How can a function with a hole (removable discontinuity) equal a function with no hole? I've done some research, and I'm hoping someone can check me. My question was this:

Assume I have the function $f(x) = \frac{(x-3)(x+2)}{(x-3)}$, so it has removable discontinuity at $x = 3$. We remove that discontinuity with algebra: $f(x) = \frac{(x-3)(x+2)}{(x-3)} = (x+2)$. BUT, the graph of the first function has a hole at $x = 3$, and the graph of the second function is continuous everywhere. How can they be "equal" if one has a hole and the other does not?

I think that this is the answer:

Because the original function is undefined at the point $x = 3$, we have to restrict the domain to $\mathbb{R} \setminus 3$. And when we manipulate that function with algebra, the final result, $f(x) = (x + 2)$ is still using this restricted domain. So even though the function $f(x) = (x+2)$ would not have a hole if the domain were all of $\mathbb{R}$, we are sort of "imposing" a hole at $x = 3$ by continuing to throw that point out of the domain.

And then just to close the loop: Removing the removable discontinuity is useful because it allows us to "pretend" that we're working with a function that is everywhere continuous, which helps us easily find the limit. But the reality is that the function $f(x) = (x +2)$ is actually NOT continuous everywhere when we restrict the domain by throwing out the point 3. Or am I now taking things too far?

Thanks in advance!

EDIT: For anyone coming across this in the future, in addition to the excellent answers below, I also found this other question about the continuity of functions with removable discontinuities helpful.

Two functions are typically defined to be equal if and only if they...

Share the same domain
Share the same codomain
Take on the same values for each input.

Thus, functions $f,g : S \to T$ for sets $S,T$ have $f=g$ if and only if $f(x) = g(x)$ for all $x$ in $S$.

For functions with holes, we typically restrict the domain by ensuring the values where the function is not defined at not included. For example, in the functions you have, you have

$$f(x) = \frac{(x-3)(x+2)}{(x-3)} \;\;\;\;\; g(x) = x+2$$

Are these equal? Yes, and no.

A function must be defined at all values of the domain. Thus, we can say $3$ is not in the domain of $f$ for sure. But we never specified otherwise the domains and codomains of these functions! Typically, unless stated otherwise, we often assume their domain to be $\Bbb R$ or $\Bbb C$, minus whatever points are causing problems - and of course, in such cases, $f \neq g$ since $f(3)$ is not defined, and thus $f$ normally has domain $\Bbb R \setminus \{3\}$ and $g$ generally has domain $\Bbb R$.

But that restriction is not necessary. For example, we could define the functions to be $f,g : \Bbb R \setminus \Bbb Q \to \Bbb R$. Notice that the domain of both functions are now all real numbers except rational numbers, i.e. the irrational numbers. This means $3$ is not in the domain of either function - and since that's the only "trouble spot," and the codomains are equal, and the values are equal at each point in the domain, $f=g$ here.

Or even more simply: we could have $\Bbb R \setminus \{3\}$ be the domain of $f$ and $g$ and again have equality! The key point in all this is that, just because $f$ or $g$ do attain defined values for certain inputs, doesn't mean they have to be in the domain.

In short, whether $f=g$ depends on your definitions of each. Under typical assumptions, $f \neq g$ in this case, but if we deviate from those assumptions even a little we don't necessarily have inequality.

One seldom writes the domain of a function, implicitly assigning the largest domain that is sensible in your context.  For much algebra (especially graphing), trigonometry, and calculus, the largest domain of interest is the real numbers, $\mathbb{R}$, and the (implicit) maximal domains of functions are subsets of $\mathbb{R}$.

Two functions are equal if they have the same domain, range, and values on each point of the domain.  (The range is a subset of the codomain.  No one cares if you expand the codomain.)  So $\frac{(x-3)(x+2)}{(x-3)}$ is not equal  to $x + 2$ because they do not have the same domain: $\mathbb{R} \smallsetminus \{3\} \neq \mathbb{R}$.

Two functions are identical if they have the same values on each point of their common domain.  The common domain of $\frac{(x-3)(x+2)}{(x-3)}$ and $x + 2$ is $(\mathbb{R} \smallsetminus \{3\}) \cap \mathbb{R} = \mathbb{R} \smallsetminus \{3\}$.  Both functions agree on this common domain, so they are identical.  "At any point you can evaluate both functions, they give the same answer, so there is no point of the common domain that can, through evaluation, show that the two functions are different."  Two unequal identical functions are unequal "before you get to evaluations" -- that is, they are unequal in their domains or ranges.

A more extreme example: $\log x$ has domain $(0, \infty)$ and $\log( -x)$ has domain $(-\infty, 0)$.  These two functions are not equal; they do not have the same domain.  Because the intersection of their domains is empty, their common domain is empty and they are (vacuously) identical.  (Here, "vacuously" means "there is literally nothing to check because the common domain has nothing in it, like a vacuum".)

When proving identities, we only use this weaker notion of equivalence of functions.

Regarding continuity:  "$f$ is continuous" means that $f$ is continuous on each point of its domain.  If you delete a point from its domain, you have relaxed the conditions imposed by continuity.  "$f$ is continuous at the point $x$" means
$$  \lim_{t \rightarrow x} f(t) = f(x)  $$
(where the indicated limit must exist).  In considering the limit, $t$ is restricted to only take values in the domain of the function.

(Most people don't talk about continuity of functions at isolated points of their domains.  With the above definition, a function is continuous at any isolated point of its domain because the limit becomes vacuous: there are no points in the domain near the isolated point except for the isolated point, so there is nothing to check.  It is not the case that the limit fails to exist.  In a rigorous setting, you would define continuity as "for all $\varepsilon > 0$, there exists a $\delta$ such that for all $t$ such that $|x-t| < \delta$, we have $|f(x) - f(t)| < \varepsilon$".  For an isolated point, once $\delta$ is small enough, the only choice for $t$ is $t = x$, and we have $|f(x) - f(t)| = 0$, which is definitely less than $\varepsilon$.)

Continuous functions are nice.  For instance, you likely have a theorem that says continuous functions are (Riemann) integrable (on non-infinite intervals of integration).  If you have a function that has a removable discontinuity, then it is identical (but not equal) to a continuous function whose domain includes the ordinate (first coordinate) of the removable discontinuity.  Any value you can get from the first function, you can also get from the second function.  This means any particular Riemann sum (that is, any particular choice of partition and sample points) that can be evaluated using the discontinuous function has the same value as the same sum using the identical, continuous function.  Riemann sums that sampled the removable discontinuity did not exist, so prevented the existence of the limit as the diameter of the partition went to zero.  The identical function sidesteps this problem by supplying the limit of the function as it approaches the removable discontinuity, so the Riemann sum using the "filled in" function exists and you can integrate it.  (Normally, one talks about an improper integral, splitting the interval of integration into pieces that avoid little intervals around the points of discontinuity, then taking limits as the little intervals shrink to zero.  If the discontinuity is removable, one can show that these two methods of sneaking up on the removable discontinuities give the same result.)


If $G$ has no proper subgroups other than the trivial group,then $G$ is a finite group. If $G$ has no proper subgroups other than the trivial group,then show that $G$ is a finite group.

totally stuck.can I get some help

Assume $G$ is infinite and $1\ne g\in G$. Then $\langle g\rangle$ is either a proper subgroup, or one has $G=\langle g\rangle\cong \mathbb Z$, which has the proper subgroup $2\mathbb Z$

See the subgroup generated by an element of $G$.


Binomial expansion question. easy! I'm trying to do the binomial expansion of
$(x-2)^{1/2}$.

How do you do it? As far as I'm aware the expansion only works for $(1+x)^n$. How could I get it in that form?

Thanks.

One way to do this is to rewrite the expression as $(1+(x-3))^{1/2}$, and then substitute $u=x-3$ and expand $(1+u)^{1/2}$ using the Binomial Series
$\displaystyle(1+u)^r=\sum_{n=0}^{\infty}\binom{r}{n}u^n$.

If what you meant to ask about was a binomial expansion, then the approach is as follows:

First, note that the binomial theorem states that
\begin{align*}
(a+b)^n = \sum_{k=0}^{\infty}\binom{n}{k}a^kb^{n-k}
\end{align*}
Thus, by substituting for the above values, we have
\begin{align*}
(x-2)^\frac{1}{2} = \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k(-2)^{\frac{1}{2}-k}
\end{align*}


Is there a polynomial which vanishes at only one point Assume $(a,b) \in \mathbb C^2$. Is there a polynomial $f(x,y)$ which vanishes only at $(a,b)$ ?
Thanks.

You could use the algebraic sledgehammer. If $V(f)$ is a point, $\sqrt{(f)}$ is a maximal ideal, which is a contradiction, since $(f)$ is always contained in a non-maximal prime ideal $(g)$, where $g$ is an irreducible factor of $f$.

Or you do it the more elementary way:

Without loss of generality assume $(a,b)=(0,0)$. For any $t \in \mathbb C \setminus \{0\}$ the polynomial $g(x) := f(x,t)$ has no root, hence is constant. The same holds for $f(t,y)$ and you can deduce that $f$ itself is constant from that.

No,

You are working on a plane, and one single equation $f(x,y)=0$ left you a 1-dimensional solutions subspace.


multivariate Ito isometry I wonder whether there exists a straightforward extension of the Ito isometry to multidimensional processes.

In the one-dimensional case the Ito isometry can be written as

$\mathbb{E}[ (\int_0^T X_t \; \mathrm{d}W_t)^2 ] = \mathbb{E}[ (\int_0^T X_t^2 \;\mathrm{d}_t) ]$.

If now $X_t$ is a vector of random variables instead, do I get something along these lines:

$\mathbb{E}[ (\int_0^T X_t \; \mathrm{d}W_t) (\int_0^T X_t^\top \; \mathrm{d}W_t^\top) ] = \mathbb{E}[ (\int_0^T X_t X_t^\top \;\mathrm{d}_t) ]$

????

Short answer: Yes, what you wrote is basically correct.
Longer anser:

See the following for a more general case, which is Lemma D.1 on the page 19 of http://www.shuangjian.info/Wasserstein_Control_HRLMC.pdf

${\bf Statement.}$
Let ${\bf B}: [0, T]\times \Omega \rightarrow \mathbf{R}^p$ be the standard $p$-dimensional Brownian motion and ${\bf M}:  [0, T]\times \Omega \rightarrow \mathbf{R}^{p\times p}$ be a matrix-valued stochastic process adapted to the natural filtration of the Brownian motion. Then
$$\mathbf{E} \left[ \left\Vert\int_0^T {\bf M}_t d{\bf B}_t\right\Vert_2^2\right] = \mathbf{E} \left[ \int_0^T \left\Vert{\bf M}_t\right\Vert_F^2 dt\right],$$
where the norm in the right hand side is the Frobenius norm of the matrix-valued process.

The implication of the above statement is your case: take ${\bf X}_t$ as the first row of ${\bf M}_t$ and let the rest of the matrix be zero all the time, then you can recover the case for $p$-dimensional vector-valued random variables:
$$\mathbf{E} \left[ \left(\int_0^T {\bf X}^{T}_t d{\bf B}_t\right)^2\right] = \mathbf{E} \left[ \int_0^T \left\Vert{\bf X}_t\right\Vert_2^2 dt\right].$$
This is basically what you wrote down.
[1]: https://i.stack.imgur.com/KNvI4.png


Rouché's Theorem for $p(z)=z^7-5z^3+12$ Let $f$ and $g$ be differentiable on a domain $D$ and suppose that $\gamma$ is a simple closed contour whose inside is contained in D.

  If $|f(z)-g(z)|<|f(z)|$ for all $z$ on $\gamma$, then $f$ and $g$ have the same number of zeros inside $\gamma$ (counted including their order).

I was reading an example of application of Rouché's Theorem, where Rouché's theorem was used to show that the polynomial $p(z)=z^7-5z^3+12$ has $0$ roots in $\{z:\mathbb{C}:|z|<1\}$.

What was done was:

Let $g(z)=z^7-5z^3+12$ and let $f(z)=12$. Then for $|z|=1$,

$|f(z)-g(z)|=|z^7-5z^3| \\ \le|z|^7+5|z| \\=1+5\\=6<12=|f(z)|$

Hence by Rouché's Theorem $p(z)=z^7-5z^3+12$ has $7$ roots in $\{z:\mathbb{C}:|z|<2\}$.

I was wondering, what is the purpose to do the step $\le|z|^7+5|z|$? Can't I just jump straight from  $|f(z)-g(z)|=|z^7-5z^3| \\=|1-5|\\=4<12=|f(z)|?$

Secondly, $|f(z)-g(z)|=|-z^7+5z^3|$, is there a reason why they used $|f(z)-g(z)|=|z^7-5z^3|$?

Thirdly, what does it mean by "(counted including their order)"? (From the definition above)

With $|z|=1$, $|f(z)-g(z)|=|z^7-5z^3| = |z^4-5|$, however the last quantity is not equal to $5-1$, as you can take any point with $|z|=1$ (eg, take $z=e^{i\frac{\pi}{4}}$, then the value is $6$). The best upper bound is the one given.

There is no difference between the $|-z^7+5z^3|$ and $|z^7-5z^3|$, since it appears under the $|.|$ (ie, |w| = |-w|).

