Video lectures on Real Analysis? One of the most annoying gaps in my math education is Real analysis.

I tried hard, but all I could find are either Harvey Mudd College lectures or MathDoctorBob. The latter are too short and the former are in horrendous format, I can barely make out what is written on the blackboard.

Ideally I'd like the lectures to cover topics such as proofs of continuity, differentiation, some main inequalities, $\limsup$ and $\liminf$, uniform/pointwise convergence and continuity, dominated and monotone convergence and maybe a bit more.

So if anyone knows if this material is available online, I'd be quite grateful.

You can try: http://www.uccs.edu/math/student-resources/video-course-archive.html

They have a Real Analysis course listed, Math 533 Real Analysis I - Fall 2007. They are good, though not that challenging.

You may need to create a [free] account to view the videos.

If you really don't want to make an account the videos start with:

http://cmes.uccs.edu/Fall2007/Math533/pages/video3.html

and go up to /video24.html. There is, for some reason, no video 1 or 2.

1 http://ocw.aca.ntu.edu.tw/ntu-ocw/index.php/ocw/cou/104S115 real analysis 1 at National Taiwan University ( Called Calculus for math majors 1)

2 http://ocw.aca.ntu.edu.tw/ntu-ocw/index.php/ocw/cou/104S210 real analysis 2 at National Taiwan University ( Called Calculus for math majors 2)


Why study quadrilaterals? My niece is in the 10th grade, and they have to do lot of theorems related to quadrilaterals. And, I was surprised to know that they have to learn by rote some theorems. This has made her feel that maths is the worst subject (though she likes probability) ! So, I was wondering , why study quadrilaterals? How can we relate this geometrical figure to the real world?

thanks

Actually, I have thought and read a lot about this topic. A brief disclaimer: I'm going to present two pictures below, both of which can be validly used and neither of which is an "objective" correct pictures. However, one has strong advantages and I won't be shying away from promoting it.

Quadrilaterals

Studying quadrilaterals is an excellent example of how definitions are used in mathematics. Most people don't appreciate this the first time around (I certainly didn't, in grade school.)

When we learn about squares, rhombi, trapezoids etc when we're young, the definitions often spring from the head of Zeus with very little connection. Most students would probably not notice what similarities connect these families.

But anyone who goes a little farther in mathematics can start seeing this "family tree":

What is to be learned here? The idea is that of a "hierarchy" of properties. Squares are special rectangles, are special parallelograms are special quadrilaterals, and etc through other paths of this picture.

Actually, you can make this picture completely symmetrical by adding in a node between kite and quadrilateral and adding a line to it from "parallelogram", but we don't have a good word for that particular shape. But that is a slight digression: the point is that all of these things fit into a coherent picture based on their properties.

Why thinking about hierarchies is useful

The main strength of exercising this kind of thinking is that it strengthens one's logical ability. For one thing, it teaches that one should view things with a lot of properties as being special cases of more general objects (like squares being special cases of parallelograms.) Another thing is that if you prove something for a quadrilateral high up, then it automatically holds for quadrilaterals lower, because they are just special cases.

Thinking in terms of hierarchies of properties like this can be found throughout mathematics. For example, when studying real analysis, I distinctly remember making diagrams of implications between pointwise convergence, a.e convergence, uniform convergence etc so that I could understand how they related to each other.

One can do the same with groups and rings in algebra. There is a little chain of implications between rings in the top of this wiki article, and I personally have an extensive one with about 90 ring types on it making a very interesting lattice.

Really, this example with quadrilaterals is one of the simplest and most accessible demonstrations of a hierarchy of properties.

An unfortunate situation at present

Unfortunately, many primary and secondary teachers still insist on teaching the "Frankenstein hierarchy" depicted below:

Thanks to presentations like this, the most important pedagogical point is pretty much obliterated. A tiny bit is retained in the parallelogram-rhombus-square-rectangle portion, but I bet many students perceive that as a "lucky break" in the organization of an otherwise terrible picture. The only way you could make it more terrible is if you had "quadrilateral" at the top, and put the other seven items in a row with individual lines going up to "quadrilateral." The result is a very shallow hierarchy with virtually no use.

So, what can you take away from the first picture?

Point out that tracing a line upward is "relaxing" requirements, and tracing a line downward is "adding" requirements. These things are interrelated by their properties, not just randomly named garbage to torture children with.
Notice that if you prove something for parallelograms, then you have automatically proven the same theorem for all rhombi, rectangles and squares. If you prove something for all kites, you've also proven the same thing for rhobmi and squares. If you prove something for trapezoids, you proved it for parallelograms as well.
Moreover, let's say you have proven something for rectangles. The next question might be "but wait, is rectangles the best I can do? Might I be able to prove this for parallelograms too?" And maybe you will be able to do it! This shows that knowing how things are interrelated can give you ideas for improving results.

There is a book about this

In fact, you might want to check this out from the library: The Classification of Quadrilaterals: a study in definition. I hope it doesn't sound too dry because it's really quite interesting. It is, if nothing else, a book with a well told story of math education.

Among other things it contains a historical overview of how the definitions evolved, and better written explanations about the pedagogical value than what I am writing here. Finally, they provide quite good analysis and evidence that supports the use of the first diagram above over the second.

Here is a brief list of works I've found useful investigating this topic. If you read anything, take a look at one of Keedy's papers and the book I mentioned above (which is also below.

Aydin, N. and Halat, E. (2009) ”The Impacts of Undergraduate Mathematics Courses on College Students’ Geometric Reasoning Stages.” The Montana Mathematics Enthusiast, ISSN 1551-3440, Vol. 6, nos.1&2, pp.151- 164
Bagni, G. T. (2001).  “An investigation of some misconceptions in High School students’ mistakes”. Learning in Mathematics and Science and Educational Technology, I, Intercollege Press Cyprus, Nicosia: pp 3-24
Craine, T.V. & Rubenstein, R.N., “A quadrilateral hierarchy to facilitate learning in geometry.” Mathematics Teacher 86 (January 1993): 30-36.
DeVilliers, M. (1994). “The Role and Function of a Hierarchical Classification of Quadrilaterals.” For the Learning of Mathematics, Vol 14 No. 1. FLM Publishing Association: pp 11-18.
Fujita, T & Keith Jones. (2007).“Learners’ understanding of the definitions and hierarchical classification of quadrilaterals: Towards a theoretical framing.”  Research in Mathematics Education, 9 (1&2): pp 3-20.
Fujita, T. (2012). “Learner’ level of understanding of the inclusion relations of quadrilaterals and the prototype phenomenon.” The Journal of Mathematical Behavior 31: pp 60-72.
Keedy, M.L. “The Fuss about trapezoids.” Mathematics Teacher 73 (October 1980):  488-489.
Keedy, M.L. “What is a trapezoid?” Mathematics Teacher 59 (7) (November 1966): 646.
Leung, I.K.C. (2008) “Teaching and learning of inclusive and transitive properties among quarilaterals by deductive reasoning with the aid of SmartBoard.” ZDM Mathematics Education. Vol 40, #6: pp 1007-1021. DOI: 10.1007/s11858-008-0159-z
Maraldo, S. “Properties of Quadrilaterals.” Mathematics Teacher 73 (1) (1980): 38-39.
Usiskin, Z. Griffin, J, Witonsky, D, Willmore, E. “The classification of quadrilaterals: A study of definition.” Information Age Publishing. (December 2007)

Memorizing theorems is not learning, and has no value whatsoever. I would say that your niece's teacher should be fired, but this so-called curriculum was almost certainly handed down by the local or regional government. Such piss-poor geometry classes, I'm told, are now the norm in those schools that even bother to teach geometry at all.


How to show that $f(x)\ge 0 \implies \int_a^x fdt \ge 0$? This is clear to me thinking about integration like area under a curve, but how can it be proven? (assuming $f$ is continuous and using the fundamental theorem of calculus, and $\forall x\in[a,b]$)

Assume that $x\ge a$ in the integral $\int_a^x f(t)dt$

Let $f(x)$ be the derivative of a function $F(x)$

Since $f(x)\ge0$, the function $F(x)$ must be weakly increasing, and since $x\ge a$ we have $F(x)\ge F(a)$

We know from the fundamental theorem of calculus that
$$
\int_a^x f(t)\,\mathrm dt=F(x)-F(a)
$$
Can you see it now?

Simply use $\int_{a}^{x} (f-0)dt \geq0$. Now split the integrals and take the second one to the RHS


How to find the limit of this function We have the function $$\dfrac{\sqrt{n^4 + 100}}{4n}$$

I think the best method is by dividing by $n$, but I have no idea what that yields, mainly because of the square root.

Dividing by $n$ is a great idea: Note that $$\sqrt{n^4 + 100}\cdot\frac 1n = \sqrt{(n^4 + 100)\cdot\frac 1{n^2}} = \sqrt{n^2 + \frac {100}{n^2}}$$

So your limit is, after dividing numerator and denominator by $n$:

$$\lim_{n\to \infty} \frac{\sqrt{n^2 + \frac {100}{n^2}}}{4} \to \infty$$

Perhaps another method I overlooked, stupidly enough:

$$ \dfrac{\sqrt{n^4 + 100}}{4n} = \dfrac{n^4 + 100}{4n^5 + 400n} = \dfrac{ n^4 + 100}{4n \cdot n^4 + 400n} = \dfrac{101}{404n} = \dfrac{1}{4n}$$

I'm not sure it is correct, I played around a bit with the algebra.


The sum of two continuous periodic functions is periodic if and only if the ratio of their periods is rational? There are several questions here on MSE about the periodicity of the sum of two continuous, periodic functions. Here's what I know so far:

It's obvious that if the ratio of the periods is rational, the sum of the functions will be periodic with a period equal to the LCM of the individual periods.
It's relatively easy to give a counterexample to show that the sum of two continuous periodic functions is not always periodic, for instance by considering $\sin(x) + \sin(\pi x)$.

My hunch is that in fact, the sum of two continuous periodic functions is periodic if and only if the ratio of their periods is rational. First of all, is this true?

Secondly, if it's true, the sufficient part  is obvious, but is there a simple proof to show the necessary part?

By a "simple" proof I mean one that is accessible to someone with no extensive formal knowledge of analysis or algebra. This is the closest proof I could find, but I can't quite follow it.

Let $f,g$ be continuous and periodic with period $p,q$ where $\alpha=\frac pq$ is irrational. Assume $f+g$ is periodic with period $r$. Let $f(x_0)=\max f$ and $g(x_1)=\max g$.
Then the set $\{\,np+mq\mid n,m\in\mathbb Z\,\}$ is dense in $\mathbb R$.
Let $\epsilon>0$. Then for some $\delta>0$, $|x-x_1|<\delta$ implies $g(x)>g(x_1)-\epsilon$.
Let $np+mq$ differ by less than $\delta$ from $x_1-x_0$.
Then $f(x_0+np)=\max f$ and $g(x_0+np)>\max g-\epsilon$.
We conclude that $\max(f+g)=\max f+\max g$.
So assume $(f+g)(x_2)=\max (f+g)=\max f+\max g$.
Then $(f+g)(x_2+nr)=\max(f+g)$ implies that also $f(x_2+nr)=\max f$ and $g(x_2+nr)=\max g$.
At least one of $\frac rp$, $\frac rq$ is irrational, say $\frac rp$ is irrational.
Then the set $\{\,x_2+nr+mp\mid n,m\in\mathbb Z\,\}$ is dense in $\mathbb R$ and we conclude that $f(x)=\max f$ on this dense set and hence throughout. Thus $f$ is constant.

In summary: The sum of two nonconstant continuous periodic functions with incommensurable periods is not periodic.

$f(x+p)=f(x); g(x+p')=g(x) \rightarrow (f+g)(x)=f(x+kp)+g(x+k'p'); k \in \mathbb Z $ . Now, $f+g$ must satisfy both periods $p,p'$, so:

Then, for a period, we need $$ x+kp=x+k'p'\rightarrow \frac {k}{k'}= \frac{p}{p'} $$.

EDIT: My claim: given $f,g$ periodic, then $h:=f+g$ repeats precisely , i.e., iff $f,g$ simultaneously repeat, i.e., $h$ must satisfy the periods of both $f,g$. This happens precisely when the ratio of their periods is Rational.


For which $k$ can $(x-1)^{k} - (x+1)^{k}$ be divided by $x$? Consider
  $$
f_{n}(x) = (x-1)^{n} - (x+1)^{n}\,.
$$
  For which $n$ can it be divided by $x$?

Explicitly I found that for $k = 1, 3$ $f_{n}(x)$ can not be divided by $x$, while for $n = 2$ It can.

Next, I used the formula

$$
a^{n} - b^{n} = (a-b)(a^{n-1}+a^{n-2}b + \dots + ab^{n-2}+b^{n-1}),
$$

but I don't see the way to show something from it explicitly.

Let $p(x)=x^k$. Since $f(x)=p(x-1)-p(x+1)$ is divisible by $x$ we see that $x=0$ is zero of $f(x)$, so $p(-1)=p(1)$, so $$(-1)^k = 1^k$$ and that is only if $k$ is even.

I would say that for all even $n$. Observe that, irrespective of the value of $n$, $(x+1)^n=x^n+....+1.$ On the other hand, $(x-1)^n=x^n-....+(-1)^n.$ The last term in this polynomial will be $+1$ for even $n$, and, thus, the difference, $ (x-1)^n-(x+1)^n$ is just a polynomial of degree $n-1$ without a constant term which can always be divided by $x.$ Again, this is true only for even $n.$


General Annuity practice question I'm taking an introductory course in finance but I don't understand how to do this question:

A used car may be purchased for 7600 cash or 600 down and 20 monthly payments of 400 each,the  first payment to be made in 6 months.  What annual effective rate of interest does the installment plan use?

