equation of tangent to a circle with zero gradient Find the equation of the tangent to the circle $$(x-3)^2 + (y-2)^2 = 1$$ at the point $(4,2)$.
I cannot show the working of the tangent equation as gradient/slope = 0.
So how can I find the gradient of tangent?
I think it is also obvious that the tangent is $x = 2$ but what about working?

The general form for a line is $ax+by = c$. This form more generally describes lines than the form $y=mx+b$, simply the functional implication of the latter form requires us to use a point at infinity in the parameterization of vertical lines.

So, a vertical line at $x=1$ can be described by $1\cdot x+0\cdot y = 1$, represented by a point in parameter space as $(a,b,c)=(1,0,1)$.

The tangent to your circle at $(x,y) = (4,2)$ has $x=4$ everywhere, so can be written as a point in parameter space as $(1,0,4)$.

For more detail, see here: http://www.mathsisfun.com/algebra/line-equation-general-form.html

Since when the line x=2 started having zero gradient, that too obviously!
Answer is y=1 and y=3.
It can be seen very easily from graph.
Mathematically any line with 0 slope is y=constant. So put y=k in equation and use condition of tangency.


Work done by a non constant force. Example 3. A 200-lb cable is 100 ft long and hangs vertically form the top of a tall building.

How much work is required to lift the cable to the top of the building?

I understood if you find the anti derivative to the definite integral of the formulas provided from 0 to 100.. you would get the same answer. But if you find the definite integral of any other interval you get different answers. For example 0 to 1. For the anti derivative of $200-2x$ which is $200x-x^2$ you get 199 vs the anti derivative of $2x$ which is $x^2$ you get 1. Both formula is stated to represent work done, but gives out different answers. So the two formulas must represent different things. For the $2x$ (anti derivative $x^2$) I imagine it to calculate the work done section by section so when you calculate say the interval of 0 to 1 you only get the answer of the work done moving a small section of the cable by 1 feet, exactly the section that only moved one feet when you pull the rope up the building and the answer you get does not equal to the whole cable moving one feet. So it does not equal to the total work done on the cable moving one feet up the building. The other formula I will need to have someone explain what it represent.

Also i do not understand what the textbook is saying at all when it stated let's place the origin at the top of the building and the x-axis pointing downward as in figure 2 not provided. We divide the cable into small parts with length $\Delta x$. If $X_i$ is a point in the $i$-th such interval, then all points in the interval are lifted by approximately the same amount, namely $X_i$. The cable weighs 2 pound per foot, so the weight of the $i$-th part is $2 \Delta x$. Thus the work done on the $i$-th part in foot-pounds is $2 \Delta x$ (force) $\times X_i $(distance). See image solution 2

And If had placed the origin at the bottom of the cable and the x-axis upward, we would have gotten see image solution 1
which give the same answer.

I don't get how orienting the x-axis upward or downward would give you different formula. I don't get what the textbook is saying. Please help

It seems you are a little confused about the physical meaning of your equations. The equation of the work done by a force $\vec F$ along a path $P$ is given by: $$W=\int_P\vec Fd\vec r$$
In the first solution, your reference frame is at the bottom of the building, with $x$-axis pointing up. If you move the chain up a distance $x$, the length of the chain is $100-x$, and the weight is $|\vec F|=200-2x$, acting downwards. But in this problem they don't ask "what is the work done by gravity?". They ask instead "what is the work done to overcome gravity?". The only difference is in the sign of the force. In the Solution 1, this force and displacement are in the same direction, so in order to lift the chain a distance $L$ you use $$W=\int_0^L(200-2x)dx$$
If you integrate to $L=1$ you just lift the chain one foot, so $99$ feet of the chain are still hanging from the building. To get the full work, just put $L=100$ and you get the answer.

In the second solution, they use the reference frame at the top of the building, pointing down. The length of the chain at some point in time is $x$, with weight $2x$. In the beginning $x=100$ feet, in the end $x=0$ feet. But now the force to overcome gravity is pointing upwards, while the $x$ axis is pointing down. Then the work done to lift the chain a distance $L$ is $$W=-\int_{100}^{100-L}2xdx=\int_{100-L}^{100}2xdx$$
If you look very carefully at the last equation, you might be able to guess that $$\int_{0}^{1}2xdx=-\int_{1}^{0}2xdx$$ is the work required to lift the last one foot of the chain.

Work $W=\int_S \mathbf F \cdot \mathbf{ds}$

Where $S$ is some path and $s$ some parameterization. See line/path integral.

To add some technicalities, $S=\langle 0, s \rangle $ $$\int \mathbf F \cdot \mathbf ds= \int \mathbf F(S(s)) \cdot \langle 0, 1 \rangle ds$$

Again, see line and path integrals or even differential forms.

$$=\int  F_y(S(s)) ds $$

Now you know the cable weighs $200 lbs$ $\times \frac{s}{100}$

$$=\int 2s ds=[s^2]_{0}^{100}=10000-0$$


Characterization of Compact Space via Continuous Function Let $(X,\mathfrak{T})$ be a topological space. We know that if $X$ is compact and $f:X\to \mathbb{R}$ be any continuous function then $f(X)$ is bounded since the continuous image of a compact set is compact and any compact subset of a metric space is bounded.
My questions are (edited after David C. Ullrich's comment below),

What is(are) the necessary and/or sufficient condition(s) on  $(X,\mathfrak{T})$ so that if for any continuous $f:X\to \mathbb{R}$ we can conclude that $f(X)$ is bounded then it would imply that $X$ is compact?

Let $(X,\mathfrak{T})$ be a topological space. If every continuous function $f:X\to Y$ is bounded for all metric space $(Y,d)$  then can we say that $X$ is compact?

Let $(X,\mathfrak{T})$ be a topological space. If there exists a metric space $(Y,d)$ such that every continuous function $f:X\to Y$ is bounded  then can we say that $X$ is compact?

A space $X$ is called pseudocompact when every real-valued continuous function on $X$ is bounded. This is a well-studied notion. Indeed a compact space, or even a countably compact space (every countable open cover has a finite subcover) is pseudocompact.

A classical theorem: for normal and $T_1$ spaces, countably compact and pseudocompact are equivalent.

The classical example of $\omega_1$, the first uncountable ordinal, is pseudocompact (and countably compact) and not compact. This shows that we cannot make the jump from countably compact to compact.

We can use Lindelöf (every cover has a countable subcover) as the gap property to compactness. So if we use at least Tychonoff spaces, a class of spaces where pseudocompactness and compactness coincide are the Lindelöf spaces.

Another class are the metric spaces, because for metric spaces countable compactness, pseudocompactness and compactness are equivalent.

A pseudocompact space has the property that every continuous map into every metric space is bounded. This follows from the equivalence for metric spaces and the fact that the continuous image of a pseudocompact space is pseudocompact.

So $\omega_1$ is also a counterexample to question 2, and a fortiori to 3 as well.

Let's say a space is RB if every continuous real-valued function is bounded.

You ask for necessary and sufficient conditions such that RB implies compact. The word "implies" is a little funny; what people often mean by it is difficult if not impossible to define precisely.

The only precise definition of "A implies B" that I know is "A is false or B is true". If we take that as the definition of "implies" then the answer to your question is that RB implies compact if and only if $X$ is compact or there exists an unbounded real-valued continuous function on $X$.

Almost surely not an answer to what you really wanted to know, but possibly the best answer that question is going to get. (The best answer that actually gives a necessary and sufficient condition.)


Does H2 depends only on abelian quotient? Consider a finite group $G$ and an abelian group $N$. Let $G$ act trivially on $N$. Is $H^{2}(G,N)\cong H^{2}(G^{ab},N)$? ($G^{ab}=G/[G,G]$ the abelianization of $G$)

I don't get group cohomology so I have no intuitions about this, and computing $H^{2}$ is a pain, so I am not sure how to even look for examples.

For trivial action, the relationship between $H^2(G, N)$ and the abelianization of $G$ is given by the universal coefficient theorem, which says that there is a split exact sequence

$$0 \to \text{Ext}^1(H_1(G), N) \to H^2(G, N) \to \text{Hom}(H_2(G), N) \to 0.$$

Here $H_1(G)$ is the abelianization. $H_2(G)$ is a group called the Schur multiplier, which can be computed using Hopf's formula, and it is generally nontrivial, so it's possible for the second term to be nontrivial even if the first term vanishes.

Explicitly, take $G = A_5$. Since the abelianization of this group is trivial, the first term vanishes, and we have

$$H^2(A_5, N) \cong \text{Hom}(H_2(A_5), N).$$

Now we need to know $H_2(A_5)$. Fortunately this was computed by our ancestors long ago: it turns out to be $\mathbb{Z}_2$. So if we take $N = \mathbb{Z}_2$ we get a nontrivial cohomology group.

If you're familiar with the relationship between $H^2(G, N)$ and central extensions of $G$ by $N$, you can also exhibit a nontrivial element of $H^2(A_5, \mathbb{Z}_2)$ by exhibiting a nontrivial $\mathbb{Z}_2$ central extension of $A_5$. There is a unique such extension called the binary icosahedral group $\widetilde{A}_5$, and it's the universal central extension of $A_5$.

Not getting group cohomology is a typical experience, I think. I expect that for many people it's a long sequence of unmotivated definitions and computations. The real meaning behind the subject is in homotopy theory and higher category theory but this requires time, experience, and the right background to appreciate.

No, these are not the same.  For example, if we choose $G=D_{4n}$ then its Abelianization is $(\Bbb Z/2)^2$, and so $H^2(D_{4n}^{ab}, \Bbb Z)=(\Bbb Z/2)^2$.  However, as shown here, the $H^2(D_{4n}, \Bbb Z)$ group is only $\Bbb Z/2$.

Side remark: in fact, non-commutative cohomology is one of the harder subjects that I've studied.  Sorry, I have to go to sleep, so it's faster to give another example than to go through the details of my initial example.


How is the Kullback-Leibler distance between probability measures well defined Let $\Delta$ be the set of all probability measures on the measurable space $(S,\mathcal{B})$. Assume that all the measures in $\Delta$ are dominated by a measure $\mu$ on $(S,\mathcal{B})$ and let $\psi_Q = \dfrac{dQ}{d\mu}$ denote the Radon-Nikodym derivative of $Q$ w.r.t $\mu$.

Given $\mu$, we can define $$K_{Q:P} = \int \psi_P \log \left (\dfrac{\psi_P}{\psi_Q} \right) d\mu.$$

It is already known that $K_{Q:P} \geq 0$. We can use either Jensen's inequality or Gibbson inequality to prove this.

In one of the papers that I was going through (this paper), the author says that

"the quanity $K_{Q:P}$ itself is well defined and finite if and only if the function $\psi_P \log \left (\dfrac{\psi_P}{\psi_Q} \right)$ is integrable w.r.t $\mu$. Integrability yields in turn the relation $\psi_Q(x) > 0$ for $P$-almost all $x \in S$, which means $P$ is absolutely continuous w.r.t $Q$."

I am not able to see how the integral being finite will imply and will be implied by the absolute continuity of the measures. Because it is possible that the integral is $\infty$.

You're right, absolute-continuity of the measures alone doesn't imply that the KL divergence exists. Here is my favorite counterexample:

  Consider the two probability measures $P,Q$ on $[0,1]$ with pdfs: $f_P=\mathbf{1}_{[0,1]}$ and $f_Q(x)=c\cdot e^{-1/x}\cdot \mathbf{1}_{(0,1]}(x)$ where $c$ is whatever constant will make $Q([0,1])=1.$ Now both $P\ll Q$ and $Q \ll P$ and $P,Q\ll$ (Lebesgue) but $D(Q \| P)$ is infinite, therefore not integrable.

But that's not what they're saying. They are saying that if the absolute-continuity condition holds and the integral is finite, then the KL divergence exists.

In summary all they mean by that is that "The KL divergence is well defined and finite iff all the components of its definition well defined and finite"

Note that the function $x\log(x)$ is well defined for all $x\in \mathbb{R}_{+}\setminus \{0\}$. However, in order to maintain right-continuity at $x=0$, it is conventional to define $0\log(0)=\lim\limits_{x\downarrow 0}x\log(x)$.

Now, $\psi_{P}\log\left(\frac{\psi_{P}}{\psi_{Q}}\right)$ is finite if and only if either of the following two conditions is met:

(i) $\psi_{P}\geq0$ and $\psi_{Q}>0$, or

(ii) $\psi_{P}=0=\psi_{Q}$ (where the convention is that $0\log\left(\frac{0}{0}\right)=0$).

In other words, what this suggests is that the support of the measure ${P}$ should be contained in the support of the measure ${Q}$. Put in another way, there cannot be an $s\in S$ for which $\psi_{P}(s)>0$ but $\psi_{Q}(s)=0$. This condition is equivalent to saying "if $\psi_{Q}(s)=0$ for some $s\in S$, then $\psi_{P}(s)=0$", which is equivalently saying that $P<<Q$.


