can be odd at most?
I have 94 out. I did it by trial and error. Did you get something else out of it or did I do it right?

If you do odd-even-even $25$ times, followed by $25$ odds in a row, you get $97$ odd sums of three consecutive numbers. The only even sum comes from an even-odd-odd in positions $75, 76, 77$.

I started by looking at just the first 10 numbers.
$$
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
$$
When adding numbers, if both of them are either even or odd, the answer will always be even. If one is even and one is odd, the answer will be odd. So, in these first 10, we see an alternating sequence when adding 3 at a time.
$$
(1 + 2 + 3)\ \%\ 2 = 0\\
(2 + 3 + 4)\ \%\ 2 = 1\\
(3 + 4 + 5)\ \%\ 2 = 0\\
(4 + 5 + 6)\ \%\ 2 = 1\\
$$
So now we can see this alternating sequence of even and odd answers. We also know there are 98 sums (since each one starts from the next natural number and we don't count 99 or 100). So just taking half of 98, 49 should be it.


Why is this 5 negative? I'm struggling with the difference between a negative number and a positive number that is being subtracted.

For example, when working on this inequality:
$$
2 - 5x \leq 7
$$

After subtracting 2 from each side, it is:
$$
-5x \leq 5
$$

Why is the 5 negative? Wasn't it a positive 5 in the original problem?

If it was negative, shouldn't the original inequality read:
$$
2 - (-5)x \leq 7
$$

Can someone explain this?

$$2-5\times x\leq7,$$
$$\Leftrightarrow$$
$$2+(-1)\times5\times x\leq7,$$
$$\Leftrightarrow$$
$$2+(-1)\times5\times x-2\leq7-2,$$
$$\Leftrightarrow$$
$$(-1)\times5\times x\leq5,$$
$$\Leftrightarrow$$
$$(-1)\times x\leq1,$$
$$\Leftrightarrow$$
$$(-1)\times (-1) \times x\geq(-1)\times 1,$$
$$\Leftrightarrow$$
$$x\geq-1,$$

So the $5$ is not negative. It is just multiplied by $(-1)$.

$2 + (-5x) \leq 7$

Take two from both sides

$(-5x) \leq 5$

You are adding a negative number, which is the same as taking a positive number. This can be thought of as the operator being held by the number on the right. Just as the two is positive, the 5 is negative.

eg:

$0 + (+2) + (-5x)$


Question on continuity in Topology $X=[0,2\pi)$ with the relative topology determined by the usual topology on $\mathbb R$, and the unit circle $S=\{(x_1,x_2)\mid x_1^2+x_2^2=1\}$, with the relative topology determined by the usual topology on $\mathbb R^2$. Define $f:X\to S$ by $f(t)=(\cos(t),\sin(t))$. Prove $f$ is continuous and prove $f$ inverse is not continuous.

I keep looking at this and for some reason solve it. Please help me

HINT: First show that the inverse image of an open interval in $S$ is either an open interval in $X$ or a set of the form $[0,x)\cup(y,2\pi)$, which is also open in $X$; from this you can conclude that $f$ is continuous. Now let $g=f^{-1}$. $[0,1)$ is an open set in $X$; is $g^{-1}\big[[0,1)\big]$ open in $S$?

f^(-1) is not even a function as df is not 1-1.


Negative modulo: why is $-3$ mod $12$ equal to $9$? I know that in general, we can represent a number $x$ as follows:

$x = qn + r$

where $r$ is the remainder, $n$ is the divisor, and $q$ is the quotient.

But suppose we try to calculate $-3 \div12$. An answer elsewhere on Math StackExchange suggests that the answer is:

$-3 = -1 \times12+9$

In other words, that $-3 \mod 12 = 9$.

But can't we represent this as:

$-3 = -2 \times 12 + 21$

And get a different answer?

It's true that $-3\equiv 21 \bmod 12$. And also that $-3\equiv 117 \bmod 12$, and and of course $-3\equiv 9 \bmod 12.$ All of these numbers are in the same congruence or residue class. Typically the smallest non-negative value is used to represent the class.