Order means multiplicity. $z^2$ has a zero of multiplicity $2$ at $0$.

It seems that you have applied Rouche's theorem for two regions, the first one is inside the unit circle |z|=1, and the second one is inside the circle |z|=2. Your calculations, regardless of a minor mistake, show that all roots of p(z) lies between these two circles. I use a form of Rouche's theorem, found here ( https://en.wikipedia.org/wiki/Rouche's_theorem ), to provide a similar but complete answer.
1_ For |z|=1 one writes $p(z)=s(z)+t(z)$, $s(z)=z^7-5z^3$, t(z)=12 and then applies triangle inequality on the counter |z|=1 to see that $|s(z)|<1+5<12=|t(z)|$. So p has no roots inside the unit circle (notice that t(z) is constant and has no roots).
2_ Mimic the method of step 1, this time for $p(z)=u(z)+v(z)$, $u(z)=-5z^3+12$, $v(z)=z^7$, |z|=2 and see that $|u(z)| < 5*8+12=52 < 128 = |v(z)|$. Thus the total roots of p(z) must lie inside the circle |z|=2.
Now it is evident from steps 1 and 2 that all roots of p(z) are between the circles |z|=1 and |z|=2.


Does this limit with a Product Exist? I have been working on
$$n \cdot \left(1-\prod_{i=1}^n (1-\frac{1}{p_i})\right)$$
and would like to know whether

$\lim \limits_{n \to \infty} (n \cdot \left(1- \prod_{i=1}^n (1-\frac{1}{p_i})\right))$

exists.

$$n-n \cdot \left(\prod_{i=1}^n (1-\frac{1}{p_i})\right) \geq n-n \cdot \left((1-\frac{1}{p_1})\right) \to \infty$$

N.S has already given a pretty direct solution. A slight twist on that which uses a nice little result about prime is that you can call in the divergence of the Harmonic Series $ H_{N}$.

$$\lim_{N \to \infty}H_{N} = \sum_{n=1}^{\infty}\frac{1}{n} = \prod_{\textrm{primes}}\frac{1}{1 - \frac{1}{p}}$$

Flipping the fraction upside down we get, abusing notation/limits,

$$ \prod_{\textrm{primes}}\left(1 - \frac{1}{p}\right) = \frac{1}{H_{\infty}}$$

Since the Harmonic series diverges it follows that $\lim_{n\to\infty} \prod_{i=1}^{n}\left(1 - \frac{1}{p_{i}}\right) = 0$ and so the ratio of the two terms $n$ and $n\prod_{i=1}^{n}\left(1 - \frac{1}{p_{i}}\right)$  can be made arbitrarily big and so the expression is dominated by the lone $n$ and also diverges.


Some basic questions about Stochastic Calculus I have a transition function for a Markov process $X_t$.  I want to find a density function for the stochastic process $Y_t := \int_0^t X_s \,ds$.  Some questions about this:

Is this the same as the transition function for the stochastic process $Y'_t := \int_0^t 1 \,dX_s$?  If not, what is the difference in real-world meaning between $Y_t$ and $Y'_t$?
In normal calculus, one typically integrates by invoking the Fundamental Theorem and taking antiderivatives.  Is there a Fundamental Theorem analog in stochastic calculus that might be useful here?
My best lead so far is to use the Feynman-Kac formula on $X_t$ to find the characteristic function of the density function of $Y_t$, then invert it.  The transition function for $X_t$ is very long and ugly, so this would be hugely complicated to execute.  I've got Mathematica as my integration calculator, so I can handle some amount of ugliness, but this strategy was too complicated for Mathematica to stomach, and it froze up.  Is there a known simpler method?

Thanks - advice on any of these three questions is appreciated.

For (1), since $dX_s \equiv X_{s+ds}-X_s$, $Y'_t = \int_0^t dX_s = X_t - X_0$.


rationalize the complex number multiplication rule For a middle school student without previous knowledge of complex number, how do one introduce the multiplication rules of complex number? i.e., if we have two real number pairs of  $(a,b)$ and $(c,d)$, why do we define a product of $(a,b) * (c,d) = (ac-bd, ad+bc)$?
I wanted to start here and introduce the $i$ and polar expression of complex number as a consequence.

You could explain it in terms of rotation.

Suppose you have a point $(a,b)$ in the plane. If you multiply it by a real number $s$ then you scale the length but preserve the direction, $(sa,sb)$. What if we want to multiply two vectors together in a way that multiplies their lengths together and adds their polar angles?

This would allow us to rotate points just by multiplying them... etc

Personally I feel that introducing $i$ and building from there is a more natural approach though.

The complex number (a, b) should actually be a + bi; where $i^2 = -1$.

Therefore, $(a,b) * (c,d) = … = (ac-bd) + (ad+bc)i =  (ac-bd, ad+bc)$.


Interpret the graph of $\frac{ax+b}{cx+d}$ as a transformation of $y=\frac{1}{x}$ As part of a problem-set I'm self-studying, I'm trying to interpret the graph of  $f(x)=\frac{ax+b}{cx+d}$ as a transformation of the graph of $y=\frac{1}{x}$, including determining what restrictions should be placed on $a, b, c,$ and $d$ for the interpretation to remain valid.
For context, the preceding problem was as follows: Sketch the graph of $f(x)=\frac{3x-7}{x-2}$ by applying transformations to the graph of $y=\frac{1}{x}$. We used polynomial long division to show $f(x)=3-\frac{1}{x-2}$, yielding the following transformations:
$\rightarrow$
$\rightarrow$$\rightarrow$
Applying the same approach to $f(x)=\frac{ax+b}{cx+d}$, I did some long division and came up with,
$$f(x)=\frac{a}{c}+\frac{bc-ad}{c(cx+d)}=\frac{a}{c}+\left(\frac{bc-ad}{c}\right)\left(\frac{1}{cx+d}\right)$$
I'm not totally confident in my interpretation of the transformations involved, but here's what I've got:

Clearly, we must have $c\ne 0$, and the first transformation of $y=\frac{1}{x}$ is $y=(\frac{1}{c})(\frac{1}{x})=\frac{1}{cx}$, which could be viewed as either a vertical or horizontal scaling. If $c>1$, it shrinks the graph, compressing by a factor of $c$. If $0<c<1$, it grows by a factor of $\frac{1}{c}$ (again, either horizontally or vertically, the graph is such that the two are equivalent). Trivially, if $c=1$, it remains as is. A similar trio of cases for negative values of $c$ would apply similar results and a reflection (and like the scaling, the reflection could be applied either vertically or horizontally).
I suspect it makes more sense to interpret the above as a horizontal transformation, which can then be followed by a horizontal shift of $d$ units, yielding $y=\frac{1}{cx+d}$. If $d>0$, everything shifts to the left, if $d<0$ it shifts to the right, and if $d=0$ it all stays put.
The next transformation scales everything vertically by $\frac{bc-ad}{c}$, and the final transformation applies a vertical shift of $\frac{a}{c}$.

Does that sound about right? I didn't come up with any restrictions on $a,b,c,$ and $d$, beyond $c\ne 0$. I feel a bit less than competent when handling horizontal graph transformations in general. Dealing with shifts is easy enough, and I can handle simple scalings, but when they all combine, and possibly involve reflections as well, I start struggling to keep my head above water.

Everything looks correct except for the order in which the horizontal transformations are applied.  The transformation
$$ y = x \qquad \leadsto \qquad y = \frac{1}{cx + d} $$
first shifts the graph left by $d$, then scales towards the $y$-axis by a factor of $c$, which means that the shift gets scaled as well.

If you like, factor:  $cx + d = c\left(x + \tfrac{d}{c}\right)$, which allows you to apply the transformations in the usual order:  shrink by factor of $c$, then shift by $-\tfrac{d}{c}$.

We should also put the restriction bc-ad ≠ 0 although this is not directly on a, b, c and d as you asked.

f(cx) should always be interpreted as a horizontal transformation (horizontal scaling)

Therefore, in the transformation f(cx),  if 0 < c < 1, the graph is scaled by a factor of c (i.e. expanded by a factor of c) and if c > 1 the graph is scaled by a factor of 1/c (i.e. shrinked by a factor of c), one cannot say "it grows by a factor of 1/c" because it would be equivalent to saying "it shrinks by a factor of c" which is incorrect in your example.

You arrived to f(cx+d) = 1/(cx+d) from f(x) = 1/x by a series of steps

  you multiplied x by c to get cx
  you added d

What you do now is to reverse the process:

  substract d (which corresponds to a horizontal shift, to the right if d<0 and to the left if d>0)
  divide by c (which is a horizontal scaling by c or expansion by c if 0 < c < 1 and a horizontal scaling by 1/c or shrinking by a factor of c if c > 1)


Equation of the line that has $x$ and $y$ intercepts at $a$ and $b$. Please can anyone help me with proving the following problem:

  Show that the line that crosses the $X$-axis at $a \neq 0$ and the $Y$-axis at $b \neq 0$ has the equation $$\dfrac{x}a + \dfrac{y}b -1=0$$

I don't have an idea of where I should start from for proving the problem.

The $x$ intercept is given by $(a, 0)$, and the $y$ intercept is given by $(0, b)$. And we know $a \neq 0, b\neq 0$, so the points are distinct.

All you need to do is substitute these values into your equation to see that equality holds: i.e., that they lie on the line defined by $$\frac xa + \frac yb - 1 = 0$$

$$(a, 0): x = a, y = 0 \implies  \frac aa + \frac 0b - 1 = 0\implies \frac aa = 1\implies 1 = 1\quad \checkmark$$

$$(0, b): x = 0, y = b \implies \frac 0a + \frac bb - 1 = 0 \implies \frac bb = 1 \implies 1 = 1 \quad \checkmark$$

Indeed: These are unique points lying on the line given by $$\frac xa + \frac yb - 1 = 0$$

Actually there are set of 4 lines. because intercepts are length and its coordinates have set of 4 coordinates.(a,0)(0,b),(a,0)(0,-b),(-a,0)(0,b),(-a,0)(0,-b).
so lines equation will be..
$$ \frac xa + \frac yb -1 =0$$

$$ \frac xa - \frac yb -1 =0$$

$$ -\frac xa + \frac yb -1 =0$$
$$ -\frac xa - \frac yb -1 =0$$


Why isn't $\frac{1}{x}$ a polynomial? Why isn't $\frac{1}{x}$ a polynomial?

Does it directly follow from definition?  As far as I know, polynomials in $F$ are expressions of the form $\sum_{i=0}^{n} a_ix^i$, where $a_i\in F$ and $x$ is a symbol.

Or is there a nicer argument involved?

Footnote: $F$ is a field of characteristic zero.

If you want a more formal "proof", you can suppose for contradiction that $1/x$ is in fact equal to some expression $a_k x^n$ of the form $ \sum_{k=0}^n a_k x^n $:

$$ \frac{1}{x} = a_0 + a_1 x + \cdots + a_n x^n$$

multiplying through by $x$ gives:

$$ 1 = a_0 x + a_1 x^2 + \cdots + a_n x^{n+1}$$

Setting $x=0$ gives:
$$ 1 = a_0 \cdot 0 + \cdots + a_n \cdot 0^{n+1} = 0$$

a contradiction, so we must have that $1/x$ is not a polynomial.

You can first use intuition from polynomials where coefficients are from the real numbers.  Being every where continuous the image of $[-1,1]$ is a compact set, which is bounded for a polynomial. But the image of the same set under  $f(x)=  \frac1x$ is unbounded. So we can see that in reals, $1/x$ is not a polynomial.

Now take any field of characteristic 0. One can see that if $\alpha$ is any root of $g(x)$ then it has to be a root of the product  polynomial $f(x)g(x)$.
 Now assuming $f(x)=\frac1x$ is a polynomial multiply with the second polynomial $g(x)=x(x-1)$. Their product is the  polynomial $(x-1)$, this product has just 1 root namely $1$ whereas $g(x)$ has two roots, 0 and 1. This contradiction should settle the issue.


volume between the planes $z = ax + by$, $z=0$ and the cylinder of radius 1 Find the volume between the planes $z = ax + by$, $z=0$ and the cylinder of radius 1, whose axis of symmetry is the z axis in the first octant $0≤x, 0≤y, 0≤z$.

I´m studying integral in 3 variables, but i stuck with this problem , i tried to solve this with doble integral i can't  found the correct integral to solve this.

The solid lies above the region $D$ in the $xy$-plane bounded by  $x^{2} + y^{2} = r^{2}$, so the volume is given by the integral $$\int\int\limits_{D} f(x,y) \ dA = \int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} f(x,y) \ dx dy$$

Its that correct? Can you help me please.

is there any restriction with $a$ and $b$?

You need to find the volume in that region,

$\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{0}^{ax+by}1dzdydx$.

When you have a triple integral with $f(x,y,z)=1$,

$\int\int\int f(x,y,z)dzdydx$ it is the volume of the region where you are  trying to solve the triple integral.