What I tried to do was equate the 7600 to the discounted value of the monthly payments, plus the 600, but I'm not sure how to solve for 'i'. This is the equation I have now: 7600 = 400 x [1-(1+i)^-20]/i + 600

You have not taken account of the delay in receiving payments.  There should be another factor $(1+i)^{-5}$ to take account of the five months with no payments.  Then solving for $i$ requires a numeric approach.  A spreadsheet will do it for you or you can use bisection.  $i=0$ is too low and you can easily find an $i$ that is too high.  Then compute it at the center point and see if it is too low or too high.

Start by letting $i$ be the monthly interest rate, not the annual rate. The answer to the question will be $(1+i)^{12} - 1$


Why formulate continuity in terms of pre-images instead of image? I wanted to discuss my intuition of why we formulate the concept of continuity in terms of pre-image of open set is open instead of images for example if we consider $f(x) = c$ where $c$ is some constant, then that should be continuous, but if we formulated continuity in terms of images it won't be continuous. Is the reason we do it that way, because pre-images will always guarantee that function in the domain can only have $1$ value?

It's not nice to formulate it in terms of images. The behavior of images varies wildly between various continuous functions. Take the interval $(-10,10)$ and the functions $f_1(x) = x$ and $f_2(x) = \sin x$. Then the image of $(-10,10)$ under $f_1$ is of course $(-10,10)$, but the image of $(-10,10)$ under $f_2$ is $[-1,1]$. With one function you get an open set, with another you get a closed set. You can of course explore the other cases of half-open/half-closed or other phenomena. You can't make any meaningful statement about images of open sets. You can't even say anything for certain about local behavior (the images of "small" open sets) because there are plentiful locally constant functions.

You might want to flip this on its head then since images of open sets doesn't work: what about images of closed sets? This also doesn't work. Consider $\mathbb{R}$ and the function $f_3(x) = \arctan x$, then the image of $\mathbb{R}$ under $f_3$ is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ which is an open set.

The beauty of pre-images is that the functions we intuitively look at as being continuous have open sets as pre-images of open sets. Perhaps the closest you can get to a statement about images is that continuous functions map compact sets to compact sets, however there are discontinuous functions which also do this: consider $[-1,1]$ and $f_4(x) = 1$ if $x>0$, $f_4(0) = 0$ and $f_4(x) = -1$ if $x<0$. Then $f_4$ maps a compact set to a compact set but it is definitely not continuous in the usual topology.

I realize this is a bit of thread necromancy, but I don't see the harm, and I'm not sure I see this perspective noted in the answers yet.  It's true that defining it in terms of pre-images works, and defining it in terms of images (without changing anything else) doesn't work.  But why doesn't it work?
This puzzled me too, when I first encountered it mumblety-mumble years ago.  The way that I ended up internalizing it is that it's related to the fact that a function has a unique output for each input, but it doesn't have to have a unique input for each output.  Two different inputs can map to the same output.  In some sense, therefore, the output of an input set—its image, in other words—"loses information" about the input set, and therefore (in a non-rigorous sense), the image can't be expected to preserve the topology of its input.
That isn't the case with the pre-image.  It may have multiple "copies," so to speak, of the information in the image—as for example with periodic functions—but each "copy" preserves the topology and corresponds, open set to open set, even for non-periodic functions.
Consider what happens when a continuous function $f:\mathbb{R} \to \mathbb{R}$ does take on the same output for multiple inputs.  So long as $f$ is (strictly) monotonic and continuous, then it will map an open set to an open set; the topology is preserved.  But if it isn't monotonic—if there's a local maximum, say, at $(x_0, f(x_0))$—then the image of some open neighborhood of $x_0$ cannot itself be open, because $f$ has to attain a maximum value of $f(x_0)$ in that neighborhood; it can't attain a maximum of "the largest real value smaller than $f(x_0)$."
Contrast this with the situation of the pre-image.  Because you can't have a single input value map to multiple output values, there's no way for an open set in the co-domain to correspond to a non-open set in the domain; you'd have to have the input "turn back on itself," which it can't do (so long as $f$ is a function).


Prove or disprove that every even integer is the difference of two squares. I don't know where to start. Let $2k$ be an even integers and $a^2 - b^2$...

I do not believe this question is correct as stated. A counterexample is the number $2$. $2$ is not the difference of squares. If it were, say $a^2 - b^2 = 2$, then $(a-b)$ would divide $2$. So $(a-b)$ must one of $1$ or $2$, since $a-b$ is positive. $a+b$ must then be $2$ or $1$ respectively. In either case, we are forced to have that $(a+b) + (a-b) = 2a = 3$. But $2$ does not divide $3$.

What is true, however, is that every integer of the form $4k$ for some integer $k$ is the difference of squares. We may assume $k$ is positive. If $k = 0$, then $a=b=0$ works. If $k$ is negative, then assuming the above, there are $a$ and $b$ so that $a^2 - b^2 = -k$. Then $b^2 - a^2 = k$.

Now, simply notice that $$(k+1)^2 - (k-1)^2 = k^2 + 2k + 1 - k^2 + 2k - 1 = 4k,$$
so $4k$ is the difference of squares.

I think we are lacking informations...

Trivial in R
If a and b belong to R then if if k positive a = sqrt(2k) and b = 0. If k negative a = 0 b = sqrt (-2k)

False in N
Starting point is the fact that every odd number is the difference if two squares.


Math Behind Creating a "Perfect" Star I am busy looking to create star paths in my app, and I was wondering how to determine the ratio between the inner radius and the outer radius of the points of a star so that the star has "straight" lines across.

I have a function that takes 3 parameters:

pointCount = 5
outerRadius = 100
innerRadius = 50

Basically it goes around the circle and alternates between a point and an inside so the star looks like this:

As you can see, the star is "bulging". What I am really trying to get is this star:

There should be some mathematical formula that can take the "outer" radius and the number of points to calculate the "inner" radius?

innerRadius = some_operation (pointCount, outerRadius)

I think you are trying to draw a pentagram. While the linked page has many properties of that shape, some of them which would be of interest to you are:
$$R = \sqrt{\frac{1}{10}\big(25 - 11\sqrt{5}\big)} \\
\rho = \sqrt{\frac{1}{10}\big(5 - \sqrt{5}\big)}$$
where $R$ is the inner radius and $\rho$ is the outer radius. This assumes that the long edges of the star are of length 1.

You could always scale these numbers if the edges are not of length 1. But as you're interested in the ratio, we have
$$\frac{\rho}{R} = \sqrt{\frac{5 - \sqrt{5}}{25 - 11\sqrt{5}}} = \frac{3 + \sqrt{5}}{2}$$
In addition to being a surprisingly clean simplification, note that the ratio is $\phi + 1$ where $\phi$ is the famous golden ratio!

Star has $n$ points. The outer points of the star are $(x_k, y_k) =(\cos \frac {2\pi*k}n, \sin \frac {2\pi*k}n)$.  $(x_k, y_k)$ connects to $(x_{k+2\%n},y_{k+ 2\%n})$.

So formula for the line  $(x_k, y_k)-(x_{k+2\%n},y_{k+ 2\%n})$ is:

$y = \sin \frac {2\pi*k}n + (x - \cos \frac {2\pi*k}n)\frac{\sin \frac {2\pi*(k+2) \%n}n-\sin \frac {2\pi*k}n}{\cos \frac {2\pi*(k+2) \%n}n-\cos \frac {2\pi*k}n}$

and the formula for line $(x_{k-1\%n},y_{k-1\%n})-(x_{k+1\%n},y_{k+ 1\%n})$ is:

$y = \sin \frac {2\pi*(k-1\%n)}n + (x - \cos \frac {2\pi*k}n)\frac{\sin \frac {2\pi*(k+1\%n) \%n}n-\sin \frac {2\pi*(k-1\%n)}n}{\cos \frac {2\pi*k+1 \%n}n-\cos \frac {2\pi*(k-1\%n)}n}$

Calculate the point of intersection (thank god for machines)

Use the distance formula to find distance between to out points and these inner point of intersect.  (Did I say thank god for machines?  Let me say it again.)

If you just want the distance I suggest you do $k = 0$.


Existence of solutions to $a^2x^2 + bx = y^2$ Is there a way to determine if there are or aren't any integer positive solutions $(x,y)$ to the equation $a^2x^2 + bx = y^2$ depending on the values of $a$ and $b$?

I tried to deal with it using Pell equations but I just couldn't work it out. Any help is appreciated.


Find the value of $(1-a_1)(1-a_2)(1-a_3)(1-a_4)$ given that $a_1,a_2,a_3,a_4$ are roots of a biquadratic equation. So the biquadratic equation is $x^4+(2-\sqrt3)x^2+2+\sqrt3=0$. Let $a_1,a_2,a_3,a_4$ be its roots. So we have to find the value of $(1-a_1)(1-a_2)(1-a_3)(1-a_4)$ .
My attempt:
So of we put $x^2=t$, and let the roots of the new quadratic equation be $a_1,a_2$. So we get that $a_1=-a_3;a_2=-a_4$. So the expression whose value we have to find will become $(1-a_1^2)(1-a_2^2)$. And now we multiply the expression to get, $1-(a_1^2+a_2^2)+(a_1a_2)^2=1-((a_1+a_2)^2-2a_1a_2)+(a_1a_2)^2$.
Now substituting the values of sum of roots and product of the same, I get the answer $5+10\sqrt3$. But the answer according to my question paper is $5$.
So please can someone point out the mistake and tell the answer?

Hint:

Just notice that
$$(x-a_1)(x-a_2)(x-a_3)(x-a_4)=x^4+(2-\sqrt{3})x^2+2+\sqrt{3}$$

About your attempt:

According to G-man (thanks) we have

\begin{align}
1-(a_1^2+a_2^2)+(a_1a_2)^2&=1-\left(-\frac{2-\sqrt{3}}{1}\right)+(2+\sqrt{3})\\
&=1+(2-\sqrt{3})+(2+\sqrt{3})\\
&=5
\end{align}
Where we have used the fact that $(a_1a_2)^2=a_1a_2a_3a_4=2+\sqrt{3}$ from the  Vieta's formulas.


Please prove: $ \lim_{n\to \infty}\sqrt[n]{\frac{1}{n!}} = 0 $ Possible Duplicate:
  $ \lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite

Please prove:
$$
\lim_{n\to \infty}\sqrt[n]{\frac{1}{n!}} = 0
$$

Hint: When writing out $n!$, you have
\[ n! = n \cdot (n-1) \cdots \left\lceil \frac n2\right\rceil \cdots 1 \]
so at least $\lfloor \frac n2\rfloor$ of the factors are larger then $\lceil \frac n2\rceil$.
So $n! \ge \lceil \frac n2\rceil^{\lfloor \frac n2\rfloor}$.

Just try to put n equal to infinity. since 'n' appears in the denominator it will tend to zero.

Please confirm the answer.


Simulate simple non-homogeneous Poisson proces Consider a Poisson process whose conditional intensity is

$$\lambda(t) = \alpha e^{-t}$$

starting at time $t=0$ for some parameter $\alpha>0$.

I would like to simulate arrival/event/failure times as efficiently as possible. (I am not sure what the standard term is for the times you get from a Poisson process.) Is there a fast way to do this?

"Amusing" downvote to an otherwise perfectly valid answer.

If one knows how to simulate a homogenous Poisson process $(N(t))$ with intensity $\alpha$ on the time interval $[0,1]$, one can then consider the process $(X(t))$ defined, for every $t\geqslant0$, by $$X(t)=N(1-\mathrm e^{-t}).$$


Is this proper notation: $\;\mathbb Z^{+-}\;$? I have a question about notation. Is it okay to represent $\;\mathbb Z^{+-}$ as $ \mathbb Z - \{0\}$ ?

Like does it make sense to manipulate both sets mathematically like that? ) i.e. $ \{0\} = \mathbb Z - \mathbb Z^{-+} $

Thanks in advance!

Using $\mathbb Z^{+-}$ would have given me little clue as to what you are denoting, were it not for the equality you use to define it. (Okay, I could have probably guessed, but why make your readers have to guess?)

I would strictly use $ \mathbb Z - \{0\}$, or better yet, $\mathbb Z \setminus \{0\}$. I'd even prefer it were spelled out in words: "all non-zero integers" than trying to guess what your notation denotes. (I have never seen the notation $\mathbb Z^{+-},\,$ to be honest.)

I don't think it makes sense to write $ \{0\}$ as $\mathbb Z - \mathbb Z^{-+} $. (I'm not clear why you'd want to avoid using $\{0\})$. In any case, using $\;\{0\}$ is much more straightforward, and easier to write and read.

$\Bbb{Z}-\{0\}$ is usually denoted by $\Bbb{Z}^*$, the set of invertible integers.


What are the first two moments of this stochastic process? The setup. Consider a doubly stochastic Poisson (i.e. Cox) process, which is a Poisson arrival process $X_t$ with stochastic intensity function $\lambda_t$, i.e., a Poisson process whose rate is itself a stochastic process.
I try to compute the expected value and the variance of $X_t$ for a variety of rate processes $\lambda_t$.
I already computed the distribution of $X_t$, which is given by $$ P(X_t=k)=\frac{1}{k!} E\left[ \left( \int_0^t \lambda_s ds \right)^k \exp \left( -\int_0^t \lambda_s ds \right) \right],$$ see for example this paper.
The Problem. I realize it is impossible to give a closed form expression for the moments of $X_t$ for arbitrary rate processes $\lambda_t$, but for some special ones it should be possible.
I already fail for the case where $\lambda_t$ is a simple Bernoulli Process, i.e., $$ \lambda_t \mbox{ i.i.d. with }, ~ P(\lambda_t=1)=\frac{1}{2}, ~ P(\lambda_t=3)=\frac{1}{2}. $$ Can anyone give me a hint on how to start? The question therefore boils down to

How can I compute $$ E\left[ \left( \int_0^t \lambda_s ds \right)^k \exp \left( -\int_0^t \lambda_s ds \right) \right] $$ if $\lambda_t$ is the stochastic process above?

