Homomorphisms into the General Linear Group Let $G$ be a finite group of order $n \geq 2$.

I want to prove that there always exists an injective homomorphism $\varphi:G \to GL_n(\mathbb R)$. Can you help?

By Cayley's theorem, we have an injective homomorphism, $$G\hookrightarrow S_n.$$ We have a natural action of $S_n$ on $\mathbb{R}^n$ by permuting the indices via: $$S_n \times \mathbb{R}^n \to \mathbb{R}^n, \ (\sigma,(x_1,...,x_n))\mapsto (x_{{\sigma}^{-1}(1)},...,x_{{\sigma}^{-1}(n)}).$$ This action induces a homomorphism $$\psi :S_n \to \mbox{Aut}(\mathbb{R}^n)=GL_n(\mathbb{R}), \ \psi(\sigma)((x_1,...,x_n))= (x_{{\sigma}^{-1}(1)},...,x_{{\sigma}^{-1}(n)}).$$

  In fact, every group action induces a group homomorphism and vice-versa: Given an action $$\rho: G \times X \to X, \ (g,x)\mapsto g.x,$$ one can get a homomorphism from $G$ via $$\phi: G \to Aut(X) , \ \phi(g)(x):=g.x$$
  I used the term "$Aut(\mathbb{R}^n)$" earlier for this reason. But in your case, since $\mbox{Aut}(\mathbb{R}^n)=GL_n(\mathbb{R})$, you might directly go to $GL_n(\mathbb{R}).$

You can easily check that the map thus defined is indeed a homomorphism. The claim follows.

I hope this helps!

Consider $n\times n$ matrices labelled by the elements of $G$ (there are $n$ of them).

Consider for every $g\in G$ the matrix $A^{(g)}= (a^{(g)}_{hk})_{h,k \in G}$ with elements
$$a^{(g)}_{hk} = \delta_{h, gk}$$
Let's check that
$$A^{(g_1)} \cdot A^{(g_2)} = A^{(g_1 \cdot g_2)}$$
Indeed we have
$$(A^{(g_1)} \cdot A^{(g_2)})_{hk} = \sum_{l} a^{(g_1)}_{hl} \cdot a^{(g_2)}_{lk}=\delta_{h, g_1 l} \cdot \delta_{l, g_2 k} = \delta_{h, g_1 g_2 k} = A^{(g_1 g_2)}_{hk} .$$

Hence we have a morphism from $G$ to $GL(n,\mathbb{Z})$.

Note that for $g\ne e$ the matrix $A^{(g)}$ has all the diagonal elements $0$ so it it not
the identity matrix. We conclude that the morphism is injective.

This morphism corresponds to the left regular representation like in the answer of @anomaly:. One can also consider the right regular representation $g\colon \ e_h \mapsto e_{hg^{-1}}$


what are the characteristics of a good error correcting code? A posteriori  information of a bit = intrinsic info. from that particular bit + extrinsic information from the code excluding that bit.

Let's say we are operating at a decent SNR per bit and expecting a decent Bit-error-rate (BER) of $10^{-5} $.

What are things that matter here?
Like for example minimum distance...
However it is clear that other things too matter.
Could someone throw some light on them.

Don't over think this.

The gist is simple. You have M bits of message and add N bits for checksums and error corrections in the data of D bits overall (D = M + N). Ideally you'd like to have N be zero. The gist is that the message efficiency is M/(M+N).

The rub here is do you design the checksums and error correction for just one error, or two errors, or three errors and so on. The more errors that you're checking for then the bigger N will get for some fixed message of size M bits.

Also it will take more bits to correct the error than it will take to just detect the error.

I guess the last consideration would be can you request that the message be resent if it is too garbled?

So for a message of size M you decide that you'll design the code to detect and correct 4 bits. But you decide that if more than 2 bits are corrupted that you'll ask for the message to be resent rather than risk that there are actually 5 or more errors in the data such that the detection/correction scheme is invalid.

If the message can't be resent then you'd have to design for more detection/correction capability.


Why is it true that every set can be classified as ordinary or extraordinary(not ordinary)? The statements of Russell's Paradox are presented below (in more or less elaborate form):

A set is called ordinary if it does not contain itself as an element.
A set is called extraordinary if it contains itself as an element.
Consider the set of all ordinary sets C.
C is either ordinary or extraordinary.
If C is ordinary then by definition of C (st.3) it should contain itself and therefore should be extraordinary. This contradicts the assumption and hence, C should be extraordinary.
But, if C is extraordinary, then it contains itself. This contradicts the very definition of C whereby it was to contain ordinary sets only. Therefore C cannot be extraordinary.

Thus, the paradox.

But.....

The problem is with the statement 4 above. It says C is either ordinary or not ordinary. This probably follows from the seemingly obvious statement:

  Every set is either of one type or not.

But, even though it seems obvious, what is the proof that it is true? In fact, there seems to be a counter-example, namely, *THE VERY SET: C. *
As statements 5 and 6 together imply, it is impossible to classify C as either ordinary or not ordinary (i.e.,extraordinary).

Shouldn't a hypothesis (here, the statement in >blockquote< ), however obvious or axiomatic it looks, be discarded as soon as a counter-example is found?

The reason people reject the definition of $C$ rather than the law of excluded middle (LEM- it has been mentioned in the comments) is because the latter seems way more intuitive: either somehting is true, or it is false; whereas the existence of $C$ relies on being able to define a "set of all objects that have property $\phi$", which is much less obvious.

However, a better reason to discard the existence of $C$ rather than the LEM, is the following: suppose for every property $\phi$, there is a set of all things that satisfy $\phi$.  Now let $P$ be any formula/property you like ($0=1$, "I am wonderwoman", "I can fly", really anything you like.). Consider $C = \{ x : x\in x \to P\}$. Then $C\in C \implies (C\in C \implies P)$ is true. However $(A\implies (A\implies B)) \implies (A\implies B)$ is an intuitionistic tautology, which means it doesn't rely on the LEM ! You can try to prove it by yourself and you'll see that nothing more than the definition of $\implies$ and modus ponens is necessary.

Therefore this implies $C\in C \implies P$, which by definition of $C$ implies $C\in C$, which together with $C\in C \implies P$, implies $P$.

Therefore $P$ is true, and I haven't used the LEM: accepting the existence of such $C$'s allows me to prove anything: that's why we discard it; so actually the paradox in question doesn't even need the LEM

This is a very odd presentation of Russell's Paradox. The introduction of the notions of "ordinary" and "extraordinary" sets actually leads to some confusion. Your statement 4 is simply not required as an explicit assumption to be discharged.

Using a variant of FOL, you would usually prove by contradiction that $\neg \exists a: \forall b: [b\in a \iff b\notin b]$ as follows:

$\exists a: \forall b: [b\in a \iff b\notin b]\space \space$ (Premise)
$\forall b: [b\in C \iff b\notin b]\space \space$ (Existential Specification, 1)
$C\in C \iff C\notin C\space \space$ (Universal Specfication, 2)
$\neg \exists a: \forall b: [b\in a \iff b\notin b] \space \space$ (Conclusion, 1, 3)

By no means then is rejecting the Law of the Excluded Middle (LEM) required to resolve Russell's Paradox (RP).


Integration of $\int (\frac{1-x}{1+x})^{\frac{1}{3}}$ I've been trying to integrate this for a long time but can't.

$$\int \left(\frac{1-x}{1+x}\right)^{\frac{1}{3}}$$

I tried assuming $\frac{1-x}{1+x}=t^3$ , also tried integration by parts but it gets stuck in a loop.

Hint. The obvious substitution here is $x=\frac{1-u}{1+u}$. Using this, we find:

$$\begin{align}
\int\left(\frac{1-x}{1+x}\right)^{\frac13}\,\mathrm{d}x
&=\int u^{\frac13}\cdot\frac{(-2)\,\mathrm{d}u}{(1+u)^2}\\
&=-2\int\frac{u^{\frac13}}{(1+u)^2}\,\mathrm{d}u.
\end{align}$$

Next, substitute $u=t^3$:

$$\begin{align}
-2\int\frac{u^{\frac13}}{(1+u)^2}\,\mathrm{d}u
&=-6\int\frac{t^3}{(1+t^3)^2}\,\mathrm{d}t.\\
\end{align}$$

This transforms the integrand to a rational function, so the rest of the problem is now just a matter of computation.

EDIT:

Since the integration of the rational function arrived at in the hint above isn't necessarily as straightforward as I initially thought, I'll go ahead and complete the solution.

First note that $$\frac{d}{dt}\left[\frac{2}{1+t^3}\right]=\frac{-6\,t^2}{(1+t^3)^2}.$$

Thus, integrating by parts we have:

$$\begin{align}
\int\frac{-6\,t^3}{(1+t^3)^2}\,\mathrm{d}t
&=\int t\left[\frac{-6\,t^2}{(1+t^3)^2}\right]\,\mathrm{d}t\\
&=\frac{2\,t}{1+t^3}-2\int\frac{\mathrm{d}t}{1+t^3}.
\end{align}$$

The integral $\int\frac{\mathrm{d}t}{1+t^3}$ can be done by partial fractions.

$$\frac{1}{1+t^3}=\frac{1}{3(1+t)}+\frac{2-t}{3(1-t+t^2)}.$$

For convenience, we may rewrite the second term as:

$$\begin{align}
\frac{2-t}{3(1-t+t^2)}
&=-\frac{-2+t}{3(1-t+t^2)}\\
&=-\frac{-4+2t}{6(1-t+t^2)}\\
&=-\frac{(-1+2t)-3}{6(1-t+t^2)}\\
&=-\frac{-1+2t}{6(1-t+t^2)}+\frac{3}{6(1-t+t^2)}\\
&=-\frac{-1+2t}{6(1-t+t^2)}+\frac{1}{2(1-t+t^2)}.
\end{align}$$

Thus,

$$\begin{align}
\int\frac{\mathrm{d}t}{1+t^3}
&=\frac13\int\frac{\mathrm{d}t}{1+t}-\frac16\int\frac{-1+2t}{1-t+t^2}\,\mathrm{d}t+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}\\
&=\frac13\ln{(1+t)}-\frac16\ln{(1-t+t^2)}+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}\\
&=\frac26\ln{(1+t)}-\frac16\ln{(1-t+t^2)}+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}\\
&=\frac16\ln{\left[(1+t)^2\right]}-\frac16\ln{(1-t+t^2)}+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}\\
&=\frac16\ln{\left[\frac{(1+t)^2}{1-t+t^2}\right]}+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}\\
&=\frac16\ln{\left[\frac{(1+t)^3}{1+t^3}\right]}+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}.
\end{align}$$

Finally, the integral $\frac12\int\frac{\mathrm{d}t}{1-t+t^2}$ can be found most easily by completing the square in the denominator. We have,

$$\begin{align}
t^2-t+1
&=\left(t^2-t+\frac14\right)+\left(1-\frac14\right)\\
&=\left(t-\frac12\right)^2+\frac34\\
&=\frac14\left(2t-1\right)^2+\frac34\\
&=\frac34\left(\frac{2t-1}{\sqrt{3}}\right)^2+\frac34\\
&=\frac34\left[\left(\frac{2t-1}{\sqrt{3}}\right)^2+1\right].
\end{align}$$

Thus, a convenient substitution would be $\frac{2t-1}{\sqrt{3}}=r\implies t=\frac{\sqrt{3}\,r+1}{2}\implies \mathrm{d}t=\frac{\sqrt{3}}{2}\,\mathrm{d}r$:

$$\begin{align}
\frac12\int\frac{\mathrm{d}t}{1-t+t^2}
&=\frac23\int\frac{\mathrm{d}t}{\left(\frac{2t-1}{\sqrt{3}}\right)^2+1}\\
&=\frac23\int\frac{\frac{\sqrt{3}}{2}\,\mathrm{d}r}{r^2+1}\\
&=\frac{1}{\sqrt{3}}\int\frac{\mathrm{d}r}{r^2+1}\\
&=\frac{1}{\sqrt{3}}\arctan{\left(r\right)}+\color{grey}{constant}\\
&=\frac{1}{\sqrt{3}}\arctan{\left(\frac{2t-1}{\sqrt{3}}\right)}+\color{grey}{constant}.
\end{align}$$

Hence,

$$\int\frac{\mathrm{d}t}{1+t^3}=\frac16\ln{\left[\frac{(1+t)^3}{1+t^3}\right]}+\frac{1}{\sqrt{3}}\arctan{\left(\frac{2t-1}{\sqrt{3}}\right)}+\color{grey}{constant}.$$

Hence,

$$\int\frac{-6\,t^3}{(1+t^3)^2}\,\mathrm{d}t=\frac{2\,t}{1+t^3}-\frac13\ln{\left[\frac{(1+t)^3}{1+t^3}\right]}-\frac{2}{\sqrt{3}}\arctan{\left(\frac{2t-1}{\sqrt{3}}\right)}+\color{grey}{constant}.$$

Finally, to obtain the integral $\int\left(\frac{1-x}{1+x}\right)^{\frac13}\,\mathrm{d}x$, we simply substitute back $t=\sqrt[3]{u}=\sqrt[3]{\frac{1-x}{1+x}}$ into the expression for the integral $\int\frac{-6\,t^3}{(1+t^3)^2}\,\mathrm{d}t$ stated above. This comes out to,

$$\int\left(\frac{1-x}{1+x}\right)^{\frac13}\,\mathrm{d}x=(1-x)^{1/3}(1+x)^{2/3}-\frac13\ln{\left[\frac{(1+x)\left(1+\sqrt[3]{\frac{1-x}{1+x}}\right)^3}{2}\right]}-\frac{2}{\sqrt{3}}\arctan{\left(\frac{2\sqrt[3]{\frac{1-x}{1+x}}-1}{\sqrt{3}}\right)}+\color{grey}{constant}.$$

Note: As an aside, it's interesting to observe that by combining the anti-derivative I found with the anti-derivative Shadock gave in terms of hypergeometric functions, we can then deduce a closed-form for the Gauss hypergeometric function $_2F_1(\frac{2}{3}, \frac{2}{3}, \frac{5}{3}, \frac{x+1}{2})$.