I think you’re overcomplicating it. It simply boils down to this:

$$b\equiv c\pmod m \iff \frac{b-c}{m}\in\Bbb Z$$

where $\Bbb Z$ contains negatives as well as positives.

Since $$\frac{-3-9}{12}=-1$$ we say $-3\equiv 9\pmod {12}$ or alternatively $-3\bmod 12 =9$.


Confusion related to palindrome language I have this confusion related to palindrome language. I have found it being mentioned if a language L is equal to its reverse, it is a palindrome language.

But lets say

L = {aaab, baaa}

Then
$L^R= \{baaa,aaab\}$

Definitely,

L = $L^R$ in this case

But the language isn't palindrome isn't it. Any insights?

"Palindrome language" is not a technical term that has a generally accepted meaning. Just from the words one could imagine several different meanings for it:

The language consisting of all palindromes over some given alphabet.
A language whose members are always palindromes (but not necessarily all the palindromes).
A language that is closed under the operation of reversing a word (that's the one you have got there).

Neither of theses definitions are, however, one that an unprepared reader can be expected to understand. But fear not! That is what definitions are for.

One can make "palindrome language" into a technical term for the duration of a book, an article, an exercise, simply by setting forth a definition that declares what it will mean. The definition can be one of the three suggested above, or something fourth and different ... so long as it defines something that you need to have a name for for your purposes.

So, if you're reading an exercise that defines a palindrome language to be an $L$ such that $L=L^R$ (which seems to be the most relevant use of the term in the first page of Google hits), then that's what "palindrome language" means for the purpose of doing that exercise. It's not as if any of the possible meanings is something that crops often enough to warrant attaching a name to it permanently.

its a word that is pronounced the same when you say it backwards. EXAMPLE:
 mom, say it backwards, it still says,mom. did, say it backwards, its still did.


If $f(x)=f(y)$ implies that $g(x)=g(y)$ is $g$ a function of $f$? Let $f: {\mathbb R}^n\to {\mathbb R}^p$ and $g: {\mathbb R}^n \to {\mathbb R}^q$ be two non-constant functions such that: $f(x)=f(y)$ implies that $g(x)=g(y)$, can we say that $g$ a function of $f$?

This is: $g(x)= h(f(x))$ for some function $h$.

Define
$$h(x) = \left\{
\begin{array}{lc}
g(z) & \mbox{ if there exists some } z \mbox{ such that } f(z)=x \\
\mbox{ anything} & \mbox{otherwise}
\end{array}
\right.$$

Show that the given condition implies that $h$ is well defined, that is it does not depend on the choice of $z$.

Then, by definition
$$h(f(x))=g(x) \forall x \in \mathbb R^d$$

I'm assuming that why you mean by "$g$ is a function of $f$" is "there exists a function $\mathfrak{F}$ such that $g=\mathfrak{F}(f)$".
$$\bigl(\forall x,y:f(x)=f(y)\implies g(x)=g(y)\bigr)\implies\exists \mathfrak{F}:g=\mathfrak{F}(f)$$

Consider $f(x)=x$. Any function $g$ having the same domain as $f$ will satisfy the first half of the above assertion. No matter what metafunction $\mathfrak{F}$ we examine, there will be functions $g$ satisfying the first half but not the second half of our assertion, so the assertion must be false.

So the answer to your question is no. Except possibly for some trivial and pathological universes of values for $x$ and $y$.


Multivariable Calculus, Help with pre-Bachelos Homework I need help with a couple of problems for homework, btw I need to finish this in less than 12 hours, I'll appreciate your help, and srry if something is wrong or is confusing I'm not native speaker.

The problems are the following:

Let $F:\ U= \mathbb{R}^3$\ {axis Z} $\subset \mathbb{R}^3$ $\rightarrow$ $\mathbb{R}^3$ defined as:

$F(x,y,z) = (\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0)$

Determine if $F$ it a solenoid field

2.Porve Green's Theorem using Stokes's Theorem

Plus: How do I calculate the mass of thin layers wich form correspond to a certain surface $\Sigma$ and a certain density function $\rho$

Let $F(x,y,z)=F(x,y)=f(x,y)\mathbf{i}+g(x,y)\mathbf{j}+0\mathbf{k}$

A solenoid field means the field is "circular", specifically its divergence is $0$