Another example that would interest me would be $$ \lambda_t \hat = \mbox{ arrival times of another Poisson process}. $$

I figured it out myself: For the process $\lambda_t$ given in the question, we compute \begin{align} P(X_t=k) &= \frac{1}{k!} \sum_{\omega \in \lbrace \omega_1, \omega_2 \rbrace} \left( \int_0^t \lambda_s(\omega)ds \right)^k \exp \left( - \int_0^t \lambda_s(\omega)ds \right) \\ &= \frac{1}{k!} \sum_{\omega \in \lbrace \omega_1, \omega_2 \rbrace} \frac{(2t)^k\exp(-2t)}{2} = \frac{(2t)^k\exp(-2t)}{k!}. \end{align} Therefore, we can compute the first two moments as follows \begin{align} E[X_t] &=\sum_{k=0}^\infty k \frac{(2t)^k\exp(-2t)}{k!} \\ &= \exp(-2t) \sum_{k=1}^\infty \frac{(2t)^k}{(k-1)!} \\ &= 2t \exp(-2t) \exp(2t) = 2t\end{align} and \begin{align} \operatorname{var}[X_t] &= \left( \sum_{k=0}^\infty k^2 \frac{(2t)^k\exp(-2t)}{k!} \right) - (2t)^2 \\ &= ... = 4t^2+2t-4t^2 = 2t \end{align}


Prove if $x > 100$ then $\frac{100}{3-2x} > -1$ If $x > 100$ then $\frac{100}{3-2x} > -1$

Pf:

Assume $x>100$

$-2x > -20$
$-2x + 3 > -200$
$\frac{1}{3-2x} > \frac{1}{-200}$

This is what I have done so far, however I'm not sure how to proceed from here on out. If anyone has any tips, or possibly and example to further push me in the right direction I'd appreciate it greatly!

Be careful! From $2>1$ you cannot deduce that $-200>-100$.

From $x>100$ you can deduce that $2x>200$ and so that
$$
2x-3>200-3=197
$$
It follows that
$$
\frac{1}{2x-3}<\frac{1}{197}
$$
and so
$$
\frac{100}{2x-3}<\frac{100}{197}<1
$$
Changing sign reverses inequalities, so
$$
-\frac{100}{2x-3}>-1
$$

By plugging in $x = 100$ in the denominator, we have

$$ \frac{100}{3-2x} > -\frac{100}{197} > -1$$


Prove $\lim_{n\rightarrow\infty}||A^n||^{\frac{1}{n}}=\inf_n||A^n||^{\frac{1}{n}}$ Let $X$ be a normed space and $A\in B(X)$ (i.e.A is continuous linear map in $X$), prove that $$\lim_{n\rightarrow\infty}||A^n||^{\frac{1}{n}}=\inf_n||A^n||^{\frac{1}{n}}$$

It is easy to see that $||A^n||^{\frac{1}{n}}\leq ||A||$

If $\| A^n \| = 0$ for some $n$, then the result is obvious since both sides are zero. Otherwise, define $(a_n)$ by $a_n = \log \|A^n\|$. Then this sequence is subadditive:

$$ a_{m+n} = \log \|A^{m+n}\| \leq \log (\|A^m\|\|A^n\|) = a_m + a_n. $$

Then by the Fekete's lemma we have

$$ \lim_{n\to\infty} \frac{a_n}{n} = \inf_{n\geq 1} \frac{a_n}{n} \in [-\infty, \infty). $$

Taking exponential to both sides yields the desired identity.

2n'th element is less or equals when nth element of this sequence => if this limit exists, it equals to the infimum of the sequence


Expressing cos 50 using x and y Given that cos15=x and sin35=y, express cos50 in x and y.
I tried solving it using the addition theorem of trigonometry:
cos50=cos (15+35)=cos15cos35-sin15sin35
                 =xcos35-ysin15
and I am stuck ;(

Some help please? This is my revision for my graduation test.
P/S The answer given is cos50=x[(1-y^2)^1/2]-y[(1-x^2)^1/2]


exercise in commutative algebra exercise: let $A$ be an integral domain in which all prime ideals are maximal (but may not be Dedekind in general).  let $P_1,  P_2$ be 2 distinct prime ideals.  prove: for any positive integer $m, n$,  $P_1^m+P_2^n=A$.

so the hint is to prove this first for arbitrary $m$ when $n=1$.  I finished this step by observing that if $P_1^m$ is contained in $P_2$,  then since they re prime ideals $P_1 \subset P_2$.  this contradicts the fact that they re distinguish and maximal.  so this was done.

but next how to use induction for $m$? I need some hint, thanks!

consider it to be true for n then multiply by P_2
the RHS is going to be P_2.

see that (P_1)^m*(P_2) + (P_2)^(n+1) is contained in (P_2).

(P_1)^m*(P_2) is contained in (P_1)^m

(P_1)^m is not contained in (P_2)

now let "a" be any such element which is in (P_1)^m but not in (P_2)
note that (a) + (P_2) is contained in (a) + (P_1)^m*(P_2) + (P_2)^(n+1) hence (P_1)^m + (P_2)^(n+1)

and from maximality of (P_2) you get (a) + (P_2) is actually A


Detect five consecutive unsorted integers It appears that the product of the differences between 3 consecutive integers in whatever order is always equal to 2. However, I can't find the pattern for 5 integers.

We could sort them and get an easier problem, but I am curious if there is a way to solve it on an unsorted 5 integers array.

Edit: Gentlemen in comments pointed out it works for non sorted 3 integers. Question remains. Is there such a property for 5 integers ?

Edit:

The product of $|a-b|$ and $|b-c|$ is always 2.

let an integers array x be $3, 1, 2$

we have : $ (|3 - 1|) \cdot (|1 - 2|)$ = 2

Is there such a relation with 5 integers ?

Take any $5$ numbers $a$, $b$, $c$, $d$, $e$ and multiply the absolute differences differences between consecutive numbers. You get
$$
|a-b|\cdot|b-c|\cdot|c-d|\cdot|d-e|
$$

if you imagine to add to each of the five numbers the same constant, it is easy to see that this product does not change, because each difference remain the same.

So, if I'm considering $5$ numbers it does not matter if they are $1,2,3,4,5$ or $14,15,16,17,18$, because the product of the differences among adiacent numbers will remain the same.

Clearly, if we take the $5$ consecutive numbers in their natural order, or in reverse order, the product will be a product of $1$'s, and thus equal to $1$.

But there are $5!=120$ ways to permute $5$ numbers and it turns out that we can have $13$ different products, namely $1$, $2$, $3$, $4$, $6$, $8$, $9$, $12$, $16$, $18$, $24$, $36$, $48$, depending on how we permute them.

Here is a table of the $13$ values, the number of corresponding permutations, and one such permutation.

$$
\begin{array}{c|c|c}
\prod & \sharp & perm.\\\hline
 1 & 2 & 12345 \\
 2 & 4 & 12354 \\
 3 & 4 & 12543 \\
 4 & 18 & 12435 \\
 6 & 20 & 12534 \\
 8 & 12 & 13542 \\
 9 & 6 & 14325 \\
 12 & 8 & 13425 \\
 16 & 12 & 15324 \\
 18 & 8 & 14352 \\
 24 & 12 & 13524 \\
 36 & 10 & 14253 \\
 48 & 4 & 24153
\end{array}
$$

For example, the line that start with $18$ means that it is possible to obtain the product $18$ with $8$ permutations, one being $14352$, which means we take the first number, then the $4$th, then the $3$rd, the $5$th and the second.

For example, for the numbers $18,19,20,21,22$ this means the order
$18,21,20,22,19$ which indeed gives
$$
|18-21|\cdot|21-20|\cdot|20-22|\cdot|22-19|=3\cdot1\cdot2\cdot3 = 18
$$

The number $T$ of different values and the maximal product $M$ that it is possible to obtain with $k$ consecutive numbers can be summarized as follows, for $k=2,\dots,10$.
$$
\begin{array}{c|ccccccccc}
k&2&3&4&5&6&7&8&9&10\\\hline
T&1& 2 & 6 & 13 & 42 & 92 & 292 & 653 & 1345\\\hline
M& 1 & 2 & 12 & 48 & 432 & 2700 & 32000 & 276480 & 4050000
\end{array}
$$

Is this a serious question?

If $a < b <c < d <e$ are 5 consecutive integers then:

e - a =4

e-b= 3

e-c=2

e-d =1

d-a=3

d-b =2

d-c=1

c-a =2

b-a =1

The product is 4x3x2x1x3x2x1x2x1=288.

Was that a serious question?


If G is a finite Abelian group, and H is a subgroup of G, How can I prove that G/H is isomorphic to another subgroup N of G?. If G is an abelian finite group, and H is a subgroup of G, How can I prove that G/H is isomorphic to another subgroup N of G?.

I'll write abelian groups additively.

You can use the character group: if $G$ is a finite abelian group, then its character group is $\hat{G}=\operatorname{Hom}(G,\mathbb{Q}/\mathbb{Z})$.

From the structure theorem of abelian groups, it follows that $G$ is the direct sum of cyclic $p$-groups. If $C$ is a finite cyclic $p$-group, then $\hat{C}\cong C$, hence there is a (non canonical) isomorphism $\varphi\colon G\to\hat{G}$.

If $H$ is a subgroup of $G$, consider $H'=\varphi(H)$, so $\hat{G}/H'\cong G/H$. Now prove that, defining
$$
N=\{x\in G:\chi(x)=0,\text{ for all }\chi\in H'\},
$$
you have $N\cong \hat{G}/H'$.

HINT:

Because $G$ is abelian, every subgroup will be normal. Therefore, it makes sense to consider $G/H$ for a subgroup $H$.

Then, use the first isomorphism theorem such that $H = ker(f)$


What is an intermediate definition for a tangent to a curve? Most students come to calculus with an intuitive sense of what a tangent line should be for a curve. It is easy enough to give a definition of a tangent to a circle that is both elementary and rigorous. (A line that intersects a circle exactly once.) Yet, when we talk about a curve, such as a polynomial, I think one must talk about infinity to give a rigorous definition. This is not a bad thing... It can help motivate a lesson on the formal definition of a tangent line at a point... But, what intermediate definition could I use to help those students who have no intuition about the matter get an idea of what we are going for before we talk about secants and such?

I have looked at some textbooks and online and I find saying a tangent line is one that "just touches" problematic-- it is the kind of phrase that is quite meaningless... Even in the setting of un-rigorous definitions. Yes, I will give examples, but I would like to do better ... I think saying that a tangent line is an arrow that points in the direction that the curve is going at that instant might make sense... Though it makes everything 'directional' and that might confuse them later.

What is the best intermediate definition you have seen?

A tangent is a line that intersects the curve once, at least if it is made short enough, but there's a direction such that if you rotate the line about the intersection point in that direction, no matter how little you rotate, it'll hit the curve again. This definition apparently goes back to Euclid in some form, and it works at inflection points, but fails in dimensions higher than $2$.  (I assume you're only talking about plane curves.)  I think we need the derivative to be continuous for it to work, and we also need it not to be constant in a neighborhood of the intersection point.

I don't see what's wrong with giving a physical definition though: it's where a particle moving along the curve would go if there were suddenly no forces acting on it (Newton's first law).  Or, even more physically, it's the line a ball would begin to travel in if you threw the ball in an arc corresponding to the curve and let it go at the point you're interested in.  This definition has the advantage that it does not depend on the background mathematics: it is (to a first approximation) an empirical fact about the universe that this concept is consistent.

according to me tangent is a line perpendicular to the radious of curvature of the curve at a given point.


$\lim_{x\to0}\frac{e^x-1-x}{x^2}$ using only rules of algebra of limits. I would like to solve that limit  solved using only rules of algebra of limits.

$$\lim_{x\to0}\frac{e^x-1-x}{x^2}$$

All the answers in How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion? do not fully address my question.

