Does this sigma function correspond to this series? I'm kind of self-taught in sigmas, and I'm not very proficient in calculus nor precalculus. Therefore, I'd like to see if the series I've thought of does correspond to this sigma function
$$\sum_{k=1}^n k\times\frac k2$$
Corresponds to this series:
$$1\times\frac 12+2\times \frac 22+3\times \frac 32 + 4\times\frac 42 ... n\times\frac n2$$
I'm wondering because I want to do a little recreational study of this series, and see if what kind of relations n has to the sums. To do that, it would be nice to compress that series to a sigma function, and I'm wondering if I did it correctly here.

Adding this as an answer to avoid excessive comments.
First point:
Yes, it is true that
$$\sum_{k=1}^n k\times\frac{k}{2}=1\times\frac{1}{2}+2\times\frac{2}{2}+...+n\times\frac{n}{2}$$
PROOF: Um... plug the numbers in. The way Sigma notation works is that
$$\sum_{k=m}^n f(k)=f(m)+f(m+1)+...+f(n-1)+f(n)$$
But we can extend this to non-indexed sets (as long as they are countable), e.g
$$\sum_{x\in\mathcal{X}}f(x)=f(x_1)+f(x_2)+f(x_3)+...~~\forall x_i\in\mathcal{X}.$$
Second Point:
Notice that I am calling this "Sigma notation" not a "Sigma function" as it really is just a way of writing things to make it shorter. You could, very formally, think of this instead as a map ("function") that takes in two inputs - another function, and a set:
$$\Sigma(f,S)=f(s_1)+f(s_2)+f(s_3)+...~~\forall s_i\in S$$
E.g,
$$\Sigma(\text{id},\mathbb{N}_{\leq m})=\sum_{k=1}^m k=\frac{m(m+1)}{2}$$
But this is frankly overkill and unnecessarily complicates a simple idea.
Third point:
Yes, it is true that
$$\sum_{k=1}^n k\times\frac{k}{2}=\sum_{k=1}^n \frac{k\times k}{2}$$
There is nothing special about this! In a very crude manner, you can just "cancel out" the summation signs on both sides to get
$$k\times \frac{k}{2}=\frac{k\times k}{2}$$
Which is obviously true, e.g $(6\times 6)/2=36/2=18$ and $6\times (6/2)=6\times 3=18.$


Calculate $c\cdot\ln{\frac{4^{25}}{5^{16}}}$ Let $f(x)=\ln{x}$ and $g(x)=x^2$. Let $c$ be any constant given $c\in(4,5)$.

Calculate $c\cdot\ln{\frac{4^{25}}{5^{16}}}$

$c\cdot\ln{5}-8$
$c\cdot\ln{4}-8$
$2(c^2\cdot\ln{5}-8)$
$2(c^2\cdot\ln{4}-8)$

  Here is an attested
  graph showing that there are indeed solutions for $c\in(4,5)$.

My Attempt:

I know that I have to use Langrange's Mean Value Theorem.

So I started as,

$$c\cdot(5^2\ln{4}-4^2\ln5)$$

$$c\cdot5^2\cdot4^2(\frac{\ln{4}}{4^2}-\frac{\ln{5}}{5^2})$$

Then tried defining,

$$h(x)=\frac{f(x)}{g(x)}=\frac{\ln{x}}{x^2}$$

then by LMVT, there must be a $d\in(4,5)$ such that,

$$h'(d)=\frac{h(5)-h(4)}{5-4}$$

$$\frac{1-2\ln{d}}{d^3}=h(5)-h(4)$$

I can subsitute this in the real equation giving,

$$c\cdot4^2\cdot5^2(\frac{2\ln{d}-1}{d^3})$$

But the $5^2\cdot4^2$ part is giving me problem. I know I can eliminate this if I had dissolved them into the $h(x)$ in the original equation. But I was not able to form a $h(x)$ without taking $5^2\cdot4^2$ commmon.

What am I missing?

Thanks for the help.

Because the question is unclear I'm going to try and articulate several versions of what I think the problem might be and then solve all of them.

problem 1:
let $f(x)=\ln{x}$, and  $g(x)=x^2$

part 1:

construct a new function $h(x)$ by applying elementary functions$(+,-,\cdot,/)$ onto $f(x)$ and $g(x)$ (which can include the multiplying by a constant at the end) such that $h(5)-h(4)=\ln{\frac{4^{25}}{5^{16}}}$

part 2:

Using MLVT find the value for $c$ when $h’(c)=\frac{h(5)-h(4)}{5-4}$

solution:

$$let: h(c)=\ln(4)\cdot \ln(5)\cdot \frac{g(c)}{f(c)}=\ln(4)\cdot \ln(5) \cdot \frac{c^2}{\ln(c)}$$

$$h’(c)=\ln(4)\cdot \ln(5)\cdot \frac{(2\cdot c\cdot \ln(c)-1)}{(\ln(c))^2}$$
$$\ln(\frac {4^{25}}{5^{16}})=\ln(4)\cdot \ln(5)\cdot\frac{(2\cdot c\cdot \ln(c)-1)}{(\ln(c))^2}$$
$$\ln(4^{25})-\ln(5^{16})=\ln(4)\cdot \ln(5)\cdot\frac{(2\cdot c\cdot\ln(c)-1)}{(\ln(c))^2}$$
$$25\cdot\ln(4)-16\cdot\ln(5)=\ln(4)\cdot \ln(5)\cdot\frac{(2\cdot c\cdot\ln(c)-c)}{(\ln(c))^2}$$
$$\frac{25}{\ln(5)}-\frac{16}{\ln(4)}=\frac{(2\cdot c\cdot\ln(c)-c)}{(\ln(c))^2}$$
$$(\frac{25}{\ln(5)}-\frac{16}{\ln(4)})\cdot(\ln(c))^2=2\cdot c\cdot\ln(c)-c$$
$$(\frac{25}{\ln(5)}-\frac{16}{\ln(4)})\cdot(\ln(c))^2-2\cdot c\cdot\ln(c)+c=0$$
$$c \approx 4.495$$
the reason why $h(x)=4^2\cdot5^2\cdot\frac{f(x)}{g(x)}$ doesn't work (which was essentially the OP's first attempt) is because the result of $h(5)-h(4)=4^2\ln{5}-5^2\ln4$ not $5^2\ln{4}-4^2\ln5$

problem 2 (this problem is assuming that d=c from the OP’s question. He said “I could state that d=c” in one of the comment threads of this web page)

$let: f(x) = \ln(x), g(x)=x^2$ Let $c$ be any constant given $c∈(4,5)$

part 1:construct a new function $h(x)$ by applying elementary functions$(+,-,\cdot,/)$ onto $f(x)$ and $g(x)$ (which can include the multiplying by a constant at the end) such that $h(5)-h(4)=\ln(\frac{4^{25}}{5^{16}})$

part 2: Given that $h’(c)=\frac{(h(5)-h(4))}{(5-4)}=\ln(\frac{4^{25}}{5^{16}})$ find when $c\cdot f’(c)=0$

the process and end result of this problem is almost identical to the last equation of problem 1. The last equation of the previous problem differs from the last equation of this problem on the last two terms on the left hand side of the last equation. These terms are multiplied by $c$.

$$(\frac{25}{\ln(5)}-\frac{16}{\ln(4)})\cdot(\ln(c))^2-2\cdot c^2\cdot\ln(c)+c^2=0$$
$$c \approx 2.118<4$$
problem 3

find all values for $c$ in the equations below when $4<c<5$
    $$c\cdot\ln5−8=c\cdot\ln (\frac{4^{25}}{5^{16}})$$
    $$c\cdot\ln4−8=c\cdot\ln (\frac{4^{25}}{5^{16}})$$
    $$2(c^2\cdot\ln5−8)=c\cdot\ln(\frac{4^{25}}{5^{16}})$$
    $$2(c^2\cdot\ln4−8)=c\cdot\ln(\frac{4^{25}}{5^{16}})$$
In the graphs that the OP has posted only the third and the fourth equations intersect in the range $4<c<5$.
Equation 3
$$2\cdot(c^2\cdot\ln5−8)=c⋅\ln (\frac{4^{25}}{5^{16}})$$
$$2\cdot c^2\cdot\ln5−16=c\cdot\ln (\frac{4^{25}}{5^{16}})$$
$$2\cdot c^2\cdot\ln5-c\cdot\ln (\frac{4^{25}}{5^{16}})−16=0$$
$$c≈4.0073$$
Equation 4
$$2\cdot(c^2\cdot\ln4−8)=c\cdot\ln(\frac{4^{25}}{5^{16}})$$
$$2\cdot c^2\cdot\ln4−16=c\cdot\ln(\frac{4^{25}}{5^{16}})$$
$$2\cdot c^2\cdot\ln4-c\cdot\ln(\frac{4^{25}}{5^{16}})−16=0$$
$$c≈4.4959$$

Based on the comments, you are asking if there is any $c\in (4,5)$ such that $c\cdot \ln \frac {4^{25}}{5^{16}}$ is equal to one of the following four expressions.  We can make a table of the values of the five at $c=4$ and $c=5$
$$\begin {array}{r|r|r} \text{Expression}&c=4&c=5\\ \hline
c\cdot\ln \frac {4^{25}}{5^{16}}&35.6254&44.5318\\
c\cdot \ln (5)-8&-1.5622&0.0472\\
c\cdot \ln(4)-8&-2.4548&-1.0685\\
2(c^2\cdot\ln(5)-8)&35.5020&64.4719\\
2(c^2\cdot\ln(4)-8)&28.3614&53.3147
\end {array}$$
All of these functions are monotonically increasing.  We can see that the last two start below $c\cdot\ln \frac {4^{25}}{5^{16}}$ at $c=4$ and end above it at $c=5$.  The intermediate value theorem tells us that there is a point in the interval where they are equal.  If we want to plug in the constants we can find the values of $c$ explicitly as the equation is no worse than quadratic.  For example
$$c\cdot\ln \frac {4^{25}}{5^{16}}=2(c^2\cdot\ln(5)-8)\\
2\ln(5)c^2-c\cdot\ln \frac {4^{25}}{5^{16}}-16=0\\c\approx 4.0073$$
An exact form is available from Alpha, but I didn't find it enlightening.


Number of even numbers with sum of digits less than 12 How many positive three digit even numbers lesser than 700 are there with all three different digits and where the sum of the three digits is 12 or less ?

Am I supposed to write solve the expression for each of the 12 cases like solution for
a + b + C=12, 11, 10...1 and then add them together?

Is there a better approach to this problem?

I would just enter the numbers 2 through 698 into a spreadsheet, use the mod function to extract the digits, and if statements to determine whether the digits are different and sum to less than 12.  Then ask it to count the results.  The copy function means you only write the things once.

9+3+0
9+2+1
8+4+0
8+3+1
8+2+2 (error)
7+5+0
7+4+1
7+3+2
7+2+2 (error)
6+5+1
6+4+2
6+3+3 (error)
5+4+3

Each can be permuted ABC, ACB, BAC, BCA, CAB, CBA, so 13x6 if I counted right.

[Oops, I failed to eliminate the numbers above 700.  We'll leave that as an exercise for the student.  Easily done by breaking the above list into two ranges.]

[Gotta learn to read more carefully!  Sum of digits LESS than 12!  Oh well...]

[I suspect the above scheme would work, though, with a little adjustment.]


Probability that $x^2+y^2 \leq 1$ for uniformly distributed random variables. $X$ and $Y$ are independent random variables uniformly distributed on $(0,1)$

Find $P(X^2+Y^2\leq1|X\leq Y)$

^conditional probability where | represents given

The final answer is $\frac{\pi}{4}$

So I know that I'm going to have to set up some sort of integral to solve this problem, but I'm confused where to start. Do I need to convert to radians first?

Note that $P(X\le Y)=1/2$ and the area of $\{(x,y)\in\mathbb{R}_+^2\ :\ x^2+y^2\le 1 \text{ and }x\le y \}$ is exactly the area of $\{(r\cos\theta,r\sin\theta)\ :\ r\le 1\text{ and }\theta\in[\pi/4,\pi/2]\}$. Therefore,

\begin{align*}P(X^2+Y^2\le1\,|\,X\le Y)&=\frac{1}{P(X\le Y)}P(X^2+Y^2\le1,\ X\le Y)\\
&=2\int_0^1\int_{\pi/4}^{\pi/2}r\,\mathrm{d}r\mathrm{d}\theta\\
&=\frac{\pi}{4}.
\end{align*}

We have $P(X^2 + Y^2 \leq1|X\leq Y)$
Note: $x = rcos\theta, y= r sin\theta$, hence $x^2 +y^2 = r^2$

We have a circle where $x \leq y$, hence a circle $\frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$ and radius $0 \leq r \leq 1$

Using this we can construct:
 $$\int_0^1\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} r^3 \mathrm{d\theta dr}$$
$$=\int_0^1 r^3 * \pi \mathrm{dr}$$
$$=[\frac{r^4 * \pi}{4}]_0^1$$
$$=\frac{\pi}{4}$$

Note: $r^3$ is because $x^2 + y^2 = r^2$ then the Jacobian of transformation is $r$.


Finite sum which sums to $x e^x$ In my analysis class, we have to prove this and another two related sums that I think I could prove if I knew this. However, I don't know how to begin solving this; any hints would be appreciated.
$$\sum_{n=0}^{\infty} \left( e^x-1-\frac{x}{1!}-\frac{x^2}{2!}-\frac{x^3}{3!}-\cdots-\frac{x^n}{n!} \right)=xe^x$$
One method of proving this mentions summation by parts, and the only reference I found in relation to that is Abel's lemma which I tried to use but doesn't help me. What other solutions are there?