Basic questions in Algebra Could someone confirm or check these out? Some of these are really simple, but I can't find a definite answer to these.

Let $R$ be a ring, $D$ an integral domain, $F$ a field, $G$ a group, and $N$ a normal subgroup of$G$

Do rings have to be closed under multiplication? My book only says multiplication needs to be associative.
Is this a proof that fields are integral domains? Let $x,y \in F - \{ 0 \}$. Assume $xy = 0 \iff xy + 1 = 1 \iff xy + xx^{-1} =1 \iff x(y + x^{-1}) = 1 \iff x^{-1} = y + x^{-1} \iff y = 0$, a contradiction, so no divisors of $0$, meaning $F$ is a $D$
Can someone briefly explain to me why the (first) Isomoprhism  theorems want to base on the Kernel so much? Without going the details, my intepretation of the theorems tells me we are just trying to cut up the group $G$ into ways we can understand it better thruogh the quotient, but I don't get why the general groundwork can't be used for arbitrary normal groups, why the kernel? My book does all these proofs, but I am not getting why the kernel is so special, why can't the isomoprhism theorems be based on some other normal groups. This is basically a case of "why" even though I know the "how".

1) Yes, rings have to be closed under multiplication.

2) That is a proof, but you can go simpler. Assume $xy = 0$. If $x = 0$, we're done. Otherwise $x \neq 0$ and so $x^{-1}$ exists. Then $y = x^{-1}xy = (x^{-1})(0) = 0$.

3) The First Isomorphism Theorem essentially states that given a homomorphism $\psi \colon G \to H$, then $G/\operatorname{ker}\psi \equiv \operatorname{im}\psi \leq H$. Basically, we use the kernel because it contains information about how $G$ is embedded into $H$. If you want the result with any other normal subgroup $N$, you'll have to find a homomorphism that has $N$ as the kernel. And there is, but it is trivial. Simply take $H = G/N$ and define $\psi(g) = gN$.

The Quotient Set G/H is made into a group by the multiplication

aHbH = abH

and

aHbH = abb^-1HbH = ab(b^-1Hb)H = abH

only if

(b^-1Hb)  is a subgroup of H, that is,  only if H is normal.,

The kernel of any groupp hom is a normal subgroup of the domain; and any normal subgroup is the kernel of some group hom.


Statistics Probability Lottery question chances The country of Statland runs a national lottery for its people. In this lottery, an urn contains $38$ balls that are individually numbered $1, 2, ..., 38$. Each week, $5$ balls are drawn from the urn without replacement. Gamblers pay a small sum to predict these five numbers and the jackpot is won by anyone who has chosen all five correctly. A runners-up prize is won by anyone who correctly selected four of the five numbers. You have decided to play the Statland lottery and have chosen your five numbers at random from the $38$ available. Give your answer to seven decimals in each part.

a)
What is the probability that you win the jackpot?
I have reasoned that it is $^{38}C_5=501942$ (combinations).

b)
What is the probability that you win a runners-up prize?

I have reasoned that it is $^{38}C_4=73815$ (combinations).

What are the correct answers and solutions?

a) To compute the probability of winning this lottery you need to divide the number of outcomes where you win by the number of total possible outcomes. There is 1 outcome where you win (where each of the drawn balls in yours) and there are $^{38}C_5 = 501942$ total possible outcome so the probability is $\frac{1}{501942}$.

b) Similarly to win the runner up prize there are $^{5}C_{4} \times {}^{(38-5)}C_{1}$ possible outcomes. This is because 4 of your 5 numbers are drawn and then some other of $38 - 5 = 33$ balls is drawn. There are still $^{38}C_5 = 501942$ outcomes and so the probability is:
$$\frac{^{5}C_{4} \times {}^{33}C_{1}}{^{38}C_5}.$$

a)
$$\binom{38}{5}^{-1}=\frac{5!33!}{38!}=\frac{1}{501942}$$
So you are correct there.

EDIT:
I completely misunderstood the b due to my lack of english skills so I'm just gonna remove that part from my answer.


Continuous Joint Distribution: Find $E(X^2Y^2)$ Question: Suppose $X$ and $Y$ have a jointly continuous distribution with joint density:

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

where $p$ is a constant for which $|p| < 1$. Find $E(X^2Y^2)$.

I tried to find $\displaystyle E(X^2Y^2) =  \iint x^2y^2f(x, y) \,dx\,dy. $

where $f(x,y)$ is the joint density, but I'm not sure how to take the integral.

Any help would be appreciated!

Thanks!

According the expression of density function of $(X,Y)$, $(X,Y)$ is joint Gaussian distributed with $\mathbb{E}[X]=\mathbb{E}[Y]=0$, $\mathbb{E}[X^2]=\mathbb{E}[Y^2]=1$ and $\mathbb{E}[XY]=\rho$. Usually, if $(X_1,X_2,X_3,X_4)$ is multivariate
Gaussian distributed with $\mathbb{E}[X_i]=0, 1\le i\le 4$, then the following formula is convenient to calculate its forth moments:
$$ \mathbb{E}[X_1X_2X_3X_4]=\mathbb{E}[X_1X_2]\mathbb{E}[X_3X_4]+\mathbb{E}[X_1X_3]\mathbb{E}[X_2X_4]+\mathbb{E}[X_1X_4]\mathbb{E}[X_2X_3]. \tag{*}$$
In (*), let $X_1=X_2=X$ and $X_3=X_4=Y$, it is easy to get the follows:
$$ \mathbb{E}[X^2Y^2]=\mathbb{E}[X^2]\mathbb{E}[Y^2]+2(\mathbb{E}[XY])^2=1+2\rho^2.$$

\begin{align}
x = {} & u + v \\
y = {} & u -v \\[10pt]
dx\,dy = {} & \left|\frac{\partial(x,y)}{\partial(u,v)}\right| \,du\,dv = 2\,du\,dv \\[10pt]
& (x^2-2\rho xy + y^2) \\[4pt]
= {} & (u+v)^2 - 2\rho(u+v)(u-v) + (u-v)^2 \\[4pt]
= {} &  2(1-\rho)u^2 + 2(1+\rho) v^2
\end{align}
\begin{align}
& \iint \exp \Big(-(x^2-2\rho xy + y^2)^2/(2(1-\rho^2))\Big) \,dx\,dy \\[10pt]
= {} & \iint \exp \Big(-(2(1-\rho)u^2 + 2(1+\rho)v^2)/(2(1-\rho^2))\Big) (2\,du\,dv) \\[10pt]
= {} & \iint \exp\left( \frac{-u^2}{1+\rho} \right) \exp\left( \frac{-v^2}{1-\rho} \right) (2\,du\,dv)
\end{align}
This can be written as a product of an integral with respect to $u$ and one with respect to $v$.

But you also have the factor $x^2y^2$. You need to write that in terms of $u$ and $v$ and then expand it and get a sum, so you then get a sum of integrals. ${}$


Finding a number when its remainder is given. I am trying to solve this problem

    W is a positive integer when divided by 5 gives remainder 1 and when divided by 7 gives remainder 5. Find W.

I am referring back to an earlier post I made. Now I am attempting to solve it that way.

We know that
$$w\equiv1(mod~5)$$
$$w\equiv5(mod~7)$$

$w=7r+5$

$w=5r+5+2r$

Since $5r+5$ is divisible by 5

$w=5(r+1)+2r$ this shows the remainder is $2r$

Now $2r$ divided by 5 gives a remainder 1 , thus giving the equation

$2r = 5k + 1$ or $r=\frac{5k+1}{2}$

Putting r back in $w=7r+5$ we get

$2w = 35k + 12$

So I guess $w= \frac{35+12}{2} = 23.5$

This is wrong and the answer is suppose to be 26. Any suggestions what I might be doing wrong ? or anything that I might be missing ?

Edit:
The problem was in calculation

$w= \frac{35+17}{2} = 26$

There is an error: $\rm\:w=7r\!+\!5\,\Rightarrow\,2w = 7(2r)\!+\!\color{#C00}{10} = 7(5k\!+\!1)\!+\!\color{#C00}{10} = 35k\!+\!\color{#C00}{17},\:$ not $\rm\:35k\!+\!\color{#0A0}{12}.$
Since $\rm\:35k\!+\!17 = 2w\:$ is even, $\rm\:k\:$ is odd, $\rm\:k = 2j\!+\!1,\,$ so $\rm\:w = (35(2j\!+\!1)\!+\!17)/2 = 35j+26.$

Remark $\ $ It is easier to do the division by $2$ before the substitution. Namely, we have $\rm\:2r = 5k\!+\!1\:$ so $\rm\:k\:$ is odd, $\rm\:k = 2j\!+\!1,\:$ thus $\rm\:r = (5k\!+\!1)/2 = (5(2j\!+\!1)\!+\!1)/2 = 5j\!+\!3.\:$ Therefore $\rm\:w = 7r\!+\!5 = 7(5j\!+\!3)\!+\!5 = 35j\!+\!26.$ Notice how the numbers remain smaller this way.

I emphasize again, it's much more intuitive if you learn about modular arithmetic (congruences). For many examples see my posts on Easy CRT (easy version of the Chinese Remainder Theorem)

$w\equiv 1 \mod5 \Longrightarrow w=5a+1$, and
$w\equiv 5 \mod7 \Longrightarrow w=7b+5$  .

Multiply the first by 7 and the second by 5:
$7w=35a+7$ , and
$5w=35b+25$.

Subtract:
$2w = 35(a-b) - 18$.

$\Longrightarrow 2w \equiv -18 \mod 35$,
$\Longrightarrow 2w \equiv 17 \mod 35$
$\Longrightarrow 2w \equiv 52 \mod 35$
$\Longrightarrow w \equiv 26 \mod 35$


Gap year to study math This is a plan in its earliest and thus least concise stage, so either bear with me or don't read the following babble (I bolded some of the important stuff):

I am a high school graduate who is about to enter a top 5 US college ("top 5" might sound arrogant and unnecessary, but I'm keeping it because I think it could somewhat affect your response), and I'm very interested in majoring in mathematics. I am not blind to other potential majors, as I am also very interested in philosophy, and to a lesser degree, certain natural sciences. However, over the past year or so I have fallen further and further in love with pure mathematics.

All that David Copperfield kind of crap: Last summer I self-studied much of basic calculus. This past year I complete AP calc AB, and for the past couple of months I have been going through How to Prove it by Velleman pretty thoroughly (really like it so far). During the rest of the summer I intend to finish HTPI and do as much of Spivak's Calculus as I can (I loved calc this year and want to try some more advanced/proof-based material).

You might be saying, "wow, this kid is basically a mathematical virgin." And you would be correct: Compared to some of the incredible people on my college's Facebook group who did calculus in 9th grade, I am highly inexperienced. But as Einstein said, "I am not a genius, I am only passionately curious," and I am the second one.

So the question: Would it be a good idea for me to take a gap year to (continue to) self-study mathematics?

On my hypothetical gap year, I would begin my self-study (as I think I learn better and possibly a little faster that way) by either continuing with calculus or starting linear algebra. Ideally I could devote, say, the first half of my year to one/both of those and the rest to basic/introductory abstract algebra, which I realize might be out of my league, but I just find it so damn intriguing.

The only reason I'm even considering this idea is because I have never really immersed myself in mathematics or had that much time to pursue it on my own, other than last summer/right now, and so I've always felt like there's a next level that I've never really experienced and probably wouldn't have time to experience in the first few years of college.

I have also spent a lot of time reading about great mathematicians of the past, and I've gotten the feeling that many of them (Grothendieck, Galois, Euler, even Newton, to some extent) learned the most in own independent studies. Now, I'm not trying to compare myself to these demi-gods, but I feel like if there's any time I could get ahead and have a chance to learn how to think like a mathematician, it is now.

So what do you think? Any personal experience in the matter? Do you think I should be worried about forgetting stuff (this is the usual concern with a gap year for mathematicians, so I thought I'd ask about it, although given that I would be doing and learning math on may gap year, it probably doesn't apply to me as much)?

It might seem kind of odd to take a gap year from learning just to learn, however, a whole year would give me a chance to, as I said before, immerse myself and learn more intensively than I'll be able to at college and beyond.

I realize that this might be better asked in the academia community, however, I would like a mathematics-specific answer before I consider this question from a university's perspective.

I also realize that this is ultimately my decision, so no need to tell me to just do what I want or what I think is best for me; the purpose of this is to help me determine what would be best for me. :)

I also realize that this question is way too long, but I'd ask you not to respond if you didn't read most of it, just to ensure that you are not misunderstanding my situation.