Hint:
$(\frac{(1-x)}{(1+x)^{1/3}}=(\frac 1{(1+x))^{1/3}}-(\frac{(x)}{(1+x))^{1/3}}   $[can you break $(\frac{(x)}{(1+x))^{1/3}}   $ more to get a suitable form?]

After edit Hint:

Try with trigonometric function.


Sum of Ito processes A stochastic processes $X_t$ is called an Ito process if $X_t$ is of the form
$$X_t = X_0 + \int_0^tu_sdB_s + \int_0^t v_sds,$$
where $B_s$ is a Brownian motion, and $u_s, v_s$ are square integrable and adapted to the filtration generated by $B_s$.

Now, suppose I have 2 Ito processes driven by two independent Brownian motions, say
$$X_t = X_0 + \int_0^tu_sdB_s + \int_0^t v_sds,$$
$$\tilde{X}_t = \tilde{X}_0 + \int_0^t\tilde{u}_sd\tilde{B}_s + \int_0^t \tilde{v}_sds,$$
where $B_s$ and $\tilde{B}_s$ are two independent Brownian motions.

We may regard $(X_t,\tilde{X}_t)$ as a two dimensional Ito processes, and by Ito's lemma $X_t + \tilde{X_t}$ is also an Ito process (one dimensional).

My question is, how can we write $X_t + \tilde{X}_t$ as an Ito process?
Specifically, I want to produce the following form
$$X_t + \tilde{X}_t = X_t + \tilde{X}_0 + \int_0^t\bar{u}_sd\bar{B}_s + \int_0^t \bar{v}_sds.$$
So that, $\bar{B}_s$ may depend on $B_s$ and $\tilde{B}_s$.
It is quite clear that $\bar{v}_t = v_t + \tilde{v}_t$. My question really concern $\bar{B}_t$ and $\bar{v}_t$.

I am sorry I don't have the answer but I just got interested in this question too. I think "m7e" is not responding something satisfactory and I came across equations where the author seems to sum Itô processes with different driving Brownian motions. Given his calculus, it looks like the following holds:

if $$dX_t = a_tdW_t^{(a)} +  b_tdW_t^{(b)} $$
then by definition $X_t-X_0=\int_0^ta_sdW_s^{a} + \int_0^tb_sdW_s^{b} $, and the author writes $$dX_t = a_tdW_t^{(a)} +  b_tdW_t^{(b)} = \sqrt{a_t^2+b_t^2} dW_t $$
where $W_t$ is brand new Wiener process.
This means that:
$$\int_0^ta_sdW_s^{a} + \int_0^tb_sdW_s^{b} = \int _0^t \sqrt{a_s^2+b_s^2}dW_s $$
and $W_t$ is only one-dimensional. I'll come back to you if I find the right theorem for this!

Usually one allows the driving Brownian motion to be multidimensional, and, correspondingly, $u$ to be vector-valued. With this convention you simply set $\bar B=(B,\tilde B)$ and $\bar u=(u,\tilde u)$.


$\omega$ is a solution of $x^2+x+1=0$, find $\omega^{10}+\omega^5+3$ I am working on a scholarship exam practice assuming high school or pre-university math knowledge. I am stuck at the question below:

  Let $\omega$ be a solution of the equation $x^2+x+1=0$. Then $\omega^{10}+\omega^5+3=.....$

My first question is how it would be possible since the discriminant of $x^2+x+1=0$ is less than $0$ so I am not sure how I can continue or start from here. The answer key provided is $2$. Please advise.

Dividing the polynomial $x^{10}+x^5+3$ by $x^2+x+1$ leaves a remainder of $2$. Plugging in $x=\omega$ yields your answer is $2$.


What Does it Really Mean to Have Different Kinds of Infinities? Can someone explain to me how there can be different kinds of infinities?

I was reading "The man who loved only numbers" by Paul Hoffman and came across the concept of countable and uncountable infinities, but they're only words to me.

Any help would be appreciated.

Suppose no one ever taught you the names for ordinary numbers.  Then suppose that you and I agreed that we would trade one bushel of corn for each of my sheep.  But there's a problem, we don't know how to count the bushels or the sheep!  So what do we do?

We form a "bijection" between the two sets.  That's just fancy language for saying you pair things up by putting one bushel next to each of the sheep.  When we're done we swap.  We've just proved that the number of sheep is the same as the number of bushels without actually counting.

We can try doing the same thing with infinite sets.  So suppose you have the set of positive integers and I have the set of rational numbers and you want to trade me one positive integer for each of my rationals.  Can you do so in a way that gets all of my rational numbers?

Perhaps surprisingly the answer is yes!  You make the rational numbers into a big square grid with the numerator and denominators as the two coordinates.  Then you start placing your "bushels" along diagonals of increasing size, see wikipedia.

This says that the rational numbers are "countable" that is you can find a clever way to count them off in the above fashion.

The remarkable fact is that for the real numbers there's no way at all to count them off in this way.  No matter how clever you are you won't be able to scam me out of all of my real numbers by placing a natural number next to each of them.  The proof of that is Cantor's clever "diagonal argument."

Just simple intuitive explanation.

How many Natural (Integers) Numbers could you count?
There are infinitely many, yet you can count them.
It's called Countable Set.

How many Real Numbers are there?
Infinitely as well (Since at least every Natural Number is a Real Number).
Yet you won't be able to count them (Intuitively, Let's say you name a number the first, then find the second, I can, for sure, find a number in between, their average which is Real Number as well).
It's called Uncountable Set.

What you are after is how we define how big is a given set.
Then you should look for Cardinality.


How to show that $\Bbb Z[x,y,z,w]/(xw-zy)$ is not a UFD I am trying to show that $R=\Bbb Z[x,y,z,w]/(xw-zy)$ is not a UFD.

Let $I=(xw-zy)$.

Let $X=x+I$, $Y=y+I$, $Z=z+I$, and $W=w+I$. My guess is that $X$ is irreducible and therefore $(X)$ is a prime ideal. Now, $XW=ZY$, so that $ZY\in(X)$. I want to show that $Z,Y\notin(X)$ and conclude that $R$ is not a UFD.

However, I cannot confirm my guess because I don't even know how $R$ looks like and how to find an argument to show $X$ is irreducible. I think it suffices to prove that $X$ is irreducible. If it is so, the argument can be applied to $Y$, $Z$, and $W$. Therefore, $Z,Y\notin(X)$.

Any help?

Since $xw = zy$ in $R$, it suffices to show that $x, w, z, y$ are all irreducible. Note that $(xy - zw)$ is a graded prime ideal in $\mathbb{Z}[x,y,z,w]$ (apply Eisenstein at the prime $w$ in the ring $\mathbb{Z}[w,y,z][x]$), so $R$ is a positively graded domain, with degree $0$ subring $\mathbb{Z}$, and if we write $\deg(a)$ for the degree of the highest nonzero component in $a$, then $\deg(ab) = \deg a + \deg b$ for any $a, b \in R$.

Now if $x = rs$ for some $r, s \in R$, then $1 = \deg x = \deg r + \deg s$, but this implies one of $\deg r, \deg s = 0$, say $\deg r = 0$. Then $r \in \mathbb{Z}$, but the only elements of $\mathbb{Z}$ dividing $x$ in $R$ are $\pm 1$ (verify this!). Thus $r$ is a unit, so $x$ is irreducible, and by symmetry, $y, z, w$ are all irreducible. Thus $xw = zy$ are two distinct factorizations into irreducibles, so $R$ is not a UFD.

By the way, $(x)$ is not a prime ideal in $R$:

$$R/(x) \cong \mathbb{Z}[x,y,z,w]/(x, xw - zy) = \mathbb{Z}[x,y,z,w]/(x, zy) \cong \mathbb{Z}[y,z,w]/(yz)$$

is not a domain. This gives another way to see that $R$ is not a UFD, as $x$ is an irreducible element that is not prime.

Consider the element $a = xw$. Then
$$ xw = xw - zy + zy \overset{mod \ I}{\equiv} 0 + zy = zy $$
so in $R$ we have $xw = a = zy$. Hence, $R$ is not a UFD.


Prove that series with $(-1)^k\cos^3(k)$ has sequence of partial sums bounded. I found that $(-1)^k\cos^3(k)$ equals $$(1-sin^2(k))cos((\pi+1)k)$$ or $$cos((\pi+3)k) + sin(k)sin((\pi+2)k) + sin(k)cos(k)sin((\pi+1)k)$$.

I wanted to flatten it into such a sum of terms that each has bounded partial sums, so the final sequence of partial sums is bounded too.

But I got the $sin(k)$ and $sin(k)cos(k)$.

I should use that $(-1)^k$ equals $cos(\pi k)$ and the trigonometric identities.

Hint: By the triple angle formula, $(\cos \theta)^3$ is a linear combination of $\cos (3\theta)$ and $\cos \theta$.

   For each $\alpha,\beta$ the partial sums of $\sum_{k=1}^\infty \cos(k\alpha+\beta)$ are bounded. (It is the difference of geometric series $\sum \exp(i(k\alpha+\beta)) - \sum \exp(-i(k\alpha+\beta))$.)

 Apply this to $\sum \cos(2k)$, $\sum \cos(2k+1)$ for the series $\sum (-1)^k \cos k$ and to $\sum \cos(6k)$, $\sum \cos(6k+3)$ for the series $\sum (-1)^k \cos (3k)$

The partial sum sequence will look like
$a_n= \sum_{k=1}^{n}(-1)^kcos^3(k)$ Now write, $4cos^3(k)=cos(3k)+3cos(k)$. Then try to use the triangle inequality.


Are the $n^\text{th}$ roots of an algebraic number algebraic? Suppose that $\alpha \in \mathbb{C}$ is algebraic over $\mathbb{Q}$, and that $n$ is a positive integer. Is it true that the roots of $x^n - \alpha \in \mathbb{C}[x]$ are algebraic over $\mathbb{Q}$?
This is certainly true when $\alpha \in \mathbb{Q}$. The case $n = 1$ is also trivial. Is the result true for arbitrary $\alpha$ and $n$? If so, how could one start a proof?

To say that $a$ is algebraic over $\Bbb Q$ is to say that there are numbers
$q_k, \ldots, q_2, q_1, q_0 \in \Bbb Q$ such that
$$
p(a) = q_k a^k + \ldots + q_1 a + q_0 = 0
$$
(where this evaluation takes place in $\Bbb C$ for the current problem).
Suppose that
$$
x^n - a = 0
$$
Then $x^n = a$. Consider the polynomial
$$
h(u) = q_k u^{nk} + \ldots + q_1 u^n + q_0.
$$
Note that all the coefficients of this polynomial are still rationals.
Let's evaluate $h(x)$ for a particular $x$ that satisfies $x^n - a = 0$:
\begin{align}
h(x)
&= q_k x^{nk} + \ldots + q_1 x^n + q_0 \\
&= q_k (x^n)^k + \ldots + q_1 (x^n)^1 + q_0 \\
&= q_k (a)^k + \ldots + q_1 (a)^1 + q_0 \\
&= 0\end{align}
hence $x$, by satisfying a polynomial with rational coefficients, is also algebraic over $\Bbb Q$.
Concrete example: from
$$
x^2 - x - 1 = 0
$$
we know that
$$
a = \frac{1 + \sqrt{5}}{2}
$$
is algebraic. What about a cube root, $u$, of $a$? Well, look at the polynomial above, with $x$ replaced by $x^3$ in every term, i.e.,
$$
x^6 - x^3 - 1 = 0.
$$
When we plug in $u$, we get
$$
u^6 - u^3 - 1 = (u^3)^2 - (u^3) - 1 = a^2 - a - 1 = 0
$$
so $u$, also, satisfies a polynomial over the rationals.

The algebraic numbers form a so-called algebraically-closed field. That means every polynomial with algebraic coefficients has an algebraic root. It follows that all the roots of a polynomial with algebraic coefficients are algebraic.


Analytical form of a joint PDF of two uniformly distributed random variables I was wondering if it is possible to write down (closed-form analytically) the joint density function of two uniformly distributed random variables ($X$ and $Y$, each on a closed interval $dX=x_2-x_1$ & $dY=y_2-y_1$) as a function of the correlation coefficient.

E.g. $f(X)=1/dX$ and $f(Y)=1/dY \Rightarrow$ support set of $f(X,Y)$ is between $(x_2,y_2)$ and $(x_1,y_1)$.