Prove $H$ is a normal subgroup of $K$ $G$ is a set.
$$G=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}0 & 1 \\-1 & 0\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}, \begin{vmatrix}0 & -1 \\ 1  & 0\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}0 & i \\ -i  & 0\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}0 & -i \\ i  & 0\end{vmatrix}\right\}$$ and $(G, \times)$ is a group.
If
$$H=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}\right\}$$ and
$$K=\left\{\begin{vmatrix}1 & 0 \\ 0  & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0  & -1\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0  & i\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0  & -i\end{vmatrix}\right\},$$
then $H$ and $K$ is a subgroup of $G$, $H$ is a normal subgroup of $K$, and $K$ is a normal subgroup of $G$.
My questions are :
a. How do I prove that $H$ is a normal subgroup of $K$?
I knew that to prove if $H$ is a normal subgroup I have to prove that each left coset of $H$ in K is a right coset of $H$ in $K.$ But I don't know how apply that to the matrices.
b. What is every element of factor group $G/H?$

as for $H$ in $K$ you could notice that $H$ is of index 2 in K so it is normal.

Longer method than Yotam's: conjugate each subgroup by each element of $K$ and $G$ and show that they are invariant. You'll have to do $4*2 + 8*4 = 40$ different conjugations, but this isn't so large as to make the computation unfeasible to do by hand.


If function $f$ satisfies given conditions, then find $\frac{1}{3} f(1,98)-f(1,99)$ Let $f$ be a function defined on $\{(m,n):$ $m$ and $n$ are positive integers $\}$ satisfying:

$$1. f(m,m+1)=\frac{1}{3}$$, for all m

$$f(m,n)=f(m,k)+f(k,n)-2f(m,k) \cdot f(k,n)$$ for all $k$ such that $m<k<n$, then find the value of

$$\frac{1}{3} f(1,98)-f(1,99)$$

Could someone give me hint as how to initiate this problem?

Hint Try to prove that $\frac{1+f(m,m+n-1)}{f(m,m+n)}=3$ Then put m=1,n=99 to yield the expression you want.


Solve complex equation $z^3=1-i$ I have attempted to solve it by converting to polar form first using $z=r\cdot e^{(i\cdot\theta)}$ and $i=e^{i\cdot(pi/2 \cdot ( 2\cdot n \cdot pi))}$.

Then I was taught to compare like terms independently for the magnitude r and angle $\theta$.

The issue I'm having is not knowing how to handle the "$1 -$ " term in front of i when I am matching terms. The similar example problem we solved in class was $z^4=i$ was much easier to solve since $i$ was by itself so comparing $r$ and the $\theta$ terms was simple. Any advice is much appreciated!

Find the norm and the angle to write out the polar form of $1-i$.

$$1-i =\sqrt{2} \exp((-\frac{\pi}{4} + 2n \pi)i) $$

Now, the problem reduces to $$z^3=\sqrt{2} \exp((-\frac{\pi}{4} + 2n \pi)i)$$

You are correct in your first step of converting $z^3=1-i$ to polar form.

$r= \sqrt{1^2+(-1)^2} = \sqrt{2}$, and $\theta= \arctan(\frac{-1}{1})=\frac{7}{4}\pi$

Thus, $z^3 = \sqrt{2} \cdot e^{i(\frac{7}{4}\pi)}$.

But remember that you can use de Moivre's Theorem in that form while taking the cube root of $(1-i)$ is just the cube root of the radius times and the exponent (for simplicity since I don't know LaTeX) is just:

$i[(\frac{7}{4} + 2n)\frac{\pi}{3}]=i[\frac{7}{12}\pi + 2n\frac{\pi}{3}]$, where $n=0,1,2$.

Thus, for this problem, we get three answers:

1) $z = 2^{\frac{1}{6}} \cdot e^{i(\frac{7}{12}\pi)}$;

2) $z = 2^{\frac{1}{6}} \cdot e^{i(\frac{7}{12}\pi + \frac{2}{3}\pi)} = 2^{\frac{1}{6}} \cdot e^{i(\frac{5}{4}\pi)}$;

3) $z = 2^{\frac{1}{6}} \cdot e^{i(\frac{7}{12}\pi + \frac{4}{3}\pi)} = 2^{\frac{1}{6}} \cdot e^{i(\frac{23}{12}\pi)}$.


Using Complex exponential definitions of sine and cosine, prove $\cos2\theta=\cos^2\theta-\sin^2\theta$ Question

Please do not just tell me the answer, please provide helpful hints and hide the answers

Using Complex exponential definitions of sine and cosine, prove $\cos\theta=\cos^2 \theta-\sin^2\theta$

All that I know is the trig identity:

$\cos2\theta=1-2\sin^2\theta$ or $\cos2\theta=2\cos^2\theta-1$

Hint:

  $\cos 2\theta + i \sin 2\theta = e^{2i\theta} =\left(e^{i\theta}\right)^2=(\cos \theta+ i\sin \theta)^2$

Then expand the right side and compare real parts

$\begin{array}{lcl}\cos(2\theta)+i\sin(2\theta) & = & e^{2i\theta} \\ & = & (e^{i \theta})^2 \\ & = &  (\cos\theta+i\sin\theta)^2 \\ & = & (\cos\theta)^2+2i\cos θ\sin θ+i^2(\sin θ)^2 \\ & = & (\cos θ)^2+2i\cos θ\sin θ−(\sin θ)^2\end{array}$

Equating real and imaginary parts gives us:

$\begin{array}{lcl}\cos(2θ)  =  (\cosθ)^2−(\sinθ)^2 \\ \sin(2θ)  = 2\cosθ\sinθ \end{array}$


insphere/circumsphere ratio of a polyhedron the same as its dual polyhedron? Is the $r/R$ ratio for any polyhedron always the same as the $r/R$ ratio of the dual of that polyhedron?

Given any polyhedron, we can find the biggest sphere that fits inside it (its insphere) and the smallest sphere outside it (its circumsphere).
I'm interested in $r/R$, which is the ratio of the radius of the insphere to the radius of the circumsphere.

Motivation: the worst-case linear distortion of a polyhedral global map projection is roughly (at least) $r/R$.
The $r/R$ ratio is one of many ways to measure how close a polyhedron is to a sphere. Some of those measures say the dodecahedron is the closest Platonic solid to the sphere.
Others say the icosahedron is the closest Platonic solid to the sphere.
Polyhedrons with larger r/R ratio are better, in the sense that they (usually) produce maps with less linear distortion. I'm surprised to discover that that ratio seem to be equal between the dodecahedron and the icosahedron (both ~ $0.7946$), and a (somewhat smaller) ratio is the same between the cube and the octahedron (both ~ $0.5773$).
(I'm not so surprised that an (even smaller) ratio is the same between the remaining Platonic solid and its dual polyhedron: $1/3$).
Can this surprising-to-me fact be generalized to some larger group of polyhedrons?
What other polyhedron have a $r/R$ ratio the same as the $r/R$ ratio of its dual polyhedron?
Do any polyhedron have a $r/R$ ratio different from the $r/R$ ratio of its dual polyhedron?

(In this answer, I do as the questioner did, calling the smallest containing sphere the circumsphere and its center the circumcenter, even if this sphere does not touch all vertices.  Similarly, I call the largest enclosed sphere the insphere and its center the incenter, even if this sphere does not touch all faces.)

Assuming that the circumcenters and incenters of both the original polyhedron and the dual all coincide, these ratios should be the same.

A convex polyhedron which contains the origin in its interior can be described either as the convex hull of a set $\{a_i\}\subseteq {\Bbb R}^3$ of vertices or as the intersection of a set of bounding half-spaces $\{S_j\}$, where each bounding half-space is of the form
$$S_j=\{x\mid x\cdot b_j\le 1\}, \qquad b_j\in {\Bbb R}^3. $$

Letting the circumsphere be centered at the origin, the circumradius $R$ will equal the maximum value of any $|a_i|$.  Also, since the distance between the origin and the plane $\{x\mid x\cdot b_j=1\}$ is $|b_j|^{-1}$, if the insphere is centered at the origin, the inradius will equal the minimum value of any $|b_j|^{-1}$.

The polar dual of a given polyhedron can be constructed by interchanging the $a_i$s and $b_j$s.  So, if the original polyhedron had circumradius $R$ and inradius $r$, and if the circumcenter and incenter of the polar dual are still at the origin, then the polar dual must have circumradius $1/r$ and inradius $1/R$, and the ratio $R/r$ of the circumradius to the inradius remains the same.

For an example, take a cuboctahedron with vertices
$$
(\pm 1, \pm 1, 0), \ (\pm 1, 0, \pm 1), \ (0, \pm 1, \pm 1).
$$
In this case, the circumsphere and insphere are both centered at the origin.  The radius $\sqrt{2}$ circumsphere touches all vertices; the insphere has radius $1$ and touches only the square faces; and the ratio $R/r$ is $\sqrt{2}$.
The dual of this polyhedron is the rhombic dodecahedron, with vertices
$$
(\pm 1,0,0), \ (0,\pm 1,0), \ (0,0,\pm 1),\ (\pm\frac12, \pm\frac12, \pm\frac12).
$$
The circumsphere and the insphere are both still centered at the origin.  The circumsphere has radius $1$ and touches only those vertices where four faces meet (the first six above); the insphere has radius $1/\sqrt{2}$ and touches all faces; and the ratio $R/r$ is still $\sqrt{2}$.

the ratio is not the same between the cube and the octahedron  they are not (both ~ 0.5773)
octahedron 0.577350269
cube       0.471404521


Sequence for which no closed form can exist I was wondering whether there exists a (computable) sequence of numbers, for which it can be proven that no closed form can exist.

Edit: By closed form I mean an expression involving only a constant number of elementary functions. So something like a sum can not occur in the expression.

Inevitably, the answer will depend on what one means by closed form.  We consider computable sequences of non-negative integers, that is, computable functions  $f(x)$ from  the non-negative integers to the non-negative integers. We will allow the least number operator $\mu$, as well as operations of ordinary arithmetic.

The least number operator may not be familiar, so we define it. Let $R(y,x_1,\dots,x_n)$ be a relation such that for all $x_1,\dots,x_n$ there is a $y$ such that $R(y, x_1,\dots,x_n)$ holds. Then $\mu y R(y,x_1,\dots,x_n)$ is the least such $y$.

As a consequence of Matijasevic's solution of Hilbert's Tenth Problem, there is a fixed polynomial $P(e, x, u_1,\cdots, u_k)$ with the following property.

For any computable function $f$, there is a non-negative integer $e=e(f)$ such that for any $x$,
$$f(x)=[\mu y( P(e, x, [y]_1,[y]_2, \dots, [y]_k)=0)]_0.$$
Here by $[w]_i$ we mean the exponent of the $i$-th prime $p_i$ in the prime power decomposition of $w$.  (The $0$-th prime is $2$.)

This gives what I would consider a positive answer to the closed form question: Every computable sequence has a closed form. Many theorems of the same general kind were known long before the work of Matijasevic, except that instead of a polynomial $P$, one had a more complicated function.

If the $\mu$-operator is not allowed, there are quite a few different workarounds.

I am going to guess that closed form numbers can be computed in polynomial time or some kind of bounded complexity (depends on your definition of closed form), but there are for sure computable sequences that take much longer to compute


Choose between Combinations & Permutations What is the most natural way to choose either permutations or combinations when encountering a particular combinatorial problem?

For example, in the following problem:

Given $n$ antennas of which $m$ are defective and  $n-m$ are functional, how many linear 'orderings' are there in which no two defectives are consecutive?

What seems to be the natural way of doing this (P or C)?

The puzzlement is understandable. You want to count sequences, which brings permutations to mind. However, the most efficient way to solve this particular problem will use "combinations."

Put the $n-m$ working antennas in a row, leaving nice big gaps between them.  If we want the bad antennas to be never next to each other, where can a bad antenna go? Into one of the gaps, or at either end.  So there are $n-m+1$ potential places for bad antennas.  We have to CHOOSE $m$ of these places.  The number of ways to do this is
$$\binom{n-m+1}{m}$$
(If $m$ is larger than $n-m+1$, then the job cannot be done at all.  We can consider that the answer is still given by the above formula, if we define $\binom{x}{y}$ to be $0$ if $y>x$.  Or else we can make the explicit restriction $m \le n-m+1$.)

It is (sadly) all too easy for a plausible sounding argument to be not correct. So one might check whether the above formula gives the right answer in "small" cases where one can count explicitly. And in addition one might see what happens for something like $n=10$, $m=3$, where you can take $7$ pennies, lay them out in a row, and try to think about where one can put $3$ dimes, so that no $2$ dimes are next to each other.

lets change the question to daily situation for understanding.
consider n students in which m are boys so we have n-m girls.how many linear orders are possible if no two boys sit consecutive?
[In our example consider 7 students in which there are 2 boys then n-m=5 so we have 5 girls]
GGGGG arrangement of n-m girls
since no boy should sit consecutively they can sit beside each girl
BGBGBGBGBGB [so there are 6 places for boys 5+1=n-m+1]
so we need m positions out of n-m+1 [we need 2 position out of 6 positions available]
possible combiantions are( n-m+1Cm)


Correstness of the mathematical expression I have a question regarding some strange mathematical expression defined in the set of positive integers - N :

g(N) - greatest power of 2 that divides N.
e.g. g(24) = 8 because 8 is divided by 2,4,8 but is not divided by 16
g(0) = ∞

Is point 2. represents correct mathematical expression? Can we use ∞ notation without using limits?