(1) Show that

$$\nabla \cdot F=0$$

Green's theorem states that given a vector field $F \in\mathbb{R}^2$

$$\oint_CF\cdot dr=\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA$$

Stokes' theorem states that given a vector field $G\in\mathbb{R}^3$

$$\oint_CG\cdot dr=\iint_S\left(\nabla \times G\right)\cdot dS=\iint_D\left(\left(\nabla \times G\right)\cdot n\right) dA$$

(2) Show that

$$\nabla \times F = \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\mathbf{k}$$

And reason why, in this case,

$$n=\mathbf{k}$$

(Plus)

Given a density function, $\rho(x,y,z)$, the mass of a thin surface is given by a surface integral

$$m=\iint_S\rho(x,y,z)dS=\iint_D\rho(x,y,z)|n|dA$$

Where the surface has center of mass coordinates $(\bar{x},\bar{y},\bar{z})$

$$\bar{x}=\frac{1}{m}\iint_Sx\rho(x,y,z)dS$$

$$\bar{y}=\frac{1}{m}\iint_Sy\rho(x,y,z)dS$$

$$\bar{z}=\frac{1}{m}\iint_Sz\rho(x,y,z)dS$$

Notes: $n$ is the normal vector, $|n|$ is the normal vector's magnitude, and $S=\Sigma$

Good luck on the exam! :)

edit:

$F$ is a $3$-dim vector field but behaves like a field in $\mathbb{R}^2$ since its $\mathbf{k}$ component is $0$ which is why we can consider the $2$-dim vector field case of Green's theorem.

To show it, take the curl of $F$ to begin Stokes' theorem

$$\nabla \times F=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f(x,y) & g(x,y) & 0\end{vmatrix}=\mathbf{i}\left(\frac{\partial}{\partial y}0-\frac{\partial g}{\partial z}\right)-\mathbf{j}\left(\frac{\partial}{\partial x}0-\frac{\partial f}{\partial z}\right)+\mathbf{k}\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)$$

$$\oint_CF\cdot dr=\iint_S\nabla\times F\cdot dS=\iint_D\left(\nabla\times F\cdot n\right)dA$$

$\displaystyle\frac{\partial g}{\partial z}=\frac{\partial f}{\partial z}=0$ since both are independent of $z$ and the partial of $0$ is $0$

$$\iint_D\left(\nabla\times F\cdot n\right)dA=\iint_D\left<0,0,\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\right>\cdot\left<a,b,c\right>dA=c\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA$$

But $n$ is a unit normal vector so if $C\in\mathbb{R}^2$ then $c = 1$ and we get Green's theorem

$$\oint_CF\cdot dr=\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA$$

I hope that answered your question

In fact, we know the answer regardless of what $C$ is

$$\frac{\partial g}{\partial x}=\frac{(x^2+y^2)-2x^2}{\left(x^2+y^2\right)^2}=\frac{y^2-x^2}{\left(y^2+x^2\right)^2}$$

$$\frac{\partial f}{\partial y}=-\frac{(x^2+y^2)-2y^2}{\left(x^2+y^2\right)^2}=\frac{y^2-x^2}{\left(y^2+x^2\right)^2}$$

$$\frac{\partial g}{\partial x}=\frac{\partial f}{\partial y}$$

Since its z-component is $0$ and $f$ and $g$ are independent of $z$, it's unnecessary to check the remaining equations.

$\therefore F$ is a conservative field.

$$c\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA=c\iint_D(0)dA=0$$

$$\oint_C F\cdot dr=\oint_C \nabla f\cdot dr=f\left(\gamma(b)\right)-f\left(\gamma(a)\right)=0$$

Since $C$ is closed so $\gamma(b)=\gamma(a)$ for any closed path $C\in\mathbb{R}^3$


Suppose $X_1,X_2$ form a closed cover of $X$ and the $U \cap X_i$ are open. Is $U$ open? Equivalently, is $X$ the pushout of the $X_i$ over their intersection?