A challenging limit problem for the level of student who knows that:
$$\begin{align*}
\lim\limits_{x\to +\infty} e^x&=+\infty\tag1\\
\lim\limits_{x\to -\infty} e^x&=0\tag2\\
\lim\limits_{x\to +\infty} \frac{e^x}{x^n}&=+\infty\tag3\\
\lim\limits_{x\to -\infty} x^ne^x&=0\tag4\\
\lim\limits_{x\to 0} \frac{e^x-1}{x}&=1\tag5
\end{align*}$$

$$
\lim\limits_{x\to 0}\dfrac{e^x-1-x}{x^2} = \lim\limits_{x\to 0}\left(\dfrac{1}{e^x+1+x}\right)\lim\limits_{x\to 0}\dfrac{e^{2x}-(1+x)^2}{x^2} = \dfrac{1}{2}\lim\limits_{x\to 0}\left(4\dfrac{e^{2x}-1-2x}{4x^2}-1\right) = 2\lim\limits_{x\to 0}\left(\dfrac{e^{2x}-1-2x}{4x^2}\right)-\dfrac{1}{2} = 2\lim\limits_{x\to 0}\left(\dfrac{e^x-1-x}{x^2}\right)-\dfrac{1}{2}.
$$

Therefore,

  $$\lim\limits_{x\to 0}\dfrac{e^x-1-x}{x^2} = \dfrac{1}{2}\,.$$

This can be extended in a natural way. Define $s_{n}(x):=\sum\limits_{k=0}^{n-1}\dfrac{x^k}{k!}$ and observe the following:
$$
\begin{eqnarray*}
\lim_{x\to0}\dfrac{e^x-s_{n}(x)}{x^n} &=& \lim_{x\to0}\left(\dfrac{1}{\sum\limits_{r=0}^{n-1}e^{rx}(s_n(x))^{n-1-r}}\right)\lim_{x\to0}\dfrac{e^{nx}-(s_n(x))^n}{x^n} \\ &&\ \\ \ \\ &=& \dfrac{1}{n}\left(\lim_{x\to0}\dfrac{e^{nx}-s_n(nx)-\frac{n^{n-1}-1}{(n-1)!}x^n-o(x^n)}{x^n}\right)\\ &&\ \\ \ \\&=&
n^{n-1}\lim_{x\to0}\left(\dfrac{e^{nx}-s_n(nx)}{(nx)^n}\right)-\frac{n^{n-1}-1}{n!}\,,
\end{eqnarray*}
$$

where, upon expansion, the polynomial $(s_n(x))^n = s_n(nx) + \frac{n^{n-1}-1}{(n-1)!}x^n + o(x^n)$. Consequently,

  $$
\lim_{x\to0}\dfrac{e^x-s_n(x)}{x^n} = \dfrac{1}{n!}\,.
$$

The foregoing arguments require the existence of the primary limits being sought after. Given that our hypothetical pupil is not armed with sophisticated $\epsilon, \delta$ analysis techniques, it is difficult to see how he might proceed in showing existence. Indeed, how does this pupil even know what a limit is, let alone the algebraic laws that follow by definition? Nevertheless, with these concerns in mind, we might at least allow our student one more atom of information: the interchange of limits for the particular class of functions we are interested in here. Thus, our pupil proceeds, somewhat mechanistically, as follows:

$$
\begin{eqnarray*}
\lim\limits_{x\to 0}\frac{e^{x}-s_n(x)}{x^n} &=& \lim\limits_{x\to 0}\lim\limits_{y\to 0}\frac{\left(e^{y}(\frac{e^{x}-1}{x})-\frac{1}{x}\Big(s_n(x+y)-s_n(y)\Big)\right)}{\Big(\sum\limits_{r=0}^{n-1}(x+y)^r y^{n-1-r}\Big)} \\ &&\ \\&=& \lim\limits_{y\to 0}\lim\limits_{x\to 0}\frac{\left(e^{y}(\frac{e^{x}-1}{x})-\frac{1}{x}\Big(s_n(x+y)-s_n(y)\Big)\right)}{\Big(\sum\limits_{r=0}^{n-1}(x+y)^r y^{n-1-r}\Big)} \\ &&\ \\&=& \frac{1}{n}\lim\limits_{y\to 0}\frac{\left(e^{y}-s_{n-1}(y)\right)}{y^{n-1}}\,.
\end{eqnarray*}$$

So, by successive application of such interchanges, our pupil deduces

  $$
\lim\limits_{x\to 0}\frac{e^{x}-s_n(x)}{x^n} = \frac{1}{n!}\,.
$$


Find the horizontal and vertical asymptotes of the curve: $\frac{(9 +x^4)} { (x^2-x^4)}=-1$ I may just be tired but I set up the limit for $\frac{(9 +x^4)}{ (x^2-x^4)}= -1$ I then factored the bottom and got $x=2$ and $x=-2$, however, the program I am using says that those are not the vertical asymptotes. Can someone give me a math jargon less explanation what I am doing wrong?

As x gets suuuper duper huge, what is going to happen? Well, let's try it with 1,000,000. Then 1,000,000,000. Then 1 x 10^12. Etc. Essentially, what you will find is that it approaches -1. That is because when x gets enormous, all that is really going to matter is the x^4 term, because it will DWARF the other terms.

We notice that there is a positive and negative x^4, so the horizontal asymptote when x is super big in the positive direction is going to be -1. The same goes for the negative direction.

The vertical asymptotes are wherever the equation is not defined for a value of x. This is the same as saying where the equation is 0 in the denominator. So, you just need to find the values of x which make the denominator 0 and the function undefined. To do this, you find the solutions to the equation in the denominator, aka the solutions to x^2 - x^4. This is because the solutions to the denominator make the denominator 0, which makes the function as a whole undefined. Hint: there are three of them, all different.

The program as a whole is interpreting y as the one horizontal asymptote and x as the three vertical asymptotes. so when you find the vertical asymptotes, put a different value of x in for each one.

Idk how you got 2 or -2; when you put them into the function, you get 25/12, a totally normal point. If you get totally stuck, desmos it.


Is mathematics considered a science Possible Duplicate:
  what is the definition of Mathematics ?

I would like to know if mathematics is considered a science? I've searched the internet and asked many people for insight to no avail.  I thought science was based on observation, experimentation and falsification.  I don't understand how mathematics can satisfy all the above.  Especially, when there are some branches of mathematics that need not actually correspond to the physical world.  Maybe I have a naive definition or unsophisticated understanding of both mathematics and science.

Thank you in advance.

Sincerely,
Maxwell

You wrote: "I thought science was based on observation, experimentation and falsification". Isn´t that philosophy ? And what is good literature without observation; trying new forms of writing, that is experimenting; and adding some non existing facts and invented stories, that is falsifying ? Yet i do not understand why you define maths as in part (one third) a falsification. History, a story told by those who won, is very often false; but why do you think maths is?


Find the limit of $\lim_{x\to \infty}(\frac{x}{x})^x+(\frac{x-1}{x})^x+(\frac{x-2}{x})^x......+(\frac{1}{x})^x$ Find the limit of $\lim_{x\to \infty}(\frac{x}{x})^x+(\frac{x-1}{x})^x+(\frac{x-2}{x})^x......+(\frac{1}{x})^x$

$\lim_{x\to \infty}(\frac{x}{x})^x+(\frac{x-1}{x})^x+(\frac{x-2}{x})^x......+(\frac{1}{x})^x$
$=\lim_{x\to \infty}\frac{1^x+2^x+3^x+.....+x^x}{x^x}$
This is in $\frac{\infty}{\infty}$ form ,so i applied L hospital rule.
$=\lim_{x\to \infty}\frac{2^x\log 2+3^x\log 3+.....+x^x(1+\log x)}{x^x(1+\log x)}$
But i am stuck here and could not solve further.Please help.

For a fixed $m$:

$$\lim_{x\to\infty}\left(\frac{x-m}{x}\right)^x= e^{-m}\tag{1}$$

So if the series converges at all, it must converge to a value at least as big as:

$$\sum_{m=0}^\infty e^{-m} = \frac{1}{1-e^{-1}}=\frac{e}{e-1}$$

Letting $f_m(x)=\left(\frac{x-m}{x}\right)^x$, then define $g_m(x)=\ln f_m(x)=x\left(\log(x-m)-\log(x)\right)$, and you get:

$$g_n'(x)=\log(x-m)-\log(x) + \frac{x}{x-m} - 1=\log\left(1-\frac{m}{x}\right) +\dfrac{\frac{m}{x}}{1-\frac mx}$$

You can use the power series for $\frac{1}{1-z}$ and $\log(1-z)$ to show that this value is positive, and hence $g_m$ is increasing for $x>m$, and this that the above series converges to the hoped-for value.

This shows that $F(x)=\sum_{0}^x f_m(x)$ is strictly increasing, and thus we have:

$$\lim_x F(x)=\sup_x F(x)\leq \sum_{m} e^{-m}$$
And also have shown above that $$\sup_x F(x)\geq \sum_{m} e^{-m}$$

So the limit is equal to $\frac{e}{e-1}$.

Aside: If you have a sequence of sequences, $a_{n,k}$, and for each $k$, you have $\lim_{n\to\infty} a_{n,k}=A_k$, it is not always true that:

$$\lim_{n\to \infty}\sum_{k=0}^\infty a_{n,k} = \sum_{k=0}^\infty A_k$$

A simple example where it isn't true is: $$a_{n,k}=\delta_{n,k}=\begin{cases}1&n=k\\0&n\neq k\end{cases}$$

Then for each $k$, $\lim_{n\to\infty} a_{n,k}=0$, and for all $n$, $\sum_{k=0}^{\infty}a_{n,k}=1$, so the limit of the sum is not the sum of the the limits.

If, for each $k$, however, $a_{1,k},a_{2,k},\dots$ is increasing, then the limit of the sums is the sum of the limit.

When x tends to infinity, the first term containing x/x is obviously equal to one. Rest all terms are of the form, "something between zero and 1 raised to power infinity". So all other terms are going to be zero.
Therefore answer should be just '1'.


The degree of the extension $F(a,b)$, if the degrees of $F(a)$ and $F(b)$ are relatively primes. Let $E$ be an extension of $F$, and let $a, b \in E$ be algebraic over $F$. Suppose that the extensions $F(a)$ and $F(b)$ of $F$ are of degrees $m$ and $n$, respectively, where $(m,n)=1$. Show that $[F(a,b):F]=mn$.

Since $[F(a,b):F]=[F(a,b):F(a)][F(a):F]$ and $[F(a):F]=n$ we have $n|[F(a,b):F]$
with the same argument we prove that $m|[F(a,b):F]$, then $mn|[F(a,b):F]$ and $mn \le [F(a,b):F]$.

My problem is with the converse, I need help.

Thank you

You should be able to prove the degree of $F(a,b)$ over $F(a)$ is at most the degree of $F(b)$ over $F$, and what you want follows from that.

Note that $F(a,b)=(F(a))(b)$. Take a basis for $F(a)/F$ and for
$F(b)/F$. What is a spanning set for $(F(a))(b)/F$?


Asymptotics of a coupled sequence In this paper, the authors make the passing remark "simple analysis reveals that" the coupled sequence

$\mu_k = \theta_k \left(\theta_{k-1}^{-1} - 1\right)$

$\theta_{k+1} = \frac{\sqrt{\theta_k^4 + 4\theta_k^2} - \theta_k^2}{2}$

is such that $\mu_k$ asymptotically equals $1 - \frac{3}{k} + \mathcal{O}(1/k^2)$, but I have been unable to prove this. I was wondering if anyone could point me in the right direction as to how I would show something like this?

This question is very similar to MSE question 2861768 and several
other questions.
The idea from these questions is to use the Stolz theorem.

For simplicity, define
$\,f(x) := (\sqrt{4x^2+x^4}-x^2)/2.\,$ Define a sequence
$\,a_{n+1} = f(a_n)\,$ where $\,a_1>0.\,$
Next, define
 $$ b_n := 1/a_n, \quad\text{and} \quad
  e(x) := \frac12 + x\sqrt{1+\frac1{4x^2}} \tag{1}$$
where
$\, b_{n+1} = e(b_n)\,$ and
$\,\lim_{ x\to \infty} e(x)-x = \frac12.\,$ By the Stolz theorem
we get $\, b_n = \frac{n}2 + O(1)\,$ and thus,
$\, a_n = \frac2{n} + O(n^{-2}).$

The inverse function of $\,f(x)\,$ is $\,f^{(-1)}(x) = x/\sqrt{1-x},\,$ and we get

$$ \mu_n := a_n\!\left(\!\frac1{a_{n-1}}
   \!-\! 1 \!\right)\!=\!
   a_n\!\left(\!\frac{\sqrt{1\!-\!a_n}}{a_n}
   \!-\! 1\!\right) \!=\! 1 \!-\! \frac32 a_n
   \!+\! O(a_n^2) \tag{2}$$
but since $\,a_n = 2/n + O(n^{-2})\,$ now
$\, \mu_n = 1 - 3/n + O(n^{-2}) \,$ which is what we wanted to prove.

ADVANCED results (only if you are interested):

Define $\,x := 1/n, \, y := \ln(x).\,$
For some constant $\,c\,$ only depending on $\,a_1,\,$
for example, if $\,a_1=1\,$ then $\,c \approx -1.2929418036,\,$
we get an expansion of $\,a_n = g(1/n) = h(2n-c-\ln(2))\,$ where $$  g(x) \approx
 2\,x + (c+y)\,x^2 +((c+c^2) + (1+2c)y + y^2)\,x^3/2 + O(x^4) \tag{3}$$
and
$$ h(x) \approx 4 / (x + \ln(x + \ln(x + \frac16 + \ln(x + \frac{29}{72} + \ln(x + \frac{437}{640} + \dots))))). \tag{4}$$

The expansion $(3)$ implies that
$$ a_{n+1}(1/a_n - 1) = 1 - 3\,x + (-1-3c-3y)\,x^2/2 + O(x^3). \tag{5}$$

Notice that $\,\frac1{n+1}=\frac{x}{x+1}\,$ and thus the recursion for $\,a_n\,$ is equivalent to $\,g(\frac{x}{1+x}) = f(g(x))\,$ and from which the coefficients of the series expansion of $\,g(x)\,$ can be solved for term by term.

MSE question 3215 has an answer to the question of the convergence rate of $\,a_n\,$ for general $\,f(x).\,$


Improving basic maths & algebraic manipulation skills I am currently doing a-levels in the UK (received an A in AS) so doing calculus (differential equations/trig integrals)/linear algebra/mechanics/statistics etc.

However I feel that a lot of my basic mathematics skills are almost non-existent (don't know how I made it this far).

Can anyone recommend any good books that provide examples and exercises to help me improve this vital skill (or set of skills)?

Thanks

It is not an easy question to answer as everyone is different and learns differently and it is not clear if you want the theory versus applied path.

However, here are some suggestions for your consideration.