The function $g$ is a function $\Bbb N \to \Bbb N \cup \{\infty\}$. This is a well-defined thing to do, but you have to be careful about what you do with the values of $g(x)$.

On one level, $\infty$ is just a symbol; it would be equally valid to define $g(0) = \clubsuit$ at which point $g$ is a function $\Bbb N \to \Bbb N \cup \{\clubsuit\}$.

However, this is only considering $\Bbb N \cup \{\infty\}$ as a set. Sets are crucial to mathematics, but a set on its own doesn't have much structure. The natural numbers have lots more structure than a set: you can add them, multiply them, work out whether one divides another, work out which of two is bigger, and so on. (And the real numbers have lots more structure!) These operations aren't defined for $\infty$, and while we can sometimes write down plausible definitions for what those operations should be, we have to be careful, because any theorems we proved about the natural numbers don't necessarily hold for $\Bbb N \cup \{\infty\}$.

(I talk about an example of this, admittedly talking about real numbers rather than natural numbers, in my answer at What is the limit of zero times x, as x approaches infinity?)

However, there is one bit of structure that definitely does extend nicely, which is the comparison relations on $\Bbb N$. For natural numbers $n$ and $m$, we know how to define whether $n < m$ or not. We can extend this definition to $\Bbb N \cup \{\infty\}$ by saying $n < \infty$ for all $n \in \Bbb N$. ("Infinity is bigger than any natural number").

This also has the advantage of making the two parts of the definition consistent: if you define

$$h(x) = \sup \{2^i: i \in \Bbb N, \exists r \in \Bbb N\text{ s.t. }2^ir = x\}$$

then $h(0) = \sup \{2^i: i \in \Bbb N\} = \infty$, and so $h(x) = g(x)$ for all $x \in \Bbb N$.

(Also, I disagree with @MPW; $0$ is a positive integer, and certainly $0 \in \Bbb N$, though I would normally call $\Bbb N$ the natural numbers rather than the positive integers, partly because of the confusion as to whether "positive" means "$> 0$" or "$\geq 0$".)

No, for two reasons:

$0$ is not a positive integer
$\infty$ is not a power of $2$

My point is that $g$ is ill-defined as stated.


A proof of 2=1 using derivative We know that
$2^2=2+2$

$3^2=3+3+3$

Similarly

$x^2=x+x+...$ ( upto $x$ times)

Now I want to differentiate both side with respect to $x$

This gives
$2x=1+1+1+.....$ ( upto $x$ times)

$2x=x$

Cancelling $x$ I have

$2=1$

Which should be false.
Where am I going wrong? Please tell. The proof has to be wrong somewhere.

You assume that $x$ is a natural number, but derivatives don't exist for functions on the set of natural numbers since any neighborhood of a natural number contains infinitely many non-natural numbers.


Is any finite sum of real numbers well-defined? I am trying to find an argument or counterexample for the following proposition.

Let $ (a_i)_{i=1}^{n} $ be a finite sequence of real numbers. Then, $ \sum \limits_{i=1}^{n}a_i $ is well defined.

By "well-defined" I mean that the sum exists and equals to the same real number no matter in which order you sum up the terms.

My guess is to use induction, but is this correct?

Thanks.

Sorry but your 'proof' is wrong. Your $P(1)$ is a statement, not a number, so it is meaningless to write "$P(1) = a_1$". Secondly, by writing "$\sum_{i=1}^{k+1} a_i = \sum_{i=1}^{k} a_i + a_{k+1}$" you are already assuming that "$\sum_{i=1}^{k+1} a_i$" is well-defined, otherwise it is invalid to write it anywhere! Thirdly, a finite sequence can have length zero. [Edit: You have only fixed the first error.]

In fact depending on your underlying logic system, the proof that such recursive definitions give well-defined results can be anywhere from simple to highly non-trivial. See this post for a sketch of the proof.

Here is what the intuitive proof be in the case of the summation notation.

  Take any $m \in \mathbb{N}$ and any sequence $(a_k)_{k \in [1..m]}$ from $\mathbb{R}$.

  Let $\sum_{k=1}^n a_k = \cases{ 0 & if $n=0$ \\ \sum_{k=1}^{n-1} a_k + a_n & if $n > 0$}$, for each $n \in [0..m]$.

[Note that I have given the definition in the above format to make it clear that it is unambiguous how to interpret the summation notation, since the cases are disjoint and one of the cases must hold for any $n \in [0..m]$. Also, the definition must be interpreted as defining "$\sum_{k=1}^n a_k$" only when the relevant sub-expressions are well-defined. For example, "$\sum_{k=1}^0 a_k$" is well-defined, and hence "$\sum_{k=1}^0 a_k + a_1$" also is, which implies that "$\sum_{k=1}^1 a_k$" is well-defined. And so on.]

Now we proceed by induction to prove that "$\sum_{k=1}^n a_k$" is well-defined for any $n \in [0..m]$.

Let $P(n)$ be the assertion that if $n \in [0..m]$ then "$\sum_{k=1}^n a_k$" is a well-defined real number.

Then $P(0)$ is true because "$\sum_{k=1}^0 a_k$" is well-defined since it is defined as "$0$", which is a well-defined real number.

Given any $n \in \mathbb{N}$ such that $P(n)$ is true:

  If $n+1 \in [0..m]$:

    "$\sum_{k=1}^{n+1} a_k$" is defined as "$\sum_{k=1}^{(n+1)-1} a_k + a_{n+1}$" since $n+1 > 0$.

    Also "$\sum_{k=1}^{(n+1)-1} a_k + a_{n+1}$" reduces to "$\sum_{k=1}^n a_k + a_{n+1}$".

    And "$\sum_{k=1}^n a_k$" is a well-defined real number because $P(n)$ is true and $n \in [0..m]$.

    Thus "$\sum_{k=1}^{n+1} a_k$" is a well-defined real number because $\mathbb{R}$ is closed under addition.

  Thus $P(n+1)$ is true.

Therefore by induction $P(n)$ is true for every $n \in \mathbb{N}$.

Thus "$\sum_{k=1}^n a_k$" is a well-defined real number for any $n \in [0..m]$.

Note that it is important to include as part of the induction predicate that the sum is a real number, otherwise you cannot perform the induction step! Also, note that the induction predicate contains the cut-off condition that $n \in [0..m]$. This is in general how we can prove statements for (arbitrary-length) finite sequences using induction.

Now I'll sketch the proof that the summation is independent of order of the terms. To state it in full generality, we first have to define permutations on finite sequences:

  We say that a function $σ$ is a permutation on a set $S$ iff $σ$ is a bijection from $S$ to $S$.

And then the theorem would be:

  Take any $n \in \mathbb{N}$ and sequence $(a_k)_{k \in [1..n]}$.

  Then $\sum_{k=1}^n a_k = \sum_{k=1}^n a_{σ(k)}$ for any permutation $σ$ on $[1..n]$.

The rough idea of the proof I'm going to give is that we first get rid of the cases of $n = 0$ and $n = 1$, and we permute the first $(n-1)$ terms in such a way that we can swap the last two terms to get the right term at the end, and then we permute the first $(n-1)$ terms again to get the desired permutation. The first permutation we use can be a simple one, such as just swapping the $σ(n)$-th term to the $(n-1)$-th position. Both the first and the last permutations do not change the value by the induction. The swap of the last two terms come from associativity and commutativity of addition on reals.

Normally, the above paragraph is sufficient for an informal proof, but below is the sketch of the rigorous proof. Suffice to say that it is worth knowing how it is done but not worth doing it every time.
$
\def\l{\left}
\def\r{\right}
\def\less{\smallsetminus}
\def\on{\restriction}
$

Let $P(n)$ be the assertion that for any sequence $(a_k)_{k \in [1..n]}$ we have $\sum_{k=1}^n a_{σ(k)} = \sum_{k=1}^n a_k$ for any permutation $σ$ on $[1..n]$.

Then $P(0)$ and $P(1)$ are trivially true.

Take any $n \in \mathbb{N}$ such that $n > 1$.

By induction we can assume that $P(n-1)$ is true. [I will mark each use of this with a "$\ast$".]

If $σ(n) = n$:

  $σ \on [1..n-1]$ ($σ$ with domain restricted to $[1..n-1]$) is a permutation on $[1..n-1]$.

  Thus $\sum_{k=1}^n a_k = \sum_{k=1}^{n-1} a_k + a_n \overset{\ast}= \sum_{k=1}^{n-1} a_{σ(k)} + a_n = \sum_{k=1}^{n-1} a_{σ(k)} + a_{σ(n)} = \sum_{k=1}^n a_{σ(k)}$.

If $σ(n) \ne n$:

  Let $c \in [1..n-1]$ such that $σ(c) = n$.

  Let $τ$ be the swap $(\ n-1 \quad σ(n)\ )$ on $[1..n-1]$. [This is the first permutation.]

  [Precisely, let $τ(i) = \cases{ σ(n) & if $i = n-1$ \\ n-1 & if $i = σ(n)$ \\ i & if $i \in [1..n-2] \less \{σ(n)\}$ }$, for each $i \in [1..n-1]$.]

  Then $\sum_{k=1}^n a_k = \sum_{k=1}^{n-1} a_k + a_n \overset{\ast}= \sum_{k=1}^{n-1} a_{τ(k)} + a_n = \l( \sum_{k=1}^{n-2} a_{τ(k)} + a_{σ(n)} \r) + a_n$

    $= \l( \sum_{k=1}^{n-2} a_{τ(k)} + a_n \r) + a_{σ(n)} = \sum_{k=1}^{n-1} b_k + a_{σ(n)}$

      where $b_k = \cases{ a_{τ(k)} & if $k \in [1..n-2]$ \\ a_n & if $k = n-1$ }$, for each $k \in [1..n-1]$.

  Note that $\langle τ(k) \rangle_{k \in [1..n-2]} \cdot \langle n \rangle$ is a permutation of $[1..n] \less \{σ(n)\}$.

  Note also that $\langle σ(k) \rangle_{k \in [1..n-1]}$ is a permutation of $[1..n] \less \{σ(n)\}$.

  Thus $b$ is a bijection from $[1..n-1]$ to $\{ a_{σ(k)} : k \in [1..n-1] \}$.

  Thus let $ρ$ be a permutation on $[1..n-1]$ such that $b_{ρ(k)} = a_{σ(k)}$ for each $k \in [1..n-1]$.

  [Precisely, let $ρ(k) = b^{-1}(a_{σ(k)})$, for each $k \in [1..n-1]$.]

  Then $\sum_{k=1}^{n-1} b_k + a_{σ(n)} \overset{\ast}= \sum_{k=1}^{n-1} b_{ρ(k)} + a_{σ(n)} = \sum_{k=1}^{n-1} a_{σ(k)} + a_{σ(n)} = \sum_{k=1}^n a_{σ(k)}$.

Therefore $\sum_{k=1}^n a_k = \sum_{k=1}^n a_{σ(k)}$.

Let $\left(a_i\right)_1^n\in\mathbb{R}$ and $A_k=\sum_{i=1}^k a_i$, then:

Addition is binary relation: $\forall a,b\in\mathbb R:a+b\in\mathbb R$
Addition is comutative: $\forall a,b\in\mathbb R:a+b=b+a$

$A_1=\sum_{i=1}^1a_i=a_1\in\mathbb R$;
$A_2=\sum_{i=1}^2a_i=a_1+a_2=A_1+a_2=a_2+A_1\in\mathbb R$;
$A_3=\sum_{i=1}^3a_i=a_1+a_2+a_3=A_2+a_3=a_3+A_2\in\mathbb R$;
$\qquad\vdots$
$A_{k}=\sum_{i=1}^ka_i=A_{k-1}+a_k=a_k+A_{k-1}\in\mathbb R$.


Suppose that $f ' (x)$ exists and $f(x)$ has two roots $x_1$ and $x_2$. Try to prove that: Suppose that  $f'(x)$ exists and $f(x)$ has two roots $x_1$ and $x_2$. Try to prove that:

  $\exists \xi \in (x_1,x_2)$ such that $f(\xi)+f'(\xi)=0$.

We cannot use the knowedge of integration.

Consider $\phi(x)=e^xf(x)$ which is  continuous and differentiable since $f$ is so.

Also $\phi ^{'}(x)=e^x(f(x)+f^{'}(x))$

Now $\phi(x_1)=0,\phi(x_2)=0$

Since between any two roots of $\phi(x)$ lies one root of $\phi^{'}(x)$,
 so there exists $\xi \in(x_1,x_2)$ such that $\phi^{'}(\xi)=0$

Hence $e^\xi(f(\xi)+f^{'}(\xi))=0$

since $e^\xi\neq 0 $  so $f(\xi)+f^{'}(\xi)=0$

Real-valued differentiable function $f(x)$ attains equal zero values at two distinct points $x_1$ and $x_2.$ According with Rolle's theorem, so exists point $x_3$ with $f'(x_3)=0.$

If $f(x_3)=0$ that $x_3$ is required point. Consider the case $f(x_3)>0.$

If $f'(x_2)=0,$ that $x_2$ is required point. Let $r_0=x_2.$ If $f'(x_2)>0,$ then we have $f(x_2-\varepsilon)<0$ for sufficiently small epsilon and a root $r_1\in(x_3, x_2-\varepsilon)$ etc. If $f'(r_k)<0,$ then let $x_r=r_k,$ wherein $x_r\leq x_2$

$f(x)+f'(x)$ is continuos in $(x_3,x_r),$ wherein
$$f(x_3)+f'(x_3)>0,\quad f(x_r)+f'(x_r)<0.$$
So there is a point $X\in(x_3, x_2)$ with $f(X)+f'(X)=0.$

Case $f(x_3)<0$ is proved.