Let's write the sum in question using summation notation:
\begin{align*}
S&=\sum_{n=0}^\infty \left(e^x-\sum_{k=0}^n\frac{x^k}{k!}\right)\\
&=\sum_{n=0}^\infty \left(\sum_{k=0}^\infty \frac{x^k}{k!}-\sum_{k=0}^n \frac{x^k}{k!}\right)\\
&=\sum_{n=0}^\infty \sum_{k=n+1}^\infty \frac{x^k}{k!}
\end{align*}
Let's switch the order of summation:
$$S=\sum_{k=1}^\infty\sum_{n=0}^{k-1}\frac{x^k}{k!}$$
But note: the value of the summand doesn't depend on $n$. Therefore, we can treat that inner sum like we have a constant summand, in which case we're just multiplying by $k=(k-1)-0+1$. So,
$$S=\sum_{k=1}^\infty k\cdot\frac{x^k}{k!}=\sum_{k=1}^\infty\frac{x^k}{(k-1)!}$$
Let $k=m+1$:
$$S=\sum_{m=0}^\infty \frac{x^{m+1}}{m!}=x\sum_{m=0}^\infty\frac{x^m}{m!}=xe^x$$
as required.


How do I map the torus to a plane? Please see my answer on Perlin noise first.

A bit of background.  Imagine a solid texture, like an actual block of sky and cloud.  If you "cut a sheet" of sky and display it as an image, you'd get something like this:

Now you want that "sheet of sky" to be repeatable, so you cut the sheet of sky by using a torus:

I'm trying to remove the distortion that appears in the resulting image:

Because clearly the portion from the inner side of the donut is "thinner" than the outer side of the donut.

So I need to map the surface of a torus to a unit square.  I'm not sure how to say this correctly but the mapping must be such that each tiny square $dS$ on the unit square must map to something non-square on the surface of the torus.

For a start, I looked at trying a mapping from u, v to spherical coordinates..

$$
\theta = 2 \cos^{-1} \sqrt{1-\zeta_x} \

\phi = 2\pi\zeta_y
$$

But that didn't seem to lead to a solution.

Everything below assumes that you have already cut two seams on the torus, so that it is homeomorphic to the unit square.

There is no local isometry (i.e. a map that preserves local distances and angles) between the torus and the unit square, or any flat domain for that matter. This is a consequence of Gauss's Theorema Egregium and the fact that the Gaussian curvature of the torus is not everywhere zero. Therefore, an infinitesimal patch in the unit square will have to get stretched, squeezed, or sheared when mapped to the torus. The Theorema Egregium is the same reason why you can't flatten a sphere without distortion, making a perfect map of the Earth impossible.

There is, however, a conformal map (i.e. a map that preserves angles but may not preserve lengths) between the torus and a flat rectangle. What this means is that you won't get the "stretching" sort of distortion you can see near the edges of your image, but the texture will look bigger in some places and smaller in others. A nice example is the first figure on CGAL's page on planar parameterization; observe that the squares remain squarish instead of being squeezed into rectangles or parallelograms. For the torus, a nice explicit formula should be possible if that's what you want.

EDIT1:

In Stereographic Projection angles are conserved ( not the sense though) but length changes do occur.

Fwiw, I suggest the following:

For portions visible from torus center O and for portions invisible from O : At first project points on torus onto a circle/sphere of inversion with radius of inversion at horizon H, centers coinciding. Again use stereographic inversion from North Pole to map/transfer to either side i.e., the top and bottom sides of an annular plate whose inner and outer radii are set by horizon circle projections.

The mapping function can be found out from the diagram:

.


Union of two vector subspaces not a subspace? I'm having a difficult time understanding this statement. Can someone please explain with a concrete example?

The reason why this can happen is that all vector spaces, and hence subspaces too, must be closed under addition (and scalar multiplication). The union of two subspaces takes all the elements already in those spaces, and nothing more. In the union of subspaces $W_1$ and $W_2$, there are new combinations of vectors we can add together that we couldn't before, like $v_1 + w_2$ where $v_1 \in W_1$ and $w_2 \in W_2$.

For example, take $W_1$ to be the $x$-axis and $W_2$ the $y$-axis, both subspaces of $\mathbb{R}^2$.
Their union includes both $(3,0)$ and $(0,5)$, whose sum, $(3,5)$, is not in the union. Hence, the union is not a vector space.

To prove that a vector(U) is a subspace of a vector space(V).
we need to prove that a+$\alpha $b,(where $\alpha$ is any scalar belonging to the field of the vector space) belongs to U.

so we will make that the two vectors make a single vector. And that is possible only when one of the vector space is a subset of the other.


How does one understand and resolve Zeno's paradox? Zeno, a follower of Parmenides, reasoned that any unit of space or time is infinitely divisible or not.  If they be infinitely divisible, then how does an infinite plurality of parts combine into a finite whole?  And if these units are not infinitely divisible, then calculus wouldn't work because $n$ couldn't tend to infinity.

Another way to think about it is a flying arrow must first travel half way to the target from where it begins ( the first task), then travel half way to the target from where it is now (the second task), then travel half way to the target (third task), etc... What you get is this...

$$\begin{array}{l}
{d_{Traveled}} = \frac{1}{2}d + \frac{1}{4}d + \frac{1}{8}d + \frac{1}{{16}}d + ...\\
\\
{d_{Traveled}} = d\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + ...} \right)\\
\\
{d_{Traveled}} = d\left( {\frac{1}{\infty }} \right) = 0
\end{array}
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$$

But suppose we wish to calculate the area below a function between $a$ and $b$ say, the bars that compose this area consist of taking a reference point on the first bar $f(a)$, multiply it by $dx$, then using the slope $f'(a)$ as a guide, "jack up" the reference point onto the top of the next bar, multiply by $dx$, jack it up, multiply by $dx$, and repeat this until we reach the final bar (L.H.S.).  The summation of all this yields the exact area.

So, it's like taking the line segment $ab$ and dividing each piece over and over again.  Per division, the sizes of the pieces are half of what they were before, but there are twice as many of them as before; but as the number of divisions tends to infinity (n tends to infinity), they diminish to almost nothing such that when added back together, they still equal the length of the original line segment $ab$.

How does one understand and resolve Zeno's paradox?

Zeno's paradox is called a paradox exactly because there is a mismatch between a seemingly logical argument that concludes that motion is impossible, and our experience in dealing with reality, which says that there is motion.

To resolve the paradox, then, you need to figure out where the argument goes wrong. Saying that in your experience motion exists does nothing to get rid of the argument. Indeed, rather than resolving the paradox, when you flap your arms or drop any balls, you emphasize the paradox!

Finally, let me also add that with his argument(s) Zeno most likely wasn't trying to conclude that motion doesn't exist, but instead offered his argument as a reductio ad absurdum against the idea that space is infinitely divisible:  If space is infinitely divisible, then [insert typical Zeno story here] motion becomes impossible. But since [flap your arms now] there is motion, space cannot be infinitely divisible.

So, you being able to drop a ball to the ground is now part and parcel of the argument!  And if you want to reject the conclusion that space is not infinitely divisible, you need to show how motion is possible in such a space. Dropping balls doesn't demonstrate such a thing, because Zeno will simply say: you were able to drop the ball exactly because space (as in: the real, physical space of the world we live in) is not infinitely divisible.

Consider the following thought experiment, which I will frame in terms that Zeno would most likely have understood.

Take an urn, and fill it with water.  It will contain a finite amount of water.

Now, transfer half the water in the urn to a second urn.

Then transfer half the water in the second urn to a third.

Repeat this process with a purely hypothetical infinite collection of urns.

How much water is present in the infinite collection of urns?  Other than the initial filling of the first urn, at no time was any water added to or removed from this thought experiment.

Is it finite because we know how much was present at the start and that quantity hasn't changed, or is it infinite / otherwise impossible to determine due to being an "infinite sum"?


Find the number of homomorphisms between $\mathbb{Z}_m$ and $\mathbb{Z}_n$ Although there are many attempts to the problem but I wasn't really able to understand those so I took some idea from examples and attempted it. Can someone just go through it please.
My attempt: Let $\sigma:\mathbb{Z}_{m}\to \mathbb{Z}_{n}$. The kernel $\ker(\sigma)$ is a normal subgroup of $\mathbb{Z}_{m}$. So $\ker(\sigma)=\langle\frac{m}{d}\rangle$, where $d|m$. Now $\operatorname{im}(\sigma$) is a subgroup of $\mathbb{Z}_n$. So, now the order of the image should be $d$ so $d|n$. Now the number of homomorphisms when the kernel is of order $d$ will be $\sum_{d|m,d|n}\varphi(d)$. From here we can conclude that the no of homomorphisms should be $\gcd(m,n)$. Is this OK? Where am I making mistakes?

Yes, there are $\gcd(m,n)$ such homomorphisms.
Another way to see it is that we need $f([1])=[a]$ where $0\le a<n$ and that
$m[a]=[ma]=0$ in $\Bbb Z_n$, that is, $n\mid ma$ or $ma\equiv 0\pmod n$.
The solution to this congruence is $a\equiv 0\pmod{n/g}$ where
$g=\gcd(m,n)$, so that $a=jn/g$
for $0\le j<g$, so there are $g$ solutions.

A homomorphism $\sigma:\mathbb Z_m\to\mathbb Z_n$ is uniquely determined by $\sigma(1)=a\in\mathbb Z_n$. This has to be such that $ma=\sigma(m)=0$, so that $a\gcd(m,n)=0$. Thus, there are $\gcd(m,n)$ such homomorphisms.


Show that the congruence equation $x^8 \equiv 16 \pmod p$ has a solution for every prime $p$. Show that the congruence equation  $x^8 \equiv 16 \pmod p$ has a solution for every prime $p$.
Let $g$ be a primitive root， then the equation is equivalent to $8
\ ind_g(x)\equiv 4 \ ind_g(2) \pmod {p-1}$. It suffices to show that $2\mid ind_g(2)$.

If $p=2$, take $x=0$.
If $p\equiv \pm 1 \pmod 8$ we have $\left(\frac{2}{p}\right) = 1$, then there exists $a$ such that $a^2 \equiv 2 \pmod p$, so $a^8 \equiv 2^4 = 16 \pmod p$.
If $p\equiv 3 \pmod 8$ we have $\left(\frac{-2}{p}\right) = 1$, then there exists $a$ such that $a^2 \equiv -2 \pmod p$, so $a^8 \equiv (-2)^4 = 16 \pmod p$.
If $p\equiv 5 \pmod 8$ then $p\equiv 1 \pmod 4$, then there exists $a$ such that $a^2 \equiv -1 \pmod p$, so $(1+a)^2 \equiv 2a \pmod p$, then $(1+a)^4\equiv 4a^2 \equiv -4 \pmod p$ so $(1+a)^8 \equiv 16 \pmod p$.

If $p=2$, see jjagmath's answer.  $x^8-16=(x^2-2)(x^2+2)(x^2+2x+2)(x^2-2x+2)=0$
must have solutions mod $p$ (an odd prime),
because if $x^2-2$ and $x^2+2$ do not have roots mod $p$, then $x^2-\dfrac2{-2}=x^2+1$ does,
because  the inverse of a quadratic nonresidue is a nonresidue
and the product of two nonresidues is a residue.
Finally, if $a^2+1\equiv0\bmod p$, then $-1\pm a$ are roots of $x^2+2x+2$,
and $1\pm a$ are roots of $x^2-2x+2$, which would be obtained with the quadratic formula.
Thus, either $x^2-2$ or $x^2+2$ has roots mod $p$, or $x^2+2x+2$ and $x^2-2x+2$ have roots mod $p$;
in any event, $x^8-16$ has roots mod $p$.


Why is $\mathbb{R}^2$ not a subset and /or a subspace of $\mathbb{R}^3$? One thing this suggests--at least to me--is that the x-y plane and $\mathbb{R}^2$ are not necessarily equivalent.  For example, I could define the following: $X = \left\{ \begin{bmatrix} x\\y\\z\end{bmatrix} x,y \in \mathbb{R} \land z = 0\right\}$.  Am I wrong to think, one, that this is a subset of $\mathbb{R}^3$?  As I write this it occurs to me that while scalar multiplication is closed under the above rules, addition doesn't pass the smell test for a subspace... so, OK, it's certainly not a subspace.  I would welcome any insight readers of this query can provide.

The elements of $\Bbb R^2$ are vectors of two coordinates; and the elements of $\Bbb R^3$ are vectors of three coordinates. (One can easily think of those vectors as $2$-tuples and $3$-tuples, for example.)

Assuming mathematics is consistent, $2\neq 3$. Therefore no element of $\Bbb R^2$ is an element of $\Bbb R^3$. It follows that $\Bbb R^2$ is not a subset of $\Bbb R^3$.

And in order to be a subspace, one first has to be a subset. So it's not a subspace either.

What you have defined as $X$ is isomorphic to $\Bbb R^2$, but just as well you could decide that $y$ is $0$, and the identification would still be natural. $X$ is a subset of $\Bbb R^3$ and indeed a subspace, but it is not $\Bbb R^2$ as a set, it is just isomorphic to it in a very obvious way.

While isomorphism is an equivalence relation, and we often think of it almost as identity, it is still not set equality which is a stricter notion.