This picture on wikipedia shows the correlation on such a rectangle graphically: http://en.wikipedia.org/wiki/Correlation_and_dependence#mediaviewer/File:Correlation_examples2.svg

Is it possible to find a closed form analytical expression of $f(X,Y)$ as a function of the correlation coefficient? Furthermore, can we express $f(X+Y)$?

In general, you are talking about copulas. There are a lot of them, but they allows you to specify a joint distribution with uniform marginals.

$\frac{1}{(d(b-a)-c(b-a))}$ for two joint uniform random variables $X$ and $Y$ ranging from $[a,b]$ and $[c,d]$, respectively. You can check by integrating the formula to $1$.


Is there an isometry $\mathbb R^3\to \mathbb R^3 $ with this property? What is an isometry $\mathbb R^3\to \mathbb R^3 $ which maps $(3,1,2) \mapsto (2,2,2)$ and $(2,2,2) \mapsto (1,1,2)$?

Hint: the first map can be handled by a simple translation.  Once you've factored out that translation, consider the 'relative positions' of the points in the second map (i.e., relative to their corresponding points in the first map) - these would be $(2,2,2)-(3,1,2) = (-1, 1, 0)$ and $(1,1,2) - (2,2,2) = (-1, -1, 0)$; can you find a rotation that maps one to the other, or prove that there isn't one?  And once you've got that, can you see how all the transformations can be composed to find a final result?

Isometries should preserve the length, but $(3,1,2)$ and $(2,2,2)$ have different length (as well as $(2,2,2)$ and $(1,1,2)$. Hence, there is no such isometry.


Logic nonsense/paradox I'm not sure if this is a paradox or a nonsense or neither of both.
Anyway this is the "problem" if we can call it like that:

A: B is True
B: A is False

How can you solve it?

It is not much different from the plain liar paradox:

A: A is false

There are various more or less contrived "philosophical" attempts to resolve it, but by far the most common resolution is to deny that the statement means anything in the first place; therefore it is also meaningless to ask for its truth value.

The formalist mathematical approach to the paradox is to note that a sentence that speaks directly and explicitly about its own truth value (or, as in your original example, a set of sentences that speak mutually about each other's truth value) cannot be formalized in the systems that are commonly used as foundations for mathematics.

I fact, it is a classical result (by Tarski) that it is not possible for a sentence in formal mathematics to express "A is true" (at the same metalevel) of any (symbolically given) sentence $A$.

(See also here for additional discussion about the unwillingness of mathematics to deal with such self-reference).

I've found most satisfactory (i.e. true) the interpretation that each sentence implicitly states that is true.

Meaningful interpretation presumes true premises. The statement A:A is false converts to 'A is true and A is false' rendering it meaningless. In that sentence we are not talking about anything that exists. We are talking about nothing.

Sadly this interpretation also invalidates many branches of mathematics and e.g. the classical Banach-Tarksi -paradox, which btw was rejected by the founders due to the same idea that mathematical truths should also corresponds to reality.


The expression is not a distribution on $\mathbb{R}$ My problem is

  Show that $$u(\varphi)=\sum_{n=0}^{\infty}\frac{1}{n!} \varphi^{(n)}(0)$$ doesn't define a distribution on $\mathbb{R}.$

I don't know where to start. I would like an answer.

thank you.

Assuming $\varphi$ has a Taylor series expansion centered at $0$, which series converges at $1$, you seem to have written $\varphi(1)$ in a complicated way.


A question regarding the didactic motive for making a proof more complicated than logically necessary. This has to do with the proof given in Thurston's The Number System, on page 61.  My notation is different, and explained below.

I'm asking about a style of presentation frequently appearing in text books.  I often see derivations, etc., written out in ways that seem unnecessarily pedantic,  lengthy and/or arcane.  Sometimes, after looking at it long enough, I can see a logical or instructive reason for writing things as they were.

This is just an example.  I'm hoping someone might explain why the derivation given would be preferred over the alternative I'm suggesting.  Details follow.

My question is: What, if any advantage or insight is given by writing:

$$ x\odot \bar{p}\odot a=x\odot p\odot \bar{a}$$
$$ =x\odot \bar{a}\odot p$$
$$ =\bar{x}\odot a\odot p$$
$$ =\bar{x}\odot p\odot a$$
$$ \implies x\odot \bar{p}=\bar{x}\odot p,$$

rather than:

$$ x\odot \bar{a}\odot \bar{p}\odot a=\bar{x}\odot a\odot p\odot \bar{a}$$
$$= x\odot \bar{p}\odot \bar{a}\odot a=\bar{x}\odot p\odot \bar{a}\odot a$$
$$ \implies x\odot \bar{p}=\bar{x}\odot p?$$

I find the first version more difficult to understand and follow.  It doesn't reveal to me any significant strategy or structure.  It's just more complicated and confusing.

The operation $\odot$ is commutative, associative and cancellative.

The notation $\alpha=\left[\![a,\bar{a}\right]\!]$ is called a dyad, and represents an equivalence class of ordered pairs, where equality between pairs $\mathfrak{p}=\left\langle p,\bar{p} \right\rangle $ and $\mathfrak{x}=\left\langle x,\bar{x} \right\rangle $ means

$$\mathfrak{p}\circeq\mathfrak{x} \iff p\odot\bar{x}=\bar{p}\odot x.$$

To make this easy, in what follows $x\odot y$ may be replaced by $x+y,$ and
 $\alpha=\left[\![a,\bar{a}\right]\!]$ may be replaced by $a-\bar{a}.$

So $\mathfrak{p}\in\left[\![a,\bar{a}\right]\!]\iff \mathfrak{p}\in\alpha$ simply means
$a-\bar{a}=p-\bar{p}\iff a+\bar{p}=\bar{a}+p$.

Our objective is to show that each $\left\langle p,\bar{p} \right\rangle$ is in exactly one dyad.

The rest of this is mostly screen-scrapes from my Mathematica notebook.

We already have a Theorem 2 giving us:

IMO, that more than justifies my suggested derivation.  This is just a short outline of the structure of the proof:

I followed Thurston's derivation in the full proof.  My suggestion for showing the first inclusion $\left[\![a,\bar{a}\right]\!]\subseteq\left[\![p,\bar{p}\right]\!],$ is take the red equation in (i); transpose it; right-$\odot$ it to the blue equation in (ii); commute $p\odot \bar{a}$ with $a$; and cancel $\bar{a}\odot a$. The 2 in the right column simply refers to the theorem I already mentioned.

I'm certain Thurston was smart enough to see that possibility.  I'm wondering why he would chose to do it they way he did.

After some reflection, I can give two arguments in support of Thruston's choice to not use the most direct and obvious proof.

The most significant is that it leads his student to spend more time on the proof than might otherwise be invested.  We might liken this to the difference in experience between driving a car through a park and walking along the same road.  When we walk, we don't get to our destination as quickly, but we observe far more of our surroundings.

A second valid motivation is to encourage us to examine different possible approaches to the same problem.  If he had presented the simplest, and most obvious proof, we would likely not explore other possibilities.  By presenting us with the less obvious and direct proof, we are left to find the obvious and direct on our own.

There are, of course other possible motives.  For one, he may have expected the student to discover something significant which has escaped me.  Another is simply, that's how he learned it, so that's how he thinks it should be taught.

The most significant drawback to not presenting the direct and obvious proof is that doing so might give the impression that the most direct solution omits something logically essential or important.  That is, we might wrongly infer that the most efficient proof is flawed.

I will add that on page 7 Thurston (rightly in my opinion) asserts that what makes the proofs of the laws of basic arithmetic difficult is not that they are complicated, but that they are simple.  That is, we aren't accustomed to reasoning about them, we are used to reasoning with them.  His goal in presenting a less obvious and efficient proof may have been to encourage his student to reason about the basic laws, and not simply to apply them as expediently as possible.


How to solve these $3$ equations for three unknowns $x$,$y$,$z$? Question:

Solve:

$xy+x+y=23\tag{1}$

$yz+y+z=31\tag{2}$

$zx+z+x=47\tag{3}$

My attempt:

By adding all we get

$$\sum xy +2\sum x =101$$

Multiplying $(1)$ by $z$, $(2)$ by $x$, and $(3)$  by $y$ and adding altogether gives

$$3xyz+ 2\sum xy =31x+47y+23z$$

Then, from above two equations after eliminating $\sum xy$ term we get

$$35x+51y+27z=202+3xyz$$

After that subtracting $(1)\times 3z$ from equation just above (to eliminate $3xyz$ term) gives

$$35x +51y-3z(14+x+y)=202\implies (x+y)[35-3z]+16y-42z=202$$

I tried pairwise subtraction of  $(1),(2)$ and $(3)$ but it also seems to be not working.

Please give me some hint so that I can proceed or provide with the answer.

Hint: Put
$$X=x+1$$
$$Y=y+1$$
$$Z=z+1$$

Then we have

$$XY=24$$
$$YZ=32$$
$$ZX=48$$

Can you take it from there?

Consider first equation :

XY + X + Y = 23
=> X ( Y + 1 ) = 23 -Y
=> X = (23-Y)/(Y+1) -(a)

Consider second equation :

YZ + Y +Z = 31
=> Y (Z+1)=31-Z
=>Y=(31-Z)/(Z+1)-(b)

Consider third equation :

XZ +Z+X = 47
=> Z(X+1)=47-X
=>Z=(47-X)/(X+1)-(c)

Now put the value of z from eq c into eq b , then simply that equation , then put the simplified value of Y from eq b into eq A , then from there you will get the value of X , then put that value into eq c you will get the value of Z and finally put the value of Z into eq b to get the value of Y .


Joint distribution of $U = X + Y$ and $V = X - Y$ I have two independent continuous random variables, $X$ and $Y$, which are uniformly distributed over the interval $[0,1]$. From this I have two further random variables, $U$ and $V$, which are defined as $U = X + Y$, and $V = X - Y$.

I am trying to figure out the density function for the joint probability distribution of $(U,V)$ but am struggling. I have calculated the density functions of U and V on their own, but do not think it makes sense to simply multiply these together, as I do not think U and V are independent. Is this assumption correct?

My university lecturer suggested sketching out the range of values over which $(U,V)$ is defined, which seems to suggest he is trying to lead me towards a more intuitive solution, but I would appreciate any explanation (analytical or otherwise) which would help me understand how to solve problems such as these.

Let $g(x, y) = (x+y, x-y)$. Then $(U, V) = g(X, Y)$.

You know that $$f_{UV}(u, v) = f_{XY}(h(u, v))|J_h(u,v)|$$

where $h = g^{-1}$, $J_h$ is the jacobian matrix of $h$

So $h(u, v) = (\frac{u+v}2, \frac{u-v}2)$ $|J_h| = \frac 12$, $f_{XY}(x, y) = f_X(x)f_Y(y)$ (because $X$ and $Y$ are independent)

Therefore $$f_{XY}(x,y) = \begin{cases} 1 & \text{on the set $[0, 1] \times [0, 1]$} \\ 0  & \text{otherwise} \end{cases}$$

hence $f_{UV}(u, v) = \frac 12$ only when $h(u, v) \in [0, 1] \times [0, 1]$. That is, only on the set $A = g([0, 1] \times [0, 1])$ which is a square with vertices on $(0, 0), (1, 1), (1, -1), (2, 0)$

So finally

$$f_{UV}(u, v) = \begin{cases} \frac 12 & \text{on the set $A$} \\ 0  & \text{otherwise} \end{cases}$$

This also tells you that no, $U$ and $V$ are not independent, because the support of their joint density is not a rectangle whose sides are parallel to the axis

$X = \left(U + V\right)/2\,,\quad Y = \left(U - V\right)/2$

\begin{align}
\color{#ff0000}{\large{\cal P}\left(U,V\right)}
&=
{\partial^{2} \over \partial U\,\partial V}
\int_{-\infty}^{V}{\rm d}V'\int_{-\infty}^{U}{\rm d}U'\,{\cal P}\left(U',V'\right)
\\[3mm]&=
{\partial^{2} \over \partial U\,\partial V}
\int_{0}^{1}{\rm d}X\int_{0}^{1}{\rm d}Y\,
\Theta\left(U - X - Y\right)\Theta\left(V - X + Y\right)
\\[3mm]&=
\int_{0}^{1}{\rm d}X\int_{0}^{1}{\rm d}Y\,
\delta\left(U - X - Y\right)\,\delta\left(V - X + Y\right)
\\[3mm]&=
\int_{0}^{1}{\rm d}X\,\delta\left(U - X - \left[X - V\right]\right)
\int_{0}^{1}{\rm d}Y\,\delta\left(Y - \left[X - V\right]\right)
\\[3mm]&=
\int_{0}^{1}{\rm d}X\,\delta\left(U + V - 2X\right)\
\Theta\left(X - V\right)\ \Theta\left(1 - \left[X - V\right]\right)
\\[3mm]&=
\Theta\left({U + V \over 2} - V\right)\Theta\left(1 + V - {U + V \over 2}\right)
{ 1 \over 2}\int_{0}^{1}{\rm d}X\,\delta\left(X - {U + V \over 2}\right)
\\[3mm]&=
\color{#ff0000}{\large%
{1 \over 2}\,\Theta\left(U - V\right)\ \Theta\left(2 -\left[U - V\right]\right)\
\Theta\left(U + V\right)\ \Theta\left(2 - \left[U + V\right]\right)}
\end{align}

This expression 'define the region':

\begin{eqnarray*}
0 < & U - V < 2
\\
0 < & U + V < 2
\end{eqnarray*}


Linear Programming Word Problems A bakery has bought 250 pounds of muffin dough. They want to make waffles or muffins in half-dozen packs out of it. Half a dozen of muffins requires 1 lb of dough and a pack of waffles uses  3/4  lb of dough. It take bakers 6 minutes to make a half-dozen of waffles and 3 minutes to make a half-dozen of muffins. Their profit will be \$1.50 on each pack of waffles and \$2.00 on each pack of muffins. How many of each should they make to maximize profit, if they have just 20 hours to do everything?