You're self-motivated; the kids you mention as a general rule pushed by parents trying to compensate for what they feel is a lack in their lives. Some will excel; some will flame out as a lot that drives them is 'being a wonder kid,' and being admired for it. At a good college, things equalize fast; and this can turn into an emotional barrier for them. I recently was at a coding bootcamp with an extreme such case: conversing at 10 with MIT faculty; lost at 22.

If you study hard, and are truly driven, things equalize fast. Take what's new and hard for you, and some in your comfort zone...and don't listen to your class-mates telling you homework was easy for them. Except for the occasional true math genius where it's no lie - and this student is unlikely to harp on it -, a strangely arrogant attitude is common among students in certain technical fields; so expect it, and shrug. I like to share advice given to me by my first hw buddy: if someone has a more elegant solution than you, it was copied from a nicer book than you had access to.  Or from the senior mathematician at my undergrad alma mater: "it is easily seen means that you see it after a sleepless night agonizing over it."  Math isn't easy for anyone.

In my grad studies, some of my classmates already had Ph.D.'s in physics prior to starting this new Ph.D. (and I have great respect for physics), and I was worried about being inadequate. As long as you are willing to work really hard, doing hw in the structured environment of a school will help you much more. Try to find a friend who to team up with, and to have friendly competitions with how to solve things more elegantly, and who to meet when you're stuck (and you will be). Working on your own is very hard (it's what I'm doing right now), and nowhere near as productive.

If you are going to a top five university for math, take a look at the courses they offer to freshmen. I can only speak for myself, but having just finished my freshman year, I took Calc(apostol), linear algebra(apostol), multivar(apostol), and abstract algebra(dummit and foote). It seems to me that the courses you are looking to take might be possible your first year. This of course alleviates the problem because you can get the best of both worlds.


Negative time value of put option I am analyzing valuation of European put options in non-Gaussian environment.

I realized, that put option can easily get negative time value, when it is deep in the money (spot price close to zero). Ultimate proof, that such negative time value exists is at limit, when spot price approaches zero. Then the option value is less, than strike as the price of the underlying can move in only one direction.

This issue (negative time value of a put) has been discussed in these pages also and opinions vary: some think negative time value is not possible, but I believe my counter-example above shows, that negative time value does exist. It is not that relevant in the Gaussian world, but can become an issue with fat tails.

This negative time value leads to a paradox: when I apply put-call parity for a put with negative time value, I get negative valuation for a call. This should be impossible, as call option holder does not have, by definition, any liabilities and therefore value must be >=0.

So, what explains this contradiction: is it so, that put-call parity only applies, when underlying return is Gaussian distributed?

You won't get any contradiction from the put call parity. On the contrary, I think the put call parity is very helpful in illustrating how puts can have negative time value.

First, we need a little confirmation on the terminology: I assume in your definitions the value of a derivative (its current price) equals its intrinsic value (what I could get if I exercised it now) plus its time value.

Then take a look at the put call parity $p_t = c_t - f_t$ in which $f_t$ is the value of a notional forward contract that has the same maturity and strike as the call and put in question. For simplicity let's assume the world follows Black Scholes assumptions (strike that, we actually don't need BL or any other model assumptions to make the following valuation of forward hold, apart from a constant risk free rate $r$), so that $f_t = S_t - Ke^{-r\tau}$ ($\tau$=time to maturity, or $T-t$). Hence
$$p_t = \color{blue}{c_t + Ke^{-r\tau} - K} + \color{red}{(K-S_t)}$$
The red part is apparently the intrinsic value, which makes the blue part the time value. Then you can see clearly that there is no guarantee that the blue part is greater than zero. Specifically, for example, it can easily drop below zero in the following two scenarios:

1). When the put is deep in the money, or equivalently when the call is out of the money. In this case $c_t\approx 0$, and $Ke^{-r\tau} - K <0$. So it's very likely they sum to a negative value.

2). When the interest rate is high or time to maturity is long. In this case $Ke^{-\tau}$ will be small and dominated by $-K$. If $c_t$ is also significantly smaller than $K$ (which happens easily when, say, $K$ is large).

The intuition is that if you were a put holder you'd want to get paid as early as possible because early payment means larger time value of money, but the put contract stipulates that you cannot get paid prior to maturity, which entails a loss from such deferral of cash flow. This is opposite to the call, in which case you as the call holder would want to defer the payment as much as possible and the call contract actually entitles you to such deferral of payment, which means a gain in time value.

The time value can't be negative.

A positive time value reflects the possibility that the price of the stock might wander in such a way as to make the option in the money at expiration. Then the option gives the holder the right to exercise the option, and a make a profit.

A negative time value would reflect the possibility that the price of the stock might wander in such a way as to make the option out of the money at expiration. However, since the option gives the holder the right to exercise the option, but not the obligation, this only results in the option being worthless. So the time value can not be negative.


Set cover: To show the number of minimum coverable combination of the power set of universe $U$ is less than $2^n$ Given a set $U=\{1,2,...,n\}$, and its power set $\mathcal S=\{S_1,...,S_{2^n-1}\}$(without emptyset $\emptyset$), a minimum coverable combination is $S_i\cup...\cup S_k=U$ and $S_i,...,S_k$ are pairwise disjoint.

For example, when $U=\{1,2,3\}$, then $\mathcal S=\{\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$ and a minimum coverable combination is $S_4=\{1,2\}$ and $S_3=\{3\}$ then $S_4\cup S_3=U$ and $S_4\cap S_3=\emptyset$.(don't count the permutations so $S_4\cup S_3$ is considered the same as $S_3\cup S_4$)

Now the question is to prove the number of minimum coverable combination of $U$ is less than or equal to $2^n$

Let $B_n$ denote the number of partitions of the set $[n]:=\{1,2,\ldots,n\}$.  This is known as the $n$-th Bell number.  The first few values are $B_0=1$, $B_1=1$, $B_2=2$, $B_3=5$, $B_4=15$, and $B_5=52$.  It can be shown that
$$B_{n+1}=\sum_{k=0}^n\,\binom{n}{k}\,B_k\text{ and }B_n=\sum_{k=0}^n\,\left\{\begin{array}{c}n\\k\end{array}\right\}$$
for all $n\in\mathbb{Z}_{>0}$, where
$$\left\{\begin{array}{c}n\\k\end{array}\right\}=\frac{1}{k!}\,\sum_{j=0}^k\,(-1)^{k-j}\,\binom{k}{j}\,j^n$$
is the Stirling number of the second kind.

It is known that
$$B_n \approx n^{-\frac{1}{2}}\,\left(\frac{n}{W(n)}\right)^{n+\frac{1}{2}}\,\exp\left(\frac{n}{W(n)}-n-1\right)\,,$$
where $W$ is the Lambert $W$-function.  That is, using $W(n)\approx \ln(n)$, we get that
$$B_n > \left(\frac{n}{\text{e}\,\ln(n)}\right)^n$$
for every positive integer $n$.  Therefore, $B_n$ grows much faster than $2^n$.


Empty intersection and empty union If $A_\alpha$ are subsets of a set $S$ then

$\bigcup_{\alpha \in I}A_\alpha$ = "all $x \in S$ so that $x$ is in at least one $A_\alpha$"

$\bigcap_{\alpha \in I} A_\alpha$ = "all $x \in S$ so that $x$ is in all $A_\alpha$"

It is the convention that $\bigcup_{\alpha \in \emptyset}A_\alpha = \emptyset$ and $\bigcap_{\alpha \in \emptyset} A_\alpha = S$.

But if $x$ is in $\bigcap_{\alpha \in \emptyset} A_\alpha = S$ then $x$ is in all $A_\alpha$ with $\alpha \in \emptyset$ and therefore $x$ is certainly in at least one $A_\alpha$ with $\alpha  \in \emptyset$. But then $x \in \bigcup_{\alpha \in I}A_\alpha$.

Can someone help me and tell me what is wrong with this? Thank you.

Some texts consider it a convention, but it is in fact a computation!

For the fixed set $S$, we are looking at the set $\mathcal P(S)$, the set of all subsets of $S$. With the operations of intersection and union (of arbitrary many subsets) the set $\mathcal P(S)$ is what is known as a complete lattice (don't worry if you don't know what that means).

First, one can argue intuitively: the more subsets of $S$ you intersect, the smaller the intersection is. Or, the fewer subsets you intersect, the larger the intersection is. So, the intersection of no subsets at all, the least amount of sets you can intersect, should be the largest subset possible. Namely, $\bigcap_{i\in \emptyset}A_i=S$. Similarly, the fewer subsets you take the union of, the smaller the union. So, the union of no subsets at all should be the smallest set possible. Namely, $\bigcup _{i\in \emptyset }A_i=\emptyset$.

Now, to make things more formal, let's define the intersection and union in $\mathcal P(S)$. The definition will be equivalent to the set-theoretic definitions but will only make use of the partial order relation of inclusion. Given a collection $\{A_i\}_{i\in I}$ of subsets of $S$, their intersection is the largest subset of $S$ that is contained in each $A_i$ (notice that this is saying that the intersection is a greatest lower bound). Similarly, the union of the family of subsets is the smallest subset of $S$ that contains each $A_i$ (notice that this says that the union is a least upper bound). Incidentally, this point of view very clearly points to a duality between union and intersection.

So now, the intersection of no subsets is the largest subset of $S$ that is contained in each one of the given subsets. There are no given subsets at all, so (vacuously) any subset $B\subseteq S$ is contained in each of the non-existent $A_i$. The largest of those is $S$, proving that $\bigcap_{i\in \emptyset}A_i=S$. Similarly, the union of no subsets is the smallest subset of S that contains each of the given subsets. No subsets are given, so any subsets $B\subseteq S$ contains each of the non-existent $A_i$. The smallest of these is $\emptyset $, thus proving that $\bigcup_{i\in \emptyset}A_i=\emptyset$.

$(x ∈ A_{\alpha} , \forall \alpha \in I )$ does not imply $( \exists  \alpha \in I \text{ such that } x \in A_{\alpha} )$, since I may be $\emptyset$


Given continuous $f_n(x)\ge 0$ are there continuous $g_n(x)\ge 0$ so that $f_n(x)g_n(x)\to +\infty$ whenever all $f_n(x)\neq 0$? Let $f_n:X\to [0,+\infty)$ be continuous functions, where $X$ is a metric space. Then do there exist continuous $g_n:X\to [0,+\infty)$ so that $f_ng_n\to +\infty$ poinwise in $\cap_n[X\setminus f_n^{-1}(\left\{0\right\})]$?

I am particularly interested in the special case where if $f_n(x)=0$ for some $n$ and $x$ then $f_m(x)=0$ for all $m\ge n$. As a special case of the special case, do such $g_n$ exist when  $f_n=f_{n-1}b_n$ for continuous $b_n\ge 0$?

Let's try for $f_ng_n\simeq n$. At the zero set of $f_n$ this of course can not be satisfied, so one has to modify this equality slightly. Multiply with $f_n$ to get a positive term
$$
f_n^2g_n\simeq nf_n
$$
and then desingularize
$$
(n^{-2}+f_n^2)g_n=nf_n
$$
So taking
$$
g_n=\frac{n^3f_n}{1+n^2f_n^2}
$$
gives a function sequence that has a chance to come close to your requirements. But as said in the other answer, without further assumption on $f_n$ it will be hard to prove pointwise divergence close to the zero set.

Not without some additional assumption on the $f_n$'s.  Consider the sequence $f_n = 0$ for all n.


"Statistics": How many of these begin and end with the letter "S"? How many distinct permutations are there of the letters
  in the word “statistics”? How many of these begin
  and end with the letter s?

The first part of the question I do understand. You have to use permutation with identical items. This is based on the number of letters.

$$\binom{10}{3,3,1,2,1} = 50400$$

Yet for the second part I am confused as to what the directions means. It says that how many letters begin and ends with letters  s so does one eliminate $2$ s and calculate this problem normally?

BONUS:
 If so using a similar example how does one find out if how many of these begin and end with the letters m for the word
 "mathematicsman"?

Oh my, such confusion! Let's try to simplify this, yet keep its essence.

Instead of statistics, let's use stats. There are 30 unique
arrangements, of which 3 have s at each end:
$$ \frac{5!}{2!\,2!} = 30\; \; \text{and} \; \; 3!/2! = 3.$$
So we can get the second number by removing s from each end
and dealing with what remains.

This means that in a random permutation. the probability of
getting an s at each end, should to 0.1.