If $X$ has a finite closed cover of subspaces $X_1$ and $X_2$ then $F \subset X$ is closed iff $F \cap X_1$ is closed in $X_1$ and $F \cap X_2$ is closed in $X_2$. One direction uses the definition of the subspace topology (left to right) and right to left uses 2 facts: firstly, a subset $A$ that is closed (in the subspace topology) in $C$ and $C$ is closed in $X$, then $A$ is closed in $X$, and secondly that a union of 2 closed subsets of $X$ is again closed in $X$.  We apply these facts to $F \cap X_1$ and $F \cap X_2$, whose union is again $F$ as the $X_i$ form a cover of $X$.

Now, we also have, dually, that $O$ is open in $X$ iff $O \cap X_1$ is open in $X_1$ and $O \cap X_2$ is open in $X_2$. This holds as

$O$ open in $X$

iff $X\setminus O$ closed in $X$

iff $(X \setminus O) \cap X_1$ closed in $X_1$ and $(X \setminus O) \cap X_2$ closed in $X_2$

iff $X_1 \setminus ( (X \setminus O) \cap X_1 ) = O \cap X_1$ open in $X_1$ and $X_2 \setminus (( X \setminus O) \cap X_2) = O \cap X_2$ open in $X_2$.


Last step of the proof of the Chinese remainder theorem. For the Chinese Remainder Theorem for rings we have:

$$
A/(I\cap J) \cong A/I\times A/J
$$

So far I have proven that there is a ring homomorphism from $\phi :A \rightarrow A/I\times
A/J $
and the $ker(\phi)=I\cap J$.
Additionally, since I and J are comaximal, $I + J = A$, implies  $I\cap J=IJ $

My question is, how does this last fact prove isomorphism in the last step of the proof?

$I + J = 1\,\Rightarrow\, i + j = 1\,$ for some $\,i\in I,\, j\in J\,$ therefore $\,i\equiv 0\pmod{I},\ i\equiv 1\pmod{J}\,$ so $\,\phi(i) = (0,1).\,$ Similarly $\,\phi(j) = (1,0).\,$ Thus $\phi\,$ is onto: $\,{\rm Im}\,\phi = A/I\times A/J,\,$ hence applying the First Isomorphism Theorem yields the result.

Remark $\ $ This  generalizes to any number of  pair-comaximal ideals $\, I + J_k = 1,\,$ since then there are elements $\ i_k+j_k = 1,\ \ i_k\in I,\, j_k\in J_k,\,$ therefore

$\qquad x := j_1\cdots j_n = (1-i_1)\cdot (1-i_n)\equiv 1^n\equiv 1\pmod{I},\,$ and $\ x\equiv 0\pmod {J_k}$

Just apply the first isomorphism theorem to complete the proof.


How to translate set propositions involving power sets and cartesian products, into first-order logic statements? As seen from an earlier question of mine one can translate between set algebra and logic, as long as they speak about elements (a named set A is the same as {x ∣ x ∈ A}).

However I've stumbled upon propositions that involve cartesian products and power sets and I'm not sure how to translate those into logic statements. For instance:

(A  × B)    =   (B  × A)    if and only if  A   =   B

or

if   A   =   ℘(A)    then    ℘(A)    =   ∅
if   A   ⊆ ℘(A)  then    ℘(A)    =   ∅
if A ∈ ℘(A)  then    ℘(A)    =   ∅

and even a combination of the two:

℘(A ×   B)  ⊆       ℘(A)    ×   ℘(B)

Note that "×" is the cartesian product symbol, and ℘ the power-set.

Can someone provide any insight on this?

Most introductory books on axiomatic set theory directly define set theoretical concepts in the language of first order logic. Once you know these definitions, all you need to do is translate, translate and translate until you are satisfied. One such book that I use can be found here: http://www.amazon.com/Theory-Continuum-Problem-Dover-Mathematics/dp/0486474844. If you buy this book, I warn you that it has errors. Fortunately, you can find a lot of the corrections here: http://comet.lehman.cuny.edu/fitting/errata/book_errors/RevisedSetBookErrors/split.html.