I'm confused with Linear Programming word problems -

I first made - ( $x =$ cost of waffles, etc.)

$C(x,y) = 1.5x + 2y$

I need help with the other equations/inequalities. I know how to graph it.

"A bakery has bought 250 pounds of muffin dough. They want to make waffles or muffins in half-dozen packs out of it. Half a dozen of muffins requires 1 lb of dough and a pack of waffles uses 3/4 lb of dough. It take bakers 6 minutes to make a half-dozen of waffles and 3 minutes to make a half-dozen of muffins. Their profit will be $1.50 on each pack of waffles and $2.00 on each pack of muffins. How many of each should they make to maximize profit, if they have just 20 hours to do everything?"

You certainly should NOT "make x= cost of waffles, etc." since that is not what is asked.  The question is "How many of each should they make" so you should let x be the number of half-dozen packs of waffles and let y be the number of half-dozen packs of muffins.
  You are told "Half a dozen of muffins requires 1 lb of dough and a pack of waffles uses 3/4 lb of dough"  so to make x packs of waffles and y packs of muffins requires x+ (3/4)y pounds of dough.  You are also told "a bakery has bought 250 pounds of muffin dough. (I am not at all clear how you make waffles out of muffin dough but apparently you can.) You cannot use more dough than you have but you can use less: $x+ (3/4)y\le 250$.

You are told that "It takes bakers 6 minutes to make a half dozen waffles and 3 minutes to make a half dozen of muffins".  So it will require 6x+ 3y minutes to make x packs of waffles and y packs of muffins.  You are also told that
"they have just 20 hours (20 hours*60 minutes/hour= 1200 minutes) to do everything".  They cannot use more than 1200 minutes but they can use less: $6x+ 3y\le 1200$.

"Their profit will be 1.50 on each pack of waffles and 2.00 on each pack of muffins" so their profit on x packs of waffles and y packs of muffins is 1.50x+ 2.00y as you say.

The problem is to maximize the profit 1.5x+ 2.0y subject to the constraints $x+ (3/4)y\le 250$ and $6x+ 3y\le 1200$.

If we denote the number of batches of muffins as "a" and the number of batches of waffles as "b," we are then supposed to maximize the profit function

P = 2a + 1.5b

subject to the following constraints:  a>=0, b>=0, a + (3/4)b <= 250, and 3a + 6b <= 1200.  The third constraint can be rewritten as 4a + 3b <= 1000.

Use the simplex method on these coefficients, and you should find the maximum profit to be $500 when a = 250 and b = 0.  So, make 250 batches of muffins, no waffles.

You use up all the dough, have 450 minutes left, and have $500 profit, the maximum amount.


How to show $\lim_{x\to 1}2^{1/(x-1)} $ does not exist? How to show $$\lim \limits_{x\to 1}2^{\frac{1}{x-1}} $$ doesn't exist?

I understand that we have to show the left-hand limit is not equal to the right-hand limit. But how to do that?

Let $f(x)=2^{\frac{1}{x-1}}$ and $x_n:=1+1/n$. Then

$f(x_n)=2^n \to \infty$ for $n \to \infty$.

If $y_n=1-1/n$, then $f(y_n)=1/2^n \to 0$ for $n \to \infty$.

$$2^{1/(1+\epsilon-1)}$$ is unbounded for $\epsilon>0$. No limit.


Solving the recurrence relation $T(n)=T(n−1)+O(n)$ for $n > 1$, $T(1)=1$. I want to solve the recurrence relation $T(n)=T(n−1)+O(n)$ for $n > 1$,  $T(1)=1$.

what is $O(n)$? is $O(n)$ equals to constant or what?

Can anyone help me solve this?

$O(n)$ is some function $g$ that is asymptotically upper bound by the function $f(n) = n$. The definition of the Big-$O$ notation says $g(n) = O(f(n))$ if there exists some constant $c \in \mathbb{R}$ such that there exists an $n' \in \mathbb{N}$ with $g(n) \leq c f(n)$ for all $n > n'$.

Let's first develop some intuition for the recurrence by substituting $T(n-1)$ with $T(n-2) + O(n)$ the right hand side. This is valid because we are just replacing $T(n-1)$ with the image of $n-1$ under $T$.

$T(n) = T(n-1) + O(n) = (T(n-2) + O(n)) + O(n)$

The goal is to express $T(n)$ in terms of $T(1)$. Now let's substitute $T(n-2)$ to get an answer in terms of $T(n-3)$.

$T(n) = ((T(n-3) + O(n)) + O(n)) + O(n)$

We see a pattern emerging. Each time we substitute $T(n-i)$ we pick up an additional $O(n)$. In general we have $T(n) = T(n-i-1) + \sum_{k=1}^i O(n)$. Since $1 = n - (n - 1)$ we set $i$ equal to $n - 1$ in our equation.

$T(n) = T(1) + \sum_{k=1}^{n-1} O(n) = 1 + \sum_{k=1}^{n-1} O(n) = O(n^2)$

We end up with the solution $O(n^2)$ since we are summing up $O(n)$ terms equal to $O(n)$.

$\mathcal{O}(n)$ denotes Big-O notation, that is it is linear. You can substitute $\mathcal{O}(n)$ with $n$ to solve the recurrence.


Construct a function which is continuous in $[1,5]$ but not differentiable at $2, 3, 4$ Construct a function which is continuous in $[1,5]$ but not differentiable at $2, 3, 4$.

This question is just after the definition of differentiation and the theorem that if $f$ is finitely derivable at $c$, then $f$ is also continuous at $c$. Please help, my textbook does not have the answer.

How about $f(x) = \max(\sin(n\pi x),0)$ or perhaps $g(x) = |\sin(n\pi x)|$?


Can you recommend a book to learn to teach math to a child? I am looking for a book which contains some ideas on introducing a child to mathematics. I am not particularly looking for a textbook to be used as part of the teaching (though feel free to mention one if it's particularly good), but for a book suggesting teaching methods and approaches (the ultimate purpose is homeschooling).

Crucially, the emphasis should be on actually teaching mathematics, not grinding through mindless exercises (in the spirit of Paul Lockhart's Mathematician's Lament).

Clarification: I am looking for material that focuses on teaching children which are already old enough to be proficient with counting and some basic arithmetic but maybe not ready yet to eat a lot of formalism, so in the 5-12 year old range?)

Math From Three To Seven by Alexander Zvonkin is an excellent book that chronicles the author's experiences running a math circle for his children and their friends from the ages 3 to 7.  It does not purport to be an instruction manual, but there are many great ideas and all activities are described in detail.  Here's a preview.

And just to give some credence to his lessons' effect: his son Dimitri is also a mathematician!

The best book I have found so far is Polya's "How to solve it."

This article has also seems like a great resource: http://www.ams.org/bookstore/pspdf/mcl-5-prev.pdf


Can someone check my work for this problem? Newton's second law of motion At $560$ metric tons, the airbus A-$380$ is the world's largest airliner. Whats the upward force on an A-$380$ when the plane is flying (a) at a constant altitude and (b) accelerating upward at $1.1 \: m/s^2$

So I know $f=ma$

So for part a I did

$f= 560 * 10^3$ kg ($9.8 \: m/s^2$)

$f=5,488,000 N$

So for part b I did

$f=560 * 10^3$ kg ($1.1 \: m/s^2$)

$f=616,000 N$

Are these answers correct?

Think about it for a moment.

You say that the force when the airplane is flying at a constant altitude is $5,488,000$ newtons, but when the airplane is accelerating upwards, the force is $616,000$ newtons.

That means that if I am pushing the plane upward so it is flying at a constant altitude, then decrease the upward force bt $4,872,000$ newtons, the airplane will suddenly start moving upward?

You answer for part a is right.

For the second you are wrong since it is accelerating at 1.1 units. The other force acting on it is gravity which is opposite so it has to oppose this and is then also accelerating at 1.1 units so the net force is m*(9.8+1.1) units where m is mass.

So the answer is 6,104,000


Solve the equation about matrix The equation is $x^2 = x$, which $x$ is a $2\times2$ matrix. Anyone can give me some hint? Thanks!

Two different hints:

1) This is a question with 2-by-2 matrices... why not just write down good ol' $x=\begin{pmatrix}a & b\\c & d\end{pmatrix}$, compute $x^2-x$, and see what happens?

2) You could also notice that for such a matrix $x$, we must have $x^2-x=0$; hence the minimal polynomial for $x$ must divide $t^2-t=t(t-1)$. What does that tell you about the matrix?

$x^2 - x = x(x-1)$ has single linear factors.

Consider the Jordan Normal Form. This shows that you can split the matrix up into single blocks with eigenvalues of 0 or 1.


How is the derivative of the CDF of a random variable $X$ its PDF? For a continuos random variable $X$, if we have its p.d.f. $f(x)$, then the cumulative densitity function (c.d.f.) of $X$, $F(x)$, is
$$F(x) = \int_{-\infty}^{x}f(t)dt$$
We also have $$F'(x) = \frac{d}{dx}F(x)= f(x)$$

Why does the last step in the following equation yields $f(x)$?
$$\frac{d}{dx}F(x) = \frac{d}{dx} \int_{-\infty}^{x}f(t)dt = \int_{-\infty}^{x} \frac{\partial}{\partial x} f(t)dt $$

This is just the Fundamental Theorem of Calculus.

A PDF (of a univariate distribution) is a function defined such that it is 1.) everywhere non-negative and 2.) integrates to 1 over $\Bbb R$.

If we define $F(x) = \int_{-\infty}^x f(t)\ dt$, then the Fundamental Theorem of Calculus gives you the desired result.

This function, $F(x)$, is called the "cumulative distribution function," or CDF. It is defined in this manner, so the relationship between CDF and PDF is not coincidental -- it is by design.

Note that your last step is incorrect -- $x$ is the independent variable of the derivative there, and it is also the upper limit of the integral (so the resulting integral will be a function in terms of $x$). You can't move the $d/dx$ inside the integral.

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
read the 1st part of proof, then you can understand but you are assumed to know mean value theorem and squezee theorem.
hope it helps.


Trigonometry equation $\sin(x)+\cos(x)-\tan(x)=0.4$ There's some way to find $x$ here ?

$$\sin(x)+\cos(x)-\tan(x)=0.4$$

Avoid squaring wherever practicable as it immediately introduces extraneous roots

Using  Weierstrass substitution we have

$$\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}-\frac{2t}{1-t^2}=0.4$$ where $t=\tan\frac x2$

On rearrangement we have a Biquadratic Equation in $t$ with no extraneous root, find the answer here

Convert $\tan x = \frac{\sin x}{\cos x}$

Now we can write $\cos x$ as $\sqrt{1-\sin^2x}$

Now you will get an equation in sin x which you can solve.


Limit of $\frac{\pi^h-1}{h}$ as h approaches zero Can someone help me find this limit here. I only know how to use L'Hospital's rule but I want to be able to evaluate this limit without using differentiation.
$$\lim \limits_{h \to 0} \frac{\pi^h-1}{h}$$

The reason I want this limit is because just like $e$ can be expressed as $\sum_{n=0}^{\infty}\frac{1}{n!}$ I want to find a way to do the same with $\pi$ so i want to find the dervivative of $\pi^x$ without having $\pi$ in the result.

There's no way we can avoid $\pi$ here. The value of this limit is $\log\pi$. Despite the method we use to get that limit, the exact value is $\log\pi$ and there's no way to avoid that. We can, however, do this without limits.

Let's say $\pi=e^{\ln \pi}$
$$\lim_{h \to 0} \frac{e^{h\ln\pi}-1}{h}$$
Now, we take @SangchiLee's advice and say $t=h\log \pi$, so that $h=\frac{t}{\log \pi}$:
$$\lim_{t \to 0} \frac{e^t-1}{\frac{t}{\log \pi}}$$
Simplify:
$$\lim_{t \to 0} \log\pi\frac{e^t-1}{t}$$
Take the $\log\pi$ out of the limit:
$$\log\pi\lim_{t \to 0} \frac{e^t-1}{t}$$
Now, the limit on the right is equal to $1$, so we have:
$$\log\pi\cdot 1=\log\pi$$

You don't need that limit to compute the derivative of $a^x$ though.  ($a$ a constant)

Since $a = e^{ \ln a}$ (log and exponential are inverse functions), you have $a^x = e^{(\ln a )x}$, so $$ \frac{d}{dx}(a^x) = \frac{d}{dx} e^{(\ln a)x} = (\ln a) e^{(\ln a) x}$$


Subspaces in $\mathbb R^{n \times n}$ If our space is $\mathbb{R}^{n \times n}$, then can we definitely conclude that

$n \times n$ invertible matrices belong to its subspace
the set of all $n\times n$ matrices of the form $k*\mathcal{I}$, where $\mathcal{I}$ is the identity matrix, and $k$ is any real number also belongs to its subspace.