Assume $m<n.$  Then on the standard basis, $\mathbb{R}^m$ is, in fact a subspace of $\mathbb{R}^n.$
The vector space $\mathbb{R}^n$:
Begin by defining $\mathbb{R}^n$ (and similarly for $\mathbb{R}^m,$ etc.,) as the $n^{th}$ Cartesian product of $\mathbb{R}.$  Doing so requires an ability to distinguish each $\mathbb{R}$ appearing as a Cartesian factor from the others.  For example, in $\mathbb{R}^3$ we distinguish each by calling them the $x,y,z$ axes.  We  nominate each factor using a subscript natural number $i=1 \dots i=n$ so $\mathbb{R}^n=\mathbb{R}_1\times\dots\times\mathbb{R}_n,$ and assume the natural ordering on the indices.  These designations are arbitrary.  We could choose any unique subscript values in any ordering, just so long as we distinguish each Cartesian factor $\mathbb{R}$ from the others.
The standard basis $\left\{\mathfrak{\hat{e}}_i\right\}$ of $\mathbb{R}^n:$
From each $\mathbb{R}_i$ we identify its real number $1\in\mathbb{R}_i$ with the symbol $\mathfrak{\hat{e}}_i\in\mathbb{R}_i.$  The $\mathfrak{\hat{e}}_i$ are called the elements (vectors) of the standard basis of $\mathbb{R}^n.$  The set $\left\{\mathfrak{\hat{e}}_i\right\}$ is linearly independent, and spans $\mathbb{R}^n$.  That is, every element of $\mathbb{R}^n$ is given as a unique linear combination of the standard basis vectors.
$\mathbb{R}^m$ as a subspace of $\mathbb{R}^n$: We construct $\mathbb{R}^m$ and its standard basis $\left\{\mathfrak{\hat{f}}_j\right\}$ using the same method with $j=1 \dots j=m<n.$  The next natural step is to identify $\left\{\mathbb{R}_i=\mathbb{R}_{j=i\le{m}}\right\}$ which gives $\left\{\mathfrak{\hat{f}}_j=\mathfrak{\hat{e}}_{j=i\le{m}}\right\}.$
Therefore, assumming standard definitions and natural identities, $\mathbb{R}^m\subset\mathbb{R}^n.$
Of course, we could use different definitions, leading to different results.  For example, a typical definition is $\mathbb{R}^{n-m}=\mathbb{R}_{i=m+1}\times\dots\times\mathbb{R}_n.$
Other than an appeal to sanity, there is nothing to stop us from identifying $\mathbb{R}^{p}=\mathbb{R}^{n-m},$ where $p=n-m,$ and $\mathbb{R}^{p}$ is constructed as above.  Indeed $\mathbb{R}^{p}$ and $\mathbb{R}^{n-m}$ are isomorphic, so there is an equivalence relation between them.


Theorems with an extraordinary exception or a small number of sporadic exceptions The Whitney graph isomorphism theorem gives an example of an extraordinary exception: a very general statement holds except for one very specific case.

Another example is the classification theorem for finite simple groups: a very general statement holds except for very few (26) sporadic cases.

I am looking for more of this kind of theorems-with-not-so-many-sporadic-exceptions

  (added:) where the exceptions don't come in a row and/or
  in the beginning - but are scattered truly sporadically.

(A late thanks to Asaf!)

Every automorphism of $S_n$ is inner if $n \neq 6$.

P.S. That $S_6$ has an `essentially' unique outer automorphism is quite a non-obvious fact.

Have a look at the Strong Law of Small Numbers:

  When two numbers look equal, it ain't necessarily so.


Range of $\frac{\log(x)}{x^{b}}$ This is from Berkeley problems. Actual question is this

  For which positive numbers $a,b,~ a >1$ does the equation $\log_{a}(x)=x^{b}$ have positive solution for $x$?

For solving this we can rewrite the expression as $\log a= \frac{\log x}{x^b}$. Hence the problem reduces to this for what values of $x$  $\log a $ lies in the range of $f(x)= \frac{\log x}{x^b}$ the solution says range of  $f(x)= (\infty,\frac{1}{be})$. Using geogebra I can verify this. But how to find range of such functions? All I know is range of $\log x$ in $(-\infty,\infty)$ and range of $\frac{1}{x^b}$ in set of all non zero reals.

To find the range of $f(x) = \frac{\log(x)}{x^b}$, it suffices to use the usual calculus techniques to confirm that:

$y = f(x)$ has a vertical asymptote at $x = 0$
there exists an $r$ such that $f(x)$ is increasing over $0 < x < r$, and that $f(x)$ is decreasing over $x > r$

Differentiate the fuction .... then find its maxima..... in this case the function is increasing for x less than $e^{1/b}$. The corresponding y for this x is ${1/be}$. Hence this the max. Value of the function. Hence the range is ${-infinity}$ to ${1/be}$.


Continuous of function. If $f$ continuous at $x_0$ and $f(x_0)>0$

then how to show there exist $\epsilon>0$ and $\delta>0$ such that $f(x)>0$ for $|x-x_0|<\delta$.

I don't know to start proving and then i use definition continuous function at $x_0$ is

$\forall\epsilon>0, \, \exists\delta>0 \, \text{such that if }|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$

Since $f(x_0)>0$ so is $\frac{f(x_0)}{2}>0$. Let $\epsilon = \frac{f(x_0)}{2}$. By continuity of $f$ we know that there exists $\delta>0$ such that,

$$|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon = \frac{f(x_0)}{2}\\
\implies -\frac{f(x_0)}{2}<f(x)-f(x_0)<\frac{f(x_0)}{2},\\
\implies \frac{f(x_0)}{2}<f(x)<\frac{3f(x_0)}{2}$$

since both sides are positive, so is $f(x)$, the end.

$f$ continuos at $x_0$, and $f(x_0) >0$.

For every $\epsilon >0$ there is a $\delta >0$ such that

$|x-x_0| \lt \delta$ implies

$|f(x)-f(x_0)| \lt \epsilon.$

Since $|f(x)-f(x_0)| \lt \epsilon$, we have

$-\epsilon \lt f(x)-f(x_0) \lt \epsilon$, or

$f(x_0) -\epsilon \lt f(x)$.

For which choice of $\epsilon$ is the LHS positive?

Recall: $f(x_0) >0.$


How can I get the limit of this problem How to prove that:
  $$\lim_{(x,y) \rightarrow (0,0)}  (x+y)\sin(\frac{1}{x})\sin(\frac{1}{y})=0$$

I want to know how to do it. I'd appreciate it a lot.

So we have, $$ f(x,y)=(x+y)\sin(\frac{1}{x})\sin(\frac{1}{y})$$
then, using the fact that $\forall u\in\Bbb{R},|\sin(u)|<1$, $$ 0<|f(x,y)|=|(x+y)\sin(\frac{1}{x})\sin(\frac{1}{y})|\leq|x+y|$$
Thus,
$$ 0<|f(x,y)|\leq |x+y|\xrightarrow[(x,y)\rightarrow(0,0)]{}0$$
By the sandwich theorem we have the answer requested.


Area between the curve and x-axis Find the area between the curve $y={2\over x-1}-1$ and the $x$-axis over the interval $[2,4]$

Would this solve the question $\int_{2}^4( {2\over x-1}-1)dx$ ?

So...

$[2\ln(x-1)-x+c]_2^4$ would be incorrect because the curve would be below the x-axis for $3$ to $4$

As I mentioned in the comment, you should divide the interval $[2, 4]$ to two parts,
\begin{align*}S = &\int_2^3\left(\frac{2}{x - 1} - 1\right) dx \;{\color{red} -}\int_3^4\left(\frac{2}{x - 1} - 1\right)dx \\
= & \left[2\log(3 - 1) - 2\log(2 - 1) - 1\right] - \left[2\log(4 - 1) - 2\log(3 - 1) - 1\right] \\
= & 4\log 2 - 2\log 3.
\end{align*}

Essentially finding the area under in the curve is integrating. So in short, yes, that would solve the question.


Is there a probability distribution with mean $1$ such that $f(x)=\frac{1}{x}f\left(\frac{1}{x}\right)$ Is there a probability distribution defined over $\mathbb{R}^{+}$ by the pdf $f$ such that,
$$\forall x > 0, f(x)=\frac{1}{x}f\left(\frac{1}{x}\right)$$
and
$$\int_0^{\infty} x~\mathrm{d}f = 1 $$

The first condition is easily satisfied by picking functions of the form

$$\left(\int_{-\infty}^{\infty} v(t)(x^t+x^{1-t})~\mathrm{d}t\right)^{-1}$$

However, in all cases, the first moment is always greater than one. Only by pushing the standard deviation to zero can it approach one.

I realize the moment can be written as $$\int_0^1 \left(x+\frac{1}{x^2}\right)f(x)~\mathrm{d}x$$

and we have the constraint

$$\int_0^1 \left(1+\frac{1}{x}\right)f(x)~\mathrm{d}x=1$$

but I'm not sure if somehow that implies it has to be greater than one...

Since you already rewrote the moment and the density constraints, let's start from that.

We have $$1 = \int_0^1 \frac{x^3 + 1}{x^2}f(x)dx$$ and $$1 = \int_0^1 \frac{x^2 +x}{x^2}f(x)dx$$.

This gives $$0 = \int_0^1 (x^3 + 1 - x^2 - x) \frac{f(x)}{x^2}$$

But it is easy to see that $\forall x \in [0,1], x^3 + 1 - x^2 - x \geq 0$, hence we have a null integral with a non-negative integrand. This implies that $f(x) = 0$ almost everywhere, a contradiction.

The penultimate expression equals 1, which is correct. That implies that

f(x) = (x+1/x^2)^-1,

works.


What is lost when we move from reals to complex numbers? As I know when you move to "bigger" number systems (such as from complex to quaternions) you lose some properties (e.g. moving from complex to quaternions requires loss of commutativity), but does it hold when you move for example from naturals to integers or from reals to complex and what properties do you lose?

The most important ones as I see it:

Naturals to integers: lose well-orderedness, gain "abelian group" (and, indeed, "ring").
Integers to rationals: lose discreteness, gain "field".
Rationals to reals: lose countability, gain "Cauchy-complete".
Reals to complexes: lose a compatible total order, gain the Fundamental Theorem of Algebra.

$\mathbb C$ is more rigid. Holomorphic functions are analytics. This means complex manifolds are polynomial-like and more akin to algebraic varieties than to real manifolds. By contrast $\mathbb R$ manifolds can be glued together using functions like $\exp(-1/x^2)$ that can $C^\infty$-smoothly transition from one function to another.


Convergence for Borell cantelli lemma Question:
Let $X_1, X_2 ... $ be RVs that are continuous, independent and equally distributed. We define $A_n= \lbrace X_n>max_{k<n} X_k\rbrace$. Prove that $P(\limsup A_n)=1$.

Thoughts:
We studied the Borell Cantelli lemma, but I'm having difficulties proving that $\sum_{n=1}^\infty P(A_n)$ doesn't converge.
Would love some help with that.
Thanks

Answer(according to the hint):
There are $n!$ ways to order the n RVs, so each permutation has a probability of $\frac 1{n!}$. Assuming that $X_n$'s is set to be largest, we can now permutate the next n-1 RVs so we have a probability $\frac {(n-1)!}{n!}=\frac 1n$ for such situation (=$A_n$). Since the Harmonic series diverges, we're through, using Borell-Cantelli.

Hint: If $n$ randomly chosen people are standing in line, how often does it occur that the tallest person is at the end of the line?

To apply the second Borel Cantelli lemma, you also have to prove that the events $\{A_n\}$ are mutually independent.

We must show two things:

$P(A_n) = 1/n$.
$(A_n)_{n \ge 1}$ is an independent collection of events.

From these two, we can use BCL2.

The short version is symmetry. $X_n$ has the same chance of being the max of $X_1, X_2, ..., X_n$ as any of the others

Continuity, independence and equal distribution are essential. See the proof here.

$A_n$'s are independent because the $X_n$'s are independent.

Consider

$A_2 = \{X_2 = \max\{X_1, X_2\}\}$

$A_3 = \{X_3 = \max\{X_1, X_2, X_3\}\}$

Since the $X_n$'s are independent, the probability that $X_3$ is the maximum among $X_1, X_2, X_3$ is independent of which is the maximum among $X_1, X_2$


Convexity and Connectedness Is it true that all convex sets are connected?  This seems intuitively true, but I would like to know if it is, then why? And if not, why not?  In other words, is showing that a set is convex sufficient to show that it is connected?

Let $C$ be a convex set. Suppose is not connected then there are to disjoint non empty open sets $U,V$ so that $U\cup V=C$. Pick points $x\in U$ and $y\in V$ since $C$ is convex the line segment between them is contained in $C$, i.e., the function $f: [0,1]\to C$ defined $f(t)=tx+(1-t)y$.
Note that $f^{-1}(U)$ and $f^{-1}(V)$ would show that $[0,1]$ is not connected a contradiction.

Consider the set X={a,b} endowed with discrete topology and a order relation. It is convex but not connected...


Boolean logic to arithmetic is it possible to convert a Boolean logic expression to an arithmetic equation?
For example, I have the following equation: T = C $\cup$(B $\cap$ A)
The above equation obeys the distributive law, so can I convert it so:
C $\cup$(B $\cap$ A) = C $\times$B + C $\times$ A?

No, rather, the distribution of union over intersection gives us:
$$C\cup (B\cap A) = (C\cup B) \cap (C\cup A)\tag{Distribution}.$$
So we can choose to prove this in logic, by showing  $$x\in [C\cup (B\cap A)] \equiv x \in [(C\cup B)\cap (C\cup A)]$$
so $$ [x\in C \lor x \in (B\cap A)] \equiv [x\in (C\cup B)] \land [x \in (C\cup A)]$$
...


What is the difference between linear mappings and linear functions? Let $V$ and $V'$ be vector spaces over a field $K$. A linear mapping $$f:V \to V'$$ is a mapping which preserves addition and scalar multiplication.

My question is: what is the difference between linear mappings and linear functions?

Depends on the definition you are using. Linear mappings are typically the same as linear functions, although in some contexts, a linear function is strictly some function in the form of $y = mx+b$.

Linear mappings are functions that map two domains and whose operations have linearity, that is the linearity of scalar multiplication and addition.