I also noticed that in the answer to your earlier question, set builder notation $\{x:\Phi(x)\}$ was used a lot. While this is not necessarily wrong, it can lead you dangerously close to Bertrand Russell's paradox. Briefly, just let $\Phi(x)$ be $x \not\in x$. Then, the entity $X = \{x:x \not\in x\}$ cannot itself be a set in the sense that it can never sit on the left hand side of the symbol $\in$ (i.e. $X \in \mathcal{S}$ is invalid syntax for any symbol $\mathcal{S}$). If it could, then either $X \in X$ or $X \not\in X$. Either case leads to a contradiction.

Thus, I will show you an alternate way of encoding set theory in first order logic. To start with, the book I referred to above studies a popular axiomatization of set theory: Von Neumann–Bernays–Gödel (NBG) set theory. This theory talks not only about sets but also about classes. The intuition behind a class is that it is an arbitrary collection of things. A set is a class that is contained in some other class. Thus, sets are just "smaller" collections of things. Now, to show you how NBG encodes set theory in first order logic, we need to be a bit more precise with our language.

I reserve the CAPITAL letters $A, B, C, ..., X, Y, Z$ with or without sub(super)scripts to denote classes. I reserve the small letters $a, b, c, ..., x, y, z$ with or without sub(super)scripts to denote sets. Appearance aside, the capital and small letters also have different usage. How so? Well, we first introduce a binary symbol $\in$. Then we impose the following condition on it: only small letters can sit on the left of this symbol while both small and capital letters can sit on its right. Hence $x \in A$ is valid while $B \in A$ is not. Thus, we have implicitly said that only sets (small letters) can be elements.

With these preliminaries, we can start defining things. I skip many technicalities to keep the exposition brief. For instance, I do not mention any axioms of set theory itself. I can come back to them later on if you want me to. For now:

We introduce a binary symbol $\subseteq$ which is used in the following way: $A \subseteq B$ is short for $\forall x(x \in A \to x \in B)$ or if you prefer to be more formal: $A \subseteq B \leftrightarrow \forall x(x \in A \to x\in B)$.
Similarly, we introduce a binary symbol $=$: $A = B$ is short for $A \subseteq B \wedge B \subseteq A$.
By $\emptyset$ is meant the set such that $\forall x(x \in \emptyset \leftrightarrow x \neq x)$ i.e. the expression $x \in \emptyset$ is short for the expression $x \neq x$.
For any set $a$, $\{a\}$ is the set defined thus: $\forall x(x \in \{a\} \leftrightarrow x = a)$ i.e. for any $x$, $x \in \{a\}$ is short for $x = a$. Similarly, $\{a,b\}$ is defined as: $\forall x(x \in \{a,b\} \leftrightarrow (x = a \vee x = b))$. Hence, $x \in \{a,b\}$ is short for $x = a \vee x = b$.
Now we can define an ordered pair in terms of set theory: for any sets $a$ and $b$, $<a,b>$ is a shorthand for the set $\{\{a\},\{a,b\}\}$. Of course, I could have defined ordered pairs in any way I wanted as long as the definition allows me to choose between the first and second element.
For classes $A$ and $B$, the class $A \times B$ is defined as the class of all ordered pairs $<y,z>$, where $y$ comes from $A$ and $z$ comes from $B$: $\forall x(x \in A \times B \leftrightarrow \exists y \exists z(y \in A \wedge z \in B \wedge x = <y,z>))$.
For a set $x$, by $\wp(x)$ we mean the set of all subsets of x i.e. $\forall y(y \in \wp(x) \leftrightarrow y \subseteq x)$ or if you want to replace the $\subseteq$ symbol, $\forall y(y \in \wp(x) \leftrightarrow \forall z(z \in y \to z \in x))$.

Now you have enough material to translate expressions with Cartesian products and power sets into expressions in first-order logic in a straightforward but cumbersome manner:

"if $a = \wp(a)$ then $\wp(a) = \emptyset$" becomes "if $\forall x(x \in a \to x \in \wp(a)) \wedge \forall x(x \in \wp(a) \to x \in a)$ then $\forall x(x \in \wp(a) \to x \in \emptyset) \wedge \forall x(x \in \emptyset \to x \in \wp(a))$" which itself decomposes to "if $\forall x(x \in a \to \forall y(y \in x \to y \in a)) \wedge \forall x(\forall y(y \in x \to y \in a) \to x \in a))$ then $\forall x(\forall y(y \in x \to y \in a) \to x \neq x) \wedge \forall x(x \neq x \to \forall y(y \in x \to y \in a))$".