What kind of example can I give to show the following?

the set of matrices similar to $A$ are not necessarily subspaces of $\mathbb{R}^{n\times n}$. We're assuming that $A$ is a fixed matrix
the set of all matrices that commute with $A$ do not fall into the subspace of $\mathbb{R}^{n\times n}$. We're assuming that $A$ is a fixed matrix as well.

Are these assumptions correct?

Let's start by the definition

  If $V$ is a vector space on a field $K$ and $W$ is a subset of $V$, then $W$ is a subspace if

  The zero vector is in $W$
  $W$ is closed under addition and multiplication by a scalar in $K$

Let us see now if the sets that you gave us are indeed subspaces o $\mathbb{R}^{n\times n}$:

The set of all invertible $n\times n$ matrices: being invertible means that $\det(A)\neq 0$. This is enough to say that this cannot be a subspace because the zero vector is the zero matrix which clearly has zero determinant and so it's not an invertible matrix
$n\times n$ matrices of the form $k\mathcal{I}$: this indeed is a subspace because it has the zero vector (just set $k=0$) and is closed under addition $$k\mathcal{I}+w\mathcal{I}= (k+w)\mathcal{I} = n\mathcal{I}$$ and it's closed under scalar multiplication which is clear even from the definition $k\mathcal{I}$
All matrices similar to a matrix $A$: this is trickier because clearly if $A$ is the null matrix than all the matrix similar to $A$ is just the zero matrix. In this case this is a subspace because it has zero matrix, adding zero matrices together get's you zero matrices, and multiplying by a scalar the zero matrix gives the zero matrix. In all other cases this is not a subspace because of the lack of zero vector.
All matrices that commute with $A$: if two matrices commute then $$[A,B]= AB-BA = 0$$ let's see if this is a subspace: the zero matrix commutes with every matrix, in fact $$[A,0] = A0-0A = 0-0=0$$ For every matrix $B$ that commute with $A$, the matrix $kB$ $$[A, kB] = AkB-kBA = kAB-kBA = k[A,B] = 0$$ commutes as well with $A$. If two matrices $B$ and $C$ commutes with $A$, then $$\begin{align}&[A,B+C] = A(B+C)-(B+C)A = \\ &AB + AC - BA - CA = \\ &(AB-BA)+(AC-CA) = \\ &[A,B]+[A,C] = 0+0=0\end{align}$$ their sum commute. So this is a subspace.

I think you should rephrase your question.

You have a given dimension space of ℝn×n, and wish to identify which of the following are subspaces?

n×n invertible matrices belong to its subspace
the set of all n×n matrices of the form k∗I, where I is the identity matrix, and k is any real number also belongs to its subspace
the set of matrices similar to A. We're assuming that A is a fixed matrix
the set of all matrices that commute with A. We're assuming that A is a fixed matrix as well.

Your previous question was a bit jumbled to understand. You're making the assumption that 1 and 4 hold true, but 2 and 3 don't. You need to justify these claims first.


What are all real solutions of equation $x^x = y^y$? What about integer solutions and rational solutions? What are all real solutions of equation $x^x = y^y$? What about integer solutions and rational solutions?

For the case when $x=y$ it is clear, what about the case when $x\ne y$?

$x^x=y^y$ is obviously equivalent to $(1/y)^{1/x}=(1/x)^{1/y}$, and that's a very popular problem at MSE (e.g. $x^y = y^x$ for integers $x$ and $y$ containing also the general parametric solution).

"OP" asked the query:
$x^y=y^x$
Above has numerical integer solution:
$(x,y)=(2,4)$


Probability that coin will fall into a square So the exercise is this:

We have and infinite chessboard and we have a coin. Every grid is of length and width $a$, whereas the coin has diameter $2 \cdot r<a$. We throw a coin into a chessboard and we want to know with what probability the coin will falll into the grid.

So let $S_{1}$=area of green rectangle, $S_{2}=S_{1}+$ area of the red border.
So the probability in my opinion is
$ P(a)= \frac{S_{1}}{S_{2}} $.

Is this in any shape or form correct?

The rounded corners of your red area are a mistake.

The best way to look at the problem is to reformulate

  The coin falls completely inside a single square.

to

  The center of the coin is at least one coin radius from any grid line.

You can then shade the area of a grid square where the coin center cannot fall -- this shaded area will look the same in every square. If the coin radius is $r$ and the grid square sides have length $d$, there's a small square of side length $d-2r$ in each square where the center can fall without the coin extending beyond the grid square.

So the probablity of staying within the square becomes
$$ \frac{(d-2r)^2}{d^2} $$
assuming that the position of the coin's center relative to the square it happens to land on is uniformly distributed.

The term completely inside the circle means that we have to exclude the areas where the coin may or may not lie within the square region. Hence the probability will be $$(d-4r)^{2}/d^{2}.$$

In addition to the above
the probability that the coin will never be inside the square will be
((d^2 -(d-2r)^2))/d^2)


How to find the average case complexity of Stable marriage problem(Gale Shapley)? Stable marriage problem
So I know that the worst case complexity will be of the order O(n^2)
and the best case is of the order O(n).
But how can I calculate the average case complexity

Average case efficiency: $$\mathcal{O} \left( n \log n \right)$$


Prove span of S is a subspace I would like to know if my proof is correct/reasonable.

I wish to prove the following: Let $V$ be a vector space over $F$. and $S$ is a subset of $V$. Prove $span(S)$ is a subspace of $V.$

I just want to know whether I am on the right track or I am completely off. Thanks

My solution:

Suppose $V$ is a vector space over some field $F$ that contains the vectors $v_{1}, v_{2}, .., v_{k}, .., v_{n}$. Take S to be the subset $\{v_{1}, v_{2}, .., v_{k}\}$

Then $span(S) = c_{1}v_{1}+c_{2}v_{2}+ .. +c_{k}v_{k}$ for some $c_{1}, c_{2},.., c_{k} \in F$. We want to show that this is a subspace of V.

Clearly, ${\bf0} \in span(S)$ when $c_{1},..,c_{k} = 0$.

Suppose ${\bf x}, {\bf y}\in span(S)$ where ${\bf x} = c_{1}v_{1}+c_{2}v_{2}+..+c_{k}v_{k}$ and ${\bf y} = d_{1}v_{1}+d_{2}v_{2}+..+d_{k}v_{k}$ for scalars $c_{1},..,c_{k}, d_{1},..d_{k} \in F$. Also let $\lambda \in F$.

Consider ${\bf x} + \lambda {\bf y} = c_{1}v_{1}+c_{2}v_{2}+..+c_{k}v_{k} + \lambda (d_{1}v_{1}+d_{2}v_{2}+..+d_{k}v_{k})$

$=(c_{1}+\lambda d_{1})v_{1}+(c_{2}+\lambda d_{2})v_{2}+..+(c_{k}+\lambda d_{k})v_{k}$.

Since $(c_{k}+\lambda d_{k}) \in F$, ${\bf x} + \lambda {\bf y} \in span(S)$. Hence $span(S)$ is a subspace of $V.$

Hints:

The span $[S]$ by definition is the intersection of all sub - spaces of $V$ that contain $S$. Use this to prove all the axioms if you must. The identity exists in every subspace that contain $S$ since all of them are subspaces and hence so will the intersection. The Associativity law for addition holds since every element in $[S]$ is in $V$. The additive inverse also exists by the argument that any element in $[S]$ is also in every subspace that contains $S$ and the inverse exists in each of these subspaces and hence in the intersection. Argue down a similar path.

This is a lot more easier if you've been taught a Theorem to characterise subspaces.


Complex numbers, polynomials Let $a$ be complex number such that $a^5 + a + 1 = 0$. What are possible values of $a^2(a - 1)$?

I have tried to find $a$. Is there any way to find it?

$$
\begin{align}
a^5+a+1&=(a^2+a+1)(a^3-a^2+1)\\
&=(a^2+a+1)(a^2(a-1)+1)
\end{align}$$

If $a$ vanishes the second factor then $a^2(a-1)=-1$.

If $a$ vanishes the first factor, then divide $a^2(a-1)$ by $a^2+a+1$. We get $$a^3-a^2 = (a-2) × (a^2+a+1)+a+2$$

Therefore $a^2(a-1)=a+2$. We know that the roots of $a^2+a+1$ are $\frac{-1\pm\sqrt{1-4}}{2}$. Then only need to add $2$ to them.

In conclusion, the possible values are

$$-1, \frac{3\pm\sqrt{-3}}{2}.$$

Groebner Basis is a general method which (among other things) allows variable elimination. In the original system of equations we want to exclude variable a (aka "x") so we obtain equation over a^2(a−1) (aka "y"), hopefully of small polynomial degree. Derivation is entirely mechanical; one just copied and pastes CAS command

 GroebnerBasis[{x^5+x+1=0, x^2(x−1)=y} ,{x,y},{x}]

into WolframAlfa or some such and witnesses

 {3 - 2 y^2 + y^3}


Trace operator in Least Squares Classification I am reading Bishop's PRML book and got a bit stuck with equation 4.15 in the part that covers application of least squares method in classification:

$E_D(\tilde{\mathbf{W}}) = \frac{1}{2}Tr\{(\tilde{\mathbf{X}}\tilde{\mathbf{W}}-\mathbf{T})^T(\tilde{\mathbf{X}}\tilde{\mathbf{W}}-\mathbf{T})\}$

What is the trace operator doing in the matrix/vector sum of least squares?  Can someone provide a simple and intuitive example that illustrates the squared error and trace operator connection?

Thank you very much in advance!

Vlad

This is simply a matrix form of the sum-of-squares error function. To see this, note that:

The rows of $\tilde{\mathbf{X}} \tilde{\mathbf{W}}$ are your predictions (one row per sample);
Thus, the entries of $\tilde{\mathbf{X}} \tilde{\mathbf{W}} - \mathbf{T}$ are the errors, where the row indicates which sample we're viewing, and the column which component of that sample;
The $i$-th diagonal entry of $\mathbf{A}^T \mathbf{A}$ is the sum of the squared elements of the $i$-th column. So, the $i$-th diagonal entry of $(\tilde{\mathbf{X}} \tilde{\mathbf{W}} - \mathbf{T})^T (\tilde{\mathbf{X}} \tilde{\mathbf{W}} - \mathbf{T})$ is the sum of the squared errors made in the $i$-th component, taken over all samples;
At the end, we apply the trace operator to sum up the diagonal entries (errors).

The factor of $\frac 1 2$ is there to make things simpler when the expression is differentiated.

The trace here is the Frobenius inner product and the function to be minimized is a squared (Frobenius) norm.


Number of combinations where A and B are together This question has been answered before, but I want an elaboration as to why:

If there are 10 people including A and B, how many arrangements in a line(I guess this is permutation since order does matter), how many of these arrangements are A and B together?

The answer is $2!9!$, but why? I don't understand the intuition because intuitively, it seems like it should be $2!8!$:

This is a word formation problem, we need two letters (A and B) to be side by side, the possible number of ways to arrange two letters side by side is $2!$, leaving use 8 other letters to choose from to create the remaining 8 spaces which is $8!$, so $2!8!$

My book says that we treat the pair as a single object, so there are $2!$ to arrange this object, and then $(n - 1)! = 9!$ to arrange the remaining people. But doesn't this include combinations where A and B are not together since $9!$ considers possibilities of where A is not seated next to B?

Why is my reasoning incorrect and why is the book reasoning correct (or why is $2!8!$ wrong and $2!9!$ correct)?

Intuition is very simple here, we look at A and B like it is one person, then we have to arrange 9 person, it is $9!$. One AB and $8$ others.

And we can say first A than B, or first B than A, so:

$$2!9!$$


A partition of 186 into five parts under divisibility constraints The sum of 5 positive natural numbers, not necessarily distinct, is 186. If placed appropriately on the vertices of the following graph, two of them will be joined by an edge if and only if they have a common divisor greater than 1 (that is, they are not relatively prime).

What, in non-decreasing order, are those 5 numbers? The answer is unique.

In order for the edges to exist for the square the way that they do, it must be that the numbers in the square are of the form $ab,ac,bd,cd$ for distinct primes $a,b,c,d$ with possible extra powers or terms being multiplied.  Further, in order for the total sum to be even and the singleton to not share any edges with the square, none of $a,b,c,d$ may be equal to $2$.  This is because if one of $a,b,c,d$ were $2$, then the sum of the square would be even and to have the total sum be even so too must the singleton be even.

The two smallest "bases" for the square then must be either $3\cdot 5, 3\cdot 7, 5\cdot 11, 7\cdot 11$ or $3\cdot 5, 5\cdot 7, 7\cdot 11, 11\cdot 3$ having sums $168$ and $160$ respectively.  If any of these had an additional prime factor larger than $2$, for example in the second base if $3\cdot 5$ was replaced by $3\cdot 3\cdot 5$, then the sum would become larger than the desired sum of $186$, so we learn that every term in the base is either a product of two small odd primes or is a $2$ times a product of small primes.

Further, the only prime which may be replaced out of $3,5,7,11$ while keeping the sum of the square less than $186$ would be switching out $11$ for $13$.  Any other switch makes the sum too large.