For the definition of linearity it still the same for functions or mappings.

But the term mapping has more general uses than than function, a mapping could be a function, functional or operator...

However, these three terms are not the same.
 A function is a term you used for mappings that map a set A to B where A and B are subsets of real of complex numbers or almost any map into $R^n$ or $C^n$ .
 A functional is a mapping that is defined from a subset of a space X into the space of real or complex numbers.
 Finally an operator is defined from a space X to a space Y, where X, Y could sequence or function spaces.

Here an examples that clearfy more the difference between these three terms,
let the function $f:R \rightarrow R $ such that  $f(x)= x$.

Let us integral this function between $0$ and $1$,  then $ \int^{1}_{0}x =\frac{1}{2}$.

You can notice that the defined integral is a functional, it mapped a function ( x ) into a real ( 1/2).

However the undefined integral of this function is a function,  $\int x =\frac{1}{2}x^2$.

Thus, the undefined integral is an operator.

You can use the term of mapping for all previous examples but the term function can be uses only for the first case.


Need help to simplify the expression involving powers $$\left(1-\frac{\sqrt{3}-i}{2}\right)^{24}$$
somehow this should be equal to :$$\left(2-\sqrt{3}\right)^{12}$$
but I can't see how...

Hint: start by squaring $(1-\frac{\sqrt{3}-i}{2})$. Remember that $(a-b)^2 = a^2-2ab+b^2$, and that $i^2=-1$.


A valid floor function trick? Given $x\in\mathbb R_+$ and $m,n\in\mathbb Z_+$, is it true that
$$\bigg\lfloor\frac{\lfloor \frac{x}{m}\rfloor}{n}\bigg\rfloor=\bigg\lfloor \frac{x}{mn}\bigg\rfloor?$$

Thanks for at least three different and convincing proofs! I'll use this result  trying to code the prime counting formula from Wikipedia $(2)$:

Given $m$, select $y$ so that
$\sqrt[3]{m}\le y\le\sqrt{m}$ and let $n=\pi(y)$. Then

$(1)\;$ $\pi(m)=\phi(m,n)+n-1-P_2(m,n)$, where

$(2)\;$ $\displaystyle \phi(m,n)=\phi(m,n-1)-\phi\Big(\frac{m}{p_n},n-1\Big)$,
$\;\phi(m,0)=\lfloor{m}\rfloor$ and

$(3)\;$ $\displaystyle P_2(m,n)=\sum_{y<p\le\sqrt{m}}
\Big(\pi\Big(\frac{m}{p}\Big)-\pi(p)+1\Big)$, $\;p$ is a prime.

Yes it is. Assume it is not, then there is an integer $N\in\mathbb{N}$ that lies between $\left\lfloor\frac{\left\lfloor\frac{x}{m}\right\rfloor}{n}\right\rfloor$ and $\left\lfloor\frac{x}{mn}\right\rfloor$.
Possible cases:

$$\frac{x}{mn}<N\leq\frac{\left\lfloor\frac{x}{m}\right\rfloor}{n} \implies \frac{x}{m}<nN\leq\left\lfloor\frac{x}{m}\right\rfloor,$$
which is a contradiction, since $\left\lfloor\frac{x}{m}\right\rfloor\leq\frac{x}{m}$.
$$\frac{\left\lfloor\frac{x}{m}\right\rfloor}{n}<N\leq\frac{x}{mn} \implies \left\lfloor\frac{x}{m}\right\rfloor<nN\leq\frac{x}{m},$$
which is also a contradiction, since there cannot lie an integer between $\left\lfloor\frac{x}{m}\right\rfloor$ and $\frac{x}{m}$.

We know that if |x-y|<1  (x>y)  then [x] = [y]

[x/m] - x/m <1 : [x/m]/n - x/mn = ([x/m] - x/m ) /n <1

So | [x/m] / n  -  x/ mn | <1  so your trick is correct


Which of these matrices are similar? $$A=\begin{bmatrix}
    1 & 1 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 2
\end{bmatrix}
\hspace{1cm} B=\begin{bmatrix}
    2 & 0 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 1
\end{bmatrix}
\hspace{1cm} C=\begin{bmatrix}
    2 & 1 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 1
\end{bmatrix}$$

I would argue as follows:

Since all matrices are triagular, each has the same set of eigenvalues. As $B$ is diagonal, we can say that both $A$ and $C$ are diagonalizable, with their diagonal matrices being $B$. It follows that $A$ and $C$ are similar to $B$. By transitivity, they are also similar to each other. Hence all matrices are similar.

However, I was told that the order of the eigenvalues (it differs among the matrices) might invalidate this reasoning, but I don't see how that could be true, given that we may asign the order of the eigenvalues arbitrarily when diagonalizing a matrix.

On second though, I could just use the theorem that says that similar matrices have the same eigenvalues, but would that line of reasoning above still be valid?

HINT

Matrices $A$ and $B$ are similar indeed $B$ is the diagonal form of $A$.

Matrix $C$ is a Jordan canonical form (i.e. eigenvalues $2$ has geometric multiplicity $=1$).


Show that a polynomial $f(x)$ over a field $k$ is irreducible if the polynomial $f(x+1)$ is irreducible I was thinking of using contradiction by assuming that that $f(x)$ is reducible but I really don't know how to continue from that idea.

Assume that $f(x)$ is reducible. Then:

$\exists \ h(x), \ g(x) \in \mathbb K[x]$ such that $f(x) = g(x)h(x)$.

Then: $f(x + 1) = g(x+1)h(x+1)$, and so reducible. A contradiction.

Let's prove the contrapositive:  That if $f(x)$ is reducible then $g(x) \equiv f(x+1)$ is also reducible.  That is logically equivalent to the statement to be proven.

Since $f(x)$ is reducible there exists some $x_0\in K:f(x_0)=0$.  Then since $1 \in K$ and $K$ is a field, $y\equiv x_0-1 \in K$.  And
$$
g(y) = f((x_0-1)+1) = f(x_0) = 0
$$ so $g(x)$ is reducible.

EDITED in response to Bill's comment:

This "proof" is invalid.  Yes, if $f(x)$ has a zero then $g(x)$ is irreducible.  But we don't know that $f(x)$ has a zero; all we know is that it is irreducible.  So "step 1.5" in the above proof is hogwash.  Sorry.


Why can chain rule not be applied recursively (reapplied to outer function)? $f(x)=\sqrt{x+1}$
$f(x)=(x+1)^{\frac 12}$
$\frac {d}{dx}f(x)=\frac {d}{dx}[(x+1)^{\frac 12}]·\frac {d}{dx}(x+1)$
$\frac {d}{dx}f(x)\neq[\frac {d}{dx}[(x+1)^{\frac 12}]·\frac {d}{dx} (x+1)]·\frac {d}{dx}(x+1)$

Why can chain rule not be reapplied to $\frac {d}{dx}[(x+1)^{\frac 12}]$ in step three even though it is also a composite function (it looks the same as $f(x)$). Is it just a rule that chain rule can only be applied once? Does Leibniz's notation explain it?

Chain rule is the following:

$$ [f(g(x))]' = f'(g(x))g'(x) $$

So, when we want differentiate things like $\sqrt{1+x}$, we do $f(x) = x^{1/2}$ and $g(x)= 1+x$ . Then, we get

$$f'(x) = \frac{1}{2\sqrt{x}} \implies f'(g(x)) = \frac{1}{2\sqrt{1+x}}$$

$$g'(x) = 1 $$

Replacing in the formula for the chain rule:

$$ [f(g(x))]' = f'(g(x))g'(x)  $$
$$ (\sqrt{1+x})' =  \frac{1}{2\sqrt{1+x}} $$

Maybe you are confused with the notation.

The chain rule can be expressed as $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$. It logically follows that  $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dv} \frac{dv}{dx}$.


Using Rolle's theorem and IVT, show that $x^4-7x^3+9=0$ has exactly $2$ roots. Using Rolle's Theorem and the Intermediate Value Theorem, show that $x^4-7x^3+9=0$ has exactly two roots.

I know how to prove that this equation has at least two real roots by using IVT, but my problem is how do I use Rolle's theorem to prove that it has at most two real roots? I've used Rolle's theorem to prove a function has one real root, but how do I do it with more than one root.

Set $f(x)=x^4-7x^3+9$. The derivative
$$f'(x)=4x^3-21x^2=x^2(4x-21)$$
has the sign of $4x-21$, so $f(x)$ is decreasing up to $\frac{21}{4}$, attains a unique minimum $f\bigl(\frac{21}4\bigr)$ and is finally increasing. By the I.V.T. and the definition of the monotony of a function, the equation has exactly two roots if this minimum is negative, none if it is positive and a double root if the minimum is $0$.

This can be generalised to any quartic polynomial $ax^4+bx^3+c$.

The derivative is $4x^3-21x^2=x^2(4x-21)$ is negative if $x< 21/4$ and positive if $x>21/4$, $f(21/4)<0$ and $lim_{-\infty}f(x)=+\infty$ $lim_{+\infty}f(x)=+\infty$. We deduce that $f$ is strictly decreasing in $(-\infty,21/4)$ and has one root in this interval, $f$ is strictly increasing in $(21/4,+\infty)$  and has one root in this interval.


If $x$, $y$, $x+y$, and $x-y$ are prime numbers, what is their sum? Suppose that $x$, $y$, $x−y$, and $x+y$ are all positive prime numbers. What is the sum of the four numbers?

Well, I just guessed some values and I got the answer.
$x=5$, $y=2$, $x-y=3$, $x+y=7$. All the numbers are prime and the answer is $17$.
Suppose if the numbers were very big, I wouldn't have got the answer.
Do you know any ways to find the answer?

Note that $x>y$, since $x-y$ is positive. Since $x$ and $y$ are both prime, this means that $x$ must be greater than $2$ and therefore odd. If $y$ were odd, $x+y$ would be an even number greater than $2$ and hence not prime. Thus, $y$ must be even, i.e., $y=2$.

Now we want an odd prime $x$ such that $x-2$ and $x+2$ are both prime. In other words, we want three consecutive odd numbers that are all prime. But one of $x-2,x$, and $x+2$ is divisible by $3$, so in order to be a prime it must be $3$. Clearly that one must be $x-2$, the smallest of the three numbers, and we have our unique solution: $x=5$ and $y=2$, and $x+y+(x+y)+(x-y)=3x+y=17$.

If $x,y > 1$ are both odd, then $x+y > 2$ is even, a contradiction.
Then $y = 2$ and by inspection $x=5$ as $x = 3$ results in $x-y = 1$.
Or, we have that $x-2,x,x+2$ are all prime meaning we have three consecutive integers modulo $3$ which is only satisfied by the triple $(3,5,7) \implies y = 5$.


Numerical methods for solving nonlinear equations I have this equation

  $x^{5}+x^{3}+3=0$

and I'm supposed to find one root with a given accuracy using the secant method. I was wondering how I can localize a solution to an interval so I can apply the formula. I'm aware that I could separate the equation into two functions and plot both of them, but I think that can be hard sometimes when I don't have access to plotting tools like Desmos and my equation gets complicated.

We take a few guesses for the function $f(x) = x^5+x^3+3$

First, we plug in $x=-1$. We find that the equation is positive, but for $x=-2$, the equation is negative.

So, we know to search in the range of $(-2,-1)$. We set $x_0=-2$ and $x_1=-1$.

Now, we use the formula for the secant method:

$$x_{i+1}=x_i-\frac{f(x_i)\cdot(x_i-x_{i-1})}{f(x_i)-f(x_{i-1})}$$

Applying this repeatedly gets us the sequence $$-2, -1, -1.0263157894736843, -1.119763138322457, -1.1034910955412918, -1.105259392177301, -1.105298653114797, -1.1052985459998306, -1.1052985460061695, -1.1052985460061695$$This is how we converge upon the solution.


Number of $0$ in great number For example,  $11111111111111100$ ends with  $2$ zeros ,when we did know the decimal representation like $100!$ also.

I would like a justified  answer for the following question . How many $0$ are in the end of decimal representation of $ 100!$?
Is there  a general process to know this number for greater number like $2^{100!}\times5^{39!}$?

It is fairly easy with factorials, a number ends with as many $0$s as the number of $5*2$ in his factorization.
It is obvious that there are more 2 than 5 in the factorization of any factorial so we only need to count how many $5$s are there in the factorization of $100!$.
These are $\frac{100}{5}+\frac{100}{5^2}=24$ since $5^3>100$

Of the 100 numbers multiplying to give 100! ten of them end in 0 (10, 20, ...100) giving 11 zeros. The only other way to get a 0 is to mutliply an even number by a number ending in 5. There are ten numbers ending in 5 (5, 15,...,95) and multiplying by any ten even numbers (except the multiples of 10) will give 10 more zeros. So that is 21 zeros on the end of 100!


Moscow State Oral Exam I have heard that during the 1960s, prospective students had to take an 'Oral Maths' exam (alongside written maths, physics and Russian literature).

I having trouble imagining what type of exam this would have been (more problem solving or merely reciting proofs?) and why anybody thought this would be a good idea or a useful indication of potential?

Was the exam purely oral, as in the candidates were given a problem they had to solve without any written work and on the spot? Or was it of a different format?

I assume it was scrapped but I am not sure when either.

Information on the internet is sparse so I thought I'd ask here, maybe some people even have first hand accounts.

Wow, how so many answers are concentrated on "Jewish problems."

The idea behind the oral exam, which was used in USSR in the places with advanced mathematical education, to understand how good the student will be able to appreciate the real mathematics. The exam was always administered after the written exam. The main point is that it is impossible to cheat at such exam, and it is simply impossible to memorize the material to convince the examiner that you actually understand math.

The structure is as follows: The student picks a random "ticket" with two theoretical questions and one problem. Then he/she is given some time to prepare. And after it the student has to present his solutions to the examiner.