First and foremost, read, study, do problems, think about theorems from various perspectives, question, explore, practice, practice and then practice!

Many questions along these lines have been asked and you should certainly review these.

MSE

how to be good at proving? (see proof books in my response)
Filling the gap in knowledge of algebra
Self-teaching myself math from pre-calc and beyond.
Self-Studying Abstract Algebra; Aye or Nay?
Motivation and methods for self-study
Complete course of self-study
How to self study Linear Algebra
Rigorous Text in Multivariable Calculus and Linear Algebra
A rigorous book on a First Course in linear algebra
$\ldots$

Web Resources

Look to the web for notes.
Look at the curriculum at various colleges and look at topical areas, syllabus, books, homeworks, et. al.
Khan Academy
Open Courseware Consortium (for example MIT)

Additional Items

Learn some programming using a Computer Algebra System

I recommend Frank Ayres First Year College Mathematics (Schaum's Outline).  Get one of the old editions (hard copy, not a scan).  Quite cheap on Amazon.

This book covers all of basic algebra, trig, analytic geometry, and has precalculus.  Since you have been exposed to the material before (but didn't learn it well), this sort of review book is perfect.  Lots of problems.  Text is very clear, simple and short.  Good examples.  And answers to all the exercises.  Just drill through that and you will solve your issue.

I was disagree with and am perplexed with several people recommending books on proof or the like.  Non responsive to what the person asked for.


Figuring out how many revolutions a bicycle tire rotates for every pedal revolution.. I'm trying to figure out this problem on my assignment, goes as follows:

A bicycle works using two gears and a chain. Turning the pedals causes the front gear to rotate and the chain then causes the rear gear to rotate which cause the bike tire to turn. The bike has a front gear with radius of 4 inches and the rear gear has radius 2 inches.

Determine the number of revolutions that the bicycle tire rotates when the pedals are rotated one revolution.

I truly have no clue.. Could someone help me out?

Any help or even clues are appreciated.

Thank you.

If you pedal one full circle, the chain moves by $2 \pi r$ ($r$ is pedal circle radius). The same length will go to the wheels (radius $R$): hence,
$$2\pi r = R \phi,\tag{1}$$
where $\phi$ is the angle (in radians) the cycle wheels rotate. From $(1)$, you get the angle the wheels rotate.

When front gear makes 1 rotation the rear gear will rotate 2 times of that. Ratio of the circumference(2×pi×R/ 2xpi×r) = 4÷2=2.
Rear gear attached  to back wheel so same rotation as rear gear


Solve the the given equation: $\sqrt{3x^2+x+5} = x-3$ We have to find the number of solution for the given equation:
$$\sqrt{3x^2+x+5} = x-3.$$
There are two solution one is

By using graph we get one solution

By squaring both sides we get no solution

I want to know which solution is correct

Hint:

The given equation is equivalent to the system:
$$
\begin{cases}
3x^2+x+5=(x-3)^2\\
x-3\ge 0
\end{cases}
$$

Note that the system has no solutions because the roots of the second degree equations :
$$
x=\frac{-7\pm\sqrt{79}}{4}
$$
are less than $3$.

And this is in accord with the fact that the graphs of the two functions
$$
y=\sqrt{3x^2+x+5} \qquad y=x-3
$$
have no common points (in your graph you  have the wrong function $x=3$).

x=-4
3*16-4+5=49
(-3-4)(-3-4)=49
So this equation was wrong first place.


What is the best book for studying discrete mathematics? As a programmer, mathematics is important basic knowledge to study some topics, especially Algorithms. Many websites, and my fellows suggest me to study Discrete Mathematics before going to Algorithms, so I want to know which Discrete Mathematics book is suitable for my needs?

Concrete Mathematics: A Foundation for Computer Science, By Donald Knuth himself!

Fundamentals of Discrete mathematical structures, 3rd Edition. It is written as per ACM-Curriculum, comprises lot of GATE level questions, and written by a Computer science Professor.


Symbolic predicate logic for "for all elements in a set except this one..." How would a statement beginning, "For all $x$ in set $S$ except $x=a$..." be translated into symbolic predicate logic? I'm somewhat of a purist in symbols and am less than satisfied with "$\forall x \in S$ except $x = a$..." or "$\forall x \in S$ where $x \ne a$...".

$$\forall x \in S\setminus \{a\}, \,\,\,\ldots$$

I would use $(\exists a\in S)(\forall x\in S)(x\neq a \Leftrightarrow \cdots)$.  It's not a direct correspondence to the English, but not everything in symbolic logic is.


How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that

$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$

for every integer $n > 0$.

I would like to know if there is another way of proving this result without using PMI. Is there any geometric solution to prove this problem? Also, are there examples of problems where only PMI is our only option?

Here is the way I have solved this using PMI.

Base Case: since $1 = 2 · 1^2 − 1$, the formula holds for $n = 1$.

Assuming that the
formula holds for some integer $k ≥ 1$, that is,

$$1 + 5 + 9 + \dots + (4k − 3) = 2k^2 − k$$

I show that

$$1 + 5 + 9 + \dots + [4(k + 1) − 3] = 2(k + 1)^2 − (k + 1).$$

Now if I use hypothesis I observe.

$$
\begin{align}
1 + 5 + 9 + \dots + [4(k + 1) − 3]
& = [1 + 5 + 9 + \dots + (4k − 3)] + 4(k + 1) −3 \\
& = (2k^2 − k) + (4k + 1) \\
& = 2k^2 + 3k + 1 \\
& = 2(k + 1)^2 − (k + 1)
\end{align}
$$

$\diamond$

Follow the rainbow!

$$ \Large \color{PaleVioletRed}1 + \color{DarkViolet}5 + \color{DodgerBlue}9 + \dots + \color{LightCoral}{(4n-3)} = n(2n-1) = 2n^2-n$$

One way using distributive laws and multiplication:

$\sum_{k=1}^n(4k-3)$

$=\sum_{k=1}^n(4k)+\sum_{k=1}^n(-3)$

$\sum_{k=1}^n(-3)=-3n$

$\sum_{k=1}^n(4k)=4\cdot\sum_{k=1}^n(k)$

Assuming n is even:

$\sum_{k=1}^n(k)=1+2+3+4...+n=(1+n)+(2+n-1)...+(\frac{n}{2}+\frac{n}{2}+1)=\frac{n(1+n)}{2}$

Assuming n is odd:
$\sum_{k=1}^n(k)=n+\sum_{k=1}^{n-1}k=\frac{2n}{2}+\frac{(n-1)(n)}{2}=\frac{n(1+n)}{2}$

Therefore it works for all n

Putting all of these together:$\sum_{k=1}^n(4k-3)=4\cdot\sum_{k=1}^n(k)+\sum_{k=1}^n(-3)=\frac{4n(1+n)}{2}-3n=2n^2+2n-3n=2n^2-n$
Q.E.D.
Or you could just know the summation identities and easily proof it.


Log of Quadratic Form - Mathematical Programming Reformulations Suppose I have the quadratic form

$$(x-\mu)^{T}Q(x-\mu)$$

where $Q$ is a positive semi-definite matrix. This form achieves a minimum at $x=\mu$.

Suppose I want to minimize instead

$$\log \left(1 +(x-\mu)^{T}Q(x-\mu) \right)$$

By monotonicity of the $\log$ function this expression, non-convex, also achcieves a minimum at $x=\mu$.

Suppose I didn't know the above and now wanted to solve this via mathematical programming

$$\text{Minimize}_{x\in\mathbb{R}}\quad \log \left(1+(x-\mu)^{T}Q(x-\mu) \right)$$

I could reformulate the problem  in the following way

$$\text{Minimize}_{x\in\mathbb{R},z\in\mathbb{R}}\quad \log \left(z \right)$$
$$\text{s.t.} \quad z\geq 1+(x-\mu)^{T}Q(x-\mu)$$
$$\text{   } \quad z\leq 1+(x-\mu)^{T}Q(x-\mu)$$

The last constraint is not a convex quadratic constraint because the quadratic is bounded from below by $z$.

My question is, isn't this last constraint redundant? Can't I throw it away? My intuition tells me that in order to minimize $\log(z)$ we need to make $z$ as small as possible. $z$ is bounded below by the quadratic form, so we need to make the quadratic form as small as possible. This alone, without the last constraint, would imply that $z=(x-\mu)^{T}Q(x-\mu)$

If I can do away with that last constraint I can do a piecewise linear approximation to the logarithm and solve the problem as a MIQCP.

If I am correct, how can I formally argue the reformulation using only the first quadratic constraint?

I'm not sure that "equivalence" between optimization models is all that well defined. If we think of a problem instance as the combination of a model (with symbolic parameters) and values for the parameters (data), then one might say two models are equivalent for a problem if, given the same data, any optimal solution to either model is also optimal in the other.

Using that definition, let (P) denote the final model (with both inequalities) and (P') the model with the second inequality removed. To show (P') is equivalent to (P), you don't need to show the feasible regions are the same, just that any optimal solution to (P) is optimal in (P') and vice versa. It's easy to prove that $(x, z)$ optimal in (P') implies $z = (x-\mu)^TQ(x-\mu)$, after which the equivalence proof is pretty trivial.

The equivalent is

$$
\begin{cases}
z^*=\text{argmin} \{\log(z)\} \\
\text{st.}\quad z\ge1
\end{cases}
$$

Which does not look like a multi-variable programming anymore.

The solution is $z=1$, which means

$$z=1 +(x-\mu)^{T}Q(x-\mu)=1$$

or

$$x=\mu$$


Find the limit of function $f_{n}(x) = nx^{n}$ Show that $\lim_{n \rightarrow \infty} nx^{n} = 0$ for $x \in [0, 1), n \in \mathbb{N}$

$\lim_{n \rightarrow \infty} nx^{n}$

$\Rightarrow \lim_{n \rightarrow \infty} n \cdot \lim_{n \rightarrow \infty} x^{n}$

$\Rightarrow \infty * 0$

$\Rightarrow 0$

Is this correct?

Now show that $\lim_{n \rightarrow \infty} \int_{0}^{1} f_{n}(x) dx = 1$

$\lim_{n \rightarrow \infty} \int_{0}^{1} nx^{n} dx$

$\Rightarrow \lim_{n \rightarrow \infty} \frac{n}{n + 1} x^{n + 1}$

(then plug in $1$ and $0$ into $x$)

$\Rightarrow \lim_{n \rightarrow \infty} \frac{1}{2}x^2$

Which is not $=1$, where did I go wrong?

As others have noted, even if $0\cdot\infty$ was defined (and it isn't), you still couldn't just break up the product inside the limit. Instead, you could apply L'Hoptial's rule to get:
$$\lim_{n\to\infty}\frac{n}{x^{-n}} = \lim_{n\to\infty}\frac{1}{-x^{-n}\log x}=0$$

$$
\lim_{n\rightarrow \infty}\left(nx^n \right)=
\lim_{n\rightarrow \infty}\left(n \over x^{-n} \right)=
\lim_{n\rightarrow \infty}\left(1 \over (-n)x^{-n-1} \right)=
\lim_{n\rightarrow \infty}\left(-x^{n+1} \over n \right)=0\;  (for\;\; 0\leq x<1)
$$


Solving the integral $\int \sqrt{C^2 - \frac{x^2}{1+x^2}}dx$ Let $C$ be a real constant, $C\neq0$. I am trying to solve the following integral:
$$
\int \sqrt{C^2 - \frac{x^2}{1+x^2}}dx
$$
If $C = \pm 1$, I known how to solve it (by using the substitution x=sinh(u)).

Thanks in advance.

Use the substitution $x = \tan \theta$ which after differentiating becomes $ \mathrm{d}x = \sec\theta \mathrm{d}\theta $. The integral now becomes:
$$ \int\sqrt{ C^2 - \frac{x^2}{x^2+1}}\mathrm{d}x = \int \sqrt{C^2 - \frac{\tan^2\theta}{1 + \tan^2\theta}}\sec\theta \mathrm{d}\theta = \int\frac{\sqrt{C^2-\sin^2\theta}}{\cos\theta} \mathrm{d}\theta$$
This form of the integral can be now solved but comming to the solution is rather tideous so here is a link of a site https://www.integral-calculator.com/ which will solve this integral in a step-by-step solution with only elementary functions in the final solution. If I manage to find a nicer solution I will post it.


What can this differential equation be used to model? So, I can model growth and decay if I start with assuming that the growth rate is constant:

$\frac{p'(t)}{p(t)}=\alpha$

and then I have

$p'(t)-\alpha p(t)=0$

A general linear differential equation, however, would have the form

$p'(t)-g(t) p(t)=h(t)$

So the growth rate is g(t).  What is h(t)?
And what type of thing is this form of equation used to model?

This question has been edited.

You can read it like this: $p'(t)= \alpha p(t)+g(t)$. Then the growth is "caused" by "reproduction" plus "immigration", for example.

I know you asked about first order, but: Second order homogeneous ODEs, e.g. $a\ddot{x} + b\dot{x} + cx = 0$, describe "unmolested" damped/driven harmonic oscillation. Introducing a function on the right hand side, e.g. $a\ddot{x} + b\dot{x} + cx = f(t)$, describes damped/driven harmonic oscillation with $f$ describing an external input. The part of the ODE which does not involve a dependent variable describes "outside influences". I am quite confident that this is the same for any order of ODE.


IGCSE/GCSE mathematics worksheet and resources I take mathematics for Cambdridge IGCSE students, Syllabus code 0580, http://www.cie.org.uk/docs/dynamic/41073.pdf ,  Please i want to know about resources available to revise the math topics,  and if there are any websites that provides the  worksheets and  lesson plans IGCSE or GCSE syllabus.