Now, we don't have many cases left to check.  If all of the terms in the square are odd, then the singleton must be of the form $2e$ for $e$ coprime to $a,b,c,d$.  In order to keep the total sum low enough, it must be a small power of $2$ or it must be $13$ (or $11$ if $13$ was switched out).  Anything larger makes the sum too big for either base.

Alternatively, if exactly one of the terms in the square is even, then the singleton must be of the form $e$ for $e$ either $1$ or a prime different than any of $2,a,b,c,d$.

As there are very few cases left to check, by finishing with brute force you can arrive at the answer and confirm that it is indeed unique.

   Factorized $3\cdot 5, 5\cdot 7, 7\cdot 11, 11\cdot 3, 2\cdot 13$, or expanded they are $15,35,77,33,26$

The solution cannot be unique.

First: $3\cdot5 $, $5\cdot7$, $3\cdot11$, $7\cdot11$ with reminder $26$

Second: $3\cdot5 $, $5\cdot7$, $3\cdot13$, $7\cdot13$ with reminder $6$


Counting numbers of fruit baskets Suppose you have $10$ apples, $12$ bananas, and $8$ peaches, and you want to divide them into $3$ baskets containing $10$ fruit each.

In how many ways can you do this, if the fruit of each type is indistinguishable but the baskets are numbered?

(This question is a modification of A combinatorial task I just can't solve,  and

I couldn't think of a simple approach to solve it other than trying to list all the possible ways of distributing the bananas in the baskets, and then the apples.

This  is not difficult  using a  CAS though  I suspect  something more
clever is expected. The species here is

$$\mathfrak{S}_{=3}
(\mathfrak{M}_{=10}(\mathcal{A}+\mathcal{B}+\mathcal{P})).$$

This yields

$$[A^{10} B^{12} P^8]
(Z(S_{10})(A+B+P)))^3 = 1980.$$

The notation $Z(S_{10})$ refers to the cycle index of the symmetric group. We also use the fact that the cycle index corresponding to the identity acting on three slots is $Z(E_3) = a_1^3.$

Recall  the recurrence by Lovasz  for the cycle  index $Z(S_n)$ of
the multiset operator $\mathfrak{M}_{=n}$ on $n$ slots, which is
$$Z(S_n) = \frac{1}{n} \sum_{l=1}^n a_l Z(S_{n-l})
\quad\text{where}\quad
Z(S_0) = 1.$$

This makes it possible to compute these cycle indices very efficiently.
We could also compute $Z(S_{10})(A+B+P)$ by iterating over partitions of the number ten into three parts, possibly zero.

Nonetheless it should be pointed out that
what we  have here is just  algebra (no qualitative  gain) however the
complexity is  better than iterating over all  $${30\choose 10,12,8} =
3784770539550$$ assignments to the  thirty slots, converting the three
consecutive blocks of ten adjacent elements into sets and tracking all
possibilities thus obtained.  Observe that $Z(S_{10})(A+B+P)$ has just
$$[z^{10}] \frac{1}{(1-z)^3}  = {12\choose 2}  = 66$$ terms.   For the
cube  we get  $$[z^{30}]  \frac{1}{(1-z)^3} =  {32\choose  2} =  496$$
terms. These values are very reasonable for a CAS.

Remark.  There  is  no  magic  here, we  are  simply  raising  the
generating function  of the possible sets  for the three  slots to the
third power and extracting the desired coefficient.

This is my best attempt:

There are 30 pieces of fruit. Split them into 3 baskets containing 10 pieces of fruit each

basket 1
30 ways for fruit 1
29 ways for fruit 2
.....
21 ways for fruit 10

basket 2
20 ways for fruit 1
19 ways for fruit 2
....
11 ways for fruit 10

basket 3
10 ways for fruit 1
9 ways for fruit 2
....
1 way for fruit 10

Looking at it from a brute force method.

Multiply each choice together gives you the number of ways to place 30 pieces of fruit into 3 baskets of 10 each.

30*29*28.....*1
= 30!


Help to improve or give a better proof that every non-negative measurable function can be written as a sequence of simple functions Let $(X,\mathcal{A})$ be a measure space. Then for any non-negative measurable function $f$, there exists an increasing sequence $f_n$ of non-negative simple functions s.t. $f_n\nearrow f$.

The proof of this result is straightforward but I can't seem to find a neat proof this.

For a $n\in\mathbb{N}$ define

$A^k_n:=\left\{
                \begin{array}{ll}
                  [k2^{-n},(k+1)2^{-n}), & k=0,1,...,n2^n-1\\
                  [n,\infty), & k=n2^n
                \end{array}
              \right.$

To me it is clear by the definition of the $A_n^k$s that they partition $\mathbb{R}_{\geq0}$, but my first snag is showing this in a neat way. They only way I can think of to show that they are disjoint is a proof by contradiction argument. Similarly use a proof by contradiction to show that $\bigcup\limits_{k=0}^{n2^n}A_n^k =\mathbb{R}_{\geq0}$. This seems like a very tedious way to show something that is plainly evident (and not sure why this needs to be verified), so can anyone suggest a neater argument to show this?

Next define $f_n:X\rightarrow\mathbb{R}_{\geq0}$, by

$f_n(x):=\sum\limits^{n2^n}_{k=0}\frac{k}{2^n}\mathbf{1}_{A_n^k}(f(x))$.

Now as $f$ is measurable and the $A_n^k$s are Borel sets, then clearly  $f_n$ is a simple measurable function as the composition of measurable maps is measurable. The next snag I have is showing that $f_n$s are an increasing sequence, the only argument I can think of is very gawky and breaks it up into cases and is as follows:

Take a $x\in X$. If $f(x)\geq n+1$ then $f_n(x)=n<n+1=f_{n+1}(x)$.

On the other hand if $f(x)<n+1$, then there is a unique (as $A_{n+1}^k$s partition $\mathbb{R}_{\geq0}$) $k^*\in \{0,1,...,(n+1)2^{n+1}-1\}$ such that $k^*2^{-n-1}\leq f(x)<(k^*+1)2^{-n-1}$ (where in fact $k^*=\max\{0\leq k\leq (n+1)2^{n+1)}|k2^{-n-1}\leq f(x)\})$, so $f_{n+1}(x)=k^*2^{-n-1}$.

Now if $k^*$ is even then $k^*=2l$ for some $l\in\mathbb{N}_0$ and so,

$\frac{l}{2^n}=\frac{2l}{2^{n+1}}\leq f(x)<\frac{2l+1}{2^{n+1}}<\frac{2l+2}{2^{n+1}}=\frac{l+1}{2^n}$,

thus $f_n(x)=\frac{l}{2^n}=f_{n+1}(x)$.

If $k^*$ is odd then $k^*=2m+1$ for some $m\in\mathbb{N}_0$ and so,

$\frac{m}{2^n}=\frac{2m}{2^{n+1}}<\frac{2m+1}{2^{n+1}}\leq f(x)<\frac{2m+1+1}{2^{n+1}}=\frac{m+1}{2^n}$,

therefore $f_n(x)=\frac{m}{2^n}<\frac{2m+1}{2^{n+1}}=f_{n+1}(x)$.

To show that $\lim\limits_{n\rightarrow\infty}f_n=f$, is very easy so I won't bother with that. So if anyone can offer any assistance with the two snags I have or provide me with an alternative proof, it will be greatly appreciated. Thanks in advance.

A slicker way: Enumerate the non-negative rationals as $\{q_n\}$ (with $q_0=0$), and let $B_n:=\{x: f(x)\ge q_n\}$.

Convince yourself that $f(x) =\sup_n q_n\cdot 1_{B_n}(x)$, for each $x$.
Define $f_m(x) = \max_{1\le n\le m}q_n\cdot 1_{B_n}(x)$, $m=0,1,2,\ldots$.

Clearly each $f_m$ is a simple function (taking only the values $q_0,q_1,\ldots,q_m$) and the sequence $\{f_m\}$ increases pointwise to $f$.

Why do you want to prove such a fact in a measure theory course? If you pressed me, I'd say that this is simply the uniqueness of the $n$th digit in a binary decimal expansion.
Fix $n$. Consider the sets $U^k_n = f^{-1}(A^k_n)$. As these partition the domain $X$, it is enough to prove that $f_{n+1} \ge  f_n$ on each of these sets. The base case is trivial. Suppose $k\le n2^n - 1$. Note then that $f_n$ restricted to $U^k_n$ is just $ \frac k{2^n} \mathbf1[\frac{k}{2^n} \le f < \frac{k+1}{2^n}).$ (Forgive me for promoting a subscript to aid readability.)
What's $f_{n+1}$ restricted to this set? Well, since $$ \frac{k}{2^n} \le f < \frac{k+1}{2^n} \iff \frac{2k}{2^{n+1}} \le f < \frac{2k+2}{2^{n+1}}$$
by the same reasoning for $f_n$, $f_{n+1}$ is a sum of two functions,
$$ f_{n+1} = \frac {2k}{2^{n+1}} \mathbf1\left[\frac {2k}{2^{n+1}} \le f < \frac {2k+1}{2^{n+1}}\right) + \frac {2k+1}{2^{n+1}} \mathbf1\left[\frac {2k+1}{2^{n+1}} \le f < \frac {2k+2}{2^{n+1}}\right)$$
and its clear now that we have the conclusion we seek, $f_{n+1} \ge  f_n$. For $k=n 2^n$, its similar: $f_{n+1}$ restricted to this set takes values $\ge \frac{k}{2^n}$.

What makes the above not too bad to write down is the fact that the partition $(A^k_{n+1})_{k}$ is a refinement of $(A^k_{n})_{k}$.


Show that f is measurable. Let $a > 0, b \geq 0$ and the function $f: \mathbb{R} \to \mathbb{R}$

$$f(x) = \left\{\begin{matrix}
1, & |x| \leq a \\
 b & |x| > a
\end{matrix}\right.$$

show that it is measurable.

I don't have clue about to show this!!!

I assume measurability with respect to the Borel $\sigma$-algebra.

You can rewrite $f$ as:
$$f(x)=\mathbf{1}_{[-a,a]}(x)+b\mathbf{1}_{\mathbb{R}\setminus [-a,a]}(x)$$

$[-a,a]$ is an interval so it is measurable. Thus $\mathbb{R}\setminus [-a,a]$ is measurable too. This simplies that the indicator functions $\mathbf{1}_{[-a,a]}$ and $\mathbf{1}_{\mathbb{R}\setminus [-a,a]}$ are measurable functions. You can then conclude by using the fact that the sum and product of measurable functions is still measurable.

I don't think it's necessarily measurable. Did you mean Lebesgue measurable? If so, then I think so.

Let $a, b \in \mathbb R$.

Define $A := \{x | |x| \le a\}$.

Consider a measure space $(S, \Sigma, \mu)$.

$1$ is constant and hence $\Sigma$-measurable. $(**)$

$b$ is constant and hence $\Sigma$-measurable.

If $A \in \Sigma$, then:

$1_A$ is $\Sigma$-measurable. $(**)$
$A^C \in \Sigma$
$1_{A^C}$ is $\Sigma$-measurable.

If $A^C \in \Sigma$, then:

$1_{A^C}$ is $\Sigma$-measurable.
$A \in \Sigma$
$1_{A}$ is $\Sigma$-measurable.

So, if $A \in \Sigma$, then

$$f = 1 \times 1_A + b \times 1_{A^C}$$

since finite sums of $\Sigma-$measurable functions are $\Sigma-$measurable and finite products of $\Sigma-$measurable functions are $\Sigma-$measurable.

If our measure space is $(\mathbb R, \mathscr B(\mathbb R), \lambda)$, then $A \in \mathscr B(\mathbb R)$ because $A$ is an interval.

$(**)$ Note: We say a function $f$ is $\Sigma-$measurable from $(S, \Sigma, \mu)$ to $(\mathbb R, \mathscr B(\mathbb R))$ if

$\sigma(f) := \{f^{-1}(B) | B \in \mathscr B(\mathbb R)\} \subseteq \Sigma$

Now, if $f$ is a constant, then

$\sigma(f) = \{\emptyset, S\} \subseteq \Sigma$

If $f = 1_B$ for $B \in \Sigma$, then

$\sigma(f) = \sigma(B) \subseteq \Sigma \tag{*}$

$(*)$ Note: $B \in \Sigma \to \sigma(B) := \{\emptyset, B, B^C, S\} \subseteq \Sigma$


Matrix derivative with respect to matrix Say I have a $ 2 \times 2$ matrix as $A$ and a $2 \times 1$ vector as $X$.
I want the derivative of the matrix product with represent to $A$:

Let $y=
\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}$

Then what is $\frac{dy}{dA}$?

According to the standard definition of Jacobian

I should be getting a $ 2 \times 4$ matrix where columns are $a,b,c,d$ but this does not agree with the material I am reading. Is the transpose involved in any of this?

Note that $A\mapsto Ax$ is linear, so everything is trivial.