Case $f(x_3)<0$ can be proved similarly.


Homeomorphism of the Disk I'm working through Massey's "Basic Course in AT."  One of the problems is prove that a homeomorphism of the closed disk maps the boundary to the boundary and the interior to the interior.  How would one prove this?  I can't seem to get this one problem.

This answer extends on Chris Eagles comment:

Let $D^n \subset \mathbb R^n$ denote the $n$-dimensional closed unit disk, that is $D^n = \{ x \in \mathbb R^n \;|\; |x|\leq 1 \}$, with boundary $\partial D^n = S^{n-1} = \{ x \in \mathbb R^n \;|\; |x| = 1 \}$ the $(n-1)$-dimensional sphere.

Let $f: D^n \to D^n$ be a homeomorphism that maps $x \in \partial D^n$ to $f(x) \in D^n \setminus \partial D^n$. Obviously $f$ induces a homeomorphism $\tilde{f}: D^n \setminus \{ x\} \to D^n \setminus \{ f(x) \}$.

Since $x \in \partial D^n$, we have that $D^n \setminus \{ x\}$ is convex and therefore homotopy equivalent to a point. On the other hand we can construct a homotopy equivalence $D^n \setminus \{ f(x) \} \simeq \partial D^n = S^{n-1}$ since $D^n$ is compact and radially convex wrt. a neighborhood of $f(x)$. Thus we get $\{pt\} \simeq D^n \setminus \{ x\} \cong D^n \setminus \{ f(x)\} \simeq S^{n-1}$, which is a contradiction by your technique of choice. For example $\pi_{n-1}(\{pt\}) \not \cong \pi_{n-1}(S^{n-1})$.

A homeomorphism $\phi:D \to D$ maps open sets to open sets. Pick a point $P$ in the interior of the disk and a disk $D(P,\delta)$ which lies inside the given disk $D$. If $P$ would be mapped on the boundary of $D$ then $\phi(P)$ would not be in the interior of $\phi(D(P,\delta))$; contradiction with the fact that $\phi(P) \in \phi(D(P,\delta))$ and $\phi(D(P,\delta))$ is open.

This means that interior points are mapped to the interior of the disk.

Pick now $Q$ on the boundary of $D$. If $\phi(Q)$ is on the interior of $D$, then by a similar argument with the one above, since $\phi^{-1}$ is also a homeomorphism it follows that $\phi^{-1}(\phi(Q))=Q$ is in the interior of $D$. Contradiction.

Therefore $\phi$ maps the boundary of the disk onto the boundary of the disk.

[edit] As mentioned in the comments, I used the standard euclidean topology, not the induced one, and that might be a problem. I think that the best solution still remains the one mentioned in the first comment:

That if the homeomorphism $\phi : D\to D$ maps $P$ to $Q$ then $D\setminus\{P\}$ and $D\setminus \{Q\}$ are homeomorphic. If $P$ is on the boundary and $Q$ is in the interior (or the other way around) we get two domains which are not homeomorphic: one is simply connected, one isn't (it has a "hole")


Contradiction of a statement? My textbook has asked me to find the "contradiction" of a given statement but I have not learned to do such a thing and googling has not yielded in any results whatsoever. Exactly what is the contradiction of a statement and how to form it, given any statement?

Edit: This is the problem in question:

  "Write the converse, contrapositive and contradiction of the statement "If ∆ABC is right angled at B, then AB²+BC²=AC²"

The answer for the contradiction of this statement is given as

  "∆ABC is right angled at B and AB²+BC²≠AC²"

What you seem to be asking, when you (your text) refers to the "contradiction" of a statement, might be better termed as the "negation" of a sentence. For example, if person A declares $p$, another person B would be contradicting A by declaring "not $p$. That is, person B is asserting the negation of $p$.

So, we have a sentence we'll call $p$.  It's negation is simply $"\lnot p".$
So if we know that $p$ is true, its negation would be false.

If the sentence (proposition), for example, is "$x$ is an odd number", it's negation would be "it is not the case that $x$ is an odd number," or equivalently, its negation would be "$x$ is an even number."

But the original post asked for the contradiction of a statement, not just a statement that "is" a contradiction.


$O$ is not dense in $\mathbb R^n$ $A$ is a matrix of order $n$, and $x$ is an $n$-dimensional column vector. The set
$$
O=\{x,Ax,A^2x,\cdots,A^mx,\cdots\}
$$
Prove that, $O$ is not dense in $\mathbb R^n$.
This claim seems to be trivial: it just looks like $\mathbb N$ in $\mathbb R$. I recalled how we verify the denseness of $\mathbb Q$ in $\mathbb R$, but it seeems that the same method doesn't work. Can anyone help?

Suppose $O=\{A^mx:m\in\mathbb N_0\}$ is dense in $\mathbb R$. There are two possibilities:

$A^T$ has a real eigenvector $y$ corresponding to a real eigenvalue $\lambda$. Then
\begin{aligned}
&\ O=\{A^mx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R^n$}\\
\Rightarrow&\ y^TO=\{\lambda^m y^Tx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R$}\\
\Rightarrow&\ Y=\{\lambda^m:m\in\mathbb N_0\}\text{ is dense in $\mathbb R$}.
\end{aligned}
$A^T=A^\ast$ has an eigevector $z$ corresponding to some non-real eigenvalue $\overline{\lambda}$. Then the $\mathbb R$-linear span of the entries of $z$ must be two-dimensional and hence it is equal to $\mathbb C$. Therefore,
\begin{aligned}
&\ O=\{A^mx:m\in\mathbb N_0\}\text{ is dense in $\mathbb R^n$}\\
\Rightarrow&\ z^\ast O=\{\lambda^mz^\ast x:m\in\mathbb N_0\}\text{ is dense in $\mathbb C$}\\
\Rightarrow&\ Z=\{\lambda^m:m\in\mathbb N_0\}\text{ is dense in $\mathbb C$}.
\end{aligned}

It remains to show that $Y$ and $Z$ aren't really dense in $\mathbb R$ and $\mathbb C$ respectively. You may continue from here.

I think the hints in the comments provide you enough help for a simple proof.
You consider two cases: the norm sequence $(||A^n x||)_{n\in\mathbb N}$ either diverges to infinity, or is eventually bounded by some real number $M$ (it could still oscillate and not converge, but we'll see that we don't care).
If it diverges, it means that any open bounded subspace of $\mathbb R^n$ (e.g., $B=\{x\in\mathbb R^n : ||x||<M\}$) contains at most a finite number of elements of $O$, which means that $O$ can't be dense in $\mathbb R^n$ (because its closure doesn't even "fill" $B$).
If the norm is (eventually) bounded ($\leq M$), you can use the same argument for $\mathbb R^n\smallsetminus \overline{B}$ (it contains at most a finite number of elements of $O$...).
In either case, $O$ cannot be dense in $\mathbb R^n$.


sequence ${1/n}$ converges in R (to 0) but fails to converge in the set of all positive real number. Why? Sequence $\{\frac{1}n\}$ converges in $\mathbb{R}$ (to $0$) but fails to converge in the set of all positive real numbers.
I don't fully understand why it fails to converge in the set of all positive reals. Is it because $0$ is not in that set?

Yes. You can also say positive real number is not complete. So even the sequence is still Cauchy (this is a property concerning the sequence itself), we don't have a limit inside the space to make it convergent.

The positive reals are the open set
$$A = (0,\infty)$$
opens sets are never compact in $\Bbb R$, and in metric spaces this is equivalent to saying that this subset is also not sequentially compact. In fact $A$ is certainly not bounded. You happened to choose a sequence in this space that has no convergent sub-sequence. So you can look at it from a topological perspective.


Is "alignment" (or something else) a better word than "direction" for a senseless direction? When I encountered the concept of direction being prior to the ordering of points on a line representing (parallel to) the direction, I thought it was a valuable distinction.  I resolved to keep this distinction in my work.  But that didn't last very long.

While taking notes on Schouten's Tensor Analysis book, I found myself writing things like: there is a sense in which a line segment determines a direction without a sense.  A sense is determined by a ray.  So in order to talk about a line having a direction in the naive sense, we have to assign a sense to a senseless line. ... ad nauseam.

It gets even worse when we talk about directed flat p-manifolds in $E_n.$

While pondering this predicament, I came upon the ideal of saying that a line determines an alignment, which when a relative ordering of points on the line is additionally specified, determines a direction.

Is the term alignment or some other, similar term ever used to denote the idea of direction without sense?

Is the term alignment used in mathematics in a way that would conflict significant with my proposed usage?

Maybe oriented is the word you are looking for?

I'd get in the word "colinear"/"colinearity".

--- rk


We can count this by considering the number of states $i$ that contribute two senators:
$$
\sum_{i=0}^{5}\binom{50}{i}\binom{50-i}{10-2i}=51590216930.
$$

I wonder at the interpretation of the question.

If there were 50 pairs of counters of different colors, I can understand,
but how can you have a committee with identical people ?

I would think the answer is simply $\dbinom{50}{10}$

Remarks

Since you were asked to use a G.F. your interpretation and that of others must be right, but I am mystified why the questioner had to insert  "the two Senators from the same state, (50 States in All) are considered identical"


Consider the ratios $$A_n=\frac{(2n+1)!}{n!^2(n+1)!}$$ The goal is to show that $$A_{1007}<1$$ To do so, compute $$A_8=\frac{2431}{4032}<1$$ and note that $$\frac{A_{n+1}}{A_n}=\frac{2(2n+3)}{(n+1)(n+2)}<1$$ for every $n\geqslant3$ since the difference of the denominator and the numerator is $$n^2-n-4\geqslant(n-3)(n+2)$$ This shows that $$A_n<1$$ for every $n\geqslant8$.

You get $(1007!)^2$ compared to $\frac{2015!}{1008!}$.  The second expression is a product of $1007$ integers.  The first expression is the product of $2014$ integers.  If you take them in carefully chosen pairs, you can have the product of each pair greater than each factor in the second expression.


Uniform continuity of $f(x)=(1-\cos(x))/\sin(x)$ on the interval $(0,1)$ I got this question:

Prove that the function $f(x)=\frac{1-\cos(x)}{\sin(x)}$ is uniformly continuous on the interval $(0,1)$

I tried to prove it directly using the definition of uniform continuity but I failed this way. Then I tried to prove it using the fact that the sum of two uniformly continuous functions is uniformly continuous by writing $f(x)=\frac{1}{\sin(x)} - \cot(x)$ and then I tried to show that both $\frac{1}{\sin(x)}$ and $cot(x)$ are uniformly continuous on (0,1) but I wasn't managed to proceed that much.

Some hints will be helpful. Thanks.

Hint: You can extend $f$ continuously to $[0,1]$ and continuous functions on closed intervals are uniformly continuous.

Extension to Simon's answer:
$$\lim\limits_{x\to 0}{\frac{1-\cos\,x}{\sin\,x}}=\lim_{x\to 0}{\frac{\frac{1}{2}x^2}{x}}=0$$
the limit exists, so you can extend $f$ from $(0,1)$ to $[0,1]$


equation of a circle ,standard form I am supposed to be finding the equation of a circle that have a center (√5,2√2) and radius of √10 by using the standard form

I already know the distance formula but I could'nt figure out how to solve this given center.


Question about sylow p-subgroup. Question: Find the number of sylow 5-subgroup of group of order 60
My attempt:
  O(G)=60=2^2•3•5

As,5|60 and 5^2 does not divide 60 so, by sylow first theorem G must have at least one sylow 5-subgroup of order 5.
Now by sylow 3rd theorem number of sylow 5-subgroup say n_5 satisfies,
n_5 = 1+5k and n_5 | 60
Both of above holds for k=0,1.
But in text book they take k= 1 and so n_5 = 6.
Why they take k=1? and not take k=0?

$n_5$ satisfies two conditions:

1) $n_5\mid |G|/5=12$ which implies $n_5\in \{1,2,3,4,6, 12\}$

2) $n_5\equiv 1 {\pmod 5}$

The two conditions are satisfied only if $n_5=6$.


Convolution, indicator function I need to calculate $(f*f)(x)$ of $f(x) = 1_{[0,1]}(x)$, which is the indicator function defined with

Calculating the integral $(f*f)(x) = \int_{0,}^{x}1_{[0,1]}(t) \cdot1_{[0,1]}(x-t) dt$ gives $(f*f)(x) = \begin{Bmatrix}
x & \mathbb{for }\space x\in[0,1] \\
0 & \mathbb{otherwise}
\end{Bmatrix}$

but the Laplace transformation of $f(x)$ gives the convolution as $(f*f)(x) = x - 2 (x-1)\theta(x-1) + (x-2)\theta(x-2)$, where $\theta(x)$ is the Heavyside theta function.

What am I doing wrong?