Thanks,
Arif

Try this link below it has some topics and more everyday.
https://nerdherdcorner.com/gcse-revision/maths-revision/


How many possible color combinations? I have a unusual shape like this.

I want to color its squares with 3 colors which no two adjacent(in vertical or horizontal) squares take the same color.

How should I solve such problems? Please give me a clear solution for given shape.

Thanks in advance :)

If you omit the 4 squares that are connected to only one other square, you are left with a rectangular 3x4 grid.

The number of 3-colourings of rectangular grids (divided by 3) is given in integer sequence A078099:

So, for the 3x4 grid, the answer is: $ 3 T(3,4) = 3 * 374 = 1122$

Now for the 4 extra squares, it's easy to see that for any of them, there are exactly 2 possibilities. So, the final answer for the specific grid is:

$ 2^4  1122 = 17952 $

overlay the pattern with a chess board to see how you can do it with two colours. Then in order to use three colours repaint one of the squares.


Given a Pell solution $(u_k,v_k)$, is there a closed form "descent" to $(u_{k-1},v_{k-1})$? Given: a solution $(u_k,v_k)$ to the Pell equation
$$U^2-dV^2=1, \qquad(\star)$$
where $d$ is a non-square integer, and $k \ge 1$ is an arbitrary integer.

There are well-known recurrences to ascend to $(u_{k+1}, v_{k+1})$, given also the fundamental solution $(u_1,v_1)$. I'm wondering if there's a closed form “reverse recurrence” which gives $(u_{k-1},v_{k-1})$, without knowing $(u_1,v_1)$ explicitly.

As a concrete numerical example, when $d=2$, we can compute $v_{k-1} = u_k-v_k$, and then compute $u_{k-1}=v_k-v_{k-1}$, which gives a solution to the negative version of $(\star)$; applied twice, you get back to the next smaller solution for the positive version. It's this kind of “descent”, in a general case, that I'm interested in.

The answer to your question is no. Knowledge of how to perform your descent is equivalent to knowledge of the fundamental solution. Given $d > 0$ not a square, the previous solution is given by multiplying the column vector by
$$
\left(
\begin{array}{rr}
s & -d t \\
-t & s
\end{array}
\right)
$$
where $$ s^2 - d t^2 = 1  $$
is the fundamental solution.
Given $(x,y)$ that solves $x^2 - d y^2 = 1,$ the previous solution is
$$ (s x - d t y, \; \;  -t x + s y); $$ but the next solution is
$$ (s x + d t y, \; \;  t x + s y); $$

Meanwhile, the matrix (either) has trace $2s$ and determinant $1.$ If you know two consecutive solutions, Cayley-Hamilton says
$$ x_{n+2} = 2 \, s \;  x_{n+1} - x_n,  $$
$$ y_{n+2} = 2 \, s \; y_{n+1} - y_n.  $$
Once again, knowledge of $2s$ tells you the fundamental solution.

If we have any solution of the equation Pell: $$x^2-dy^2=1$$
   Having a form: $(x_1;y_1)$    The following solution we can always be obtained by the formula:  $$x_2=x_0x_1+dy_0y_1$$   $$y_2=y_0x_1+x_0y_1$$    This formula gives all solutions. In order to choose the direction of growth is necessary to change the sign of the substitution solution. Instead " $+$ " write " $-$ "

$(x_0;y_0)$ - the first solution of the Pell equation.


Is there a generic function to find the area of any shape? Given the coordinates of all points making up a two-dimensional shape and the order in which they are connected, is there any generic formula that will give the area of that shape?

EDIT: The list of points would be a list of all of the vertices.

If you mean arbitrary sets of points in the plane, the answer is probably "no" unless you regard the defintion of Lebesgue measure as such a formula.  However, you write about "the order in which they are connected" and that makes me suspect you may have in mind a sequence of points on the boundary with line segments between them as components of the boundary.  In that case, maybe the shoelace formula will help you.

And if you mean polygons whose corners are integer points, there is Pick's theorem.

I believe so, using shoelace theorem. If you line up the co-ordinates starting anywhere but always going counter-clockwise and ending with the first co-ordinate you can multiply the previous x value and after y value and then subtract vice versa and divide by 2. Just search it up....


how to show $c=0$,$f:(0,\infty)\rightarrow\mathbb{R}$ is differentiable, $\lim_{x\rightarrow\infty}f(x)=1$, $f:(0,\infty)\rightarrow\mathbb{R}$ is differentiable, $\lim_{x\rightarrow\infty}f(x)=1$, $\lim_{x\rightarrow\infty}f'(x)=c$   we need to show $c=0$

well, I tried like this $|f(x)-1|<\epsilon\forall x>M$, where $M$ is very large, $|f'(x)-c|<\epsilon\forall x>M$,  what more I can say?thank you.

Since $\lim_{x \to \infty} \frac{f'(x)}{1}$ exists, by L'Hospital

$$\lim_{x \to \infty} \frac{f(x)}{x}=\lim_{x \to \infty} \frac{f'(x)}{1}=c \,.$$

But the first limit is $0$.

Alternately Use the MVT on the interval $[x,2x]$. You have

$$\frac{f(2x)-f(x)}{x}=f'(c_x)$$

Now, when $x \to \infty$ we have $f'(c_x) \to f(c)$ and the LHS gos to $0$.

If c isn't 0 then f is increasing/decreasing and can't converge to a limit.


Simple conditional probability problem Question states: Consider two events A and B in a sample space S.
• Assume that Pr(A) = 1/2 and Pr(B|A') = 3/5. Determine Pr(A∪B).

Not quite sure how to answers this question. Here's how I started:

So Pr(A) = 1/2 so Pr(A') = 1/2.

Then Pr(B|A) = 3/5.

Pr(A∩B) = Pr(A) * Pr(B|A) = 1/2 * 3/5 = 0.3

Pr(A∩B) = 0.3

But how do I get (A∪B)?

Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B)

but I don't know the Pr(B)? What am I missing here? Thanks for any insight.


Two ways of saying span Suppose $V$ is a vector space, and $v_1,...,v_n$ spans $V$. Is that equivalent to saying that $V=\operatorname{span}\{v_1,...,v_n\}$?

Yes, the two statements are equivalent.
Here is a definition quoted from Friedberg:

A subset S of a vector space V generates (or spans) V  if span(S) = V. In this case, we also say that the vectors of S generate (or  span) V.

Friedberg, Stephen H.. Linear Algebra (p. 31). Pearson Education. Kindle Edition.

You run into issues if you try to define what it means to span in another way. For example, we commonly define a basis as follows:

A basis of a vector space V is a set $(v_1,...,v_n)$ of vectors that is independent and also spans V.

Artin, Michael. Algebra (Page 88). Pearson Education. Kindle Edition.
But now consider that $S = \{(1,0), (0,1) \}$ is an independent set and spans $\mathbb{R}^2$.
If we only require that $V \subseteq \text{Span } S$ in order for $S$ to span $V$, then we get that $S$ is an independent set that spans every subspace of $\mathbb{R}^2$, and so is a basis for every subspace of $\mathbb{R}^2$, which is not correct.

Saying that $V$ is spanned by $v_1,v_2,...,v_n$ means that every vector in $V$ can be written as a linear combination of $v_i$. Hence $V$ is a subspace of $\text{span}\{v_1,...,v_n\}$.
For example, $V=\{(k,0):k\in\mathbb R\}$ is spanned by $(1,0),(0,1)$ over the field of real numbers. But, $V\subset\text{span}\{v_i\}=\mathbb R^2$.

Note: In case you add the condition $v_1,...,v_n\in V$, then $\text{span}\{v_i\}\subseteq V$ by closure property of $V$. Then it can be concluded that $V=\text{span}\{v_i\}$.


Explanation of trivial zeros of the Riemann Zeta Function Why do negative even numbers plugged into the Zeta function produce a zero? The Riemann Hypothesis implies that the non-trivial zeros are connected to the primes, so how does that fit with negative even numbers?

The functional equation of the zeta function, needed to extend anatically that function to $\;\Bbb C\setminus\{1\}\;$ and one of the most astonishingly beautiful equations in mathematics, is

$$\zeta(s)=2^s\pi^{s-1}\,\sin\frac{\pi s}2\,\Gamma(1-s)\,\zeta(1-s)$$

Well, now for $\;s=-2n\;,\;\;n\in\Bbb N\;$ , you get $\;\zeta(-2n)=0\;$ ...

For example, the zeta function at $-2$ is $1^2 + 2^2 + 3^2 + ...$ which is trivially equal to 0.

Similarly, the zeta function at $-4$ is $1^4 + 2^4 + 3^4 + ... $ which is also trivially equal to zero.

It is trivial to see that the same holds for the function at $-6$, $-8$, ... etc. For that reason, these are considered the trivial zeros of the Zeta function.


The number $2^{2^n} + 2^{2^{n - 1}} + 1$ can be expressed as the product of at least $n$ prime factors Prove that the number $2^{2^n} + 2^{2^{n - 1}} + 1$ can be expressed as the product of at least $n$ prime factors, not necessarily distinct.

Doing what the hint has suggested, I have done the same where $n>1$, and took $\bmod 7$. I then found that $2^{2^n} + 2^{2^{n - 1}} + 1=7k$.

However, I got stuck when trying to simplify the case of $n=3$, since $$2^8+2^4+1=273=3\cdot 7\cdot 13$$

I do not understand where the 13 is from, since I thought that it would be something like $$2^{2^n} + 2^{2^{n - 1}} + 1 \equiv 0 \pmod {2^n-1}$$

We start with $x^4+x^2+1=\left(x^2+1\right)^2-x^2=\left(x^2-x+1\right)\left(x^2+x+1\right)$.  Thus, we can proceed by induction (on $n$) that $2^{2^n}+2^{2^{n-1}}+1$ has at least $n$ prime factors.  For $n=1$, there is nothing to prove.  Suppose the statement is true for $n=k$ for some $k\in\mathbb{N}$.  Then, for $n=k+1$,
$$2^{2^{k+1}}+2^{2^k}+1=\left(2^{2^k}-2^{2^{k-1}}+1\right)\left(2^{2^k}+2^{2^{k-1}}+1\right)\,.$$
By the induction hypothesis, $2^{2^k}+2^{2^{k-1}}+1$ has $k$ prime factors.  Furthermore, $2^{2^k}-2^{2^{k-1}}+1>1$ has at least one prime factor.  Hence, $2^{2^{k+1}}+2^{2^k}+1$ has at least $k+1$ prime factors.  We are done.

You could substitute $2^{2^{n-2}}$ for a letter, say $a$ and prove that $a^4+a^2+1$ is divisible by $a^2+a+1$ using polynomial long division.
More info here: https://en.m.wikipedia.org/wiki/Polynomial_long_division

This way, you can prove that there is another factor added for every increase in an $n$ value.


Can a function be increasing *at a point*? From what I understand we say that a function is increasing on an interval $I$ if
$$
x_1 < x_2 \quad\Rightarrow\quad f(x_1) < f(x_2).
$$
for all $x_1,x_2\in I$. I understand that some might call this strictly increasing and allow for equality when saying increasing.

We know that when a function is differentiable on the interval $I$, then if $f'(x) > 0$ on $I$ then $f$ is increasing on $I$. But here we have to be careful because if $f(x) = x^3$ then $f$ indeed is increasing on $\mathbb{R}$ even though $f'(0) = 0$. So we can't conclude that if $f$ is increasing, then $f'(x)$ must be positive on the interval.

Also, even though $f(x) = x^2$, then $f$ is increasing on $[0,\infty)$ even though $f'(0) = 0$.

My question is

  Does it make sense to talk about a function being increasing at a point?

The only definition I can think of making is to say that $f$ is increasing at a point $a$ if $f'(a) > 0$. But the problems with saying this is that then the function given by $f(x) = x^3$ is not increasing at $0$ even thought it is increasing on all of $\mathbb{R}$.

I am aware (from here) of the example with
$$
f(x) = \begin{cases}x^2\sin\left(\frac{1}{x}\right) + \frac{x}{2} & \text{if }x\neq 0 \\ 0 & \text{if } x = 0\end{cases}.
$$
This function is differentiable at $0$ with $f'(0) = \frac{1}{2} > 0$. For $x\neq 0$ we have $f'(x) = 2x\sin(1/x) - \cos(1/x) + 1/2$. And so for all $x_k = \frac{1}{2k\pi}$ we have $f'(\frac{1}{2k\pi}) = -\frac{1}{2} < 0$. That is, there is not interval around $0$ where $f$ is increasing. And so it shouldn't really make sense to say that $f$ is increasing at $0$.

One example I can think of where it maybe should make sense to take about a function being increasing at a point is when when we take the derivative of function $s(t)$ that gives the position of a particle. The derivative is then the velocity $v(t) = s'(t)$. And maybe this is just a language issue, but here it sounds reasonable saying that if $v(a)>0$ then the velocity is positive at time $t=a$ and so the rate of change is positive at $a$. This is almost saying that the position is increasing at that point.

Again, my question is just if it makes sense to talk about a function being increasing (or decreasing) at a point. If so, how is this defined?

A function $f$ is increasing at $x$ if $f(t)\gt f(x)$ for every $t\gt x$ close enough to $x$ and $f(t)\lt f(x)$ for every $t\lt x$ close enough to $x$. More rigorously, one asks that there exists $\varepsilon\gt0$ such that, for every $(t,s)$ such that $x-\varepsilon\lt t\lt x\lt s\lt x+\varepsilon$, $f(t)\lt f(x)\lt f(s)$.