There is a very nice Russian book by Tkachuk (I link a Russian text) about preparation to entrance exams with examples and general discussion. I am not sure though that it was translated into English.

Here is an example of a question: We have a quadrangle, which has all four sides equal. Does it mean that it is a rombus?

In the former Soviet Union MIR Publishing helped to develop a number of books to aid in the study of such exams. Most of these books are now public domain. An example of one can be found here. http://mirtitles.org/2013/11/22/problems-in-higher-mathematics-minorsky/
These MIR problem books can be found for various technical subjects and are useful study aids.


Acceptable Definition for $\sqrt{a}$? Is this an acceptable definition for $\sqrt{a}$, where $a \in \mathbb{R}$?

If $a\geq 0, \sqrt{a} = b \in \mathbb{R}$ s.t. $b\geq 0, b^2 = a$.

I'm proving some theorems involving $\sqrt{a}$, in the context of introductory real analysis.

Let $a, b \in \mathbb{R}$. $\sqrt{a}$ is the square root of $a$ and is defined as $f(a) = b$ such that $f:a \to a\times a = b$. Notice the product then can never be negative, because $\mathbb{R}$ is reflexive.


Problem in proving that $\mathbb{A}^2$ is not homeomorphic to $\mathbb{P}^2$ Let $k$ be an algebraically closed field. All spaces are equipped with the usual Zariski topologies.

All the proofs of this fact that I've seen rely on the fact that two lines in $\mathbb{P}^2$ intersect but this doesn't necessarily hold in $\mathbb{A}^2$. I'm stuck on proving that this property is a "Zariski-topology invariant" (i.e preserved by homeomorphism). All the proofs use this fact without proving it, so I assume it is trivial, but I do not know how to prove it.

Does someone have a hint on how to prove it?

It would be enough for me to prove that lines in $\mathbb{A}^2$ are sent to projective lines to complete the proof, or that the image of an algebraic curve is an algebraic projective curve. But I cannot prove any of these. Any help?

There exist two disjoint irreducible  closed subsets  both containing more than one point   in $\mathbb A^2_k$ but not in $\mathbb P^2_k$ (Bézout).

Well, there are two things.

Homeomorphism preserves compactness (can you see why? homeomorphisms being one-to-one preserves unions, inclusions and opennes).
$\mathbb{P}^2$ is compact, while $\mathbb{A}^2$ is not.


If $f_n \to f$ in measure, then there is a subsequence such that $f_{n_k}\to f$ almost everywhere Suppose we are on a finite measure space and $f_n\to f$ in measure. I want to show there is a subsequence such that $f_{n_k}\to f$ a.e.

I know that $f_n\to f$ in measure $\iff$ $\int_X\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu\to0$. So if $f_n\to f$ in measure, then by Fatou's lemma, $\int_X\text{lim inf}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu\leq\text{lim inf}\int_X\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu=0$. Since $\text{lim inf}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}\geq 0$, this means $\text{lim inf}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}=0$. Can I infer from this that there is a subsequence such that $\frac{|f_{n_k}(x)-f(x)|}{1+|f_{n_k}(x)-f(x)|}\to 0$ and so $f_{n_k}\to 0$ almost everywhere?

No, your argument does not give a convergent subsequence. (You can only get a subsequence depending on x). Here is a valid proof: Choose $n_k$ such that $\mu \{|f_n -f| >2^{-k}\} < 2^{-k}$ for $n \geq n_k$. Since $\sum_k \mu \{|f_{n_k} -f| >2^{-k}\} < \infty$ it follows the set of points x for which $|f_{n{_k}}(x)-f(x)| > 2^{-k}$ has measure 0 which shows that $f_{n{_k}} -f \to 0$ a.e..

Is your function $f$ non-negative? Without that, you can not apply Fatou's lemma. The answer to your question is yes, I think you may find this post helpful Do we need to have a subsequence such that $\lim_{k\to\infty}\left\|x_{n_k}\right\|=\liminf_{n\to\infty}\left\|x_n\right\|$?


Prove that from the equalities, $\frac{x(y+z-x)}{\log x}=\frac{y(x+z-y)}{\log y}=\frac{z(y+x-z)}{\log z}$ follows $x^yy^x=y^zz^y=z^xx^z$. Problem :

Prove that from the equalities, $$\frac{x(y+z-x)}{\log x}=\frac{y(x+z-y)}{\log y}=\frac{z(y+x-z)}{\log z}$$ follows $$x^yy^x=y^zz^y=z^xx^z$$.

My approach :

$$\frac{x(y+z-x)}{\log x}=\frac{y(x+z-y)}{\log y}=\frac{z(y+x-z)}{\log z} = k $$

$$ \Rightarrow xy +zx -x^2 = k\log x  .....(i)$$

$$\Rightarrow xy +zy -y^2 = k\log y ...(ii)$$

$$\Rightarrow zy +zx -z^2= k\log z ....(iii)$$

How to proceed further please guide thanks

Then , $k(y\log x+x\log y)=k(xy^2+xyz-x^2y+x^2y+xyz-xy^2)=2kxyz$

Again , $k(z\log y+y\log z)=k(xyz+z^2y-zy^2+y^2z+xyz-z^2y)=2kxyz$.

So, $k(y\log x+x\log y)=k(z\log y+y\log z)\implies x^yy^x=y^zz^y$ , as $k\not=0$.

Similarly you can prove the second equality..

$$x^yy^x=y^zz^y=z^xx^z$$
$$\Rightarrow  \left\{\begin{array}{ll}
                 x^{y-z}=(\frac{z}{y})^x \\
                 y^{x-z}=(\frac{z}{x})^y \\
                 z^{y-x}=(\frac{x}{y})^z
                \end{array} \right. $$
Take the log of both sides
$$\Rightarrow  \left\{\begin{array}{ll}
                 (y-z)\log{x}=x\log{\frac{z}{y}} ...(1) \\
                 (x-z)\log{y}=y\log{\frac{z}{x}} ...(2)\\
                 (y-x)\log{z}=z\log{\frac{x}{y}} ...(3)\\
                \end{array} \right. $$
From (2)
$$\log{y}=\frac{y}{x-z}\log{\frac{z}{x}} ...(4)$$
subtitute (4) into (1)
$$(y-z)\log{x}=x\log{z}-x(\frac{y}{x-z}\log{\frac{z}{x}} )$$
$$\Rightarrow (z^2-xz-yz)\log{x}=(x^2 -xz-xy)\log{z}$$
$$\Rightarrow \frac{z(x+y-z)}{\log{z}}=\frac{x(y+z-x)}{\log{x}}$$

Similarly, we get the
$$\frac{z(x+y-z)}{\log{z}}=\frac{x(y+z-x)}{\log{x}}=\frac{y(x+z-y)}{\log{y}}$$


Show that any conjugate pair of complex numbers (with non-zero imaginary part) cannot be the spectrum of any 2x2 matrix with real, nonnegative entries My professor showed me this in her office today but I didn't like her method and wanted to use another method.

So, I computed the characteristic polynomial of some arbitrary $2 \times 2$ matrix with nonnegative entries $a,b,c,d$, and solved for its two eigenvalues using the quadratic formula.

Then I equated the conjugate pair of complex numbers (with non-zero imaginary part) to the two eigenvalues, and am trying to derive some contradiction at this point.  But, I'm stuck.

My professor had used some complexification process and showed that the realizing matrix must have certain trigonometric entries (she mentioned rotation matrices), and then derived an impossibility.

Thanks in advance,

For a $2 \times 2$ matrix $A$, the characteristic equation is $x^2 - \text{Tr}(A)x + |A| = 0$. If this equation has complex roots, then the discriminant is negative. But if you assume A = [a,b;c,d], then $\text{Tr}(A)^2 - 4|A| = (a-d)^2 + 4bc \geq 0$!!

Hint: Calculate the discriminant of the characteristic polynomial of
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
after some easy calculations to be $$\frac{1}{4}\left((a-d)^2+4bc\right).$$

Remark: Complex-conjugated eigenvalues may only exist if $bc<0$ regardless of the signs of $a$ and $b$.


Why are direct proofs often considered better than indirect proofs? As the title indicates, I'm curious why direct proofs are often more preferable than indirect proofs.

I can see the appeal of a direct proof, for it often provides more insight into why and how the relationship between the premises and conclusions works, but I would like to know what your thoughts are concerning this.

Thanks!

Edit: I understand that this question is quite subjective, but that is my intention. There are people who prefer direct proofs more than proof by contradiction, for example. My curiosity is concerning what makes a direct proof preferable to such individuals. In the past, I've had professors grimace whenever I did an indirect proof and showed me that a direct proof was possible, but I never thought to ask them why a direct proof should be done instead. What's the point?

I just did a quick lookup and it suggested that the two flavors of indirect proof was contraposition and contradiction.  What I'm about to say is criticizing contradiction, because contraposition seems fine to me.

Imagine you have a 1000 statement direct proof.  Then every step along that way is provable.  Maybe somebody reads your proof and realizes that an observation you made halfway through is exactly the idea they need to solve a problem they have.  Mathematical history has many examples of lemmas that are more famous than the theorems they originally supported.

By contrast, a 1000 statement proof by contradiction starts out with two hypotheses that are inconsistent.  Everything you're building is a logical house of cards that is intended to collapse at the end. Nothing you wrote can be counted on outside that framework without a separate analysis.

If it truly takes both hypotheses to get you to the result, then so be it.  But I was rightfully dinged by my professors when I wrote a proof by contradiction that could easily be modified into a direct proof by contraposition.

Even if direct proof is preferable, it's not always possible.  How could you prove, for example, that primes are infinitely many, except by showing that any finite number we assign to them entails the existence of yet another prime (Euclid Elements, IX, 20)--[edit] or contradicts something else known to be true, e.g. the divergence of the harmonic series (Euler)?

Proof of any kind of unlimited multitude, it seems, has got to be indirect, i.e. by contradiction.


For $x\in\mathbb R\setminus\mathbb Q$, the set $\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$ is dense on $[0,1)$ Let $x\in \mathbb{R}$ an irrational number. Define $X=\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$. Prove that $X$ is dense on $[0,1)$.

Can anyone give some hint to solve this problem? I tried contradiction but could not reach a proof.

I spend part of the day studying this question Positive integer multiples of an irrational mod 1 are dense
and its answers. Only one answer is clear and give clues to solve the problem. This answer is the first one. However, this answer does not answer the question nor directly, nor the proof follows from this answer.

This answer has some mistakes, he use that $[(k_1-k_2)\alpha]=[k_1\alpha]-[k_2\alpha]$ which is not true. Consider $k_1=3, k_2=1, \alpha=\sqrt{2}$ we have $[(k_1-k_2)\alpha]=2\not= 3=[k_1\alpha]-[k_2\alpha] $. We only can assure that $[k_2\alpha]-[k_1\alpha]-1\leq [(k_2-k_1)\alpha]\leq[k_2\alpha]-[k_1\alpha]$.

Who answered said something interesting about additive subgroups of $\mathbb{R}$, but unfortunately the set $X=\{nx-[nx] : n\in \mathbb{N} \}$ is not a subgroup. Considering the additive subgroup $G=\langle X \rangle$, if we prove the part (a) of the link, we get that indeed $G$ is dense on $\mathbb{R}$ but we can not conclude that $X$ is dense on $[0,1)$.

I think this problem has not been solved.

Thanks!

Ok, since you've asked and it doesn't fit into a comment, there you go. I'll do it on a circle since it's slightly easier to explain and I'll leave it to you to complete it in the case of an interval. let's say you have a circle of length $1$. you take 'steps' along the circle of an irrational length, let's say counter-clockwise. you'll never hit the same spot twice so for any fixed $\epsilon > 0$ you'll eventually find two 'steps' $a_n$ and $a_m$ such that $0 < |a_n - a_m| < \epsilon$. the distance from $a_n$ to $a_m$ is the same as between $a_{n-m}$ and $a_0 = 0$ and so on. therefore if you let $k:= n-m$ and you only consider each $k$-th step you'll be going around the circle travelling a distance smaller than $\epsilon$ hence if you divide your circle into arcs of equal lengths greater than $\epsilon$ (but just slightly, say smaller than $2 \epsilon$) you'll have to land in each one of those in order to make your way all around the circle (because your steps are to small to jump over them). Every point of the circle is in at least one of those intervals which means that for each point of the circle you can find a number $a_j$ in your sequence that is closer than $2 \epsilon$ to it. Now conclude taking smaller and smaller $\epsilon$'s.

edit: oh, just note that I'm taking the distance along the circle, not the euclidean one

$f \colon x \rightarrow x-\lfloor x \rfloor$

$F=\{f(nx),n \in \Bbb{Z}\}$

$F \neq \emptyset$ and $x \ge 0$ $\exists d=\inf(F)$

We can prove that $d=0$ (suppose a $\gt$ 0 and ...)

hence $\forall$ y $\gt$ 0 $\exists$ n $\in$ $\Bbb{N}$ , $y \gt f(nx) \gt 0$

$a,b \in [0;1[$
if $b-a \gt 0$ $\exists n \in \Bbb{N}$, $b-a \gt f(nx) \gt 0$

$\exists p \in \Bbb{N}$ , pf(nx) $\gt$ a (1) , if we suppose that p is the smallest number who verifies (1) we have that (p-1)fn(x) $\lt$ a therefore
 pf(nx) $\lt$ a+f(nx) $\lt$ b

therefore pf(nx) $\in$ [a;b] and pf(nx)=f(pnx)

we can conclude about the density

First time I use this way of writing I hope I have been clear


Positivity of power function. Prove that
           $6^a-7^a+2\cdot 4^a-3^a-5^a\ge0$ for $-\frac{1}{2}\le a\le0$.