If you really want to write the Jacobian in coordinates, it's
$$
J=\begin{pmatrix}
x1 & x2& 0& 0\\
0& 0&x1&x2
\end{pmatrix}
$$
because, as you may check yourself, we have
$$
Jvec(B)= Bx
$$
for any matrix $B\in\mathbb{R}^{2,2}$ with vectorization $vec({B})\in\mathbb{R}^4$

The Jacobian, the way you wrote it, really only works after vectorizing w.r.t. the components of $A$. More generally, given a function $f\colon X\to Y$, you can think of the derivative as a $(\operatorname{rank} Y, \operatorname{rank}  X)$-tensor, which allows one to keep the matrix structure of the input here.
In this case, since the input is a matrix, i.e. a rank-$2$ tensor, and the output is a vector, i.e. a rank-$1$ tensor, you can think of $dy/dA$ the (1, 2) tensor with components
$$\Big(\frac{\rm d y}{\rm d A}\Big)^k_{ij}=\frac{\partial y_k}{\partial A_{ij}} \leadsto D = \frac{\rm d y}{\rm d A} = \begin{bmatrix} \begin{bmatrix} x_1 & x_2 \\ 0& 0\end{bmatrix},  \begin{bmatrix}0& 0 \\ x_1 & x_2\end{bmatrix}\end{bmatrix}$$
This tensor encodes a linear map given by the corresponding tensor contraction
$$D\cdot X = \sum_{ij} D^{k}_{ij} X^{ij}$$
In particular, note that for $f(A)=Ax$ we have the first order Taylor expansion
$$ T_f^{(1)}(A) = f(0) + Df(0)\cdot A =\begin{bmatrix}0 \\ 0\end{bmatrix}+ \begin{bmatrix}A_{11}x_1 + A_{12}x_2 \\ A_{21}x_1 + A_{22}x_2\end{bmatrix} = f(A)$$
I.e. the function is in fact equal to its first order Taylor expansion, as expected, since it's a linear function.


Probability of winning a tennis match with a two point difference? Let $A$ and $B$ be two tennis players, playing against each other.
A win of the game is considered when one of the two players is leading by two points (e.g. $A$ scores $5$ points vs. $B$ scoring $3$ points).
The probability of a winning a point when $A$ is serving is $\frac{3}{5}$ the probability of a winning a point when $B$ is serving is $\frac{2}{3}$

$A$ serves first and they take turns after each serve. What is the probability of $A$ winning the game.

My progress so far:

I realise that A wins the whole game when the following sequence occurs: $WLW$ (where $W$ denotes a win and $L$ denotes a loss). However, also a game win can occur when we have $WLLLW$ or $WLLLLLW$ etc. If $A$ wins a point and $B$ wins a point, the score is essentially re-set. From here on, I am unsure how to continue...

P(wins first point) = 3/5

P(wins second point) = 1/3

so he can win those two points with probability 1/5 and win in first two points - but if he wins the first point, then loses the second point, with probability 2/5 - he is back to square one - if he loses the first point and wins the second point (prob 2/15) - he is back to square one - if he loses two points - he lost (prob 4/15)

so after two points, he either wins, loses, or is back to square one with the same probability of winning - therefore
P(winning) = M = P(wins in two points) + M x P(back to a draw after two points)

P(winning) = M = (3/5)(1/3) + (3/5)(2/3)M + (2/5)(1/3)M + 0 x(2/5)(2/3)

M = 1/5 + (2/5)M + (2/15)M

M = 3/15 + (6/15)M+ (2/15)M

M = 3/15 + (8/15)M

(7/15)M = 3/15

M = 3/7

Answer - 3/7

Let E be the event sania winning the match, then $P(E)
                                        =0.63$ (given)
Since $P(\text{Sania loosing the match } )= P(\text{not } E)=P(E)
                                 =1-P(E)=1-0.63
                                 =0.37$


Finding a seating arrangement of $4$ different people in $n$ rounds I have a practical problem. I want to arrange a speeddate event with $24$ or $32$ people in $7$ rounds. I have a room with sufficiently many tables and I want $4$ people per table and each round must consist of $4$ different people per table.

Is it possible to find such an arrangement?

It is impossible for 24 and 32 people. I'll demonstrate how it is impossible for 24 people (the case for 32 is similar)

Suppose there are 24 people. This means that every round there will be 6 tables of 4. In total, there are 24C2 ways to choose a pair of people from this group of 24, and every such pair must have sat together at a table in one of the rounds. Each table contributes a maximum of 4C2 new pairs, so we have the inequality

4C2 * 6 * 7 >= 24C2

which evaluates to be 252 >= 276, which is a contradiction.

What do you wish to keep constant? Do you want a speed dating with exactly 7 rounds? Or do you want the minimum number of rounds for 24 or 32 people?


Evaluating $\lim\limits_{z \to  0} \frac{z\cdot \cos(z)}{\sin(z)}$ I'm reading Boas' chapter on functions of a complex variable, and she's talking here about finding residues. However, I don't understand the evaluation of the limit(below). How is it that we can take $\cos(0)$ out of the limit while $\sin(z)$ and $z$ stay within the limit?

$$
\lim_{z \to 0} \frac{z\cdot\cos(z)}{\sin(z)}=\cos (0)\cdot \lim_{z \to 0}\frac{z}{\sin(z)}=1\cdot 1=1
$$

You can take stuff out of the limits which are smooth (or even continuous) and non-zero about that point. Here $\cos(0) = 1$, but $z(0)$ and $\sin(z)$ although smooth, evaluate to zero.

The reason we can do this is the product rule of limits. We can actually write an intermediary step including $lim_{z\rightarrow0}\cos(z)$ which evaluates to $\cos(0)$. We can't take the other two parts $z(0)$ and $\sin(z)$ as they evaluate to zero and we'd be left with a 0/0.

I think you CAN take $z$ or $sin(z)$ out of the limit, however, you will not be able to solve the resulting equation. For example, if you take z out of the limit, you get:

$$
\lim_{z \to 0} \frac{z\cdot\cos(z)}{\sin(z)}=0 \cdot \lim_{z \to 0}\frac{\cos(z)}{\sin(z)}=0\cdot \infty=???
$$

Boas was just smart enough to arrange the equation in a way that will become solvable...


Linear Programming Problem? You are about to take a test that contains computation problems worth 6 points each and word problems worth 10 points each. You can do a computation problem in 2 minutes and a word problem in 4 minutes. You have 40 minutes to take the test and may answer no more than 12 problems. Let x represent the number of computation problems and y the number of word problems. Write the objective function to determine how many of each question type you should answer to maximize your score. Assume you answer all the problems attempted correctly.Which are constraints for this problem?

It's not system of equations, it's linear programming.

Technically this is a poorly posed question. What the variables ought to represent are the number of computation and word problems you attempt, not, as it says, the number of problems, period.

In any case, every linear programming problem has the same structure: you specify the variables; then write down a function of the variables, called the objective function, which represents the quantity you are trying to optimize (maximize or minimize); then you write down your constraints, which are usually inequalities but can also include equations; then you solve the system of constraint inequalities, which gives the "feasible region"; and finally you find the optimum value of your objective function by looking at its values on the corners of the feasible region.

The first part of the problem is done for you: they told you (if imprecisely, as I said at the beginning) what the variables are.

So the next step is to write the objective function. In your problem, you're trying to maximize the score on your test. But you get 6 points for every computation problem you get right and 10 points for every word problem. So you'd like to maximize $$6x+10y$$

That is the objective function.

Next, you write your constraints. There are two constraints in your problem: a constraint on the time you can spend (no more than 40 minutes) and a constraint on how many problems you can attempt (no more than 12).

The first constraint can be written $$2x+4y\leq40$$
because it takes you 2 minutes to attempt a computation problem and 4 minutes to attempt a word problem.

The second constraint can be written $$x+y\leq12$$

To get started I write down the first constrain, which is described in the sentence: "You have 40 minutes to take the test and may answer no more than 12 problems."

$2x+4y \leq 40$

Is this inequality understable for you ? If so, then try the next.

greetings,

calculus


A ﬁrst-term expansion of the solution of the ODE Get a ﬁrst-term expansion of the solution of $$εy''+y'+ y = 0, \text{where } y(0) = 0, y'(0) = 1$$ that is valid for large values of $t$.

I tried the usual approach with two time scales $t$ and $s=ε^a t$. Assuming $$y \sim y_0 + ε^b y_1$$ I get stuck immediately after writing it out since I can't see any obvious choice of $a$ and $b$.

Added: here is how I wrote it out: $$ε (\partial^2_t+2ε^a\partial_t \partial_s+ε^{2a}\partial^2_s)(y_0 + ε^b y_1) + (\partial_t + ε^a \partial_s)(y_0 + ε^b y_1) + (y_0 + ε^b y_1)=0$$
If we set $a=1$ and $b=1$, then $O(1)$ gives $$\partial_t y_0 + y_0 = 0$$ which can not satisfy our initial conditions.

So we want some other terms to be in $O(1)$ but both possible choices, namely $a = -1/2$ and $a = -1$, give us $1/ε$ in front of some terms.

$$\epsilon y''+y'+y=0 \quad \begin{cases} y(0)=0\\y'(0)=1\end{cases}$$

NOTE :

In special case $\epsilon=0$ the solution of $y'+y=0$ is $y=ce^{-t}$

The condition $y(0)=0$ implies $c=0$ and the solution is trivial : $y(x)=0$

So, with the specified conditions, the solution of $\epsilon y''+y'+y=0$ tends to $y(x)=0$ when $\epsilon\to 0$. But the condition $y'(0)=1$ is ignored.

BRUT FORCE SOLVING METHOD :

With the eigen values $\epsilon r^2+r+1=0 \quad\to\quad r=\frac{-1\pm\sqrt{1-4\epsilon}}{2\epsilon}$ the general solution is :
$$y=c_1 e^{r_1t}+c_2 e^{r_2t} \quad
\begin{cases}
r_1=\frac{-1+\sqrt{1-4\epsilon}}{2\epsilon} \sim -1-\epsilon\\
r_2=\frac{-1-\sqrt{1-4\epsilon}}{2\epsilon} \sim -\frac{1}{\epsilon}
\end{cases}$$
With conditions $y(0)=0=c_1+c_2$ and $y'(0)=1=r_1c_1+r_2c_2$ leading to $c_1=-c_2=\frac{\epsilon}{\sqrt{1-4\epsilon}}$
$$y=\frac{\epsilon}{\sqrt{1-4\epsilon}} \left( e^{\frac{-1+\sqrt{1-4\epsilon}}{2\epsilon} t}- e^{\frac{-1-\sqrt{1-4\epsilon}}{2\epsilon} t}\right)$$
$$y \sim \epsilon\left( e^{-t}- e^{-\frac{t}{\epsilon}}\right)$$
Bringing it back into the ODE and into the two conditions shows that it agrees for the first approximate.

This approch makes clear the difficulty when it is assumed that $y(t)\sim y_0(t)+\epsilon^b y_1(t)$ :

First, we have $y_0(t)=0$ .  Second, we have $\begin{cases} b=1\\ y_1(t)=e^{-t}\end{cases}$ . This is OK up to this point.

But the term $-\epsilon e^{-\frac{t}{\epsilon}}$is missing. This is acceptable for $\epsilon>0$ and $t$ large because if $\frac{t}{\epsilon}$ is large, the missing term $\epsilon e^{-\frac{t}{\epsilon}} \sim 0$.

On the other hand, keeping only the term $\epsilon e^{-t}$ doesn't satisfy the condition $y'(0)=1$.

Nevertheless, on the common sens of perturbation, I think that an acceptable answer to the question is :
$$y(t)\sim\epsilon e^{-t}$$


The Area of Trapezium is given by $A=\frac{1}{2}(4-x^2)(2x+4)$ The Area of Trapezium is given by: $$A=\frac{1}{2}(4-x^2)(2x+4)$$ Find the Maximum area of Trapezium.

Hi, can anyone help me with this question. I know we differentiate the equation, but i don't know what to do next. Can anyone help me.

Thanks.

The maximum of the curve (in this case of the area of the trapezium) can be found through the zeroes of the derivative
$$A'=-x(2x+4)+(4-x^2)=0.$$
This equation will lead to two solutions $x_1,x_2$ (potential maximums or minimums). What you have to do next is to demonstrate which one is a maximum. You can do it by analyzing the second derivative, $A''$ in the pints found before, $x_1,x_2$. If $A''(x_1)<0$ then $x_1$ is a maximum. In the opposite case, $A''(x_1)>0$ $x_1$ is a minimum (and same for $x_2$). I hope it is clear enough!


Showing system contains peroidic orbit Consider the IVP for the $2$-dimensional dynamical system ($X=[0, \infty )^2$) $$\dot{x_1}=a-x_1-\frac{4x_1x_2}{1+x_1^2}$$ $$\dot{x_2}=bx_1 \bigg( 1- \frac{x_2}{1+x_1^2} \bigg)$$ for all $t \in I$, and $\textbf x (0)=(x_{10}, x_{20}) \in [0, \infty )^2$, with $a,b >0$.

Show that for certain $a$, $b$, and $c$, there exists an invariant region
$$\{f(x_1, x_2) \in X : 0 < x_1 < a, \, \, 0 < x_2 < c \}$$
that contains a periodic orbit for the dynamical system.

My attempt:

So we have a rectangle and we have to choose $a,b,c$ such that the flow always remains within the rectangle. Let's consider each edge and corner of the rectangle. Imagine our graph to be $x_1$ as the horizontal axis and $x_2$ as the vertical.

At corner $(0,0)$: $\dot x_1=a$, $\dot x_2 =0$ so at this point, flow is directed $x_1$ direction so this is inside the region.