No notion of differentiability is needed. Consider for example the function $f$ defined by $f(t)=2t$ for every rational $t$ and $f(t)=t$ for every irrational $t$. Then $f$ is increasing at $x=0$ and only at $x=0$ while $f$ is nowhere differentiable.

The truth is that it doesn't really make sense to say that a function is increasing or decreasing at point. However, due to culture and "bad lexicon,"
people usually say that a function is increasing at point when they really mean to say that the point is an interval in which the function is increasing. Most people who know math are aware of this, so they usually aren't bothered by this language. For students learning calculus for the first time this is unfortunate. But English speakers say many things that don't really make sense like driving on parkways and parking in driveways, etc. Also, you could try arguing with someone that a so-called combination lock should really be called a permutation lock. Please don't take everything you hear too seriously.


Why does $x^2+47y^2 = z^5$ involve solvable quintics? This is related to the post on $x^2+ny^2=z^k$. In response to my answer on,

$$x^2+47y^2 = z^3\tag1$$

where $z$ is not of form $p^2+nq^2$, Will Jagy provided one for,

$$x^2+47y^2 = z^5\tag2$$

$$ (14p^5 + 405p^4q + 3780p^3q^2 + 13410p^2q^3 + 11550pq^4 - 14647q^5)^2 + 47  ( p^5 - 270p^3q^2 - 2520p^2q^3 - 8115pq^4 - 8344q^5)^2 = (3p^2 + 28pq  + 81q^2)^5\tag3 $$

As noticed by Elaqqad, the cubic polynomials I used for $(1)$ involve the discriminant $d=-47$ and I assumed it would be same with the (irreducible) quintic polynomials used by Jagy for $(3)$. Then I wondered if they were solvable in radicals as well. (I knew Ramanujan played with a solvable quintic with $d=-47$.) It turns out they are.

This Magma calculator computes the Galois group and the command is:

  Z := Integers(); P < x > := PolynomialRing(Z); f := 14*x^5 + 405*x^4 + 3780*x^3 + 13410*x^2 + 11550*x - 14647; G, R := GaloisGroup(f); G;

Testing both polynomials, it shows the group has order 20 and hence is solvable.

  Q: Given $x^2+dy^2=z^k$ where $z\neq p^2+dq^2$, is it true that if

  $$\big(P_1(x)\big)^2+d\big(P_2(x)\big)^2=\big(P_3(x)\big)^k$$

  then the equations $P_1(x) = P_2(x) = P_3(x) = 0$ are solvable in radicals?

P.S. Or is this $5$th parameterization special only because the class number $h(-47) = 5$? One way to check would be to solve $x^2+47y^2 = z^\color{red}7$ analogous to $(3)$ (Will, care to oblige?) and see if it involves solvable septics.

I am not finding a short online summary of the composition of two binary quadratic forms, as done by Dirichlet. Also, the edition of Cox that I have has a typo, corrected in the second edition (2013), here it is.

Given $\gcd(a,a',B) = 1,$ define
$$ X = xz-Cyw,  $$
$$ Y = axw + a'yz + B yw,  $$ then
$$ (a x^2 + B xy + a'C y^2) (a' z^2 + B zw + aC w^2) = aa'X^2 + B XY + C Y^2   $$
which you ought to check!

Here are the binary forms (primitive) of discriminant $-284$

Discr  -284 = 2^2 * 71  class  number  7

 all
     284:  < 1, 0, 71>
     284:  < 3, -2, 24>
     284:  < 3, 2, 24>
     284:  < 5, -4, 15>
     284:  < 5, 4, 15>
     284:  < 8, -2, 9>
     284:  < 8, 2, 9>

The first few primes integrally represented by $3x^2 + 2xy+24y^2$ are
$$     3,     29,     89,    103,    109,    151,    157,    191, $$ and below, we show how to represent each $p^7$ once we have $x,y.$

compared with $-71$ primitive, where this time a form represents the prime $2$

Discr  -71 = 71  class  number  7

 all
      71:  < 1, 1, 18>
      71:  < 2, -1, 9>
      71:  < 2, 1, 9>
      71:  < 3, -1, 6>
      71:  < 3, 1, 6>
      71:  < 4, -3, 5>
      71:  < 4, 3, 5>

ummmm, $h(-71) = h(-284) = 7.$  Since $4 \cdot 3^7 - 284 = 92^2,$ the principal form is $\langle 1, 92, 2187\rangle.$ The class group is cyclic, everything is a power of  $\langle 3, 92, 729 \rangle$ under Dirichlet's version of Gauss composition. All I am doing is repeatedly multiplying by $3 x^2 + 92 xy + 729 y^2,$ the rules for composition eventually give the quadratic form $\langle 2187, 92, 1 \rangle$ with variables which are homogeneous degree seven in the original $x,y.$ Oh, any form that represents $1$ is $SL_2 \mathbb Z$ equivalent to the principal form. At the very end, I show how to write $t^2 + 71 z^2 = (3 x^2 + 92 xy + 729 y^2)^7. $ I am showing the whole gp-pari session, there is nothing difficult once we get that fortunate expression for the coefficients of a generator of the group.

a=3; a1=3; b=92; c=243; z = x; w = y;

zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;

? a=3; a1=3; b=92; c=243; z = x; w = y;
? zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? z
%3 = x^2 - 243*y^2
? w
%4 = 6*y*x + 92*y^2
?

a1 = 9; c = 81; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? a1 = 9; c = 81; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? z
%6 = x^3 - 729*y^2*x - 7452*y^3
? w
%7 = 27*y*x^2 + 828*y^2*x + 6277*y^3
?

a1 = 27; c = 27; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;

? a1 = 27; c = 27; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? z
%9 = x^4 - 1458*y^2*x^2 - 29808*y^3*x - 169479*y^4
? w
%10 = 108*y*x^3 + 4968*y^2*x^2 + 75324*y^3*x + 376280*y^4
?
?
a1 = 81; c = 9; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;

? a1 = 81; c = 9; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? z
%12 = x^5 - 2430*y^2*x^3 - 74520*y^3*x^2 - 847395*y^4*x - 3386520*y^5
? w
%13 = 405*y*x^4 + 24840*y^2*x^3 + 564930*y^3*x^2 + 5644200*y^4*x + 20889961*y^5
?

 a1 = 243; c = 3; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;

?  a1 = 243; c = 3; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? z
%15 = x^6 - 3645*y^2*x^4 - 149040*y^3*x^3 - 2542185*y^4*x^2 - 20319120*y^5*x - 62669883*y^6
? w
%16 = 1458*y*x^5 + 111780*y^2*x^4 + 3389580*y^3*x^3 + 50797800*y^4*x^2 + 376019298*y^5*x + 1098952052*y^6
?
?

 a1 = 729; c = 1; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;

?
?  a1 = 729; c = 1; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
?
? z
%18 = x^7 - 5103*y^2*x^5 - 260820*y^3*x^4 - 5931765*y^4*x^3 - 71116920*y^5*x^2 - 438689181*y^6*x - 1098952052*y^7
? w
%19 = 5103*y*x^6 + 469476*y^2*x^5 + 17795295*y^3*x^4 + 355584600*y^4*x^3 + 3948202629*y^5*x^2 + 23077993092*y^6*x + 55417244077*y^7
?

2187 * z^2 + 92 * z * w + w^2

( 3 * x^2 + 92 * x * y + 729 * y^2)^7

?
? 2187 * z^2 + 92 * z * w + w^2
%20 = 2187*x^14 + 469476*y*x^13 + 46911879*y^2*x^12 + 2892076488*y^3*x^11 + 122889105423*y^4*x^10 + 3807263630268*y^5*x^9 + 88688782583499*y^6*x^8 + 1578039270279536*y^7*x^7 + 21551374167790257*y^8*x^6 + 224815110103695132*y^9*x^5 + 1763324345027822661*y^10*x^4 + 10084047184857263688*y^11*x^3 + 39747900724268273397*y^12*x^2 + 96660945131267433924*y^13*x + 109418989131512359209*y^14
?
?
? ( 3 * x^2 + 92 * x * y + 729 * y^2)^7
%21 = 2187*x^14 + 469476*y*x^13 + 46911879*y^2*x^12 + 2892076488*y^3*x^11 + 122889105423*y^4*x^10 + 3807263630268*y^5*x^9 + 88688782583499*y^6*x^8 + 1578039270279536*y^7*x^7 + 21551374167790257*y^8*x^6 + 224815110103695132*y^9*x^5 + 1763324345027822661*y^10*x^4 + 10084047184857263688*y^11*x^3 + 39747900724268273397*y^12*x^2 + 96660945131267433924*y^13*x + 109418989131512359209*y^14
?
? 2187 * z^2 + 92 * z * w + w^2 - ( 3 * x^2 + 92 * x * y + 729 * y^2)^7
%22 = 0
?

t = w + 46 * z

t^2 + 71 * z^2

?
? t = w + 46 * z
%23 = 46*x^7 + 5103*y*x^6 + 234738*y^2*x^5 + 5797575*y^3*x^4 + 82723410*y^4*x^3 + 676824309*y^5*x^2 + 2898290766*y^6*x + 4865449685*y^7
?
?
?
? t^2 + 71 * z^2
%24 = 2187*x^14 + 469476*y*x^13 + 46911879*y^2*x^12 + 2892076488*y^3*x^11 + 122889105423*y^4*x^10 + 3807263630268*y^5*x^9 + 88688782583499*y^6*x^8 + 1578039270279536*y^7*x^7 + 21551374167790257*y^8*x^6 + 224815110103695132*y^9*x^5 + 1763324345027822661*y^10*x^4 + 10084047184857263688*y^11*x^3 + 39747900724268273397*y^12*x^2 + 96660945131267433924*y^13*x + 109418989131512359209*y^14
?
? t^2 + 71 * z^2  -  ( 3 * x^2 + 92 * x * y + 729 * y^2)^7
%25 = 0
?
?
? t
%26 = 46*x^7 + 5103*y*x^6 + 234738*y^2*x^5 + 5797575*y^3*x^4 + 82723410*y^4*x^3 + 676824309*y^5*x^2 + 2898290766*y^6*x + 4865449685*y^7
?
? z
%27 = x^7 - 5103*y^2*x^5 - 260820*y^3*x^4 - 5931765*y^4*x^3 - 71116920*y^5*x^2 - 438689181*y^6*x - 1098952052*y^7
?

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

There is a standard calculation procedure.  Suitable for any degree.

We make the conversion.

$$x^2+ay^2=(p^2+as^2)(k^2+an^2)(t^2+ar^2)=z^5$$

And then solve the system of equations.

$$\left\{\begin{aligned}&z^2=k^2+an^2=t^2+ar^2\\&z=p^2+as^2\end{aligned}\right.$$

In the first equation to parameterize the substitution in the second.  And then, we decide and we obtain the parameterization of all solutions.

Not clearly only one. Need to find other solutions?


Dependence of posterior probability on parameters Suppose that there are 5 blood types in the population, named type 1 through type 5, with probabilities p1, p2,. . . , p5. A crime was committed by two individuals. A suspect, who has blood type 1, has prior probability p of being guilty. At the crime scene blood evidence is collected, which shows that one of the criminals has type 1 and the other has type 2.

Find the posterior probability that the suspect is guilty, given the evidence. Does the evidence make it more likely or less likely that the suspect is guilty, or does this depend on the values of the parameters p, p1,. . . , p5? If it depends, give a simple criterion for when the evidence makes it more likely that the suspect is guilty.

What I got so far is that P(guilty|evidence) = P(evidence|guilty)P(guilty)/(P(evidence|guilty)P(guilty)+P(evidence|not guilty)P(not guilty)).

P(guilty) = p

P(not guilty) = 1 - p

P(evidence|guilty) = 1/2 because one criminal has blood type 1

P(evidence|not guilty) = 1/5 because there are 5 blood types.

So I get P(guilty|evidence) = 0.5p/(0.5p + 0.2(1-p)) and I do not know if this is correct.

$\mathit G$: event that the suspect with blood type 1 is guilty
$\mathit E$: event that  the suspect's blood type matches one of the blood types found at the crime scene

We want to calculate $\mathbf P(G\vert E)$

Using Baye's theorem,

$\mathbf P(G\vert E) = \frac{\mathbf P(G)\mathbf P(E\vert G)}{\mathbf P(E)} $

$\mathbf P(G) = p$

$\mathbf P(E) = \mathbf P(G)\mathbf P(E\vert G) + \mathbf P(G^c)\mathbf P(E\vert G^c) $

$\mathbf P(E\vert G)$ is the probability that the suspect's blood type matches one of the blood types found at the crime scene i.e. type 1 and 2, given that the suspect with blood type 1 is guilty. Or simply that the probability that the suspect's blood type was found at the crime scene given that he is guilty. If the suspect is guilty, his blood type with definitely be at the crime scene. So this probability amounts to one.
$\mathbf P(E\vert G) = 1$

$\mathbf P(E\vert G^c)$ is the probability that the suspects blood type matches the ones found in the crime scene, given that he is not guilty. In other words, the probability that the suspect has the same blood type as one of the criminals. This would amount to $\frac{p_1}{p_1 +\,p_2}$

This gives us the following result,

$\mathbf P(G\vert E) = \frac{p}{p\, +\,(1-p)\cdot\frac{p_1}{p_1 +\,
 p_2}} $

What I understand as the evidence is one of the criminals has type 1 and the other has type 2. So if our suspect having type 1 blood is guilty the probability of this evidence occurring is 1.