In the simulation program below, I have avoided the messiness
of dealing with character strings in R, by substituting numbers
for letters in stats (1 represents s). A million random
permutations ought to give 2-place accuracy, so the answer
substantially matches the theoretical value.

 stats = c(1,1,2,2,3)
 n = length(stats)
 m = 10^6; x = numeric(m)
 for (i in 1:m) {
  perm = sample(stats,n)
  x[i] = (perm[1]==1 & perm[n]==1) }
 mean(x)
 ## 0.099806

Now let's use mamam instead of mathematicsman. Again here, we can
use the standard method find $\frac{5!}{3!\cdot 2!} =10$ as the number of unrestricted, distinguishable permutations. If we ignore two of the (indistingusihable) m's, then we
have the 3 arrangements of ama. So the probability
a random permutation has m's at both ends should be $3/10 = 0.3.$ The simulation below
confirms this.

 mamam = c(1,1,1,2,2)
 n = length(mamam)
 m = 10^6; x = numeric(m)
 for (i in 1:m) {
   perm = sample(mamam,n)
   x[i] = (perm[1]==1 & perm[n]==1) }
 mean(x)
 ## 0.300217

I think it is (3 choose 1) times (9 choose 2,3,1,2,1).
My reasoning is that you force an 's' at the front of your word in three different ways and then shuffle the rest of the letters around.


Joint Density Function Example f(x,y)= $\frac 1y$  for 0 < x < y <1

f(x,y)= 0 elsewhere

find P(x+y>.5):

$\iint$ $\frac 1y$ dxdy

I basically can't figure out the limits, since x and y are not only dependent on each other but also the probability statement itself.

$\displaystyle{%
{\rm f}\left(x,y\right)
=
{\Theta\left(x)\Theta\left(y - x\right)\Theta(1 - y\right) \over y}}$

\begin{align}
&\color{#ff0000}{\large\int_{-\infty}^{\infty}{\rm d}x
\int_{-\infty}^{\infty}{\rm d}y\,{\rm f}\left(x,y\right)}
=
\int_{-\infty}^{\infty}{\rm d}y\,{\Theta\left(1 - y\right) \over y}
\int_{-\infty}^{\infty}{\rm d}x\,
\Theta\left(x\right)\Theta\left(y - x\right)
\\[3mm]&=
\int_{-\infty}^{\infty}{\rm d}y\,{\Theta\left(1 - y\right) \over y}
\Theta\left(y\right)\int_{0}^{y}{\rm d}x\,
=
\int_{-\infty}^{\infty}{\rm d}y\,\Theta\left(1 - y\right)\Theta\left(y\right)
=
\int_{0}^{1}{\rm d}y =
\color{#ff0000}{\large 1}
\end{align}

\begin{align}
\color{#ff0000}{\large{\rm P}\left(x + y > 5\right)}
&=
\int_{-\infty}^{\infty}{\rm d}x
\int_{-\infty}^{\infty}{\rm d}y\,{\rm f}\left(x,y\right)
\Theta\left(x + y - 5\right)
\\[3mm]&=
\int_{-\infty}^{\infty}{\rm d}y\,{\Theta\left(1 - y\right) \over y}\,
\Theta\left(y\right)
\int_{0}^{y}{\rm d}x\,\Theta\left(x + y -5\right)
\\[3mm]&=
\int_{-\infty}^{\infty}{\rm d}y\,{\Theta\left(1 - y\right) \over y}\,
\Theta\left(y\right)
\overbrace{\int_{y - 5}^{2y - 5}{\rm d}x\,\Theta\left(x\right)}
^{\left(~=\ 0\ \mbox{since}\ y - 5\ <\ 2y - 5\ <\ 0\
         \mbox{when}\ 0\ <\ y\ <\ 1~\right)}
=
\color{#ff0000}{\large 0}
\end{align}

This result is quite obvious since ${\rm P}\left(x > 1\right) = 0$
and ${\rm P}\left(y > 1\right) = 0$ such that ${\rm P}\left(x + y > 2\right) = 0$.

In general
$$
\left\langle\phi\right\rangle
\equiv
\int_{0}^{1}{{\rm d} y \over y}\int_{0}^{y}{\rm d}x\,\phi\left(x,y\right)
$$

The region given is $ 0 < x< y<1$. In particular, this gives $0<x<y$

so we have $\int\int f(x,y) dx dy = \int_0^1\int_0^y \frac{1}{y} dx dy + \int_0^1\int_y^1 0 dx dy$

We can evaluate this, and we get

$\int_0^1\int_0^y \frac{1}{y} dx dy + \int_0^1\int_y^1 0 dx dy = \int^1_0 \frac{1}{y} *(y-0)dy + 0 = \int_0^1 1 dy = 1$

(edit) So to answer the probability statement for $P(x+y>0.5)$

$x+y>\frac{1}{2} \Leftrightarrow x > \frac{1}{2}-y$, so we are interested in the intersection of the regions $0<x<y<1$ and $1/2-y < x<1$. Combined, these give the region $\frac{1}{2} -y < x< y$ and $1/2 < y<1$

$$\int_0^1 \int_{1/2-y}^1 f(x,y) dx dy = \int_.5^1\int_{1/2-y}^y \frac{1}{y}dxdy$$

$$= \int_.5^1 \frac{1}{y}(2y-1/2) = \int_.5^1 2-\frac{1}{2y}$$
$$= 2*.5 -\frac{1}{2}(\ln(1) - \ln(.5))$$
$$= 1 - \ln(\sqrt{2})$$


For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7 For $n \in \mathbb{N}$, show that $4^n + 10 \times 9^{2n-2}$ is divisible by 7.

I'm not sure how to do this proof so any help would be appreciated.

Induction can certainly be used, but a more direct method will also work as follows - (The trick here is to get to 7 as far as possible, so 9 = 7+2, etc)

$4^n + 10 \times 9^{2n-2}$

= $4^n + 3 \times (7 + 2)^{2n-2}$  (mod 7)

= $4^n + 3 \times 2^{2n-2}$ (mod 7)

= $4^n + 3 \times 4^{n-1}$ (mod 7)

= $4^{n-1}(4 + 3)$ (mod 7)

= $4^{n-1}(0)$ (mod 7)

= $0$ (mod 7)

Edit:-  In fact, because the expression is even, $14$ divides it for all $n$


find the least a, for which two equations have a common root Could you help me out please. I have two equations: $2x^2-3x+1=0 $  and  $ 2x^2-(a+3)x+3a=0$

I need to find the least $a$ for which these two equations have a common root.
At a first glance I thought it'd be easy, just creating an equation with these two, then creating a function for $a$ and then just a small derivative knowledge. But, unfortunately it seems that it's not as simple as I think, because I've been getting very strange answers.

It'd be wonderful if you could help me out here, just can't concentrate enough maybe.

No derivative knowledge is needed. A common zero of your two polynomials is a root of their difference. So it must be a root of the equation $ax=3a-1$. It is easy to see that $a\ne 0$. So any common root must be equal to $3-1/a$. Substitute in the first equation and solve for $a$.

Because this is homework, we omit the rest of the calculation. But after a while you should get a quadratic in $a$.

Comment: For various reasons, it is nice to put off dividing as long as possible. Since $a\ne 0$, we can rewrite the first equation as $a^2x^2-3a^2x+a^2=0$. Then we can substitute $3a-1$ for $ax$. This yields
$$2(3a-1)^2 -3a(3a-1)+a^2=0,$$
and then simplification is pleasant and quick.

The resultant of the two polynomials is
$\left| \begin{array} {cccc}
2 & -3 & 1 & 0 \\
0 & 2 & -3 & 1 \\
2 & -a-3 & 3a & 0 \\
0 & 2 & -a-3 & 3a
\end{array} \right| .$

Equating this determinant to $0$ we get $2 (2a-1) (5a-2)=0$, giving $a = \frac 1 2$ and $a = \frac 2 5$.


indefinite integral using trig functions. I am unable to solve this integration. i dont understand this integration py using trig identities. any help?

$$\int \frac{dx}{\sqrt{-x^2+25}} , \quad \frac{-\pi }{2}\le t\le\frac{\pi}{2}$$
use the substitution :$$x=5\sin t$$

and also my question is that what is $$\sqrt{-x^2+25}$$ in terms of $t$.

Its just a practise question.

Hint:

$$
\sqrt{25-x^2}=\sqrt{25-25 \sin^2 t}=5\sqrt{1-\sin^2 t}
$$

Now you can do it?

Let's compute

If $x=5\sin(t)$ then $dx=5\cos(t)dt$.

On the other hand,
 $$\sqrt{-x^2+25}=\sqrt{-{(5\sin(t))}^2+25}=\sqrt{-25\sin^2(t)+25}$$
by trig identities we find that $\sin^2(t)=1-\cos^2(t)$ so
 $$\sqrt{-25\cdot\sin^2(t)+25}=\sqrt{\ -25+{\cos t}^2 +25}=\cos(t)$$
I guess you'll be able to compute that integral.


Adjoint Operators and Inner Product Spaces My linear algebra textbook gives the definition of the Adjoint Operator and then says,

  You should verify the following properties:

  Additivity: $(S + T)^* = S^* + T^*$
  Conjugate homogeneity: $(aT)^* = \overline{a}\,T^*$
  Adjoint of adjoint: $(T^*)^* = T$
  Identity: $I^* = I$, where $I$ is the identity operator on $V$.

I've stared at the pages for a couple hours now. How do you verify this?

Here's my attempt at a proof for Adjoint of Adjoint: (T*)* = (T*v, w)* = (v, Tw)* = (Tv, w) = T

Is that correct reasoning?

BTW, this is NOT homework. Just reading for pleasure.

Thanks!

The adjoint of a transformation is defined as the unique transformation $T^{*}$ so that $$ \langle Tx, y \rangle = \langle x , T^{*}y \rangle $$ for every $x$ and $y$.

So to prove any of your equalities above you simply need to show that the transformation you want to be the adjoint satisfies this property. Generally you can do this using the properties of an inner product.

For example, to prove $(T^{*})^{*} = T$ you need to show that for any $x,y$ $$ \langle T^{*}x , y \rangle = \langle x , Ty \rangle.$$
This follows because $\langle T^*x,y\rangle= \overline{\langle y , T^*x \rangle} = \overline{\langle Ty, x \rangle} = \langle x , Ty \rangle$.

The adjoint of a transformation is defined as the unique transformation $T^\ast$ so that
$$\langle Tx,y\rangle = \langle x,T^\ast y\rangle$$
for every $x$ and $y$. So

$$\langle T^{\ast\ast} x, y\rangle = \langle x, T^\ast y\rangle = \overline{\langle T^\ast y, x\rangle} = \overline{\langle y, Tx\rangle} = \langle Tx,y\rangle.$$

And we'll get that $\langle T^{\ast\ast}x, y\rangle = \langle Tx, y\rangle$ implies that $T^{\ast\ast} = T$.


Viscous fluid boundary condition Consider an incompressible viscous ﬂuid of kinematic viscosity
$ν$ , dynamic viscosity
$µ$
and density
$ρ$ . A viscous boundary layer is located over a solid surface at $y = 0$ and $x > 0$. The ﬂow in the boundary layer must match with the constant inviscid bulk velocity $U$. The steady two-dimensional boundary-layer equations are $$\frac{∂ u}{ ∂ x} + \frac{∂ v}{ ∂ y} = 0$$
$$ρ
\bigg( u
\frac{∂ u }{∂ x}
+v
\frac{∂ u }{∂ y} \bigg)=
µ \frac{∂ ^2u }{∂ y^2}$$ where $u(x,y)$ is the horizontal velocity and $v(x,y)$ the vertical velocity.

(i) State the boundary conditions on $u$ and $v$ along $y = 0$ and $x > 0$.

(ii) State the far-ﬁeld condition on $u$ as $y→∞$.

It says "The ﬂow in the boundary layer must match with the constant inviscid bulk velocity $U$" so does this mean that the resultant of the $u$ and $v$ velocity components should be equal to $U$?

Then for (i) $u(x,y=0)=v(x,y=0)=U/\sqrt2$?

For (ii), is the far field condition just $u \rightarrow 0$ as $y \rightarrow \infty$?

'It says "The ﬂow in the boundary layer must match with the constant inviscid bulk velocity U'

It means that as you approach the boundary from within the boundary layer, the velocity must approach the bulk velocity U.  It does NOT mean that the velocity within the boundary layer is constant.


the Gaussian integers are isomorphic to $\mathbb{Z}[x]/(x^2+1)$ I am trying to prove that $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$.

My initial plan was to use the first isomorphism theorem.  I showed that there is a map $\phi: \mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$, given by $\phi(f)=f(i).$  This map is onto and homorphic.  The part I have a question on is showing that the $ker(\phi) = (x^{2}+1)$.

One containment is trivial, $(x^2+1)\subset ker(\phi)$.  To show $ker(\phi)\subset (x^2+1)$, let $f \in ker(\phi)$, then f has either $i$ or $-i$ as a root.  Sot $f=g(x-i)(x+i)=g(x^2+1).$
How can I prove that $f \in \mathbb{Z}[x]\rightarrow g \in \mathbb{Z}[x]$?