At corner $(0,c)$: $\dot x_1=a$, $\dot x_2 =0$ so at this point, flow is directed $x_1$ direction so this is inside the region.

At line $x_1=0$ so on points $(0,l)$ with $l \in (0,c)$: $\dot x_1=a$, $\dot x_2 =0$ so at this point, flow is directed $x_1$ direction so this is inside the region.

At corner $(a,c)$: $$\dot x_1 =-\frac{4ac}{1+a^2}<0, \, \, \forall a,c$$

$$\dot x_2 =ba \bigg( 1- \frac{c}{1+a^2} \bigg)$$ with $ba>0, \, \, \forall a,b$ but we need the sign of $\dot x_2$ to be $\leq 0$ and this is dependent on the sign of the following (so we WANT the following): $$ 1- \frac{c}{1+a^2} \leq 0$$ so then the flow at this corner will go negative $x_1$ direction or down within the region. Either way, flow will still be in the region.

At line $x_2=0$ so on points $(q,0)$ with $q \in (0,a)$: $\dot x_1 =a-q >0, \, \, \forall a,q$, and $\dot x_2 =bq>0$. So the flow is directed in the region.

At line $x_1=a$ so on points $(a,p)$ with $p \in (0,c)$: $$\dot x_1 = - \frac{4ap}{1+a^2}<0, \, \, \forall a,p$$
$$\dot x_2 =ba \bigg( 1- \frac{p}{1+a^2} \bigg)$$ We are unsure on the sign of this but regardless the flow will remain in the region since $\dot x_1 <0$.

At line $\dot x_2 =c$, so on points $(t,c)$ with $t \in (0,a)$: $$\dot x_1 = a-t -\frac{4tc}{1+t^2}<0 $$

$$\dot x_2 =bt \bigg( 1- \frac{c}{1+t^2} \bigg)$$ We are unsure on the signs of both of these but we WANT $\dot x_2 \leq 0$ and then regardless of the sign of $\dot x_1$, the flow will remain in the region.

So we need the flowing inequalities to hold: $$1- \frac{c}{1+a^2} \leq 0$$ $$1- \frac{c}{1+t^2} \leq 0$$ Choose $c=66$ and $a=8$ and then they hold.

Next we need our fixed point to be out of the region and this happens when $$1 + \frac{a^2}{25} < c$$ and this is true for our chosen values of $a$ and $c$. Hence by Poincare-Bendixon theorem, there exists a periodic orbit.

IS THIS CORRECT? PLEASE COMMENT!!

I think we have to use this:

Poincare-Bendixon: Consider $\dot x = f(x)$ with $f \in C^1(X)$ (continuous on first differential) and $X \subset \mathbb R^2$. If the forward orbit, $O^+(x_0)$ (for $x_0 > X$) is
contained within a closed bounded set $D$ containing no fixed points, then
$\omega (x_0)$ contains a periodic orbit.

Generalized Poincare-Bendixon: Let the hypothesis of the above hold, except now, let $D$ contain fixed
points $p_1, ..., p_N$. Then $\omega (x_0)$ is one of the following;

1) a fixed point $p_i$.

2) a periodic orbit.

3) a finite number of fixed points $p_i$ , and a countable number of
homoclinic/ heteroclinic orbits between $p_i$ and $p_j$.

Hint:

Find the directions of the vector field along the sides of the rectangle $[0,a]\times[0,c]$, for some carefully chosen $c$. For example, along the side $x_1=0$ you have $\dot x_1=a$ and $\dot x_2=0$, showing that the vector field points inside the rectangle. Similarly, for $x_2=0$ you have $\dot x_1=a-x_1$ and $\dot x_2=bx_1$, and again the vector field points inside the rectangle (you see the pattern).

After that study the vector field in a neighborhood of the equilibrium point.

A minor comment about the wording: you should really write "forward invariant" instead of "invariant", and "equilibrium point" instead of "fixed point".

to make life easier, i am going to pick $a = 5, b = 1$ so that we have one rest point $x_1 = 1, x_2 = 2.$ we will make a change of variable
$$\xi_1 = x_1 - 1, \xi_2 = x_2 - 2 \text{ and } x_1 = 1+\xi_1 , x_2 = 2+\xi_2.$$  the governing equations for $\xi_1, \xi_2$ are:

$$\dot \xi_1 = 5 - (1+\xi_1 ) - \frac{4(1+\xi_1)(2+\xi_2)}{1+(1+\xi_1)^2}\\
\dot \xi_2 = (1+\xi_1)\left( 1- \frac{2+\xi_2}{1+(1+\xi_1)^2}\right) $$ which simplifies to  $$\dot \xi_1 = \frac{-2\xi_1-4\xi_2+2\xi_1^2-\xi_1^3-4\xi_1\xi_2}{1+(1+\xi_1)^2}\\
\dot \xi_2 = \frac{(1+\xi_1)(2\xi_1-\xi_2+\xi_1^2)}{1+(1+\xi_1)^2}$$
the associated linear system has the coefficient matrix $\pmatrix{-1&-2\\2&-1}$ with the eigenvalues $-1\pm 2i$ suggesting that the fixed point is a stable spiral.

i will now now do transformation to polar coordinates $$\xi_1 = r\cos \theta, \xi_2 = r \sin \theta. $$

i will come back to this tomorrow if you don't get to it by then.


Is this the standard definition of cumulative distribution function? A cumulative probability function is usually defined as:

$$F_X(x)=P(X\leq x)$$

But is it ever also sometimes defined as follows?
$$F_X(x)=P(X< x)$$

Is there a particular reason why the second one would be problematic, or incorrect, or less general?

The 2nd definition is less general, used by some Hungarian mathematicians.

Note if you use the 1st definition, you CDF is right-continuous; if you use the 2nd definition, your CDF is left-continuous and related definition and theorem will change a little bit.

Ok, just remember use the 1st one and that CDF always right-continuous.


Etymology of transpose of morphisms in an adjunction Let $F: \mathbf{C} \to \mathbf{D}$ be left adjoint to $G : \mathbf{D} \to \mathbf{C}$, witnessed by the family of bijections between hom-sets, natural in objects $X, Y$: $$\operatorname{Hom}_{\mathbf{C}}(X,GY) \overset{\psi_{X,Y}}{\longrightarrow} \operatorname{Hom}_{\mathbf{D}}(FX,Y).$$

I've noticed that among some authors, the image $\psi_{X,Y}(f)$ of a map $X \overset{f}{\to} GY$ is called the transpose of $f$. Does this have anything to do with a categorical realization of transpositions in linear algebra? Is there a canonical motivating example for adjoint functors involving transposes of linear operators, involving (say) adjoint representations of Lie groups?

The etymology of the word "adjunction" is the characterization of the adjoint, or transpose, of a linear map between inner product spaces by the formula $\langle Av,w\rangle=\langle v,A^T w\rangle$. I guess that the use of "transpose" to indicate passing a morphisms across adjunction comes out of this history. But this seems to be only a psychological, not a technical, explanation. In particular, it's difficult to realize transposes in linear algebra as adjunctions in category theory.

Let $V$ and $W$ be finite free $R$-modules. Since the transpose reverses multiplication of matrices, evaluating the transpose $f^T$ of a map $f: V \to W$ at an element $(w_1, \dots, w_n)$ of $W$ amounts to precomposing the functional $[w_1, \dots, w_n]$ by $f$. So $f^T$ is just the image of $f$ under $\operatorname{Hom}_{R\text{-}\mathbf{Mod}}(-,R),$ which we'll call $(-)^{T}$.

With a little thought and a not-so-little amount of verification and diagram-chasing, one can see that $(-)^{T}$ is right adjoint to its opposite functor $\big((-)^{T} \big)^{\operatorname{op}}$ as functors $$\left((-)^{T}\right)^{\operatorname{op}} : R\text{-}\mathbf{Mod} \leftrightarrows R\text{-}\mathbf{Mod}^{\operatorname{op}} : (-)^{T}.$$

In particular, this restricts on finite free $R$-modules to an isomorphism of categories.


Solve for $x$ in $3^x=5^2$ by logarithm. Was solving using properties of logarithm but got stuck at the equation $x\log 3=\log5+\log5$

Hint: $\log 3$ and $\log 5$ in your equation are just constants. How would you solve for $x$ if they were replaced by say, $1$ and $2$?

$$3^x=5^2\implies \log3^x=\log5^2\implies x\log3=2\log5\implies x=\frac{2\log5}{\log3}$$
I hope that helps :)


Calculate the unit normal vectors to the both sides of a plane Calculate the unit normal vectors to the both sides of a plane passing through three points with coordinates (1,0,1), (1,1,-1) and(-1,1,1).

My answer is [$\sqrt{6}/6 , \sqrt{6}/3 , \sqrt{6}/6$] and for the other side I just have to multiply by -1 which gives [$-\sqrt{6}/6 , -\sqrt{6}/3 , -\sqrt{6}/6$]

I would like to check whether I'm correct and for calculating the normal, does it matter which coordinate i choose to form the vectors??

Call $A=(1,0,1)$, $B=(1,1,-1)$, $C=(-1,1,1)$.

Each of the two vectors you found give $0$ in scalar product with $A-B$, $B-C$ (and $C-A$) and the modules of those vectors are $1$ in both cases (if you call the first $\vec{v}$, the second is $-\vec{v}$).

Without pen and paper next to me, I'd say you're right: you found two unit vectors, $\pm\vec{v}$, normal to the plane thru $A,B,C$.

P.S.: Changing system of coordinates may alter single components but NOT modules and orthogonality, which are indeed invariants.


Discrete Random Variable Probabilities? I have the following discrete random variable $\mathbb{X}$ over $\Omega = \{1,2,3,4,5\}$ with the pmf:

$f_{\mathbb{X}}(1) = P(\mathbb{X} = 1) = 0.25$

$f_{\mathbb{X}}(2) = P(\mathbb{X} = 2) = 0.4$

$f_{\mathbb{X}}(3) = P(\mathbb{X} = 3) = 0.1$

$f_{\mathbb{X}}(4) = P(\mathbb{X} = 4) = 0.15$

$f_{\mathbb{X}}(5) = P(\mathbb{X} = 5) = 0.1$

and I want to find $P(\mathbb{X} = \mu)$

I can easily find $\mu$ as follows:

$$\mu = \sum_{i = 1}^5 xf_{\mathbb{X}}(x) = 1(0.25)+2(0.4)+3(0.1)+4(0.15)+5(0.1) = 2.45$$

but, how do you find $P(\mathbb{X} = 2.45)$?

Do I simply round down since $\mathbb{X}$ is discrete? Do I interpolate? Any clarification would be great!

You need to assume a functional form for the distribution:

A simple quadratic fit gives:  $f(X) = 0.34 - 0.0335714 x - 0.00357143 x^2$
and then $f(2.45) = 0.236313$.


How to write this term compactly? How can I write this term in a compact form where $a$ only appears once on the RHS (in particular without cases)?

$T(a) =
  \begin{cases}
    a^2  &,\text{ if $a \leq 0$}\\
    2a^2 &,\text{ if $a > 0$}\\
  \end{cases}$

I have already thought about $T(a) = \max\{\sqrt{2}a,|a|\}^2$ or $T(a) = \frac{3+\text{sgn}(a)}{2}a^2$, but in both cases $a$ appears twice.

After discussing the problem with a brilliant friend we came up with the following solution:

$T(a) = \left[\Re\left((\sqrt[4]{2}-i)\sqrt{a}\right)\right]^4$

However, I am still up for further suggestions!

Another class of solutions to this problem is the "trivial substitution class".

Examples

$$T(a) = 2\int_0^a x\cdot(\max\{x,0\}+1)dx$$

or even more trivial

$$T(a) = \max\{\sqrt{2}x,|x|\}^2\bigg|_{x=a}$$


$f>0$ is a integrable function, How to prove that $\int_a^bf(x)dx>0$ Here we're talking about Riemann integrable real functions defined on $[a,b]$. Even though this is a simple question I didn't find it here, if it's duplicated I'm sorry...

I could prove this if $f$ was continuous, or
$$
f\geq0 \rightarrow \int f\geq0.
$$

Trying to prove this one I stucked in a point that if I could prove the following

  if for every partition of $[a,b]$,
  $$
\inf \{f(x)|x\in[x_{t_i},x_{t_{i+1}}]\}=0
$$
  for all intervals of the partition then $f=0$,

then I could prove the initial thing... But I couldn't do this neither. Anyone can help me with that?

This is also true in Lebesgue sense. We can prove it as follows: Let $A=\{x\mid f(x)>0\}$. For each $n\in\mathbb{N}$, let $A_n=\{x\mid f(x)\geq \frac{1}{n}\}.$ Note that $A=\cup_n A_n$. Since $A$ has non-zero measure, there exists $n$ such that $A_n$ has non-zero measure. Now $\int f \geq \int_{A_n} f \geq \frac{1}{n}m(A_n)>0$, where $m(A_n)$ denotes the Lebesgue measure of $A_n$.

Since $f(x) \gt 0$, then $\exists m \in \Re$ s.t. $f(x) \ge m \gt 0$.

Compose a constant function $\phi$ s.t.

$\forall x \in [a,b]$, $\phi(x) = m$.

Then

$\int_a^b f(x) \ge \int_a^b \phi(x) = (b-a)*m \gt 0$.