P(evidence|guilty)=1

And if not guilty the evidence has probability $2p_1p_2$ (both blood types occurring and order not important)

P(evidence|not guilty)=$2p_1p_2$

Hence we get $\frac{p}{p+2p_1p_2(1-p)}$


Solve $ \int_0^a x^4 \sqrt{a^2-x^2}dx$ Solve $$ \int_0^a x^4 \sqrt{a^2-x^2}dx.$$

Substitution: $x = a\sin t \Rightarrow dx = a\cos t dt$

$$ \int_0^a x^4 \sqrt{a^2-x^2}dx = a^6 \int_0^{\pi/2}\sin^4t\cos^2 tdt = \frac{a^6}6 \frac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(3)} = \frac{a^6 \pi}{288}.$$

When evaluating this integral in Wolfram Alpha, I get another result. Also plugging in values and checking if they correspond to the answer I would get with my result, give contradictions. Where is my mistake?

Thanks.

EDIT: I found my mistake. Just had  a little brain fart: $\Gamma(3) = 2!$ and not $3!$. Thus we get $$\frac{a^6}{12}\cdot \frac{3\pi}{8} = \frac{a^6 \pi}{32},$$ which is correct. Thank you for your time and responses!

Hint: Substitute $$\sqrt{a^2-x^2}=xt\pm a$$ the so-called Euler Substitution.


Can you distribute the nth derivative via the product rule? I just want to check a particular step I am trying out. Normally I would prove an nth derivative formula via induction, but I am wondering if there is an easier way to potentially make this work. Is it true for some possible manipulation of this formula for analytic functions f(x) and g(x) for $ x \in {\Bbb R}$ that:

$$ \frac{d^n}{dx^n}f(x)g(x)= \frac{d^{n}f(x)}{x^n}g(x)+f(x)\frac{d^{n}g(x)}{x^n}?$$

The generalisation you are seeking is Leibniz' formula:
$$\frac{d^n}{dx^n}(f(x)g(x))
=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{d^k f(x)}{dx^k}\frac{d^{n-k} g(x)}{dx^{n-k}}.$$

In comments, you asked why it never holds.  It sometimes holds (just not always).  Consider the two following (simple) examples:

$f(x)$ could be anything, and $g(x)=1$.

In this case, $\frac{d^{n}}{dx^n} f(x)g(x)=\frac{d^{n}}{dx^n} f(x)\cdot 1 = \frac{d^{n}f(x)}{dx^{n}}(1)+f(x)\cdot 0$ and the formula works.

Suppose that $f(x)=g(x)=x^2$ so that $f(x)g(x)=x^4$ and take $n=4$.

By easy computation $\frac{d^4}{dx^4} x^4 = 24$.

But, using your formula you would get:

$\frac{d^4 (x^2)}{dx^4} x^2 + x^2 \frac{d^4 (x^2)}{dx^4} = 0\cdot x^2 + x^2 \cdot 0 = 0$ which disagrees with what you know the derivative should be (24).


Solving $\sin(\omega t)=- \frac 1 2$ Solve the given trigonometric equation:
   $$\sin(\omega t)=- \frac 1 2$$

Here is my attempt:

$$\sin(\omega t)=- \frac 1 2 = \sin \biggr (\pi +\dfrac{\pi}{6}\biggr )$$

Which yields

$$\omega t = \dfrac{7\pi }{6}$$

or

$$\sin(\omega t)=- \frac 1 2 = \sin \biggr (2\pi -\dfrac{\pi}{6}\biggr )$$

$$\omega t = \dfrac{11\pi}{6}$$

Is my assumption correct?

Regards!

From the unit circle definition of sine, we see that

$$\omega t = 2n\pi - \dfrac{\pi}{2} \pm \dfrac{\pi}{3}$$

and we can rephrase that as we wish.

Yes it is right but recall to add the $2k\pi$ term with $k\in \mathbb{Z}$, indeed we have

$$\sin(\omega t)=- \frac 1 2\iff \omega t=\dfrac{7\pi }{6}+2k\pi\,\lor\,\omega t=\dfrac{11\pi }{6}+2k\pi$$


What is the difference between parametrization and change of variables? I am embarrassed to ask this, but really need to, in order to clarify my confusion. I am taking multi-variable calculus and I am confused as to the difference between when I should be parametrizing and when I am making a change of variables.

My question is really motivated by a question such as this:

  Compute $ \int \int_{E} e^{-4x^2-9y^2}\,dxdy $ where $E$ is the Ellipse
  $4x^2+9y^2 \le 25$

When I see a question like this, the first thing I think is, ok parametrize the curve, so I simply did:

$x = \frac{5}{2}cos\theta \space$and$ \space y = \frac{5}{3}sin\theta$

and then proceeded to to substitute and do:

$ \int \int_{E} e^{-25}\,rdrd\theta $

getting the wrong result. Actually as I type this, I think I may have answered my own question. Have I got the wrong idea because, I have essentially turned a double integral with two variables into a 1 variable thing? And use an incorrect  $r$?

What is a good rule of thumb to keep in mind the difference of when I am making a change of variables and when I am just parametrizing?

In your case, you can change the variables since your surface ($2D$) is represented by $2$ variables (you could call this "parameterization" but "change of variables" is more appropriate). $(x,y) = (\frac52r \cos \theta, \frac53 r\sin \theta)$, $r$ goes from $0$ to $1$ and $\theta$ goes from $0$ to $2\pi$, calculate the Jacobian, and go from there.

You want to parametrize an $n$-dimensional object when it is being represented by more than $n$ variables. This is because you want to work with $n$ variables. For example, take $S$ to be the surface of the unit sphere with $xyz$-cordinates, and you try to calculate $$\iint_S f(x,y,z)dS$$
Then, using the parameterization $(x,y,z) = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$, the integral becomes
$$\int_0^{2\pi}\int_0^\pi g(\theta,\phi)\frac{dS}{d\theta d\phi}d\theta d\phi$$
where $\frac{dS}{d\theta d\phi} = \|T_\theta\times T_\phi \|$

paramaterizing is with a single integral

change of variables is a double integral

so for this problem you're meant to use a change of variables, with r and theta.
your x and y just need to be multiplied by r. and then chose your boundaries and integrate


In which categories are epimorphisms equivalent to surjections? For example, in the category of sets a morphism is an epimorphism iff it is surjective. This is true in the category of groups with homomorphisms, the category of topological spaces with continuous functions. However this is not true in the category of monoids, rings, etc. (For more examples see https://en.wikipedia.org/wiki/Epimorphism#Examples)

Is there a classification of categories where the two are equivalent? Or are there some properties of categories where they're equivalent? Obviously we must be working in a concrete category to talk about "surjective". But is there anything else we can say about the category?

If you want to talk about "surjection" your category must have a functor $C \stackrel{U}{\to} \text{Set}$ that specifies the underling set of every object.

You question can be reformulated as follows: when does $U$ reflect epimorphisms?

I do not know when such a characterization exists, but if $U$ is faithful, then it reflects epimorphisms.

I do believe that all the cases you can think of are due to this phenomenon, the forgetful functor in those cases is faithful.


Show $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ I have a homework problem which asks me to show
$$ \lim\limits_{x \to 0}\frac{e^x-1}{x} = 1 $$

Any help is appreciated.

By the definition of the derivative and if we denote $f(x)=e^x$ then
$$\lim_{x\to0}\frac{e^x-1}{x}=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=f'(0)=e^0=1$$

You can use l'Hopitals rule. Notice that if we substitute directly we get $\frac{0}{0}$, an indeterminate form. So we can differentiate the numerator and denominator separately then take the limit. Then we have

$$\lim_{x\to 0} \frac{e^x-1}{x}=\lim_{x \to 0}\frac{e^x}{1}=e^0=1$$

Though I've doubt you've seen this yet, we don't even need that because we can use the Taylor series for $e^x$.

$$\lim_{x \to 0}\frac{e^x-1}{x}=\lim_{x\to 0} \frac{\sum_{n=0}^\infty\frac{x^n}{n!}-1 }{x}=\lim_{x \to 0} \frac{1+\sum_{n=1}^\infty \frac{x^n}{n}-1}{x}=\lim_{x \to 0}\frac{\sum_{n=1}^\infty \frac{x^n}{n!}}{x}=\lim_{x\to 0} \sum_{n=1}^\infty \frac{x^{n-1}}{n!}$$

But then we have

$$\lim_{x\to 0} \sum_{n=1}^\infty \frac{x^{n-1}}{n!}=\sum_{n=1}^\infty \frac{0^{n-1}}{n!}=1$$

Though seeing that last step with $0^0=1$ is a bit trickier. Not necessarily the way to do it but another way of thinking of it for those that know the Taylor Series.

EDIT: You wouldn't even need to use the fact that $0^0$ is $1$ as

$$
\lim_{x\to 0} \sum_{n=1}^\infty \frac{x^{n-1}}{n!}=\lim_{x\to 0} 1+\sum_{n=2}^\infty \frac{x^{n-1}}{n!}=1+ \sum_{n=2}^\infty \frac{0^{n-1}}{n!}=1
$$


About a definition of skew field and field, with example $\{0\}$ I use following definitions from Corso di geometria-Stoka Marius-CEDAM-1995:

  let be $(C,+)$ a commutative group and $(C,\cdot)$ a semigroup $$(C,+,\cdot) \text{ is ring iff }\begin{cases}
 \forall x\in C, \forall y\in C, \forall z\in C: (x+y)\cdot z=(x\cdot z)+(x\cdot y)\\
 \forall x\in C, \forall y\in C, \forall z\in C:  x\cdot (y+ z)=(x\cdot y)+(x\cdot z)
\end{cases}$$
  let be $(C,+,\cdot)$ a ring $$(C,+,\cdot) \text{ is commutative ring iff }
 \forall x \in C, \forall y\in C: (x\cdot y)=(y\cdot x)
$$
  let be $(C,+,\cdot)$ a ring $$(C,+,\cdot) \text{ is ring with 1 iff }
 \exists x=:1 \in C,\forall y \in C: (x \cdot y)=y=(y\cdot x)
$$

I thinked,  are the following definitions of skew field and field possible?

  let be $(C,+,\cdot)$ a ring with 1 $$(C,+,\cdot) \text{ is skew field iff } \forall x \in C\setminus{0},\exists y \in C: (x \cdot y)=1=(y \cdot x)$$
  let be $(C,+,\cdot)$ a skew field $$(C,+,\cdot) \text{ is field iff } \forall x \in C,\forall y \in C: (x \cdot y)=(y\cdot x)$$

If yes, is $(\{0\}, +, \cdot)$ a field?

A skew-field (or skew field) is a ring in which the equations ax=b and ya=b with a≠0 are uniquely solvable. In the case of an associative ring (cf. Associative rings and algebras) it is sufficient to require the existence of a unit 1 and the unique solvability of the equations ax=1 and ya=1 for any a≠0. A commutative associative skew-field is called a field. An example of a non-commutative associative skew-field is the skew-field of quaternions, defined as the set of matrices of the form

(a−bb¯a¯)
over the field of complex numbers with the usual operations (see Quaternion). An example of a non-associative skew-field is the Cayley–Dickson algebra, consisting of all matrices of the same form as above over the skew-field of quaternions. This skew-field is alternative (see Alternative rings and algebras). Any skew-field is a division algebra either over the field of rational numbers or over a field of residues Fp=Z/(p). The skew-field of quaternions is a 4-dimensional algebra over the field of real numbers, while the Cayley–Dickson algebra is 8-dimensional. The dimension of any algebra with division over the field of real numbers is equal to 1, 2, 4, or 8 (see [Ad], and also Topological ring). The fields of real or complex numbers and the skew-field of quaternions are the only connected locally compact associative skew-fields (see [Po]). Any finite-dimensional algebra without zero divisors is a skew-field. Any finite associative skew-field is commutative (see [Sk], [He]). An associative skew-field is characterized by the property that any non-zero module over it is free. Any non-associative skew-field is finite-dimensional [ZhSlShSh]. A similar result applies to Mal'tsev skew-fields [Fi] (see Mal'tsev algebra) and to Jordan skew-fields [Ze] (see Jordan algebra). In contrast to the commutative case, not every associative ring without zero divisors can be imbedded in a skew-field (see Imbedding of rings).

Associative skew-fields are also known as division rings, in particular if they are finite dimensional over their centre


How do I calculate this integral:$\int_{0}^{1}\ln^2 \left| \sqrt x-\sqrt{1-x} \right|dx$? Is there someone who can show me how to calculate this integral :

$\int_{0}^{1}\ln^2\left|\sqrt{x}-\sqrt{1-x}\right|dx$ ?

I substituted $x=\cos ^2 t$ and after some cleanup ended up with the expression

$$-2 \int^{\pi/2}_{\pi/4}\log ^2(\sqrt2 \cos u)\cos (2u)du=-2 \int_0^1 \frac{\left(b^2-1\right) \log ^2(b)}{\sqrt{2-b^2}} \, db$$

Which mathematica, in its infinite wisdom, evaluates as

$$\frac{1}{4} (4 C+2+\pi  (1+\log (2)))$$

Where C is the Catalan number.

In order to develop the integral, you should make this steps,

1 Use logarithmic properties

$\int\ln^2 \left| \sqrt x-\sqrt{1-x} \right|dx$ = $\int\ 2ln \left| \sqrt x-\sqrt{1-x} \right|dx$ = $2\int\ln \left| \sqrt x-\sqrt{1-x} \right|dx$

2 Now I would suggest you to this variable change, x=(\sin x)^2, so our new integral, using trigonometrical identities, is,
   $4\int\ln\left|(\sin x -\cos x )\right|\ sin x\ cos x dx$

3  Here use the formula, $\int u*dv$ = $u*v - \int v*du$
where u=$\int\ln(\sin x -\cos x )$
      $dv=sin x \cos x dx$,
Then, you have to solve a trigonometrical integral, which can be easily solved by the universal change of variable