Let me elaborate on Daniel Fischers comment. You have a ring homomorphism $\phi: \mathbb Z[x]\to\mathbb Z[i]$ given by $x\mapsto i$. Take $f \in \ker \phi$. By the division algorithm,
$$ f = q\cdot(x^2+1) + r,$$
where $\deg r < 2$ and $q,r\in\mathbb Z[x]$, since $x^2+1$ is monic. Applying $\phi$ to this equation yields
$$ 0 = \underbrace{\phi(q)\cdot(i^2+1)}_0 + \phi(r).$$
Since $r$ is of degree $<2$, we can write $r = ax+b$ for some $a,b\in\mathbb Z$. Then $\phi(r)=0$ gives
$$ai+b=0.$$
This equation in $\mathbb Z[i]$ implies $a=b=0$, so we have $r=ax+b=0\in\mathbb Z[x]$ and therefore
$$
f = q\cdot (x^2+1) \in \langle x^2+1\rangle.
$$
We conclude that $\ker \phi \subseteq \langle x^2+1\rangle$.

You can define your isomorphism as follows

$\phi:\mathbb{Z}[i]\rightarrow \mathbb{Z}[x]/\langle x^2+1 \rangle$

$\phi(a+bi)=(a+bx)+\langle x^2+1 \rangle$

If you compute $[a+bx+\langle x^2+1 \rangle]\times [c+dx+\langle x^2+1 \rangle]$

We get $ac+(ad+bc)x+bdx^2+\langle x^2+1 \rangle$. We reduce by long division to get,

$(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.

Therefore,

$\phi([a+bi][c+di])=(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.

Which is exactly what you want.

The map is obviously bijective and a ring homomorphism.


Why does the Binomial Theorem use combinations and not permutations for its coefficients? I have been trying to understand the Binomial Theorem formula. I can see that it works.

What I don’t understand is how or why using combinations finds the coefficients.

  What I mean is, isn’t each coefficient actually a permutation?

In the sense, that a combination isn’t concerned with the order. Yet the coefficient seems to reflect the ways a selection of items can be ordered.

It seems like a contradiction.

A simple explanation would be greatly appreciated. As I am not a mathematician.

Many thanks.

The reason combinations come in can be seen in using a special example. The same logic applies in the general case but it becomes murkier through the abstraction.

Consider

$$(a+b)^3$$

If we were to multiply this out, and not group terms according to multiplication rules (for example, let $a^3$ remain as $aaa$ for the sake of our exercise), we see

$$(a+b)^3 = aaa + aab + aba + baa + abb + bab + bba + bbb$$

Notice that we can characterize the sum this way:

$$(a+b)^3 = (\text{terms with 3 a's}) + (\text{terms with 2 a's}) + (\text{terms with 1 a}) + (\text{terms with no a's})$$

(You can also do the same for $b$, the approach is equivalent.) Well, we see from our weird expansion that we have every possible sequence of length $3$ made up of only $a$'s and $b$'s. We also know some of these terms are going to group together, as, for example, $aba = aab = baa$.

So how many summands are actually equal? Well, since they all have the same length, two summands are equal if and only if they have the same number of $a$'s (or $b$'s, same thing). And we also know that every possible sequence of length $3$ and only $a$'s and $b$'s are here.

So we can conclude that

$$\begin{align}
(\text{# of terms with 3 a's}) &= \binom{3}{3} = 1\\
(\text{# of terms with 2 a's}) &= \binom{3}{2} = 3\\
(\text{# of terms with 1 a}) &= \binom{3}{1} = 3\\
(\text{# of terms with no a's}) &= \binom{3}{0} = 1
\end{align}$$

Thus, we conclude:

There will only be one $aaa = a^3$ term
There will be $3$ $aba=aab=baa=a^2b$ terms.
There will be $3$ $abb = bab = abb = ab^2$ terms.
There will be $1$ $bbb=b^3$ term.

Thus,

$$(a+b)^3 = \sum_{k=0}^3 \binom{3}{k}a^k b^{3-k}$$

and in general, for positive integers $n$,

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

In short, the reason we use combinations is because the order does not matter, because we will get terms like $aab, baa, bab$ which are all equal in the expansion. Since multiplication is a commutative operation over the real numbers, then, we can say they're equal. Thus, the number of terms of that "type" (characterized by how many $a$'s or $b$'s they have) is given precisely by the number of sequences of length $n$ ($n=3$ in our example), made of only $a$ and $b$, that has exactly $k$ $a$'s (or $b$'s).

Of course this all relies on the central premise that multiplication commutes in the reals and thus ensures that the order of the factors does not matter. That suggests that it does not always hold in situations where multiplication does not commute - for example, the multiplication of a type of numbers known as quaternions is not commutative, and thus the binomial theorem does not hold there as it does here (since there $ab$ need not equal $ba$).

The nature of this commutativity, or the lack of it, and the consequences of each is better divulged in a discussion on abstract algebra, and this tangent is long enough as it is.

The use of combinations in the coefficients DOES convey order, but it is indirect.  Thus, it would make sense for combinations to be present. Here's why:

Consider: $(a+b)^3 = (a+b)(a+b)(a+b)$ Note: There are 3 factors, this is important

Now, how many $a^2b$ terms are there? 3

This 3 can be thought of as the result of a repeated permutation of 2a's and one b, aab, aba, aab

This 3 is also a result from combinations.
i.e. From the 3 factors of $(a+b)$, which 2 am I going to choose for the a. If you chose the first 2 factors, then the term will be aab, if you chose the first and third factors, the term would be aba. However choosing the first and third factor, is the same as choosing the third and first factor and that's why it is a combination.

Conclusion: The combination involved is to do with the factors of the entire binomial $(a+b)$, not the individual a and b's! Just switch your perspective slightly :)


A proof that $\mathbb{Q}^+$ is not finitely generated. Is it correct? Is $\mathbb{Q}^+$ finitely generated? Justify your answer.

This is an abstract algebra exercise I've tried to solve, but I don't know whether my answer (proof) is right or not.

    Proof
    So, let us suppose that $\mathbb{Q}^+$ is finitely generated and $\mathbb{Q}^+=\langle q_1,\dots , q_n \rangle$ , where $q_1, \dots , q_n$ generators and $n\in \mathbb{N}$. The "mean value" of $q_1 \dots , q_n$ is in $\mathbb{Q}$. Thus there exist $k_1, \dots , k_n$ such that $$\frac{q_1+\dots +q_n}{n}=k_1q_1+\dots +k_nq_n$$ or $$\left(k_1-\frac{1}{n} \right)q_1 +\dots +\left(k_n-\frac{1}{n}\right)q_n=0 \in \mathbb{Q}^+$$ Now, every $\left(k_i-\frac{1}{n} \right)$, where $i=1,\dots,n$ is not an integer (something we would have wanted), as $\frac{1}{n} \notin \mathbb{N}$ and that leads us to contradiction. Thus, $\mathbb{Q}^+$ is not finitely generated.

But, here comes my question. Are we sure that any rational number can be written uniquely as a word in $q_1,\dots,q_n$? Where are the possible mistakes in my proof? Thank you in advance?

No, there is no guarantee that the representation is unique. The mean of a sequence of rational numbers can even be an element of the sequence, so nothing is precluding the mean from being expressible as an integer linear combination of generators, just with different coefficients than the ones you give. In a general group, there's no guarantee that an expression of an element in terms of generators is unique, even if it is a minimal generating set.

The way to show it is not finitely generated is to look at the denominators. Every element expressible in terms of a finite number of elements has a denominator that divides the product of all of the denominators of the generators, hence if we pick a prime that does not divide that product as a denominator then the element is not expressible in terms of the generators.

As mentioned, you have not justified that every rational number is a unique $\mathbb{Z}$-combination. However, you may assume WLOG that $q_1, \dots, q_n$ are independent over $\mathbb{Z}$ by pruning dependencies if this is not the case. Multiplying by $n$ gives you $nk_i - 1 = 0$ for all $i$ now that you have assumed independency. This means $k_i = 1/n,$ contradiction. You also forgot to dispose of the trivial case $n=1.$


Why is $\lim\limits_{x\to-\infty}\log [1 + \exp(-x)]+x=0$? Could someone explain why

$$\lim\limits_{x\to-\infty}\log [1 + \exp(-x)]+x=0$$

Here is a one-line (full) explanation: $$\log(1+\mathrm e^{-x})+x=\log((1+\mathrm e^{-x})\cdot\mathrm e^{x})=\log(\mathrm e^{x}+1)\underset{x\to-\infty}{\longrightarrow}\log(0+1)=0$$

I think you may have meant the exponential function, which is Exp[-x].

See: http://www.wolframalpha.com/input/?i=Limit%5BLog%5B1%2BExp%5B-x%5D%5D%20%2B%20x%2C%20x%20-%3E%20Infinity%5D&t=crmtb01

Do you see why it is infinity, just by inspection?

The exponential approaches zero, Log[1] is zero and the remaining terms grows to infinity.

If this is acceptable, you should accept the answer.


Reducing Pcp (Post's correspondence problem) to mPcp Recently I have been studying Post's correspondence problem ($Pcp$), and I have stumbled upon a problem where I need to find a reduction from $Pcp$ to a modified version, $mPcp$. This modified version requires any matching sequence of "dominoes" $p_{i_1}, p_{i_2}, ..., p_{i_m}$ to start with the first "domino", $p_1$.

I am very well aware of the fact that there  exists  a  match  for
an  instance $I=\langle p_1, ..., p_n\rangle$ of $Pcp$  if  and  only  if  there  exists  a match  for  one  of  the  instances $I_1, I_2, ..., I_n$ of $mPcp$ where $I_j=\langle p_j, p_{j+1}, ..., p_n, p_1, ..., p_{j-1}\rangle$ for $j=1, ..., n$. In other words, if and only if $I$ is a yes-instance for $Pcp$, there is a way to rotate the list of "dominoes" in a cyclic manner such that the starting piece in the found match for $Pcp$ is put in the first position.

However, it is not clear to me how this observation could lead to a mapping from an instance $I$ of $Pcp$ to the "correct" instance $I_j$ of $mPcp$ that can be carried out in a finite amount of steps for any $I$, thus serving as a suitable reduction.

Could anybody point me in the right direction? Thank you very much.

http://infolab.stanford.edu/~ullman/ialc/slides/slides14.pdf

To summarise what is said in the attached pdf:

Let the dominos be (u1,v1), (u2,v2),...(un,vn).
(* is some symbol not already in the alphabet) For all the u's, put a * after them, and for all the v's, put a star before them, like such:
(u1*,v1), (u2,v2),...(un,*vn)

Now in the MPCP if we wanna start with the first term, we put a * before u1 like such:
(u1,v1), (u2,v2),...(un,*vn)

So now you the first domino MUST start, as it is the only one which has the first symbol matching. If there is a solution to this, it must start with the first domino.

Now to make the endings match, we add a domino:
(@,*@)
where @ is anything not in the alphabet.

Now MPCP is yes iff PCP is yes.

Put the *s before (or after) each symbol, not the words. So for d=(01101,001) add (0*1*1*0*1*,*0*0*1) but no (01101*,*001). If d is a starting domino add  (*0*1*1*0*1*,*0*0*1), too.


Need help figuring out error on Improper Integral Resolution my friends. I upfront apologize for using any unusual terms below, I'm not used to writing math in english. It's a rather simple integral that, when put on wolframalpha, we find out its sum does not converge.

$$\int_0^3 {{1 \over {x - 1}}dx} $$

The thing is I came across one resolution of it that shows  ln(2) as final answer, and I can't find in which step he got something wrong:

$$\int_0^3 {{1 \over {x - 1}}dx}  = \mathop {\lim }\limits_{a \to {1^ - }} \int_0^a {{1 \over {x - 1}}dx}  + \mathop {\lim }\limits_{b \to {1^ + }} \int_b^3 {{1 \over {x - 1}}dx} $$

$$\mathop {\lim }\limits_{a \to {1^ - }} \left( {\ln \left| {a - 1} \right| - \ln \left| { - 1} \right|} \right) + \mathop {\lim }\limits_{b \to {1^ + }} \left( {\ln 2 - \ln \left| {b - 1} \right|} \right)$$

$$\mathop {\lim }\limits_{\scriptstyle a \to {1^ - } \atop
  \scriptstyle b \to {1^ + }}  {\rm{ }}\left( {\ln \left( {1 - a} \right) - \ln \left( {b - 1} \right)} \right) + \ln 2$$

$$\mathop {\lim }\limits_{\scriptstyle a \to {1^ - }  \atop
  \scriptstyle b \to {1^ + } }  \ln \left( {{{1 - a} \over {b - 1}}} \right) + \ln 2 = \mathop {\lim }\limits_{x \to 1} \ln \left| {{{1 - x} \over {x - 1}}} \right| + \ln 2$$

The limit on the left was solved using L'Hospital rule which returned an Ln(1)=0, so, final answer would be Ln(2).