I tried to do it by first derivative test but derivative become more complicated (same with 2nd derivative for convexity).
Here is it's plotting.

For  $a\in[-\frac12,0)$ and $x>0$ let $$g_a(x)=x^a-(x+1)^a-(x+3)^a$$
Then  $$f(a)=g_a(4)-g_a(3)$$
By the Mean Value Theorem this equals $g_a'(\theta)=ag_{a-1}(\theta)$ for some $\theta\in [3,4]$.  We have
$$\begin{align}\frac{g_{a-1}(\theta)}{\theta^{a-1}}&=1-(1+\tfrac1\theta)^{a-1}-(1+\tfrac3\theta)^{a-1}\\&\le 1-(1+\tfrac13)^{a-1}-2^{a-1}&\text{(negative exponent $\Rightarrow$ maximize $\tfrac1\theta$)}\\
&\le 1-(\tfrac43)^{-3/2}-2^{-3/2}&\text{(bases $>1$ $\Rightarrow$ minimize neg. exponent)}\\
&=\tfrac18\left(8-3\sqrt3-2\sqrt 2\right).\end{align} $$
We have $5<3\sqrt 3<6$ and $2<1\sqrt 2<3$, therefore the sign of the expression in parenthesis is not immediately clear (without using a calculator). But the conjugates $8+3\sqrt3-2\sqrt 2$, $8-3\sqrt3+2\sqrt 2$, $8+3\sqrt3+2\sqrt 2$ are all definitely positive (in fact $>8+2-6=4$).
The product of all four conjugates however is
$$\begin{align}&(8-3\sqrt3-2\sqrt 2)(8-3\sqrt3+2\sqrt 2)(8+3\sqrt3-2\sqrt 2)(8+3\sqrt3+2\sqrt 2)\\=\:&\bigl((8-3\sqrt 3)^2-8\bigr)\bigl((8+3\sqrt 3)^2-8\bigr)\\=\:&(83-48\sqrt 3)(83+48\sqrt 3)\\=\:&83^2-3\cdot 48^2\\
=\:&{-23}.\end{align} $$
We conclude $8-3\sqrt 3-2\sqrt 2<0$, so that
$$f(a)=ag_{a-1}(\theta)\ge a\theta^{a-1}\tfrac18\left(8-3\sqrt3-2\sqrt 2\right)>0 $$
for $-\frac12\le a<0$.

$f(a) = 6^a - 7^a + 2 \cdot 4^a - 3^a - 5^a$.

Value of function

$f(-1/2) = 0.00572$, and $f(0) = 0$.

First derivative:

$2^{2 + 2 a} Log[2] - 3^a Log[3] - 5^a Log[5] + 6^a Log[6] - 7^a Log[7]$.

Evaluated at $a = -1/2$ and $0$:

${d f(a) \over d a} |_{a=-1/2} = 0.02825$ and ${d f(a) \over d a} |_{a=0} = -0.0896$.

Second derivative:

${d^2 f(a) \over d a} = 2^{2 a+3} \log ^2(2)-3^a \log ^2(3)-5^a \log ^2(5)+6^a \log ^2(6)-7^a \log ^2(7)$

which is always negative in the range $-1/2 \leq a \le 0$.  We know this because the second derivative is negative at each end of the range and the third derivative is always negative (as can be seen on a plot).

QED.


How to write neither ___ nor ___ in First Order Logic? If X exists, it is neither Y nor Z.

Would this be written as X ⇒ ¬Y ∨ ¬Z  ?

Or would it be X ⇒ ¬(Y ∨ Z)  ?

\begin{equation}
(\exists x \in S)(\exists y, z \in S) \Rightarrow x \neq y \wedge x \neq z
\end{equation}

Where S is some arbitrary set


Geometric Interpretation of Total Derivative? Say that:

$$z = xy$$

So:

$${\partial z \over \partial x} = y$$

and

$${\partial z \over \partial y} = x$$

If we plot in 3D space the 2D surface corresponding to eq1, than take a point on that surface, the tangent with respect to the x axis is y, and the tangent corresponding to the y axis is x.

Do the total derivatives ($dz \over dx$ and $dz \over dy$) have a similar geometric interpretation?

Here are two ways to think about the total derivative of a function $z$.

If the gradient vector field is $$\nabla z = \begin{pmatrix}\frac{\partial z}{\partial x} \\ \frac{\partial z}{\partial y}\end{pmatrix},$$ the total derivative is a "covector field" given by the transpose of the gradient,
$$dz = \begin{pmatrix}\frac{\partial z}{\partial x}\ \ \frac{\partial z}{\partial y}\end{pmatrix}.$$
What information does this encode?  Directional derivatives.  If $v = \begin{pmatrix}v_1 \\ v_2\end{pmatrix}$ is a vector field, then matrix-multiplying $(dz)(v)$ is the function that tells you the directional derivative of $dz$ along $v$ at each point.
Remember from calculus that the derivative is the slope of the best affine-linear approximation to a function.   In other words, if you zoom in close to $x_0$, then $f$ begins to look very much like $f(x_0) + f'(x_0)(x-x_0)$.  In this context, $dz$ is the "best linear approximation" to $z$ at $(x,y)$.  If you zoom in close to $(x_0,y_0)$, the map $z$ looks very much like the affine-linear map $$x_0y_0 + \begin{pmatrix}\frac{\partial z}{\partial x} \ \ \frac{\partial z}{\partial y}\end{pmatrix}\begin{pmatrix}x-x_0 \\ y-y_0\end{pmatrix}.$$ The first term is $z(x_0,y_0) = x_0y_0$, and again we are matrix-multiplying in the second term.

The total derivative may be used in dynamic geometry. See Geogebra.  The addition of time to geometry was a part of Newton's calculus.  The static $(x,y,z)$ coordinate system expanded to $(x,y,z,t)$.  The static coordinates expanded to $[x(t),y(t),z(t)]$.  One derivative of interest is $dz/dt$.  The independent variable is time which is plotted along the $x$-axis.  The dependent variables $(x,y,z)$ are plotted individually along the vertical axis.  Also, in geometry $z$ may be plotted as the radius of an expanding circle $x(t)^2 + y(t)^2 = z(t)^2$.  This adds time to the Pythagorean  theorem.  In special relativity $z=ct$ where $c$ is the speed of light.  This geometry is the light cone. Newton was interested in the orbits of planets.  This requires keeping track of the directions of motions with vectors. Fortunately the motion of the earth is mostly in a plane.  In this case the object in plane geometry is the ellipse for closed orbits.  It is a hyperbola or parabola for open orbits.  Another geometric item of interest is the curvature. For a circle it is $1/r$ and for a sphere it is $1/r^2$. The $1/r^2$ is interpreted as the gravitational force. Euler proved that the second derivative with respect to time is also the curvature.  This second derivative is interpreted as the acceleration. Thus Newton's second law states that temporal curvature must be equal to spatial curvature using local masses and a universal constant as parameters.  This was the dynamic geometry that was used before Einstein created his geometry of gravity, known as General Relativity. Quantum mechanics required the development of a completely different type of geometry.


Non-constant continuous map from $\mathbb{R}P^2$ to $\mathbb{R}$ Give an example of non-constant continuous map from $\mathbb{R}P^2$ to $\mathbb{R}$.

My attempt: Since $\mathbb{R}P^2$ is closed unit disc with antipodal points of the boundary identified. I will restrict the projection map from $\mathbb{R}^2$ to $\mathbb{R}$ to closed unit disc. then can i use this to get the required map.

Indentify $\mathbb{R}^2$ by $\mathbb{C}$ and consider the map $\nu\colon \mathbb{C}\longrightarrow \mathbb{R}$ which takes $z$ to its norm $\nu(z)=|z|$.


Solution to $y^{\prime \prime} -xy = 0$ I have a differential equation of the form

$y^{\prime \prime} - xy = 0$,

which I'm told has a solution of the form

$y(x) = p(x) e^{-q(x)}$.

I've been trying to solve for q(x) by differentiating y(x) and plugging it into the ODE directly, which led me to find

$p^{\prime \prime}(x) - 2p^{\prime}(x)q^{\prime}(x) + p(x)(q^{\prime}(x)^2 - q^{\prime \prime}(x)) = xp(x)$,

but I'm not sure where to go from here, or if this is even the right direction. I've been given a hint that to solve for $q(x)$, I'll need to think about the behaviour of $p(x)$ for large $x$. Since $p(x)$ looks like an amplitude to an oscillation, I feel like it will go to $0$ over time.

I would appreciate any help on how to to get a solution of this form.

EDIT: for large enough $x$, I think we can make the approximation that $y^{\prime \prime}(x) << y^{\prime}(x)^2$, although I'm not sure yet exactly how this could help.

You have $2$ functions parametrizing one solution. Thus you can add one external condition on the relation of $p$ and $q$. One relatively natural condition is that the "amplitude" $p$ varies much more slowly than the "phase" $q$, meaning that the derivatives of $p$ are dominated by the derivatives of $q$. This then suggests to extract from the dominant terms
$$
q'(x)^2=x
$$
so that the remaining equation is
$$
p''(x)-2p'(x)q'(x)-p(x)q''(x)=0
$$
Here again one can concentrate in a first approximation on the dominant terms
$$
-2p'(x)q'(x)-p(x)q''(x)=0\implies p(x)^2q'(x)=C
$$
In total this gives the first order of the WKB approximation method.

Well, if all you want to do is solve for $q(x),$ then that's just the logarithm of $p(x)/y(x).$ That is, if you rewrite your equation as $$\frac{y(x)}{p(x)}=e^{-q(x)},$$ whence, by raising both sides to the power $-1$ we get  $$\frac{p(x)}{y(x)}=e^{q(x)},$$ from which it follows that $$q(x)=\log\left(\frac{p(x)}{y(x)}\right).$$


Direct Product of Cyclic Groups I have the direct product: $G_{1} \times G_{2}$ and I know it is a cyclic group. I want to prove $G_{1}$ and $G_{2}$ are themselves cyclic groups. Any help getting started?

Your group $G_1 \times G_2$ is cyclic, i.e. there exists $g \in G_1 \times G_2$ which generates the big group. Be aware that $g = (g_1, g_2)$ for some $g_1 \in G_1$ and $g_2 \in G_2$. So why not trying to prove your statemant with these two "obvious" candidates?

Here are some thoughts. Working with the integers $\mod p$ where $p$ is a prime number, if $G$ is a cyclic group of order $p-1$, and $x$ is a primitive root ($x$ has multiplicative order $p-1$), and $G=[x,x^2,x^3,...x^{p-1}]$, then every non-zero element $e$ is contained in $G$. This is the "bigger group". Let $a$ and $b$ be relatively prime integers such that $ab=p-1$. If $y$ is an element with multiplicative order $a$, and $z$ is an element with multiplicative order $b$, the cyclic group of order $a$ is $A=[y,y^2,y^3,...y^{p-1}]$ and the cyclic group of order $b$ is $B=[z,z^2,z^3,...z^{p-1}]$ are the "smaller groups" and hence subgroups of $G$. It follows the multiplication of every element in $A$ with every element in $B$ will yield the group $G$. That is Group $A$ multiplied by group $B$ is group $G$, or $AB = G$. Another consequence is that every non-zero element $e$ is the product of an element in $A$ and $B$. In other words, each non-zero element $e$ has the form $y^nz^m$ for integers $n$ and $m$ where $y$ and $z$ are defined earlier.


What does "$\to$" mean in "$a^3+b^3=(a+b)^3\rightarrow ab(a+b)=0$"? I'm working on this problem and I'm having difficulty understanding the solution. In the solution, it states $a^3+b^3=(a+b)^3\rightarrow ab(a+b)=0$. What do these equations mean? It doesn't seem to fit the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$ if the arrow meant subtracting.

I know that right arrows can define a function, but I don't see how $(a+b)^3\rightarrow ab(a+b)=0$ could be a function. If it is, then I don't know how it is applicable to the problem, in which case it would be great if someone could explain the rest of the solution.

P.
S. I don't see this as asking two questions if that's not allowed. One is just a prerequisite to the other.

Assume that
$$
a^3  + b^3  = \left( {a + b} \right)^3
$$
Since
$$
\left( {a + b} \right)^3  = a^3  + b^3  + 3a^2 b + 3ab^2
$$
it follows that
$$
3a^2 b + 3ab^2  = 0
$$
thus
$$
3ab\left( {a + b} \right) = 0
$$
so that
$$
ab\left( {a + b} \right) = 0
$$

the arrow means it follows:  just calculate (a+b)^3 and subtract a^3+b^3
 what remains is 0


Implementing Neumann boundary condition for elasticity problem using the finite element method I am solving the following mixed boundary elasticity problem with the finite element method on the unit circle, using a triangle mesh with 3 nodes on each element:

$ \frac {\partial}{\partial x_j}\sigma_{ij}(u(x_1,x_2))=0 $, $(x_1,x_2) \in \Omega$

$ u_1(x_1,x_2)=u_1^{an}(x_1,x_2)=\frac{1-\nu}{2G(1+\nu)}\sigma_0x_1$ on $\Gamma$

$\sigma_{2j}(u(x_1,x_2))n_j(x_1,x_2)=t_2^{an}(x_1,x_2)$ on $\Gamma$

where:

$ t_2^{an}(x_1,x_2)=\sigma_{21}^{an}(x_1,x_2)n_1+\sigma_{22}^{an}(x1,x2)n_2(x_1,x_2) $ is the 2nd component of the traction vector $ t^{an}=(t_1^{an}, t_2^{an}) $

$ \sigma_i^{an}(x_1,x_2)=\sigma_0 \delta_{ij} $, $i,j=1,2$

$\sigma_0=1.5 *10^{10} N/m^2$ , $ \nu=0.34 $ , $G=3.35*10^{10}N/m^2$

With the following weak form:

$\int_\Omega \sum_{i.j}^2 \sigma_{ij}(u) \epsilon_{ij}(v)= \int_\Gamma v2*t2 $

I have already implemented it correctly for the pure Dirichlet problem. (so the global stiffness matrix is also good for this problem), but I do something wrong when implementing the Neumann boundary condition and I cannot figure out what. (I still get the right solution for u1 where I have the Dirichlet condition).