Looking carefully at the definition of the integrand (for $x\in\mathbb{R}$), you can see that $f\ast f$ is $0$ for $x\notin[0,2]$, and that for $x\in[0,2]$
$$
f\ast f(x) = \int_{\max(0,x-1)}^{\min(1,x)} dt = \begin{cases} \int_{0}^{x} dt = x & x\leq 1 \\
\int_{x-1}^{1} dt = 2-x & x\geq 1
 \end{cases}
$$
As mentioned in a comment above, you can alternatively get this by seeing that you are computing the probability density function of $D=\mathcal{U}_{[0,1]}\ast\mathcal{U}_{[0,1]}$, where $\mathcal{U}_{[0,1]}$ is the uniform distribution on $[0,1]$. This is the distribution of a random variable $Y=X_1+X_2$, where the $X_i$'s are independent uniform r.v.s on $[0,1]$: in particular, $Y$ take values in $[0,2]$, and it is "easy" to see that its density is symmetric around $1$ and that the probability to be around $1/2$ is bigger than the probability to be around $0$ or $1$. (on a side note, $Y$ follows a triangular distribution on $[0,2]$ with mode $1$).

The support of the convolution of a function with support $[0,1]$ with itself should have support $[0+0,1+1] = [0,2]$. The convolution integral is non-zero for $x$ in $[1,2]$. If you


orthogonalization and orthonormalization I have to construct a diagonalizable and orthogonal matrix starting from this quadratic form $Q(x_1,x_2,x_3)=-2x_1x_3+2x_1x_3-2x_2x_3$ in order to reduce it in canonic form with a variables change.

I've constructed the associated symmetric matrix $\begin{bmatrix}0 & -1 & -1\\ -1 & 0 & -1\\  1&  -1& 0\end{bmatrix}$ with autovalues $\lambda_1=-1$ and $\lambda_2=2$, respectively by algebraic multiplicity $m(-1)=2$ and $m(2)=1$.
Then, I've found their respective autovectors: for $\lambda_1=-1$ we have $\bar{a}=l\begin{bmatrix}1\\ 1\\ 0\end{bmatrix}+k\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}$, and for $\lambda_2=2$ we have $\bar{b}=t\begin{bmatrix}1\\ -1\\ 1\end{bmatrix}$.

Now, to construct the orthonormal matrix i have to prove that autovectors are orthogonal two by two. If i'd had three autovectors like $l\begin{bmatrix}a\\ b\\ c\end{bmatrix}$, $k\begin{bmatrix}d\\ e\\ f\end{bmatrix}$ and $t\begin{bmatrix}g\\ h\\ i\end{bmatrix}$ I would have just prove (for example) that $l\cdot k\begin{bmatrix}a\\ b\\ c\end{bmatrix}\begin{bmatrix}d\\ e\\ f\end{bmatrix}=0$, but now my first autovector is a sum $l+k$. So i ask you:

1) How can i orthogonalize my two autovectors $\bar{a}$ and $\bar{b}$? Could you show me the algebraic passages?

After this, i know that to orthonormalize a vector i can write $\sqrt{a^2l^2+b^2l^2+c^2l^2}=1$, but now (for the second time) my autovector is a sum $l+k$. So i ask you:

2) How can i orthonormalize my first autovector $\bar{a}$?

Thanks for any help!

Thanks for your answers. So, trying to apply Gram-Schmidt to the vectors $\begin{bmatrix}1\\
1\\ 0\end{bmatrix}$ and $\begin{bmatrix}-1\\ 0\\ 1\end{bmatrix}$i write that $v_1=\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}$, $v_2=\begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}$. Now:

$w_1=v_1=\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}$

$w_2=v_2-\frac{v_2\cdot w_1}{w_1\cdot w_1}\cdot w_1=\begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}-\frac{\begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}\cdot \begin{pmatrix}1\\ 1\\ 0\end{pmatrix}}{\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}\cdot \begin{pmatrix}1\\ 1\\ 0\end{pmatrix}}\cdot \begin{pmatrix}1\\ 1\\ 0\end{pmatrix}=\begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}+\frac{1}{2}\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}\\ \frac{1}{2}\\ 1\end{pmatrix}$.

Then, i normalize the vector $\begin{pmatrix}1\\ -1\\ 1\end{pmatrix}$:

$\sqrt{t^2+t^2+t^2}=\sqrt{3t^2}=1\rightarrow t_1=\frac{\sqrt{3}}{3}\Rightarrow ||\bar{b}||=t_1\begin{bmatrix}
1\\ -1\\ 1\end{bmatrix}=\begin{bmatrix}\frac{\sqrt{3}}{3}\\\frac{-\sqrt{3}}{3} \\ \frac{\sqrt{3}}{3}\end{bmatrix}$.

So, my orthonormal matrix is $P=\begin{bmatrix}1 & -\frac{1}{2} & \frac{\sqrt{3}}{3}\\  1& \frac{1}{2} & -\frac{\sqrt{3}}{3}\\  0& 1 & \frac{\sqrt{3}}{3} \end{bmatrix}$.

But now, since $P=P^{T}=P^{-1}=\begin{bmatrix}1 & 1 & 0\\  -\frac{1}{2}& \frac{1}{2} & 1\\  \frac{\sqrt{3}}{3}& -\frac{\sqrt{3}}{3} & \frac{\sqrt{3}}{3}\end{bmatrix}$ and being the diagonal matrix $D=\begin{bmatrix}-1 & 0 & 0\\  0& -1 & 0\\  0& 0 & 2 \end{bmatrix}$, i obtain (for the condition $\bar{X^{T}}A\bar{X}=\bar{X^{T}}PDP^{T}\bar{X}=\bar{Y^{T}}D\bar{Y}$) that the change of variables is $\bar{Y}=P^{T}\bar{X}=\begin{bmatrix}1 & 1 & 0\\  -\frac{1}{2}& \frac{1}{2} & 1\\  \frac{\sqrt{3}}{3}& -\frac{\sqrt{3}}{3} & \frac{\sqrt{3}}{3}\end{bmatrix}\bar{X}$.

In the light of all this I ask you all whether the procedure is correct and, with it, the result.
Thanks again for any help.

Con "dei passaggi" intendi forse che avrei dovuto calcolare anche $w_3=v_3-\frac{v_3\cdot w_1}{w_1\cdot w_1}-\frac{v_3\cdot w_2}{w_2\cdot w_2}=\begin{pmatrix}1\\ -1\\ 1\end{pmatrix}-\frac{\begin{pmatrix}1\\ -1\\ 1\end{pmatrix}\cdot \begin{pmatrix}1\\ 1\\ 0\end{pmatrix}}{\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}\cdot \begin{pmatrix}1\\ 1\\ 0\end{pmatrix}}\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}-\frac{\begin{pmatrix}1\\ -1\\ 1\end{pmatrix}\cdot \begin{pmatrix}\frac{-1}{2}\\ \frac{1}{2}\\ 1\end{pmatrix}}{\begin{pmatrix}\frac{-1}{2}\\ \frac{1}{2}\\ 1\end{pmatrix}\cdot \begin{pmatrix}\frac{-1}{2}\\ \frac{1}{2}\\ 1\end{pmatrix}}\begin{pmatrix}\frac{-1}{2}\\ \frac{1}{2}\\ 1\end{pmatrix}=\begin{pmatrix}1\\ -1\\ 1\end{pmatrix}$?

Una volta fatto questo, i vettori normalizzati si ottengono calcolando per ognuno di essi ad es. $e_1=\frac{\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}}{||\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}||}=\frac{\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}}{\begin{pmatrix}l_1\\ l_1\\ 0\end{pmatrix}}$ (con $l_1=\frac{1}{\sqrt2}$) oppure posso normalizzarli separatamente (cioè a dire $\sqrt{l^2+l^2}=1$, il che vuol dire che nella matrice diagonalizzante ortogonale andrò ad inserire direttamente $\begin{pmatrix}l_1\\ l_1\\ 0\end{pmatrix}$)?

Se la risposta fosse il primo caso, come si svolge quel rapporto matriciale?


Why is $i^3$ (the complex number "$i$") equal to $-i$ instead of $i$? $$i^3=iii=\sqrt{-1}\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)(-1)}=\sqrt{-1}=i
$$

Please take a look at the equation above. What am I doing wrong to understand $i^3 = i$, not $-i$?

Since $i^2=-1$ by definition, $i^3=i^2\cdot i=-i$.

$\sqrt{a}\sqrt{b}=\sqrt{ab}$ is only guaranteed for positive real $a$ and $b$.

Someone posted earlier that it is i x i x i which is -1 x i, ie -i (but the post seems to have been deleted!

But that's all there is to it. It doesn't matter what i represents, the algebra will be consistent. But to prove it, substituting the value of i, ie (√-1), you get

√-1 x √-1 x √-1
= -1 x √-1
= -(√-1)
= -i

However it may be worth mentioning that i is not a complex number as has been suggested - it is an irrational number. A complex number is of the form (x +iy)


What is the solution of $\cos(x)=x$? There is an unique solution with $x$ being approximately $0.739085$. But is there also a closed form solution?

The equation in question is a transcendental equation. Apart of guessing, numerical or analytical methods, there is no way of solving the equation without using another transcendental function, and therefore argue in circles.
In this case, denote $g(x)=\cos x -x$, see that its derivative is negative with countable many zeros, and therefore $g$ is strictly decreasing, yielding that there is at most one solution to $g(x)=0$. Since $g(0)g(\pi/2)<0$ there is such a solution. Arbitrary precise approximations can be found using Newton, bisection, or false position method.
As user Myself commented, it is a challenge (not so hard) to prove that the sequence $x_{n+1}=\cos x_n, x_0 \in \Bbb{R}$ converges to the unique solution to $\cos x=x$.
Another related problem which I encountered last week when trying to help one of my friends for an exam is to find all continuous functions $f : \Bbb{R} \to \Bbb{R}$ with the property that $f(x)=f(\cos x)\ \forall x \in \Bbb{R}$.

this can be really easy with the help of graphs. Draw the graphs of y= cosx and x=y. Intersection points on the graph will show the solutions.


Finding $x$. The summation of the floor of the equation. I would appreciate if somebody could help me with the following problem

Q:Finding $x$. The summation of the floor of the equation.

$$\sum_{i=1}^{2013}\left\lfloor\frac{x}{i!}\right\rfloor=1001$$

Factorial increases very quickly, so because of the floor, most of the terms are zero. If we solve the equation without floors and only five first terms we get:

$$x+x/2+x/6+x/24+x/120= \frac{103x}{60}= 1001 \Leftrightarrow x \approx 583 $$

Now plugging this back into the original equation we see only five first terms are non-zero and sum gives 999. Increasing $x$ by one to $584$ gives the correct answer.

We have the series of the exponential function:

$$\sum_{n=0}^\infty\left\lfloor\frac{1}{n!}\right\rfloor=e$$

In our case, take the X outside, as a constant, and you get that:

$$x * \sum_{i=1}^\infty\left\lfloor\frac{1}{i!}\right\rfloor=e-1$$

Note, the -1 appears when you take the first element, i=0 out of the sum. As i goes >> 9, each element of the sum goes very fast to 0, so going to infinity or 2013, makes no difference. Thus, the result goes:

$$x = \frac{1001}{e-1}$$
Which is approximately 582.55


Combination problem distributing In how many ways 8 candies can be distributed be among 3 children?

I have done by using combination method $8C4$ but the answer is given 45.
someone has done in this way:

008 in 3 ways
710 in 6 ways
611 in 3 ways
620 in 6 ways
530  in 6 ways
521 in 6 ways
440 in 3ways
431 in 6 ways
422 in 3   ways
332 in 3 ways

in total 45 ways.

Why my formulation is not right.  And how can I do this math within the formula I have writen? I mean within short time done.

Original Answer

Seeing as you asked me to look at this problem for you I'll give my own answer. I get that your main concern here is understanding and not just being given the correct solution. But, correctness has a context. Your formulation, although lengthy has none the less given the correct answer.

If you don't understand how to prove the stars and bars formula, instead I'll show you a method using the basic idea of the stars and bars diagram to hopefully understand what is going on. From this you may go on to understand the proof.

The two bars stand between the three children. Start with the two children on the left with no sweets and the child on the right with all of them. Leave the left most bar at the left and move the right hand bar, past one star at a time, to the right. (This first one is equivalent to (0, 0, 8))

$$\ |\ |\ \bigstar\bigstar\bigstar\bigstar\bigstar\bigstar\bigstar\bigstar$$

The following is what it looks like after two steps to the right (which is equivalent to (0, 2, 6))

$$\ |\ \bigstar\bigstar\ |\ \bigstar\bigstar\bigstar\bigstar\bigstar\bigstar$$

You can do this 8 times. So the first step has 9 combinations.

Then put the two bars after the first star and again leave the left hand bar where it is and move the right hand bar, one at a time to the right. (This one is equivalent to (1, 0, 7))

$$\ \bigstar |\ |\ \bigstar\bigstar\bigstar\bigstar\bigstar\bigstar\bigstar$$

You can do this 7 times. So the second step has 8 combinations.

You should be seeing a pattern by now.

Each step decreases by one. The last step has the two bars on the far right with the right hand bar having nowhere else to go, finishing with 1 combination. Thus the pattern gives you a triangular number.