As you can see, a full translation like that is definitely possible but it is very very tedious! For example, try translating "$A \times B = B \times A$ if and only if $A = B$". It is not fun!

There will be differences of opinion about notation, but I like to use:

The Cartesian Product of sets $A$ and $B$ is given by $ A\times B\$ such that:

  $\forall x,y:[(x,y)\in A\times B\iff x\in A\land y\in B]$

The power set of set A is given by $\mathcal P(A)$ such that:

  $\forall x: [x\in \mathcal P(A)\iff \forall y:[y\in x\implies y\in A]] $


Interpretation of parametric differentiation "How does parametric differentiation make sense both mathematically and graphically?"

I mean the method is simple to understand since one only has to differentiate both equations and divide (generally), but what does it actually mean to differentiate 2 questions and then divide them to find the differential of one with respect to another?

I've worked out the idea for the mathematical aspect and realized that $da/db$ is just a ratio so it's just the idea that $a:b = 2:3$ and $b:c = 6:11$ and if one is required to find $a:c$, he'd divide both. I think same thing holds for differentiation. But then what's the purpose of the notion familiarly stated with derivatives "derivative of one with respect to another?"

But I was struggling with the graphical interpretation, I got ideas that the graph would become a curved surface in 3D and the tangent will become a plane or that it might be related to partial derivatives and such but I think this is beyond me (class 12 student in India). A detailed explanation is welcome though.

Here's one from a snippet of a book,
“ x and y are given as functions of a single variable e.g. x =f(t) and y=g(t) are 2
functions of a single variable. In such a case x and y are called parametric functions or parametric equations and t is called the parameter. To find dy/dx in case of parametric functions, we first obtain
relationship between x and y by eliminating the parameter t and then we differentiate it with respect to x. But, it is not always convenient to eliminate the parameter. Therefore dy/dx can also be by the following formula,

$$
\frac{d y}{d x}=\frac{d y / d t}{d x / d t}
$$

I have also found this alongside with it,
“To prove it, let Δx and Δy be the changes in x and y respectively corresponding to small change is Δt in t. Then,
$$
\frac{\Delta y}{\Delta x}=\frac{\Delta y / \Delta t}{\Delta x / \Delta t} \Rightarrow \frac{d y}{d x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\frac{\lim _{\Delta t \rightarrow 0} \frac{\Delta y}{\Delta t}}{\lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}
$$
So I guess, this takes care of the mathematical aspect (though I'm still not convinced with it) but what about the graphical interpretation?

The "proper" way to do the things you did above. Consider the function $u(x)$. Now lets assume that $x$ is actually again a Function or something that depends on $parameter$. So in your case lets say $x = x(v)$. Now when we want to calculate the derivative of $u$ we with respect to $v$, where we need the chain rule:
$$
\frac{d}{dv} u(x(v_0)) = \frac{du}{dx}(x(v_0)) \cdot \frac{dx}{dv}(v_0).
$$
Now, let's not evaluate the derivative, then the above expression becomes
$$
\frac{du}{dv} = \frac{du}{dx} \frac{dx}{dv}.
$$
Observe, that this last expression is a function, and so are $\frac{du}{dx}$ and $\frac{dx}{dv}$ are both functions!

Now, in the beginning of the development of calculus (e.g. the 17 century). A lot of things were not known, especially about limits and convergence and that stuff. So, when people, namely Newton and Leibniz, developed this idea of taking a derivative, they used a lot of intuition and good will. Leibniz very much argued in a similar spirit as you, say, hey look, lets denote by $du$ a very very small difference of $u$ and by $dx$ lets denote a very very small difference of $dx$, then even so these difference will super close to zero (but never really) zero, then the fraction $\frac{du}{dx}$ actually could be something meaningful and let's called it the derivative.