This is what I do when implementing the Neumann boundary condition:

I go over each triangular element:

-> I extract the coordinates of each node of the element:
   (x1,y1), (x2,y2), (x3, y3)
   (the elements are numbered in anticlockwise sense)

-> if (x1,y1) and (x2,y2) are on the boundary:

I calculate the outward unit normal:
$$ n2=\frac{x_1-x_2}{\sqrt{(y_2-y_1)^2+(x_1-x_2)^2}} $$
I update the force vector F in the place corresponding to $C_e^12$ and $C_e^22$ ($v_2=\sum_{B=1}^3 N^BC_e^B2$) by adding :
$$ \sigma_0*n_2*\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}*0.5 $$

-> if (x1,y1) and (x3,y3) are on the boundary:

$ n2=\frac{x_3-x_1}{\sqrt{(y_1-y_3)^2+(x_3-x_1)^2}} $
update with
$ + (-1)*\sigma_0*n_2*\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}*0.5 $

-> if (x2,y2) and (x3,y3) are on the boundary:

$ n2=\frac{x_2-x_3}{\sqrt{(y_3-y_2)^2+(x_2-x_3)^2}} $
update with:
$ +\sigma_0*n_2*\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}*0.5 $

The methodology seems fine. The issue could be due to bugs in your code. Check your code carefully and see if there are any bugs.


How can I prove this using a Combinatorial Proof? Can someone explain this to me? How can I approach this?

Let $k \geq 1$ be an integer and consider a sequence
$n_1,n_2,\ldots,n_k$ of positive integers. Use a combinatorial proof
to show that.

$$
 {{n_1} \choose 2} + {{n_2} \choose 2} + \cdots + {{n_k} \choose 2}
         \leq
   {{n_1 + n_2 + \cdots + n_k} \choose 2}$$

Consider the complete graph on $n_i$ vertices. How many edges does this graph have?

Elaborating on the hint, consider a graph formed by the disjoint union of complete graphs of sizes $n_1,n_2,n_3,\dots,n_k$

It will look something like this:

How many edges appear in the graph made from the disjoint union of complete graphs of size $n_1,n_2,\dots,n_k$?

If you are unable to immediately answer that, consider reading this related question.

What does the right side of the inequality count and how does that relate to our scenario here when considering graphs?

   The LHS, $\binom{n_1}{2}+\binom{n_2}{2}+\dots+\binom{n_k}{2}$, is the number of edges in a graph on $n_1+n_2+\dots+n_k$ vertices made up of the disjoint union of complete graphs of sizes $n_1,n_2,\dots,n_k$ respectively.  The RHS, $\binom{n_1+n_2+\dots+n_k}{2}$, is the largest number of edges in a graph on $n_1+n_2+\dots+n_k$ vertices and is in particular greater than or equal to the number of edges in the graph described in the first sentence.  Q.E.D.

This should hopefully mostly be a combinatorial proof: the $LHS$ of the equality is the same as (let's call it $S_{left})$:

$$\frac{n_1!}{2!(n_1-2)!} + \frac{n_2!}{2!(n_2-2)!} + \frac{n_3!}{2!(n_3-2)!} + \ ... \ + \frac{n_k!}{2!(n_k-2)!} = S_{left}$$
$$\frac{n_1!}{(n_1-2)!} + \frac{n_2!}{(n_2-2)!} + \frac{n_3!}{(n_3-2)!} + \ ... \ + \frac{n_k!}{(n_k-2)!} = 2S_{left} \tag{1}$$

We know that $\frac{n!}{(n-k)!} = n(n-1)(n-2)...(n-(k-1))$. E.g. $\frac{10!}{(10-2)!} = \frac{10(9)(8)...(2)(1)}{8(7)(6)...(2)(1)}$ will just become $10(9)$ because everything else cancels out. Use this fact for $(1)$:

$$n_1(n_1-1) + n_2(n_2-1) + n_3(n_3-1) + \ ... \ + n_k(n_k-1) = 2S_{left} $$
$$n_1^2 + n_2^2 + n_3^2 + \ ... \ + n_k^2 - (n_1 + n_2 + n_3 + \ ... \ + n_k) = 2S_{left} \tag{2}$$

Now let's move on to the $RHS$, let's call it $S_{right}$:

$$\frac{(n_1 + n_2 + n_3 + \ ... \ + n_k)!}{2!(n_1 + n_2 + n_3 + \ ... \ + n_k - 2)!} = S_{right}$$
$$\frac{(n_1 + n_2 + n_3 + \ ... \ + n_k)!}{(n_1 + n_2 + n_3 + \ ... \ + n_k - 2)!} = 2S_{right} \tag{3}$$

What is $n_1 + n_2 + n_3 + \ ... \ + n_k$? Well it's just a sum of an arithmetic series so it's equal to $\frac{k(n_1+n_k)}{2}$, use that for $(3)$ (we will use it later for $(2)$ as well):

$$\frac{\Big(\frac{k(n_1+n_k)}{2}\Big)!}{\Big(\frac{k(n_1+n_k)}{2} - 2\Big)!} = 2S_{right} \tag{4} $$

Using the same logic here as we did for terms in $(1)$, we can see that $(4)$ is just the sum for $k$ terms times the sum for $k-1$ terms (in the same way that a term in $(1)$ was $n(n-1)$). Change $(4)$ to:

$$\Big(\frac{k(n_1+n_k)}{2}\Big)^2 - \frac{k(n_1+n_k)}{2} = 2S_{right} \tag{5}$$

Use the definition of an arithmetic series on $(2)$ and then compare to $(5)$:

$$n_1^2 + n_2^2 + n_3^2 + \ ... \ + n_k^2 - \frac{k(n_1+n_k)}{2} = 2S_{left} \tag{6} $$
$$\Big(\frac{k(n_1+n_k)}{2}\Big)^2 - \frac{k(n_1+n_k)}{2} = 2S_{right} \tag{7}$$

Use your equality to set $$2S_{left} \le 2S_{right} \tag{8}$$ $S_{left} \le S_{right}$ is the same as $(8)$ so it's valid for our comparison. Then use $(6)$ and $(7)$ in the equality: $$ n_1^2 + n_2^2 + n_3^2 + \ ... \ + n_k^2 - \frac{k(n_1+n_k)}{2}\le \Big(\frac{k(n_1+n_k)}{2}\Big)^2 - \frac{k(n_1+n_k)}{2}$$

$$ n_1^2 + n_2^2 + n_3^2 + \ ... \ + n_k^2\le \Big(\frac{k(n_1+n_k)}{2}\Big)^2 $$

Note that $\Big(\frac{k(n_1+n_k)}{2}\Big)^2 = \sum_{i=1}^{k} n_i^3$, and $n_1^2 + n_2^2 + n_3^2 + \ ... \ + n_k^2 = \sum_{i=1}^{k} n_i^2$.

So obviously for $k \ge 1$ and positive $n_i$ (for negative $n_i$ this is not true because the cubic series will become negative):

$$ \sum_{i=1}^{k} n_i^2 \le \sum_{i=1}^{k} n_i^3$$

Which completes the proof for the first question.

Regarding the second question: in a graph with $n$ vertices, the first vertex has $n-1$ vertices to draw edges to, the second vertex has $n-2$ vertices to draw edges to etc. Maybe you can figure it out by this? Alternatively, how many combinations of $2$ vertices connected together can a graph with $n$ vertices have? Where order doesn't matter because $v_1 \leftrightarrow  v_2$ is the same as $v_2 \leftrightarrow  v_1$.


how can I show that the translate of an open set is open and that the translate of a closed set is closed Let $P$ be an open set of real numbers and let $C$ be an closed set of real numbers.  Let $x$ be a real number.  How can I show that $P + x$ is open and that $C + x$ is closed?

It's been a while since I worked with topology and I'm sort of stuck.

Since $f:\mathbb{R}\to\mathbb{R} , f(u) =u+x$ is homeomorphism


If $X$ and $Y$ are countable, then $X \times Y$ is countable. (Analysis 1 by Tao) Exercise 8.1.8 Use Corollary 8.1.13 to prove Corollary 8.1.14.
Corollary 8.1.13. The set $\mathbb{N} \times \mathbb{N}$ is countable.
Corollary 8.1.14. If $X$ and $Y$ are countable, then $X \times Y$ is countable.

For the proof of Corollary 8.1.13, the book shows that $\mathbb{N} \times \mathbb{N} = A \cup B$, where $A = \{(n,m) \in \mathbb{N} \times \mathbb{N} : 0 \le m \le n\}$ and $B = \{(m,n) \in \mathbb{N} \times \mathbb{N} : 0 \le n \le m\}$.
I also know that $X$ and $Y$ are countable, there exists bijections from $\mathbb{N}$ to $X$ and $\mathbb{N} $ to $Y$. To finish the proof, I need to find the bijection from $\mathbb{N} \times \mathbb{N}$ to $X \times Y$, but I do not know how to get this.
Any help would be appreciated.

Denote by $f:\Bbb N \to X$ , $g:\Bbb N \to Y$ the bijections from $\Bbb N$ to $X,Y$.

Then $h : \Bbb N \times \Bbb N \to X \times Y$ which is defined by $h(n,m)= (f(n),g(m))$ is the bijection you are looking for.

As this was downvoted, I will provide another argument. In the following argument I will allow countability to include finite and countably infinite.

Before starting, a useful lemma is the following:

If $f:X\rightarrow Y$ is injective and $Y$ countable then $X$ is countable.

Proof:

Suppose $Y$ is finite. Observe, $f:X\rightarrow f(X)$ is bijective and since $f(X)$ is finite, so is $X$. Hence $X$ is countable. Suppose now that $Y$ is countably infinite. Then $f(X)\subseteq Y$ is finite or countably infinite (countable). In either case, we admit a bijection from $X$ to $f(X)$, in which case $X$ is either finite or, if $f(X)$ is countably infinite, we obtain a bijection $g:f(X)\rightarrow \mathbb{N}$ and as a result, considering $f:X\rightarrow f(X)$, we obtain a bijection, $g\circ f: X\rightarrow \mathbb{N}$.

Now, with regards to your problem,

Suppose $X$ and $Y$ are both countable.

Case 1: Suppose $X$ is finite and $Y$ is countably infinite.

In this case, one may define an injection, $f: X \rightarrow \mathbb{N}$ (Exercise: Find such a injection) and a bijection, $g:Y\rightarrow \mathbb{N}$. Therefore, define a function $h: X\times Y \rightarrow \mathbb{N} \times \mathbb{N}$  by $h(x,y)$ $=$ ($f(x),g(y))$. It is straightforward to check that this function is injective.

Case 2: Suppose $X,Y$ are both countably infinite. In which case, we are able to obtain bijective maps, from $X$ to $\mathbb{N}$ and $Y$ to $\mathbb{N}$. Therefore, repeating what we did in case 1 gives us the desired conclusion.

Other cases follow similarly.


2 dice are rolled, and it is revealed that one of the numbers rolled was a 4. P(other number rolled was a 4)=? The above question was present on brilliant with slight change as written below. I was able to correctly solve it and get the probability=2/11.
Question on brilliant: Two fair dice are rolled, and it is revealed that (at least) one of the numbers rolled was a 4. What is the probability that the other number rolled was a 6?
Note: You are not told which of the numbers rolled is a 4.
I am getting the answer to the question I have asked as 1/11 (both 4's). Is that correct? Just wanted to confirm.

You already have the correct answer $1/11$ and have responded to
@Riemann in your comment, giving the correct method.
I'm a great fan of simulation, especially to verify my
answers to such problems, guarding against a lapse in
logic. So here's my simulation in R, which is consistent
with your $1/11.$
I simulated rolling two dice a million times, counting
the number of 4s at each roll of a pair of dice.
This gives me a vector nr.4 with a million entries
(each of which may be 0, 1, or 2). The simulation gives
the answer $0.0913 \pm 0.0006,$ consistent with $1/11 = 0.0909.$
set.seed(812)  # for reproducibility
nr.4 = replicate(10^6, sum(sample(1:6, 2, rep=T)==4))
table(nr.4)/10^6
nr.4
       0        1        2
0.695074 0.277077 0.027849

mean(nr.4[nr.4>=1] == 2)
[1] 0.09133036     # aprx P(both 4|one or more) = 1/11
1/11
[1] 0.09090909
2*sd(nr.4[nr.4>=1] == 2)/1000
[1] 0.0005761576   # aprx 95% margin of sim error for answ

Notes on R code (if interested):

If you repeat this bit of code starting with
my set.seed statement, you will get exactly my results; for
your own fresh simulation, pick a different seed, or omit
the statement and R will pick an unpredictable seed.

The procedure sample simulates rolling two dice.

The logical vector nr.4 >= 1 has a million entries (TRUEs
and FALSEs).

I look at the number of instances with nr.4==2
which also have nr.4 >= 1. The bracket notation [ ] can
be read as 'such that'. The mean of a logical vector is
the proportion of its TRUEs.

The last statement allows me to get a 95% margin of simulation
error. With a million iterations one can usually expect two
places of accuracy.

If the two throws are independent, then the probability of a "6"  on one throw is 1/6 no matter what the other throw was.
If you need a detailed analysis there are 36 results of two 6 sided dice:
{(1, 1), (1, 2), (1, 3), ... (6, 4), (6, 5), (6, 6)}.
Of those, exactly 6 have a "4" in the first place:
{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
one of which has a "6" in the second place.  Therefore, the probability that when rolling two dice, if one die is a "4", the other is a "6" is 1/6.