Total = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45

Addendum 1: Stars And Bars Proof

Stars and Bars Theorem $n\leq k$: We have $n$ containers and $k$ identical items to distribute. If you start by insisting that each container has at least 1 item in it, you may as well start by putting 1 item into each container. You have no choice with this. Hopefully you can see that, counting wise, this is the same as starting with $k-n$ items and allowing empty containers.

So, lets start with $k$ items, $n$ containers and insist at least one in each container. In the following argument this allows having gaps only between stars. If we tried to start with having empty containers, you have a problem with bars accumulating at the edges and the argument doesn't work.

Consider the following diagram with $k=11$ and $n=4$. The 3 bars represent the boundaries between separate containers, where $3=4-1$. We can't put bars at the left and right hand edges, otherwise that would mean we had a container at the right or left or both ends with no items in so the outside edges are automatically boundaries of the first and last containers.

\begin{align*}
|\ \bigstar\bigstar\bigstar\bigstar\bigstar\bigstar\ |\ \bigstar\bigstar\ |\ \bigstar\bigstar\bigstar &= (0, 6, 2, 3) \text{ this is not allowed } \\
\bigstar\ |\ \bigstar\bigstar\bigstar\bigstar\bigstar\ |\ \bigstar\bigstar\ |\ \bigstar\bigstar\bigstar &= (1, 5, 2, 3) \text{ this is allowed }
\end{align*}

Think of this as allocating the gaps between stars to the bars and note also that you can't have more than one bar in a single gap, otherwise we wouldn't be able to apply the standard combinations formula. So, we have in general $k$ stars, $k-1$ gaps between them to choose from and $n-1$ bars to distribute in them, which gives ${{k-1}\choose{n-1}}$ ways.

As above, insisting on at least 1 in each container, is the same as having $k-n$ items and allowing 0 in any container. So, if we distribute $k+n$ items into $n$ containers with at least 1 in each container, this is the same, counting wise, as distributing $k$ items into $n$ containers allowing 0 in any container.

Thus we have $${{k+n-1}\choose{n-1}} = \frac{(k+n-1)!}{((k+n-1)-(n-1))!(n-1)!} = \frac{(k+n-1)!}{k!(n-1)!} = {{k+n-1}\choose{k}}.$$

There is a formula for working this out called the related to the binomial coefficient it is: n!/(n-k)!k!, where n is the number of sweets and k is the number of children

Using the formula there are 112 possible combinations of sweets and children


Use integration by substitution I'm trying to evaluate integrals using substitution. I had

$$\int (x+1)(3x+1)^9 dx$$

My solution: Let $u=3x+1$ then $du/dx=3$

$$u=3x+1 \implies 3x=u-1 \implies x=\frac{1}{3}(u-1) \implies x+1=\frac{1}{3}(u+2) $$

Now I get $$\frac{1}{3} \int (x+1)(3x+1)^9 (3 \,dx) = \frac{1}{3} \int \frac{1}{3}(u+2)u^9 du = \frac{1}{9} \int (u+2)u^9 du \\
= \frac{1}{9} \int (u^{10}+2u^9)\,du = \frac{1}{9}\left( \frac{u^{11}}{11}+\frac{2u^{10}}{10} \right) + C$$

But then I get to this one

$$\int (x^2+2)(x-1)^7 dx$$

and the $x^2$ in brackets is throwing me off.

I put $u=x-1\implies x=u+1,$ hence $x^2+2 =(u+1)^2 +2 = u^2+3$. So
$$ \int(x^2+2)(x-1)^7\, dx = \int (u+1)u^7 du = \int (u+u^7) du = \frac{u^2}{2}+\frac{u^8}{8} + C $$
Is this correct or have I completely missed the point?

Nice work, (though a wee bit hard to read).

Let's look a bit more closely at the second integral.

Your substitution is correct: $$u = x-1 \implies x = u+1,\;dx = du, \\ x^2 + 2 =(u+1)^2 + 2$$

You got a bit mixed up when substituting into the factor $x^2 + 2$, to express it as a function of $u$.

Substituting, we get  $$\begin{align}\int (x^2 + 2)(x-1)^7\,dx & = \int ((u+1)^2 + 2)u^7\,du \\ \\
&= \int(u^2 + 2u + 3)u^7\,du \\ \\
&= \int(u^9 + 2u^8+ 3u^7)\,du\end{align}$$

I trust you can take it from here.


What is the maximum number of the pieces that a watermelon can be cut into 10 knife cuts? (circle cutting problem extension) I saw this problem in a game "Brain out" level 85. In this game, the answer is 1024, simply 10th power of 2. But I learned circle cutting problem in school and by solving recurrence formula, I could get explicit formula for the problem.
I think this problem also require recurrence formula similarly and the answer is not just 1024. What is the true answer about this game stage?

The original watermelon is (presumably) a convex shape. Performing a straight cut with a convex shape results in wo convex shapes. If you can arrange $n$ convex shapes in a way such that the cutting plane intersects each part, this gives you $2n$ convex shapes. So, if you are allowed to rearrange the pieces before applying your next knife cut, you can produce $2^k$ pieces with $k$ cuts.

If you are not allowed to rearrange the pieces, the answer is strictly lower for $k>3$ cuts of a convex watermelon in $\Bbb  R^3$.

In fact, if we count cuts through different pieces as different cuts (even when performed in the same movement), the answer is much simpler: After $k$ cuts, we have $k+1$ pieces.

If the expected answer is $1024$, it appears (though could perhaps not be inferred from the problem statement) that rearrangement of pieces is allowed.

i think that answer's actually right since it doesn't have any limitations like the watermelon cannot be rearranged. But if you weren't allowed to rearrange it and some force were holding the watermelon pieces together as you cut, the answer would be $176$. It's the same as the cake number problem and follows the formula $(n^3+5n+6)/6$


A proof of the principle of mathematical induction Is there a definition of the natural number in which someone might be able to prove "The Principle of mathematical induction"?

It is a common misconception that induction over natural numbers can somehow be justified (in a non-circular way). This is simply impossible, and there are two parts to the reason.

Firstly, note that to even talk about induction we need to work within a meta-system. And any reasonable meta-system will already have access to the natural numbers, in the sense that it has an inbuilt assumption that there is a collection $N$ of objects and operations $+,\times$ on $N$ that satisfy first-order PA (or something equivalent to it). PA already includes the induction schema. It just makes absolutely no sense to ask for a justification of induction over natural numbers without first defining natural numbers, but that cannot be done without already using assumptions with PA essentially built into them.

In short, it is like asking whether there is a proof that the English language is a language. Well, in English we have already agreed upon what "the English language" refers to, which is indeed a language!

For example, ZFC set theory does not seem to have the natural numbers as basic objects, but note the axiom of infinity (Inf). It is totally meaningless unless we already believe in the existence of a set that is structure-isomorphic to the natural numbers, besides believing in the other ZFC axioms. Furthermore, it is nothing more than a set-theoretic version of induction! Its modern form states that "an inductive set exists":

  (Inf) $∃I ( ( ∀x ( ¬∃y ( y∈x ) ⇒ x∈I ) ∧ ∀x ( x∈I ⇒ ∀y ( ∀z ( z∈y ⇔ z∈x ∨ z=x ) ⇒ y∈I ) ) )$

The other axioms such as specification and union permit ZFC to 'implement' arithmetic operations on the set $N$ given by (Inf), as in any standard textbook, and ZFC can indeed prove that $N$ satisfies induction in terms of the conventional axiomatization of PA.

But of course you can't possibly believe that (Inf) is true unless you already believe in ZFC plus the existence of a set that is inductive (intuitively some set containing all and only the sets that can be obtained from the empty-set by a finite natural number of the successor operation $x \mapsto x \cup \{x\}$).

Secondly, one might still ask whether induction is superfluous, in the sense that any proof using induction can be converted to a proof without induction. No, for any reasonable interpretation of the question. Of course, one could be silly and replace induction by a slightly different axiom (or axiom schema), but as with ZFC's axiom of infinity it would be obvious to any logician that the core notion of induction cannot be justified non-circularly.

The mathematical basis is as follows. PA$^-$ is stronger than PA minus the induction schema. But PA$^-$ is strictly weaker than PA and has a model that everyone agrees is structurally different from natural numbers. Specifically, let $P$ be the collection of polynomials with integer coefficients such that the coefficient of the highest-degree term is non-negative. Then $P$ with the usual polynomial arithmetic and the ordering defined via $x<y \overset{def}\equiv x \ne y \land \exists z\ ( x+z=y )$ satisfies PA$^-$ but does not satisfy induction. This is because PA proves $Q \overset{def}\equiv \forall n\ ( \exists m\ ( n = 2m \lor n = 2m+1 ) )$ but $P$ does not satisfy $Q$, and so $P$ does not satisfy the axiom given by induction that was used in the proof of $Q$ over PA.

A working list of the Peano Axioms defining the natural numbers $\mathbb{N}$ is:

In the set of the natural numbers there is an element named 1.
For every natural $n$ there is another natural, named successor, denoted by $n+1$.
There are no natural numbers $n$ for which $n+1 = n$.
If $n+1 = m+1$ for two natural numbers $n$ and $m$, then $n=m$.
If $S$ is a subset of the naturals satisfying that $1\in S$ and if $k\in S$ then $k+1\in S$, then $S = \mathbb{N}$.

Since these are axioms, there is nothing to prove. Notice how the last axiom can be rewritten as:

If there is a function $P:\mathbb{N}\rightarrow \{T,F\}$ satisfying that $P(1) = T$ and that if $P(k) = T$ then $P(k+1) = T$, then $P(n) = T$ for every natural $n$.

Which is precisely the Induction Principle.

Consider now the natural numbers defined as ordinals (notice that with these axioms we require the empty set to exist): define $1 = \{\emptyset \}$ and $n+1 = n \cup \{n\}$, and by definition $\mathbb{N} = \{1,2,3,\dots\}$ (one can include $0 = \emptyset$ if it so desires). Notice that under the definition as ordinals, $\mathbb{N}$ is a well-ordered set with respect to the inclusion, namely $n\subset n+1$.

We prove the fifth Peano axiom: suppose we have $S$ a subset of the naturals satisfying that $1\in S$ and if $k\in S$ then $k+1\in S$. If $S = \mathbb{N}$, we are done. Suppose $S \neq \mathbb{N}$, then the set $T = \{n\in\mathbb{N} : n\notin S\}$ is a non-empty subset of a well-ordered set $\mathbb{N}$, meaning that it has a least element, say $m$. Since $1\in S$ we have $m \neq 1$ so $\emptyset,\{\emptyset\}\in m$. By construction $m$ is a well-ordered set and contains only a finite number of sets, thus it contains a maximal set $k$. Again by construction, we have that $m = k\cup \{k\}$ so $m = k+1$. Since having $k\in T$ contradicts that $m$ is the least element in $T$, we have that $k\in S$ and thus by hypothesis $m = k+1\in S$, a contradiction.


Group action decomposes $X$ into distinct orbits Define the group action as $g\cdot x:=g^{-1}xg.$ Let $G=A_5$, and
$X=\{\sigma\in A_5:=\sigma=(a,b,c,d,e)\}.$ Show that the group action on X decomposes $X$ into two distinct orbits.

There are 60 elements in $A_5,$ so I assume that we need to use Burnside's theorem. But I am not sure how to use it to show that G decomposes $X$ into two distinct orbits.

Hint: show that the subset of permutations whose signature is 1 and the subset whose signature is $-1$ are preserved by the action.

You're basically looking for the conjugacy classes of all elements of the form $\sigma = (a, b, c, d, e)$.

In any $S_n$, all elements of the same cycle length are conjugate to each other.

We know that the conjugacy class of an element of $A_n$ splits into two (from $S_n$) if the element has a cycle decomposition of odd length where every length is unique.

Here, we know that $\sigma = (a, b, c, d, e)$ which is of length 5. Hence, its conjugacy class splits into two.


Product of a exponential and discrete Distribution Let $U \sim e(1)$ and $V$ a discrete random variable independent of $U$ such that
$p_V(v)=1/2$ if $v \in \{-1,1\}$ and $p_V(v)=0$ otherwise.

Problem:
Let $W=UV$. Find the distribution function of $W$ $\forall w\in \mathbb{R}$.

My try:
\begin{align}
P(W\leq w)&=P(UV\leq w)=P(U\leq w \mid V=1)P(V=1)+P(-U\leq w \mid V=-1)P(V=-1)\\
&= P(U\leq w)\frac{1}{2}+P(U\geq -w)\frac{1}{2}\\
&=P(U\leq w)\frac{1}{2}+(1-P(U\leq-w)\frac{1}{2}\\
&=\left(\int_0^we^{-u} \, du\right)\frac{1}{2}+\left(1-\int_0^{-w} e^{-u} \, du \right) \frac{1}{2}\\
&=(1-e^{-w})\frac{1}{2}+\frac{1}{2}e^w = F_W(w)
\end{align}

The correct answer in my book is the following
$$F_W(w)=\frac{1}{2}(1-e^{-w})+\frac{1}{2}, w>0$$ and
$$F_W(w)=\frac{1}{2}e^{w}, w\leq 0$$

Can anyone tell me what i am doing wrong?