Today, we do not argue like this any more (at least if you are a mathematician). But Leibniz invented by this an amazing symbolism that you see in play above. Because starting from $\frac{du}{dv}$ simply by multiplying by $1 = \frac{dx}{dx}$ you actually get the chain rule effortless (as hell!). Now lets keep this idea, that $\frac{du}{dx}$ is actually a fraction, then you can also get your expression
$$
\frac{du}{dx}\frac{dx}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}}
$$
and with a little bit of magic this completely unjustified manipulation of symbols actually gives the right answer (in the right circumstance!).

Maybe so much for now. Is it helpful? Not sure if I actually addressed  the question properly.

What you're doing is applying the chain rule. You have (following your example), $$u=f(v),$$ where $v=g(x).$ Then of course $$u'_x=f'_vv'_x.$$

Well, in particular, for your example, since $v=\sqrt{\cos x},$ it follows that $\sin x=\sqrt{1-\cos^2x}=\sqrt{1-v^4},$ so that $$u=\log \sqrt{1-v^4},$$ explicitly.


prove that $ a^2 + b^2 + c^2 \ge 2\left( {a^3 b^3 + a^3 c^3 + c^3 b^3 + 4a^2 b^2 c^2 } \right)$ I have:

  let $a$, $b$ and $c$ be non-negative real numbers with sum $2$. Prove that
  $$a^2  + b^2  + c^2  \ge 2\left( {a^3 b^3  + a^3 c^3  + c^3 b^3  + 4a^2 b^2 c^2 } \right)$$

I should determine whether this is a convergent or divergent integral. The problem is that I don't know how to start.

This inequality is come from?  This inequality can use this

$$\Longleftrightarrow \frac{(a^2+b^2+c^2)(a+b+c)^4}{16}-2(\sum{a^3b^3}+4a^2b^2c^2)=\frac{3abc(a+b+c)}{4}\sum{(a-b)^2}+\sum\frac{c(2a+2b+c)(a^2+b^2-c^2)^2}{16}+\frac{(a-b)^2(b-c)^2(c-a)^2}{2}+\frac{a^2b^2c^2}{4}$$

You can use the KKT condition: http://en.wikipedia.org/wiki/Karush-Kuhn-Tucker_conditions

As a hint, here's a physicist's argument: given the symmetry among the 3 variables, the optimal value is likely to occur at the high-symmetry scenario when a=b=c. And you'll see that the inequality holds.


What is the difference actually between big O and small o in numerical methods? I am studying numerical methods for ODE. I am confused about the "o"s there.
What is the difference actually between big O and small o in numerical methods? I have read somewhere that the big O means the convergence order. Am I right? If so, what about small o then?

The big $O$ notation is usually used in evaluating the complexities of algorithms. A function $f$ is said to be $O(g)$ for a function $g$ if there exist constants $k$ and $C$ such that $f(n) < kg(n) + C$ for all large enough $n$. For example, if you make an algorithm that takes $2n^2+172n + 1$ steps to complete for an input of size $n$, you say that it takes $O(n^2)$ steps, as the $n^2$ factor dominates the polynomial for large $n$.

The small $o$ means something else. It is used to evaluate how quickly a function drops towards $0$ when $x$ approaches some number. A function $f$ is said to be $o(g)$ around $a$ if the limit $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$$
exists. For example, the function $\sin x$ is $o(x)$, but not $o(x^2)$.


How to prove that $7^{15} + 7^{16} + 7^{17} - 1$ is divisible by $10$? This was a question on my math exam. We weren't able to use calculators so proving by manually calculating the exact value would take too long.
In the end I ignored this question to save time but I'm still curious.


Best way to approach a Multiple Choice exam I have an exam tomorrow with 100 multiple choice questions. Each question has 4 options, only 1 is correct.

If I answer the question and get it right, I get 1 mark.
If I leave it blank, I get 0.
If I answer incorrectly, I lose 0.25 marks.

How should one approach the test? Mathematically/statistically are my odds better if I guess if I don't know the answer?

Imagine if you had $4$ questions you don't understand.

If you leave them blank, you will get a mark of $0$.
If you guess all of them, the expected result is getting $1$ right and $3$ wrong. This will get you an expected mark of: $$1-3(0.25)=0.25$$

Therefore it is better to guess than to leave questions blank.

If you can eliminate at least one choice on a question, then you should guess from the remaining choices.