Likelihood Ratio Test statistic to test $H_0$ vs $H_1$ Let ${Y_1,...,Y_n}$ be independent random variables and $Y_i$~$N(\beta x_i, 1)$ where $x_1,...,x_n$ are fixed known constants, and $\beta$ is an unknown parameter.

I'm looking to find the p-value or rejection region for  the test

$$H_0: \beta=0 \quad \text{vs} \quad H_1:\beta\ne0$$

The Likelihood Ratio Test statistic $\Lambda$ is

$$e^{-1/2\left(\frac{\sum_{i=1}^n y_i x_i}{\sum_{i=1}^n x_i^2}\right)^2\sum_{i=1}^n x_i^2}$$

After setting $\Lambda < k$ and then solving the inequality

$\sqrt{-2\text{log}(k)}<\sum x_iy_i/\sqrt{\sum_{i=1}^n x_i^2} <-\sqrt{-2\text{log}(k)}$

(Someone told me that my last step is incorrect)

The algebra looks correct to me except you have the signs flipped on the last line. It should be
$-\sqrt{-2\text{log}(k)}<\sum x_iy_i/\sqrt{\sum_{i=1}^n x_i^2} <
\sqrt{-2\text{log}(k)}$


Keeping a dynamic-dimensioned square centered in a rectangle For some reason this problem which seemed fairly easy on the surface has given me a lot of trouble. I have a square who's side length changes based on an independent variable. The square's upper left corner is at the origin, and the square's lower right corner is at $(l_f,-l_f)$, where $l_f$ is the length of the square.

Our rectangle must always have a width to heigth ratio of $1280/720$. When the square is at it's initial length, the length of the square and the height of the rectangle should  be the same. Additionally, the square must always be directly in the middle of the rectangle (that is, the center of the square and the center of the rectangle coincide). And finally, the ratio of the square's length and the rectangles lengths must get linearly smaller as the square's length decreases, and linearly larger as the square's length increases. I need the equations for $(x_i,y_i)$ and $(x_f,y_f)$, where $x_i$ and $y_i$ are the coordinates of the top left corner of the rectangle, and $x_f,y_f$ are the coordinates of the bottom right corner of the rectangle,  based on the following variables.

$d = $ an independent variable, in the domain $(0, \infty)$, initial value of $1$

$l_i = $ constant (initial size of square...for my problem it is 250)

$l_f = l_i\cdot d \rightarrow$ (where $l_f$ is the length of the square)

My Attempt

I believe my attempt at getting the y limits is correct.
$$y_i = \frac{l_f}{2} - \frac{l_i}{2}$$ $$y_f = \frac{l_f}{2} + \frac{l_i}{2}$$

The $x$ coordinates are the hard ones now. Because the initial height of the rectangle and the length of the square must be the same, we need a corrective term to add onto the $x$ coordinates. The following term worked (not sure if it's neccessary in the end though). $\frac{1}{2} \cdot l_f(\frac{1280}{720} - 1)$. Then, if the independent variable changes from it's default value of 1, we need another term since the width and height of the rectangle can't continue to scale at the same rate, otherwise the 1280x720 ratio is not maintained. I thought this term would do the trick $\frac{1280}{720}(\frac{l_f}{2} - \frac{l_i}{2})$, but using these terms does not give me the correct answer. Using these terms, I get the following:

$$x_i = \left(1 + \frac{1280}{720}\right) \left(\frac{l_f}{2} - \frac{l_i}{2}\right) - \frac{1}{2} \cdot l_f\left(\frac{1280}{720} - 1\right)$$
$$x_f = \left(\frac{l_f}{2} + \frac{l_i}{2}\right) - \frac{1}{2} \cdot l_f\left(\frac{1280}{720} - 1\right) - \frac{1280}{720} \left(\frac{l_f}{2} - \frac{l_i}{2}\right)$$

Like I said this doesn't work though. Let me know if you have any tips, thanks!

If I'm understanding everything correctly, here's what I think you're looking for:
\begin{align*}
y_i &= \dfrac{-l_i}{2} - \dfrac{l_f}{2} \\
y_f &= \dfrac{-l_i}{2} + \dfrac{l_f}{2} \\
x_i &= \dfrac{l_i}{2} - \dfrac{\frac{1280}{720}l_f}{2} \\
x_f &= \dfrac{l_i}{2} + \dfrac{\frac{1280}{720}l_f}{2} \\
\end{align*}
Here's what I did. Initially, the centre of the square/rectangle is at $(l_i/2, -l_i/2)$. Note that linearly scaling the square/rectangle by a factor of $d$ will not change this centre point. Now to get the corners, we simply have to either add or subtract half the dimensions of the rectangle. Initially, the rectangle has a height of $l_i$, so scaling it by $d$ gives us a new height of $l_f$. Since we know the aspect ratio, it follows that the new width must be $\frac{1280}{720}l_f$.

From the answers you gave to my comments, I figured you mixed up the $y$-direction in your question. You say that the rectangle is centered at $(l_f/2, -l_f/2)$, but in your calculation you use $(l_f/2, l_f/2)$ to obtain the $y$-boundaries of your rectangle. (Note that you get $y_f=l_f$ for $d=1$ instead of $y_f=-l_f$ as you told me in a comment.)

Suppose the formulas for the $y$-boundaries are correct in your question and the center of the square is in fact at $(l_f/2, l_f/2)$ so everything fits together, I would guess you want the following $x$-boundaries for the rectangle:

$$\begin{align*}
x_i &= \frac{l_f}2 - \frac{1280}{720} \cdot \frac{l_i}{2} \\
x_f &= \frac{l_f}2 + \frac{1280}{720} \cdot \frac{l_i}{2}
\end{align*}$$


Is there a function $f$ with $f'(x) = e^x * f(x)$? Edit : I'm ''derivation'' and ''integration'' beginner. So I don't have any techniques to solve equations like that.This why I think there have to be a clever trick to solve this. a), b), c) was okay. You can see my solution below. I want to emphasize that I don't want to see the solution. Maybe someone can show me the techniques for $b)$? I found a $f$ in b), c) only with try and error. If you show me the trick, I would  try again $c)$ and $d)$ by my own.

Are there any functions $f$ with following equations:

$ a) f'(x) =e^x + f(x)   $  a) is clear now. Thank you for all of your help

$ b) f'(x) =e^x \cdot f(x) $ solved BUT only with try and error. what is the beginner technique ( like I said: Im a beginner ) to find $f$?

$ c) f'(x) =e^x \cdot f(x)^2 $ solved BUT only with try and error. what is the beginner technique to find $f$?

$ d) f'(x) =e^{f(x)} $

Remark : $f$ is not a constant function.

Update: $a ) f:= x \cdot e^x $. Then we use the multiplication rule for derivation. $f'$ = $x \cdot e^x + 1 \cdot e^x = x \cdot e^x + e^x$.

Update2 b) $f:= c \cdot e^{e^x}.$ Then we can use the chain rule: $f' = c \cdot e^{e^x} \cdot e^x$.

Update3 c) $f:= - \frac{1}{e^x +c}$. Then use chain rule and quotient rule: $ f' = e^x \cdot  \frac{1}{(e^x +c)^2}$

$$f′(x)=e^x+f(x)\Longleftrightarrow f'-f =e^x\Longleftrightarrow(fe^{-x})'=1\Longleftrightarrow fe^{-x}=x-c\Longleftrightarrow f(x)= e^x(x+c)$$

In order to be sure whether such a function exist you would need to solve the differential equations. b),c) and d) can be solved by separation of variables. a) is on the form $$y^\prime(x) + P(x)y(x) = Q(x), \ \ P(x) = -1,  \ \ Q(x) = e^x$$
and can thus be solved by the method of integrating factor https://en.wikipedia.org/wiki/Integrating_factor.


How does rewriting $x^2 -y^2$ as $(x+y)(x-y)$ avoid catastrophic cancellation? Why is rewriting $x^2 -y^2$ as $(x+y)(x-y)$ a way to avoid catastrophic cancellation?

We are still doing $(x-y)$. Is it because the last operation in the second form is a multiplication?

$x^2-y^2$ will be only as precise as $x^2$ is. If $x$ and $y$ are close to each other, the computational error might be larger than the result.

In the second case, less loss of significance occurs.

Consider e.g. $x = 9000.2$ and $y = 9000.1$, where the floating-point precision is 5 significant digits. $x^2-y^2$ is actually equal to 1800.03.

Let's try to compute it in $x^2-y^2$ way: $x^2 = 81003600.04 = 81003000$; $y^2 = 81001800.01 = 81001000$, so $x^2-y^2 = 2000$, which is quite far away from the precise answer.

So let's try to compute it in $(x+y)(x-y)$ way: $x+y = 18000.3 = 18000$; $x-y = 0.1$, so $(x+y)(x-y) = 1800$, which is significantly closer to the precise answer (and is actually as close as possible with 5 significant digits).

The loss of significant digits is approximately the same for both computations. The second approach will sometimes give you an extra significant digit, but that's not what I would call a catastrophic cancellation.

The real advantage of the second approach is that it avoids unnecessary arithmetic overflow, if $x^2$ and/or $y^2$ exceed the largest representable number. (Or arithmetic underflow, if they are smaller than the smallest representable positive number.)


Number of homomorphism Recently, I was reading on group operations and came across a question like determine the number of group homomorphism for two groups. I don't specifically recall the example but how can one determine the number of homomorphism? Wouldn't that depend on the order of the group? and bigger the order, more would be the no. of homomorphism. The example that I have in mind is rather simple, determine the number of homomorphism from $\mathbb{Z_{8}}\rightarrow \mathbb{Z_{10}}$?

Figuring out the number of homomorphisms from a cyclic group is relatively easy; once you know what happens to the generator, you immediately know what everything else in the group is mapped to.  All that remains from there is to check which of those give you a homomorphism.

In this case, the question we need to ask is if $f:\mathbb{Z}_8 \to \mathbb{Z}_{10}$, what values of $f(1)$ could give us a homomorphism?

Keeping in mind that we need to have $8[f(1)]\equiv 0\pmod{10}$ (why?), the only valid choices are $f(1) \equiv 5$ and $f(1) \equiv 0$. You can verify that both of these define a valid homomorphism.  Thus, there are two homomorphisms in this instance.

order of ${Z_8}=8$,  order of ${Z_{10}}=10$. Both are finite cyclic groups. Hence, to check number of Homomorphism between them is gcd of their orders, $\text{gcd}(8,10) = 2$, Hence the number of homomorphism function is $2$. This is not coincident.

One of The property of homomorphism states ${\phi{(g^n)= (\phi(g))^n}}$, such a homomorphism is completely specified by the image of 1. That is, if, 1 maps to a then x maps to xa.
Another property states that if ${|g|}$ is finite, then ${|\phi(g)|}$ divides ${|g|}$, so ${|a|} ={1,2}$ as then ${a={0,5}}$, i.e., ${1 \mapsto 0}$ or ${1 \mapsto 5}$ .These are the only two homomorphism is possible.


When is something "obvious"? I try to be a good student but I often find it hard to know when something is "obvious" and when it isn't. Obviously (excuse the pun) I understand that it is specific to the level at which the writer is pitching the statement. My teacher is fond of telling a story that goes along the lines of

  A famous maths professor was giving a lecture during which he
  said "it is obvious that..." and then he paused at length in thought, and then
  excused himself from the lecture temporarily. Upon his return some fifteen minutes later he said "Yes, it is obvious that...." and continued the lecture.

My teacher's point is that this only comes with a certain mathematical maturity and even eludes the best mathematicians at times.

I would like to know :

Are there any ways to develop a better sense of this, or does it just come with time and practice ?
Is this quote a true quote ? If so, who is it attributable to and if not is it a mathematical urban legend or just something that my teacher likely made up ?

Like Florian, I really like Gowers' definition of obvious. Of course this is a very personal definition. A proof that instantly springs to mind for one person may not spring to mind for another. I am not really sure what there is to say at this level of generality beyond that.

Really phrases like "it is obvious that..." and "clearly..." are bad habits. In a mathematical argument they are the places you should look at first for possible errors.

Perhaps another story will be illuminating: a professor of mine once made an assertion in lecture that I didn't quite see instantly. I asked him "is that obvious?" and he replied "yes." I asked him "is it obvious that that's obvious?" and, after a short pause, he replied "no."

I have never experienced a non obvious "obviously" in a textbook in mathematics. If it wasn't obvious at first sight it became obvious after some investigations.

To me, what is really obvious is a claim that is true and can be easy proved, but that might not be that obvious from the wrong perspective.

Anybody can fail to see what is obvious and anybody with appropriate knowledge can agree after some thinking. So there is no way to develop a sense of it, but having knowledge about the subject.


trigonometry identity I have some problem with proving this identity:

$$2\left(1+\cos\alpha \right)-\sin^2\alpha=4\cos^4\frac{\alpha}{2}$$

I tried to start from the right side rewritting it to $(2\cos^2(2\frac{\alpha}{4}))^2$ but it's not working.

Replacing $\sin^2(x)=1-\cos^2(x)$ gives
$$
\begin{align}
2(1+\cos(\alpha))-\sin^2(\alpha)
&=1+2\cos(\alpha)+\cos^2(\alpha)\\[6pt]
&=(1+\cos(\alpha))^2\\
&=\left(2\cos^2\left(\frac\alpha2\right)\right)^2
\end{align}
$$
using the identity $\cos(\alpha)=2\cos^2\left(\frac\alpha2\right)-1$.

begin by writing sin a = 2sin a/2 cos a/2 in left side ,easy by factoring you will have the answer