You wrote:
\begin{align}
& P(W\leq w) = P(UV\leq w) \\[10pt]
= {} &P(U\leq w \mid V=1)P(V=1)+P(-U\leq w \mid V=-1)P(V=-1).
\end{align}

What is $P(U\le w\mid V=1)$?

It is $\begin{cases} 1-e^{-w} & \text{if } w\ge0, \\ 0 & \text{if } w<0. \end{cases}$

What is $P(-U\le w\mid V=-1)$?

It is $\begin{cases} e^w & \text{if } w\le0, \\ 1 & \text{if } w>0. \end{cases}$

You need to consider the cases $w\ge0$ and $w<0$ separately.

If you find the density function, you will be able to write it more simply by using the notation $|w|.$

Hint: An exponentially distributed random variable has positive support.


Number of conjugates of an $m$-cycle $\sigma$ in $S_n$ I am currently reading the following from Dummit and Foote can someone explain to me why the number of conjugates of $\sigma$ is:

$$\frac{n(n\ -\ 1)...(n\ -\ m\ + \ 1)}{m}\ ?$$

Two permutations in $S_{n}$ are conjugate if and only if they have the same cycle type, so really the question is how many $m$-cycles there are in $S_{n}$. Well, choose your $m$ elements from $n$ in $n\choose{m}$ ways. But we can rearrange those $m$ elements in $m!$ ways. Finally, each of our cycles has been represented $m$ times by a cyclic rearrangement, so we divide by $m$. So our final count is $$\frac{n!}{(n-m)!m}$$
as required.

The number of combinations of $n$ distinct things taken $m$ at a time is your starting point, for which the binomial coefficient comes in handy: $n!/m!(n-m)!$ Since each combination will represent more than just one $m$-cycle, for every choice of the 1st spot of an $m$-cycle there will be $m-1$ choices for the 2nd spot, $m-2$ choices for the 3rd spot, etc. You then arrive at $(m-1)! \times n!/m!(n-m)!$


A closed form for $\sum_{k=0}^n \frac{ (-1)^k {n \choose k}^2}{k+1}$ Mathematica gives  $$\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{k+1}= ~_2F_1[-n,-n;2;-1],$$ where $~_2F_1$ that is Gauss hypergeometric function. Here the question is: Can one find a simpler closed form for this summation. Recently, the absolute summation for this has been discussed at MSE:

A binomial summation: $\sum_{k=0}^{n} \frac{{n \choose k}^2}{k+1}$

We seek to evaluate

$$\sum_{k=0}^n \frac{(-1)^k}{k+1} {n\choose k}^2.$$

This is

$$\frac{1}{n+1} \sum_{k=0}^n (-1)^{k} {n\choose k} {n+1\choose k+1}
= \frac{1}{n+1} \sum_{k=0}^n (-1)^{k} {n\choose k} {n+1\choose n-k}
\\ = [z^n] (1+z)^{n+1} \frac{1}{n+1}
\sum_{k=0}^n (-1)^{k} {n\choose k} z^k
\\ = [z^n] (1+z)^{n+1} \frac{1}{n+1} (1-z)^n
= \frac{1}{n+1} [z^n] (1+z) (1-z^2)^n.$$

Now if $n=2m$ we get

$$\frac{1}{n+1} [z^{2m}] (1+z) (1-z^2)^n
= \frac{1}{n+1} [z^{2m}] (1-z^2)^n
\\ = \frac{1}{n+1} [z^{m}] (1-z)^n
= \frac{1}{n+1} (-1)^m {n\choose m}.$$

On the other hand when $n=2m+1$ we find

$$\frac{1}{n+1} [z^{2m+1}] (1+z) (1-z^2)^n
= \frac{1}{n+1} [z^{2m+1}]z  (1-z^2)^n
\\ = \frac{1}{n+1} [z^{2m}] (1-z^2)^n
= \frac{1}{n+1} [z^{m}] (1-z)^n
= \frac{1}{n+1} (-1)^m {n\choose m}.$$

We thus have even or odd the closed form

$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{n+1} (-1)^{\lfloor n/2\rfloor}
{n\choose \lfloor n/2\rfloor}.}$$

The second case could have been done by inspection given the first.
This result matches the comment by @SangchulLee.


How to know if equation can be solved by factorising before trying? So, I have core 1 test tomorrow and there is a lot of solving of quadratic equations without calculator and my weakest point is the time I waste in trying to factorise and equation but then it ends up it doesn't factorise or the opposite way in which I use equation while I could've done factorising. Is there any way you can know if equation factorise or not?

Check the discriminant. If it is a perfect square, the radical expression in the quadratic formula becomes an integer (I'm assuming you've cleared fractions before beginning the problem, and all coefficients are integers after that).

Recall that the discriminant for $ax^2 + bx + c = 0$ is $\Delta = b^2-4ac$, and that the solution(s) are $$x = \frac{-b\pm\sqrt{\Delta}}{2a}$$

I'm also assuming you're only interested in factoring if you can do it without using radicals. For example, you could factor $x^2-2$ as $(x+\sqrt{2})(x-\sqrt{2})$, but I'm guessing that doesn't interest you.

For example,
x2+6x+10
x2+6x+10
could only have m,nm,n as (1,10)(1,10) or (2,5)(2,5); since neither of those add up to 66 you can't factor this with easy factors.

The other way is to find b2−4acb2−4ac. If that is a perfect square, then the equation can be factored nicely. If not, then at least you are halfway toward finding the roots using the quadratic formula.


Summation of convergent series Sum the series
$$\sum_{n=0}^{\infty}\left(\frac{2}{2n+1} - \frac{1}{n+1}\right)$$
Is there some general method for summing such series?

Hint:

\begin{equation}
 \sum_{n=0}^{\infty}\left(\frac{2}{2n+1} - \frac{1}{n+1}\right)
=
2 \sum_{n=0}^{\infty}\left(\frac{1}{2n+1} - \frac{1}{2n+2}\right)
=
- 2 \sum_{n=1}^{\infty}(-1)^n \frac{1}{n}
\end{equation}

So this is the alternating harmonic series, rest is up to you


How many ring can $(\mathbb{Z},+)$ upgrade to? (Can we define $(-1)\times(-1)=-1$?) We know that $(\mathbb{Z},+)$ is an abelian group. And under usual multiplication, $(\mathbb{Z},+,\times)$ become a ring. Is there another binary operation $\square$ on $\mathbb{Z}$ that can make $(\mathbb{Z},+,\square)$ to be a ring?

PS: The question comes with a more philosophical argument: junior high students often ask why we should define $(-3)\times(-5)=15$, why not define $(-3)\times(-5)=-15$ or just another number, say $(-3)\times(-5)=155555$? I think a good motivation and reason for operations with negative numbers defined like so, is because only when we defined like that, the integers $\mathbb{Z}$ with $+$ and $\times$ can have associativity, distribution law and so on(i.e. forms a ring). If we didn't choose to define so, then we don't have such good properties!

Yes, there is a different ring structure on $(\mathbb{Z}, +)$: letting $\cdot$ denote the usual product, define $*$ as $$a*b=-(a\cdot b).$$

Why does this work? Well, let $(G, \oplus, \otimes)$ be any ring, and $\alpha$ any automorphism of $(G, \oplus)$. Then we can "push" $\otimes$ through $\alpha$ to get a new operation, $\otimes_\alpha$, such that $(G, \oplus,\otimes_\alpha)$ is also a ring. Namely, $\otimes_\alpha$ is defined as $$a\otimes_\alpha b=\alpha^{-1}(\alpha(a)\otimes\alpha(b)).$$ In our case, we have the automorphism $\alpha:x\mapsto -x$, and $\alpha(a)\cdot \alpha(b)=a\cdot b$, so this yields the operation described above.

In fact, $\cdot$ and $*$ are the only operations which yield a ring structure: suppose $k$ is a unit of $\odot$ with $k\not\in\{1, -1\}$. WLOG, $k$ is positive - but then $1\odot 1+1\odot 1+...$ ($k^2$ times) equals $k$, but no integer has this property if $k>1$.

It's now enough to show that any ring expansion with unit $1$ is the usual one, and with unit $-1$ is the "twisted" version described above; and this isn't too hard.

I think you can answer this to your students :

Let $\mathbb{Z}[1/2] = \{ \frac{a}{2^b}, a \in \mathbb{Z},b \in \mathbb{N}\}$.  With the usual operation it is a (commutative) ring.

But you can define another multiplication $\ast$ on it :
$$a \ast b = 2 ab$$
This way $1/2$ is the multiplicative identity, and $(\mathbb{Z}[1/2],+,\ast)$ is a  (commutative) ring where

$$(-1) \ast (-1) = 2$$


Let $\frac{a}{b},\frac{c}{d}\in \mathbb{Q}$. prove that the product is well-defined We have that $\frac{a}{b}\cdot \frac{c}{d}=\frac{a'}{b'}\cdot\frac{c'}{d'}$, which can be rewritten as $\frac{ac}{bd} = \frac{a'c'}{b'd'}$ and again as $(ac)(b'd') = (bd)(a'c').$

If $\frac{a}{b} = \frac{a'}{b'}$ and $\frac{c}{d} = \frac{c'}{d'}, $ then $ab' = ba'$ and $cd' = dc'.$

$= (ab')(dd') = (ba')(dd')$ and $(cd')(bb') = (dc')(bb')$

$= (ab')(dd')(cd')(bb') = (ba')(dd')(dc')(bb')$ when multiplying the two equations.

Here is where I got stuck. I was attempting to get the final equation equal to $(ac)(b'd') = (bd)(a'c')$ so as to prove that the multiplication is well-defined, but I am unsure of how to get there from where I am or whether I'm thinking about this wrong to begin with.

You are correct that you need to prove that $(ac)(b'd')=(bd)(a'c')$, but you're making it too complicated. By definition,

$$\frac{a}{b} = \frac{a'}{b'} \: \: \text{and} \:\: \frac{c}{d} = \frac{c'}{d'}, $$

means that $ab' = ba'$ and $cd' = dc'$. By multiplying these two equations together (ie left times left, right times right) you simply get $(ac)(b'd')=(bd)(a'c')$, which is what you're looking for.

You can simply say:
$$\frac{a}{b}\cdot\frac{c}{d}=\frac{a}{b}\cdot\frac{c}{d}$$
which is always verified.
For the hypothesis done $\left (\frac{a}{b}=\frac{a_1}{b_1}, \frac{c_1}{d_1}=\frac{c}{d}\right)$, we arrive at:
$$\frac{a}{b}\cdot\frac{c}{d}=\frac{a_1}{b_1}\cdot\frac{c_1}{d_1}$$


How to sketch a graph for fourier series I had to find fourier series for $f(x) = x$, $-\pi < x < \pi$. I found that the Fourier series for $f$ is

$$\sum_{n=1}^{\infty}(-1)^{n+1}\cdot\frac{2}{n}\cdot\sin(nx).$$

Now I have to sketch the graph on $\left[-3\pi,3\pi\right]$. How do I do this?

The fourier series is infinite, you can only graph a partial sum of the series for your interval. My guess is that it will look something like this in the first 3 terms:

See how the graph is starting to resemble f(x) = x (for x between -Pi and Pi) by the third  sum. And also notice that it is periodic with period 2*Pi

By Fourier series , you are representing a function by an infinte sum of sinusoids. You cannot plot all of them. What you can plot is the amplitude of those sinusoids by taking index n on x-axis and magnitude on y-axis.


How much would it cost to try every possible burger combination? I was at a restaurant that allows you to build your own custom burger. I got bored and started to work out how many possible combinations of burger there could be. After figuring that out and sharing the number with a friend he wondered how much it would cost you if you were to order every possible combination. This is where I got stuck.

Ignoring all the options that don't affect the price here's the basic menu:

Choose either one (\$7.99) or two (\$9.48) patties.
Choose from 7 cheeses \$0.89 each
Choose from 10 hot toppings \$0.99 each
Choose from 3 premium toppings \$1.49 each
Choose from 4 cold toppings \$0.39 each

I've been mulling it over in my head for a few days and I can't really figure out a good way to go about it. The best thing I've come up with so far is to pick a category and find out the total to try all combinations in that category using $\sum _{i=1}^m \binom{m}{i} c i$ where $c$ is the cost per item and $m$ is the total number of items to choose from. Then taking that and multiplying it by all the other possibilities created by the other categories.

Please forgive me if I'm missing something obvious or screwed something up, my formal schooling doesn't extend much beyond algebra and that was a decade ago.

Here's an alternative way to justify Thomas Andrews' calculation without appealing to averages:

Order all burgers in pairs containing one selection and its exact opposite. The total cost of each such pair is $$7.99 + 9.48 + 7\cdot 0.89 + 10\cdot 0.99 + 3\cdot 1.49 + 4\cdot 0.39 $$

To count the number of different pairs, identify each pair by its one-patty member. There are $2^{7+10+3+4}$ different one-patty burgers, so this is also the number of pairs.

So easy - too much work done above
$$2  \times   2^7  \times  2^{10}  \times   2^3  \times   2^4$$
Because except for the burger size (pick one or the other), you have to answer yes or no for each of the other add-ons, so $2^7$ means that you could have no cheese, one cheese, two cheeses, etc... up to $7$ cheeses... and so on.

Answer: $33,554,432$ ways