Can anyone see where mistakes were made?

Thanks in advance!

$$\mathop {\lim }\limits_{\scriptstyle a \to {1^ - } \atop
  \scriptstyle b \to {1^ + }}  {\rm{ }}\left( {\ln \left( {1 - a} \right) - \ln \left( {b - 1} \right)} \right) + \ln 2$$

I not aware of that notation being considered acceptable. You can take the limit as the input approaches a point, but that is different from taking two different limits of two different variables. In the comments, @MikeEarnest mentions http://mathonline.wikidot.com/limits-of-functions-of-two-variables Note that here, the definition of "limit" translates easily, with one $\delta$ and one $\epsilon$. Now, we could interpret your notation as being the limit as (a,b) approaches (1,1), but there are several problems with that. First, there is no guarantee that this limit exists (and, indeed, in this case it does not), and second, you have a approaching 1 from below and b approaching 1 from above. It would be quite difficult to translate this into the limit as we approach (1,1).

  $$\mathop {\lim }\limits_{\scriptstyle a \to {1^ - }  \atop
  \scriptstyle b \to {1^ + } }  \ln \left( {{{1 - a} \over {b - 1}}} \right) + \ln 2 = \mathop {\lim }\limits_{x \to 1} \ln \left| {{{1 - x} \over {x - 1}}} \right| + \ln 2$$

Here, any pretense that this can be interpreted as the limit as we approach (1,1) has to be abandoned. For the limit as we approach (1,1) to exist, it must be the same regardless of what path we use. You have limited us entirely to the a = b = x path, which is not valid.

Basically, what you're doing is taking this:

  $$\int_0^3 {{1 \over {x - 1}}dx}  = \mathop {\lim }\limits_{a \to {1^ - }} \int_0^a {{1 \over {x - 1}}dx}  + \mathop {\lim }\limits_{b \to {1^ + }} \int_b^3 {{1 \over {x - 1}}dx} $$

and then you're replacing both a and b with the same variable. If you look at the graph of $\frac{1}{x-1}$, there's a singularity at 1. If you approach this singularity at the same rate from both sides, the values cancels out and the integral is finite. On the other hand, if you approach the left side faster than the right side, then you get the positive amounts faster, and you get positive infinity. If you approach the right side faster, you get negative infinity. It's not valid to pick one of these and call it the limit; if these limits aren't equal, then the limit doesn't exist. It's the same principle that says that if the sum of the limit of two series is not the limit of their sums, if their limits are not finite


Quick and painless definition of the set of real numbers I am looking for a simple way to describe the underlying set of the real numbers without getting into cauchy sequences or dedekind cuts. Furthermore, I want the description to not rely on some notion of equivalence (like how one can use the notion of coprime to give the rationals unique representatives). I think the following works:

$\mathbb{R} = \{\text{All decimal expansions | does not end in repeating 9s}\}$.

My question is am I forgetting about something or does this do the trick?

Edit: Indeed, this is not closed under the usual definition of $+$ and $\cdot$ but we can redefine these operations to "round up" when necessary. I can verify the field axioms on my own time; I am merely seeing if anyone can spot a subtlety that I missed or if my description truly does give unique representatives of the reals.

Edit 2: (Some context) The posters have given some great ways to define the reals if you want to spend a decent amount of time on it. I want to give the reals as a set of decimal expansions without any fancy notion of equivalence so that I can move on and do linear algebra. For example, if I were teaching a first year course, the field axioms would take a whole lecture and be super boring and Cauchy sequences/dekekind cuts would have students lining up to drop the course.

It will do the trick, but it will be messy and painful to prove the field axioms. Moreover, the definition in this way is somewhat arbitrary. Why choose base $10$? There is a more natural way to do it which is the Bourbaki definition of the reals. It relies on the completion construction of a uniform space. The completion of a uniform space is precisely the set of minimal Cauchy filters (so in particular, there is no equivalence involved). The reals are the completion of the rationals, and the rationals can be given the structure of a uniform space (basically, since it is topological group), thus the reals can be defined to have underlying set the set of all minimal Cauchy filters in $\mathbb Q$.

The simplest way: The real numbers are a set with operations +, -, *, / and relations < = > which follow the axioms of arithmetic, and where every nonempty set with an upper bound has a least upper bound.

Equivalence classes of Cauchy sequences are IMO the easiest way to prove that the real numbers exist, and they come kind of natural when you start talking about limits (just construct a sequence that should have the limit $2^{1/2}$ but doesn't in the rational numbers because the number doesn't exist).

(Going to decimals with unlimited number of digits will get you into huge trouble just defining the product of two numbers).


A problem about $\sigma (n)$ If $\sigma(n)$ is odd, then $n$ is a square or the double of a square.

I don't know how to begin with it... Any suggestions?

Use the multiplicativity of $\sigma$. Let $p$ be a prime divisor of $n$. Write $n=p^e m$, with $p$ not dividing $m$. Then $\sigma(n) = \sigma(p^e) \sigma(m)$. If $\sigma(n)$ is odd, what can you say about $\sigma(p^e)$ and so about $p$ and $e$? Consider the case $p=2$ and $p$ odd separately, as usual.

Or just use the multiplicativity of $\sigma$ and solve the problem for prime powers.

Let $n=p_1^{a_1}\ldots p_k^{a_k+1}$ be the prime factorization of $n$. Thus $\sigma(n)=(a_1+1)\ldots(a_k+1)$. If this is odd, then every $a_i$ is even.


Doubt in finding general term of the given sequence The following image has both the problem and its solution. I have a doubt in the solution, the details of which I have included below the image.

(Assume the terms in given series are generated from a polynomial)

Here, the author has assumed $T_n$ as an arbitrary cubic equation (Step indicated by the RED box).

My doubt is, why has he assumed it as a cubic equation and not a quadratic or biquadratic equation or any other degree equation?

Please do not use Newton's forward interpolation rule as I don't know such rules. Please explain in simple terms and properties.

Kindly clarify my doubt.

The author is assuming the original terms of the series are values from a polynomial at increasing integral values. Thus, when a sequence of differences causes all of the terms to be the same at some level $m$, the sequence $T_n$ represents a polynomial of degree $m$. For more information, see Theorem $1$ in Difference Tables of Sequences. In your case, taking the sequence of the third order differences based on those second order differences causes all of the values to be $6$. This means the original sequence represents a cubic equation.

Also, although the author doesn't use this, and the linked page only hints at it, the sequence of constant values are $a_n(n!)$ where $a_n$ is the coefficient of the $x^n$ term. Thus, in this case, $6 = a_3(3!) \implies a_3 = 1$. The author could have used this to determine that $a = 1$ more directly.

(Assuming the terms in given series are generated from a polynomial)

When you take the difference, you reduce the degree by $1$.

As an example, for a quadratic $T_n = n^2$, the first order differences would be in A.P: $$(n+1)^2 - n^2 = 2n+1$$

And for a cubic $T_n = n^3$, the first order differences gives you a quadratic:
$$(n+1)^3-n^3=3n^2+3n+1$$

Now if you take the difference on above quadratic, you get a sequence in A.P.


Determine if a function is uniformly continuous by looking at its graph? Today my professor introduced the idea of uniform continuity. However, I had difficulty visualizing what the graphical interpretation of a function being uniformly continuous is. Is it possible to determine if a function is uniformly continuous by looking at its graph?

Can anyone give an intuitive interpretation of uniform continuity?

Generally it is impossible to judge whether a function is uniformly continuous only by looking at the graph.

Note that even differentiable uniformly continuous functions can have unbounded derivatives (namely unbounded slopes). One immediate example is $f(x)=x\sin(1/x)$ on $(0,1]$, see here for a proof.

Of course, if defined on a compact set (say a bounded closed interval) then it trivially suffices to check whether the function is continuous.

As far as I can tell, I can't find a good intuitive interpretation of uniform continuity.

Yes, the slopes of the tangents are bounded. At least if you are dealing with a differentiable function, but this gives you the intuition.


Formulate the integer linear programming problem to maximize the expected aggregate action for the team of gymnastics For future Olympics, the Gymnastics Federation can choose
between 6 contestants, and you need 3 of them to participate both in the balance beams
as in parallel bars. The Olympic committee requires that 4
people at each event. The following table shows the average ratings of
each gymnast and their coach assumed that their performance would be similar in the future
competition. Formulate the integer linear programming problem to choose the
gymnasts who participate in each exercise in such a way as to maximize the
expected aggregate action (sum of each gymnast's score) for the team.

Gymnasts
Balance beams
Bars

1
88
79

2
94
83

3
92
85

4
75
87

5
87
81

6
91
86

My try:
Maximize:  $$88x_1+79x_2+94x_3+83x_4+92x_5+85x_6+75x_7+87x_8+87x_8+87x_9+81x_{10}+91x_{11}+86x_{12}$$
restricted to: $$\sum_{i=1}^6x_{2i}=4$$
$$\sum_{i=1}^6x_{2i-1}=4$$
For $x_i\in\{0,1\}$ for all $i\in\{1,2,3,4,5,6\}$
I'm not sure how to simplify the restriction of having 3
gymnasts to participate both in the balance beams as in parallel bars. Any suggestions would be great!

The easiest way to do it would be to actually have three sets of variables:

$x_1, x_2, \dots, x_6$ to indicate that a contestant is participating in the first event only;
$y_1, y_2, \dots, y_6$ to indicate that a contestant is participating in the second event only;
$z_1, z_2, \dots, z_6$ to indicate that a contestant is participating in both events.

You'll need to modify your existing constraints to use $x_i + z_i$ to test if a contestant is participating in the first event, and $y_i + z_i$ for the second. To make sure that at most one of these is true for a contestant, you'll need to add the constraint $x_i + y_i + z_i \le 1$ for all $i$. Finally, the condition you had trouble with can now be written as $$z_1 + z_2 + z_3 + z_4 + z_5 + z_6 = 3.$$

You need to set $x_i \le 1$ to finish off your LP formulation. Strictly you have an IP (integer programming) and depending on how you're solving the LP, you can usually specify that the values must be integers in the software. In any case this type of problem may well give you integer values for the $x_i$. If not, then you need to search for integer points near your optimal solution.
It worries me that you don't seem to have addressed the constraint that three must compete in both events...


A question about linear transformations mapping straight lines to straight lines Obviously, a linear transformation over a space maps all straight lines to other straight lines.

My question is: is the converse true? That is, if we're looking at a space after some transformation and we're observing that all straight lines were mapped to straight lines, does that imply a linear transformation? (Lines are understood as finite and infinite collections of points that are aligned, well, in a line)

EDIT: The comments mention affine transformations which fullfil the above condition but are not linear since they also involve translation. I think I'll reask then: Is every transformation that maps ALL lines to lines necessarily affine?

not always but converse will  be true iff T(x) that is a function of x .You can suppose y=T(x) represents the line that Passes through origin


Find all values of $p$ for which the series $\sum \frac{n}{(n^2-1)^p}$ converges Use the integral test to find all values of p for which the series converges
$$\sum a_n =\sum \frac{n}{(n^2-1)^p}$$

The interval is chosen such that $a_n$ is defined $\implies n \geq 2$.
(a) Condition for the integral test:
For the integral test to work, the function has to be decreasing and positive. As $a_n$ is a function of $n$, let's consider the function $$f(x) = \frac{x}{(x^2-1)^p}$$.
$$f'(x)= (x^2-1)^{-p-1}(x^2(-2p+1)-1)$$
two critical points exist in the reals $x=1, x=-1$.Testing  $f'$ in the interval of interest $[2, \infty)$,
As we want $f$ to be decreasing, so $P$ is such that $$p > \frac{x^2-1}{2x^2}$$
$\forall x \in [2, \infty)$, $x^2 -1 > 0 \implies f(x)>0$, $\forall p$.
.
(b) the integral
$$\int_{2}^{\infty} x(x^2-1)^{-p}= \frac{1}{2} \lim\limits_{t \rightarrow \infty} \left[\frac{1}{1-p} (x^2-1)^{1-p} \right]_2^t = \frac{1}{2} \lim\limits_{t \rightarrow \infty} \left( \frac{(t^2-1)^{1-p}}{1-p} - \frac{3^{1-p}}{1-p} \right)$$
We can observe that the integral is indeterminate when $p=1$
$$ \int f <\infty \implies   \lim\limits_{t \rightarrow \infty}  \frac{(t^2-1)^{1-p}}{1-p} <\infty \implies \lim\limits_{t \rightarrow \infty}  (t^2-1)^{1-p} <\infty \implies 1-p \leq 0 \implies p\geq 1 $$
To conclude $\sum a_n < \infty$ iff ${p \geq 1}$ and $p>\frac{x^2-1}{2x^2}$
.
I would like to know if my argumentation is correct, appropriate for the integration test?
Much appreciated for your input/help

$\int \frac{n}{\left(n^2-1\right)^p} \, dn=\frac{1}{(2-2 p) \left(n^2-1\right)^{p-1}}$

so the condition for the convergence is $p>1$

If $p=1$ the integral diverges because $\frac{n}{n^2-1}$ is asymptotic to $\dfrac{1}{n}$ as $n\to +\infty$


Precise definition of the support of a random variable $\newcommand{\F}{\mathcal{F}} \newcommand{\powset}[1]{\mathcal{P}(#1)}$
I am reading lecture notes which contradict my understanding of random variables. Suppose we have a probability space $(\Omega, \mathcal{F}, Pr)$, where

$\Omega$ is the set of outcomes
$\F \subseteq \powset{\Omega}$ is the collection of events, a $\sigma$-algebra
$\Pr:\Omega\to[0,1]$ is the mapping outcomes to their probabilities.

If we take the standard definition of a random variable $X$, it is actually a function from the sample space to real values, i.e. $X:\Omega \to \mathbb{R}$.

What now confuses me is the precise definition of the term support.

According to Wikipedia:

  the support of a function is the set of points where the function is
  not zero valued.

Now, applying this definition to our random variable $X$, these lectures notes say:

  Random Variables – A random variable is a real valued function defined
  on the sample space of an experiment. Associated with each random
  variable is a probability density function (pdf) for the random
  variable. The  sample space is also called the support of a random
  variable.

I am not entirely convinced with the line the sample space is also callled the support of a random variable.

Why would $\Omega$ be the support of $X$? What if the random variable $X$ so happened to map some element $\omega \in \Omega$ to the real number $0$, then that element would not be in the support?

What is even more confusing is, when we talk about support, do we mean that of $X$ or that of the distribution function $\Pr$?

This answer says that:

  It is more accurate to speak of the support of the distribution than
  that of the support of the random variable.

Do we interpret the support to be

the set of outcomes in $\Omega$ which have a non-zero probability,
the set of values that $X$ can take with non-zero probability?

I think being precise is important, although my literature does not seem very rigorous.

I am not entirely convinced with the line the sample space is also called the support of a random variable

That looks quite wrong to me.

  What is even more confusing is, when we talk about support, do we mean that of $X$ or that of the distribution function $Pr$?

In rather informal terms, the "support" of a random variable $X$ is defined as the support (in the function sense) of the density function $f_X(x)$.

I say, in rather informal terms, because the density function is a quite intuitive and practical concept for dealing with probabilities, but no so much when speaking of probability in general and formal terms. For one thing, it's not a proper function for "discrete distributions" (again, a practical but loose concept).

In more formal/strict terms, the comment of Stefan fits the bill.

Do we interpret the support to be

- the set of outcomes in Ω which have a non-zero probability,
- the set of values that X can take with non-zero probability?

Neither, actually. Consider a random variable that has a uniform density in $[0,1]$, with $\Omega = \mathbb{R}$.
Then the support is the full interval $[0,1]$ - which is a subset of $\Omega$. But, then, of course, say $x=1/2$ belongs to the support. But the probability that $X$ takes this value is zero.

A random variable is defined as a function that maps outcomes to numerical quantities, typically real numbers, i.e., X: Ω ↦ R.

The “domain of a random variable” is the set of possible outcomes. In the case of the coin, there are only two possible outcomes, namely heads or tails. Since one of these outcomes must occur, either the event that the coin lands heads or the event that the coin lands tails must have non-zero probability. For an unbiased coin p(H)=p(T)=1/2.

Consider another random variable Y, be the number of heads. So toss a coin twice produce y = 0, 1 and 2 with p(Y) = ¼, ½ and ¼ respectively.

The “support of a real-valued function f ” is the subset of the domain containing those elements which are not mapped to zero.  Let f(x) be the p.d.f. of normal distribution with support x∈R , so that f(x)≠0.


How to prove that compact subspaces of the Sorgenfrey line are countable? How to prove that every compact subspace of the Sorgenfrey line is countable?

Let $C$ be a compact subset of the Sorgenfrey line (so $X = \mathbb{R}$ with a base of open
sets of the form $[a,b)$, for $a < b$). The usual (order) topology on $\mathbb{R}$ is coarser (as all open intervals $(a,b)$ can be written as unions of Sorgenfrey-open sets $[a+\frac{1}{n}, b)$ for large enough $n$, so are Sorgenfrey-open as well) so $C$ is compact in the usual topology as well. This means in particular that $C$ is closed and bounded in the usual topology on $\mathbb{R}$.

Suppose that $x_0 < x_1 < x_2 < \ldots $ is a strictly increasing sequence in $C$, and let $c = \sup \{x_n: n =0,1,\ldots \}$, which exists and lies in $C$ by the above remarks. Also let $m = \min(C)$, which also exists by the same.

Then the sets $[c,\rightarrow)$ and $[m, x_0)$ (if non-empty), $[x_n, x_{n+1})$, for $n \ge 0$ form a disjoint countable cover of $C$, so we cannot omit a single member of it (we need $[x_n, x_{n+1})$ to cover $x_n$, e.g.), so there is no finite subcover of it that still covers $C$.
This contradicts that $C$ is compact.

We conclude that $C$ has no infinite strictly increasing sequences. Or otherwise put: $C$ in the reverse order (from the standard one) is well-ordered.

And so we have shown that every compact subset of $C$ corresponds to a well-ordered subset of $\mathbb{R}$ (by reversing the order, and note that the reals are order isomorphic to its reverse order). And all well-ordered subsets of $\mathbb{R}$ are (at most) countable (this follows from several arguments, including one using second countability, e.g.).

not true. The compact subsets of Sorgenfrey line must be finite sets. If it is infinite countable set even it can not be compact in the ususal topology on R


What is $\frac{1}{2} \int {e^{\frac{t}{2}}dt}$ equal to? Would using substitution be helpful to get rid of the exponent of the variable?
I tried substituting "$u$" in but it did not seem to help finding the integral.

Yes, it would be helpful.

$$u = t/2$$
$$du = \frac{1}{2}dt \Longrightarrow dt = 2du$$

And so:

$$\int e^{t/2} dt = 2\int e^{u}du$$

Of course, the exponential function is its own derivative, and it's own integral up to an integration constant.


Is $\infty + (\infty/\infty)$ indeterminate? I know $ (\infty/\infty)$ is indeterminate, but it can't be less than $0$.

So can you assume $\infty + (\infty/\infty)$  is determinate because $\infty + n$ where $n\ge 0$ is still $\infty$ ?

The equation this question is based off of is $$\lim_{n \to \infty} \frac{n \log n + n}{\log n}.$$

This is in the context of big O notation. Would this be form be valid to use to determine the numerator's function is big Omega of the denominator? Or should l'hopitals rule be used to find a determinate and defined limit?

You can use the fact that if $\lim\limits_{x\to c}f(x)= \infty$ and $\liminf\limits_{x\to c}g(x)\ge 0$, then eventually $f(x)+g(x)\ge f(x)-1\to \infty$.

This is a very interesting question... I believe you are correct. Below is a proof.

Let $lim_{x \rightarrow a}[f(x)/g(x)]$ be some arbitrary indeterminate function such that $lim_{x \rightarrow a}[f(x)/g(x)]=\infty/\infty$ where $x\in \mathbb{R}$ and $a$ is a finite real number.
Also let $lim_{x \rightarrow b}[h(x)]=\infty$ where $x\in \mathbb{R}$ and $b$ is a finite real number.
As $x\rightarrow a$, we know that $f$ and $g$ either (i) grow at the same rate, (ii.) $f$ grows faster than $g$, or (iii.) $g$ grows faster than $f$.

Case i.

If $f$ and $g$ grow at the same rate, then $lim_{x \rightarrow a}[f(x)/g(x)]=L$ where $L\in \mathbb{R}$.
Then, $lim_{x \rightarrow b}[h(x)]+lim_{x \rightarrow a}[f(x)/g(x)]=\infty+L=\infty$.

Case ii.

If $f$ grows faster than $g$, then $lim_{x \rightarrow a}[f(x)/g(x)]=\infty$.
Then, $lim_{x \rightarrow b}[h(x)]+lim_{x \rightarrow a}[f(x)/g(x)]=\infty+\infty=\infty$.

Case iii.

If $g$ grows faster than $f$, then $lim_{x \rightarrow a}[f(x)/g(x)]=0$.
Then, $lim_{x \rightarrow b}[h(x)]+lim_{x \rightarrow a}[f(x)/g(x)]=\infty+0=\infty$.

Hence, in any case, $lim_{x \rightarrow b}[h(x)]+lim_{x \rightarrow a}[f(x)/g(x)]=\infty+(\infty/\infty)=\infty$

In this proof we are using limits to make sense of your statement, but I don't think you NEED limits to make sense of the statement. $\infty$ is simply a quantity increasing without limit... Some folks are saying $\infty/\infty$ is undefined; this is not true. It is indeterminate because the answer EXISTS, only it cannot be determined in the current form. By adding $\infty$ to the expression you obtain another expression that is suddenly in determinate form.


Which is bigger $\frac{1}{{\sqrt{1001}}+ {\sqrt{1000}}}$ or $\frac {1}{10}$. What is the best way to find the inequality? I know the answer but my solution is messy.

For the time being, leave the relation $@$ undefined.
$\frac1{10}@\frac1{\sqrt{1001}+\sqrt{1000}}$
$\sqrt{1001}+\sqrt{1000}@10$
Compute $\sqrt{1001}$ and $\sqrt{1000}$.
Call their sum $x$.
$x@10$
Should $@$ be $\lt$, $=$ or $\gt$?


Finding density functions from conditional distribution I'm currently taking a statistics course, but I'm having trouble with a specific concept, and hope this is a good place to ask.

Essentially, for random variables $y_{1},y_{2}$, how do you get from the conditional distribution function

$f(y_{1}|y_{2})=something$

to the joint density function

$f(y_{1},y_{2})$

or the marginal density function? Is there an general algorithm/relationship for converting between these?

Thank you


Symmetry in ordinary differential equations Suppose I am  given an ode $${dy\over dx}={1\over x^2}f(xy)$$ where $f$ is some arbitrary function.

How then does doing the following help solve the equation? :

First I have a vector field $$V=x\partial_x-y\partial_y$$

I see that this corresponds to the transformation $$(x,y)\to (x\exp(p), y\exp(p))$$ where $p$ is a parameter.

Then I sought invariant coordinates $q_1,q_2$ of the vector field such that $$V(q_1)=1, V(q_2)=0$$

In this case I could take $q_1=xy$ and $q_2=\log|x|$. Though there might be other choices that are more suitable?

Please help!

The vector V is one of the Lie point symmetries of the ODE. By knowing the maximal Lie algebra you can use other choices too. Also, because of the existence of the arbitrary function f(x,y), by group classification, you can find specific forms of the function f that give more symmetries and more integrating factors.

Sorry, I couldn't follow your hint. So here is a different idea which hopefully simplify the original equation.
$$
u(x):=x\,y(x)$.
$$
Then
$$
y(x)=\frac{u(x)}{x},
$$
so we obtain
$$
u'x-u=f(u).
$$
From this
$$
\frac{du}{dx}x=u+f(u),
$$
consequently
$$
\frac{1}{u+f(u)}du=\frac{1}{x}dx,
$$
if $f(u)\neq -u$. (If it is, then $\frac{du}{dx}x=0$, and thus $u$ is a constant function.) Now integrating both sides we get
$$
\int \frac{1}{u+f(u)}du=\ln(|x|).
$$


Distance between two hyperplanes I have two parallel hyper planes $$a^Tx=b_1,a^Tx=b_2$$ where $a \in \mathbb{R}^n, x \in \mathbb{R}^n ,b \in \mathbb{R}$ and I want to find the distance between the two.
I have read that the distance between the two hyperplanes is also the distance between the two points $x_1$ and $x_2$ where the hyperplane intersects the line through the origin and parallel to the normal vector $\vec a$. These points are given by $$x_1=\frac{b_1}{\|a\|^2_2}a$$ and $$x_2=\frac{b_2}{\|a\|^2_2}a$$
Then the distance is $|x_1-x_2|$ but I don't really understand how we got $x_1$ and $x_2$.

Let $x_1$ be any point in the first hyperplane and consider the line $L$ that passes through $x_1$ in the direction of the normal vector $a$. An equation for $L$ is given by $x_1 + at$ for all $t\in\mathbb{R}$. Now find the intersection of $L$ and the second hyperplane:

$$
a^T(x_1 + a t) = b_2 \iff t = (b_2 - a^T x_1)/a^Ta = (b_2 - b_1)/a^T a
$$

Therefore the intersection point is $x_2 = x_1 + a(b_2 - b_1)/a^Ta$. The distance between these two points is the distance between the hyperplanes:

$$
\|x_1 - x_2\| = \frac{|b_2 - b_1|}{a^Ta}\|a\| = \frac{|b_2-b_1|}{\|a\|}
$$

Points $x_1$ and $x_2$ are the images of the coordinate system's origin in an orthogonal projection on the two given planes.


What does $2^x$ really mean when $x$ is not an integer? We all know that $2^5$ means $2\times 2\times 2\times 2\times 2 = 32$, but what does $2^\pi$ mean? How is it possible to calculate that without using a calculator? I am really curious about this, so please let me know what you think.

When $x \in \mathbb{N}$

You were probably taught that “exponentiation is repeated multiplication”:

$$b^x = \underbrace{b\times b\times b\times\cdots\times b}_{x\text{ times}}$$

From this simple definition, you can observe two properties:

$b^{x+y} = b^x \cdot b^y$
$b^{xy} = \left(b^x\right)^y $

For example:

$2^{3+4} = 2^7 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = (2 \cdot 2 \cdot 2) \cdot (2 \cdot 2 \cdot 2 \cdot 2) = 2^3 \cdot 2^4$
$2^{3 \cdot 4} = 2^{12} = 2^3 \cdot 2^3 \cdot 2^3 \cdot 2^3 = \left(2^3\right)^4$

We can then definite exponentation over more general sets of numbers in a way that these two properties continue to hold.

When $x \in \mathbb{Z}$

From the above rule for addition of exponents, we obtain a rule for subtraction of exponents:  $b^{x - y} = {b^x \over b^y}$, because then $b^{(x - y) + y} = b^{x-y} \cdot b^y = {b^x \over b^y} \cdot b^y = b^x$ as expected.  This lets us expand the domain of exponents to include zero and negative integers:

$$b^0 = b^{y-y} = {b^y \over b^y} = 1,\; b \ne 0$$
$$b^{-y} = b^{0-y} = {b^0 \over b^y} = {1 \over b^y},\; b \ne 0$$

When $x \in \mathbb{Q}$

If you assume that the multiplicate property of exponents holds for rationals, then $\left(b^{1 \over n}\right)^n = b^{{1 \over n} \cdot n} = b^1 = b$.  So $b^{1 \over n}$ is a number whose $n$th power is $b$.  In other words,

$$b^{1 \over n} = \sqrt[n]{b},\; b \ge 0$$

And $b^{m \over n} = \left(b^{1 \over n}\right)^m = (\sqrt[n]{b})^m$.

For example, $4096^{5/12} = \left(\sqrt[12]{4096}\right)^5 = 2^5 = 32$.

When $x \in \mathbb{R}$

I still haven't answered your question of what $2^\pi$ means.  But at this point, we can calculate $2^x$ for $x$ aribitrarily close to $\pi$.

$2^3$ = 8
$2^{3.1} = 2^{31/10} = \sqrt[10]{2^{31}} \approx 8.574187700290345$
$2^{3.14} = 2^{314/100} = \sqrt[100]{2^{314}} \approx 8.815240927012887$
$2^{3.141} = 2^{3141/1000} = \sqrt[1000]{2^{3141}} \approx 8.821353304551304$
$2^{3.1415} = 2^{31415/1000} = \sqrt[10000]{2^{31415}} \approx 8.824411082479122$
$2^{3.14159} = 2^{314159/10000} = \sqrt[100000]{2^{314159}} \approx 8.824961595059897$

As $x$ approaches $\pi$, $2^x$ approaches a limit, which is approximately $8.824977827076287$.  For the sake of making $2^x$ continuous, we define $2^{\pi}$ to be equal to this limit.

(Note that there's nothing special about decimal fractions.  I could have used the sequence $[3, {22 \over 7}, {333 \over 106}, {355 \over 113}, \ldots ]$ of best rational approximations, but that would have been less obvious.)

However, taking the trillionth root of huge powers of a number isn't very practical for calculation.  A more useful method is to use logarithms.

$\log_c y$ is defined as the number $x$ such that $c^x = y$.  From the two basic properties of exponentation, you can obtain the identities:

$\log_c (ab) = \log_c a + \log_c b$
$\log_c (b^x) = x \cdot \log_c b$

And from the latter, you get $$b^x = c^{x \cdot \log_c b}.$$ This means that if you have an exponential and logarithm function for one value of $c$, you can calculate them for any value for $b$.

Typical choice of $c$ are:

2, for convenience in working with computers
10, the base of our number system, giving "common logarithms"
$e \approx 2.718281828459045$, the base of the "natural logarithm" ($\ln$), for its convenient properties in calculus.

So, if you wanted to calculate $2^{\pi}$, you'd actually calculate $10^{\pi \cdot \log_{10} 2}$ or $e^{\pi \cdot \ln 2}$.  And that would typically be done with the assistance of a logarithm table or a slide rule.

When $x \in \mathbb{C}$

In Calculus, you'll learn about Taylor series, and the well-known ones for $e^x$, sine and cosine:

$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \dots = \sum_{k=0}^{\infty} \frac{x^k}{k!}$
$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \cdots = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!}$
$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \frac{x^8}{40320} - \cdots = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k+1)!}$

What happens when you plug $x = i \theta$ into the Taylor series for $e^x$?

\begin{align}
e^{i \theta} & = 1 + i \theta + \frac{(i \theta)^2}{2} + \frac{(i \theta)^3}{6} + \frac{(i \theta)^4}{24} + \frac{(i \theta)^5}{120} + \frac{(i \theta)^6}{720} + \frac{(i \theta)^7}{5040} + \frac{(i \theta)^8}{40320} + \cdots \\[10pt]
& = 1 + i \theta + i^2 \frac{\theta^2}{2} + i^3 \frac{\theta^3}{6} + i^4 \frac{\theta^4}{24} + i^5 \frac{\theta^5}{120} + i^6 \frac{\theta^6}{720} + i^7 \frac{\theta^7}{5040} + i^8 \frac{\theta^8}{40320} + \cdots \\[10pt]
& = 1 + i \theta - \frac{\theta^2}{2} - i \frac{\theta^3}{6} + \frac{\theta^4}{24} + i \frac{\theta^5}{120} - \frac{\theta^6}{720} - i \frac{\theta^7}{5040} + \frac{\theta^8}{40320} + \cdots \\[10pt]
& = \left( 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24} - \frac{\theta^6}{720} + \frac{\theta^8}{40320} - \dots\right) + i \left(\theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \frac{\theta^7}{5040} + \cdots \right) \\[10pt]
& = \cos\theta + i \sin\theta
\end{align}

This is called Euler's formula, and it lets us extend exponentiation to the complex numbers:

$$e^{x+iy} = e^x \cdot e^{iy} = e^x (\cos{y} + i \sin{y})$$

Euler's identity would be another use of exponents outside of integers as there are some complex numbers used in the identity.  Euler's formula explains how to evaluate such values which does help in some cases to evaluate the function.


How to differntiate $\int_{0}^{2\pi} u(re^{i\theta}) d\theta$? Suppose $u$ is a twice continuously differentiable function on $a< |z|<b,  \  z\in \mathbb C,$ which is harmonic that is, it satisfies $u_{rr}+\frac{1}{r}u_r + u_{\theta \theta}=0.$ (If we put $z=x+iy,$ then $x=r \cos \theta, y =r \sin \theta, r=|z|$ (polar coordinates)).

Define $f(r):= \frac{1}{2\pi} \int_{0}^{2\pi} u(re^{i\theta}) d\theta$

  My Question is: Is $f(r)$ diffrentiable?  Can we expect $\frac{d^{2}f}{dr^{2}} +\frac{1}{r} \frac{df}{dr}=0$?

My vague attempt: $f'(r)= \frac{df(r)}{dr}=  \frac{1}{2\pi} \int_{0}^{2\pi} \frac{d}{dr}[u(re^{i\theta})] d\theta;$ (can we interchanged derivative and integrals; do I need to use chain rule;$\frac{d}{dr}[u(re^{i\theta})]$ = ? )

Edit: Does there exist $g\in L^{1}(\mathbb T)$ such that $|\frac{\partial} {\partial r}h(r, \theta)| \leq g(\theta)$? (If this this the case then we $f(r)$ is diffrentiable and we can take derivative inside the integral sign [Folland Real Analysis: page 56. Theorem 2.27])

You can interchange the integral over $\theta$ and the differential by $r$. Then you obtain:

$f'(r) = \frac{1}{2 \pi} \int_0^{2 \pi} \frac{du}{dr}(re^{i \theta}) d \theta$.

Now you can substitute the harmonicity condition $u_r = -r(u_{rr}+u_{\theta \theta})$:

$f'(r) = - \frac{r}{2 \pi} \int_0^{2 \pi} (u_{rr}(re^{i \theta}) + u_{\theta \theta}(re^{i \theta}))d \theta$.

The integration over $u_{\theta \theta}$ will vanish due to the periodicity of $u_{\theta}$ and the $u_{rr}$ term can be evaluated with the chain rule (it is easy). Now consider $|f'(r)|$ and use the identity $|\int_0^{2 \pi} A(\theta) B(\theta,r)d \theta| \leq \int_0^{2 \pi}  |A(\theta)| d\theta\int_0^{2 \pi}  |B(\theta,r)| d \theta$.

You will get a finite value for the maximum estimation since $u$ is assumed to be twice differentiable.


Conditional Probability Question: What is the probability that the plant will be alive when you return? Before leaving for a two week trip to Hawaii, you ask a neighbor to water a plant that you have at home. You are 85% certain that the neighbor will remember to water the plant. You also estimate that without water, the plant will die with probability 90%, and with water, it will survive with probability 90%.
What is the probability that the plant will be alive when you return?

I am thinking I have to solve P[alive]=P[alive|water]*P[water]+P[alive|no water]*P[no water] but I'm not sure if this is the correct interpretation.

I have P[alive|water]=.9; P[water]=.85; P[no water]=.15 but I am having trouble solving P[alive|no water]. I would assume I would need to solve P[alive and no water]/P[no water] but I'm not sure. Any help would be appreciated. Thanks

What you did is correct. Now
$$P(\mbox{alive} | \mbox{no water}) = 1- P(\mbox{die} | \mbox{no water}) = 1 - 0.9 = 0.1$$
and you can conclude the rest.

To produce the probability of plant being alive, use the numbers you have.

(Certainty of Being Watered)(Plant will survive with Water)+(Certainty of Not being Watered)(Plant surviving without Water)


Is "$X$, but $Y$" logically equivalent to "$X$ and $Y$"? While reading about Mathematical Logic in a book, I found the following,

  Conjunction. If $X$ is a statement and $Y$ is a statement, the
  statement "$X$ and $Y$" is true if $X$ and $Y$ are both true, and is false
  otherwise. ...

  ...Interestingly, the statement
   "$X$, but $Y$" is logically the same statement as "$X$ and $Y$", but
  they have different connotations (both statements affirm that $X$
  and $Y$ are both true, but the first version suggests that $X$ and
  $Y$ are in contrast to each other, while the second version suggests
  that $X$ and $Y$ support each other). Again, logic is about truth,
  not about connotations or suggestions.

It is the second part which I can't understand. Precisely my questions are,

How does the statement "$X$, but $Y$" is interpreted?
How does both statements affirm that $X$
and $Y$ are both true?
What is meant by saying that, "$X$, but $Y$" suggests that $X$ and
$Y$ are in contrast to each other, and "$X$ and $Y$" support each other?

"Her judgement is firm, but fair" is both firm and fair with the connotation that judgement is not often both.   The qualities are held in contrast.

"This ice-cream is soft and creamy" is both soft and creamy, with the connotation that this is better than ice-cream being either alone.   The qualities reinforce each other.

Logically X and Y is the same as X but Y. There is no difference between them. There is no Objective difference. But psychologically, they influence your Subjective conclusion.

Let me bring some examples of X but Y:

"Fermat's claim was discovered after his death in the margin of a book, but he provided no proof"

X= "Fermat's claim was discovered after his death in the margin of a book"

Y= "He provided no proof"

Here, after hearing X, you expect Fermat to provide the proof too.

Z= "The proof is provided"

While, such idea is rejected by Y.

X satisfies Z but Y contradicts Z.

"There is an elementary proof for this theorem but it involves millions of steps"

X="There is an elementary proof for this theorem"

Y="This proof involves millions of step"

Here after hearing X, you expect Z

Z="There is a basic proof for this theorem"

But, role of Y is to reject such idea. Again X satisfies Z and even make some people to expect Z and Y contradicts Z. Y comes to clarify it.

"Her judgement is firm, but fair"

X="Her judgement is firm"

Y="Her judgement is fair"

Z="She might have biased judgement"

X satisfies Z and Y rejects Z