Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Example of a normal extension. Can you give an example of a Normal extension which is not a splitting field of some polynomial.? I know that splitting field of a polynomial is always a normal extension but i am looking for the converse. I am sure that the normal extension will have to be infinite.
|
I guess it depends on how one defines "normal" for infinite extensions. Let $$K={\bf Q}(\sqrt2,\sqrt3,\sqrt5,\sqrt7,\dots)$$ Then $K$ is normal over the rationals, in the sense that any polynomial irreducible over the rationals with a zero in $K$ splits into linear factors over $K$. But it's not a splitting field, in that there is no single polynomial $f$ such that $K$ is obtained from the rationals by adjoining the roots of $f$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/993274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Commuting matrices and exponential function Let $A,B$ be $n\times n$ commuting matrices, that is $AB=BA$.
I also know that $\exp(Bt)=X(t)X(0)^{-1}$ where $X$ is the fundamental matrix function.
How can I show that $A\exp(Bt)=\exp(Bt)A$?
|
For any $n$,
$$A\cdot(I+Bt+\frac12B^2t^2+\frac16B^3t^3+\ldots +\frac1{n!}B^nt^n)=(I+Bt+\frac12B^2t^2+\frac16B^3t^3+\ldots +\frac1{n!}B^nt^n)\cdot A $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/993510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to express $z^8 − 1$ as the product of two linear factors and three quadratic factors
Verify
$$(z-e^{i \theta} ) (z - e^{-i \theta} ) ≡ z^2 - 2\cos \theta + 1$$
Hence express $z^8 − 1$ as the product of two linear factors and three quadratic factors, where all
coefficients are real and expressed in a non-trigonometric form.
For the first part I just expanded the LHS and showed its equal to RHS.
Roots of $z^8-1$:
$$z= e^{i \frac{\pi k}{4}}$$
Where $k=0,1,2,3,4,5,6,7$
Ok since they want non-Trignometric so :
I know two roots, which are obvious:
$$z=1,-1$$
$$z^8-1=(z-1)(z+1)(z^6+z^5+z^4+z^3+z^2+z+1)$$
They want two linear factors which I believe I have found: $(z+1)$ and $(z-1)$
How do I make
$$(z^6+z^5+z^4+z^3+z^2+z+1)$$
to three quadratic factors? And they mentioned Hence , so I have to use the identity I verified. Please help.
|
Recall the two versions of the fundamental theorem of algebra:
*
*Any complex polynomial can be factored into a product of linear terms
*Any real polynomial can be factored into a product of linear and quadratic terms (and the quadratic terms have no real roots)
In this case, we can find the factorization as a complex polynomial into linear terms easily; the trick then to recover the factorization as a real polynomial is to multiply the linear terms together appropriately to convert pairs of complex linear terms into real quadratic terms.
In particular, you want to group the conjugate roots together, and multiply them as described by the identity you list.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/993682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
to find the smallest and largest number of equivalence relation in a set Let $s$ be a set of $n$ elements. The number of ordered pairs in the largest and smallest equivalence relation on set $s$ are $n^2$ and $n$.
I am able to understand the largest set of equivalence relation, but in case of smallest set of equivalence relation it could be an empty set..so according to me it is 0.
Am I missing something?
|
$S$ is a set of $n$ elements.
$S$ is non-empty and we know, every non-empty set always contains the identity relation.
So , It can't be empty relation , I suppose since empty relation is invalidated , that explains why smallest equivalence relation on set $S$ is $n$.
i.e Empty relation is transitive and symmetric on every set but not equivalence.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/993781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
How to find the sum of the series $\sum_{k=2}^\infty \frac{1}{k^2-1}$? I have this problem :
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}$$
My solution
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\frac{1}{3})+(\frac{1}{6}-\frac{1}{4})+...]$$
I think that the sum should be $\frac{1}{2}$ since the limit of :
$$-\frac{1}{2}[-1+\frac{1}{k}+...+\frac{1}{n}] = \frac{1}{2}$$
But that wrong. Any ideas?
|
Let $S_n=\displaystyle\sum_{k=1}^n\frac{1}{k^2-1}=\sum_{k=1}^n\frac{1}{2}\left[\frac{1}{k-1}-\frac{1}{k+1}\right]$
$\displaystyle\hspace{.2 in} =\frac{1}{2}\left[\bigg(\frac{1}{1}-\frac{1}{3}\bigg)+\bigg(\frac{1}{2}-\frac{1}{4}\bigg)+\bigg(\frac{1}{3}-\frac{1}{5}\bigg)+\cdots+\bigg(\frac{1}{n-2}-\frac{1}{n}\bigg)+\bigg(\frac{1}{n-1}-\frac{1}{n+1}\bigg)\right]$
$\hspace{.2 in}\displaystyle=\frac{1}{2}\left[1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right]$,
so $\displaystyle S=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{1}{2}\left[1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right]=\frac{3}{4}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/993890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
$ (1+\sin{x})^{\cos{x}} + (1+\cos{x})^{\sin{x}} > 3x $ How do I show that, for $ 0 < x < \dfrac{\pi}{4} $ (first quadrant), the inequality $ (1+\sin{x})^{\cos{x}} + (1+\cos{x})^{\sin{x}} > 3x $ is valid?
I've tried Bernoulli's, but it took me to a false inequality (though all restrictions were respected). I actually thought of using Jensen's, but I don't know where to begin.
|
Use:
$$\cos x > 1/\sqrt{2},\quad \sin x > x - x^3/6>0$$
So that we have:
$$(1+\sin{x})^{\cos{x}} + (1+\cos{x})^{\sin{x}} >
1+\left(1+\frac{1}{\sqrt{2}}\right)^{x-x^3/6}$$
Now, we can use the series expansion of $(1+a)^x$:
$$(1+a)^x = 1+ \log(1+a)x+\frac{1}{2}\log^2(1+a) x^2 + \ldots =
\sum_{n=0}^\infty \frac{1}{n!}\log^n(1+a) x^n$$
Since all members of the series are positive for $a>0$, we can write:
$$1+\left(1+\frac{1}{\sqrt{2}}\right)^{x-x^3/6} >
2+\log(1+1/\sqrt2)x>\frac{x}{2}+2>3x$$
With the last being true since $\pi/4<4/5$, the point where $3x=2+x/2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Calculation of polynomial in the finite field I'm trying to understand the McEliece cryptosystem and I'm looking to this paper http://www.mif.vu.lt/~skersys/vsd/crypto_on_codes/goppamceliece.pdf
On page 26 they are calculating syndrome and somehow they got $1 + a^{10}z$ from $
(a^8+ a^4+ a^{10}) + (a^7+ a^{11} + a)z$
Could somebody please explain me how is that possible?
|
At the start of the section on page 25 there's a reference to Section 2.6.2. In Section 2.6.2, it's stated that $a$ (actually $\alpha$) is a primitive element satisfying $a^4+a+1 = 0$. So $a^4 = -a-1 = a+1$ (the final equality is because $-1 = 1 \mod{2}$).
Substitute $a^4 = a+1$ into $(a^8+a^4+a^{10})+(a^7+a^{11}+a)z$ and reduce. You'll find that it's equal to $1+(a^2+a+1)z$. Note that $a^{10}=a^2+a+1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Least possible number of squares with odd side length An $n\times(n+3)$ rectangular grid ($n>10$) is cut into some squares, with all cuts being along the grid lines. What is the least possible number of squares with odd side length?
[Source: Russian competition problem]
|
OLD AND SURELY WRONG INTERPRETATION OF THE PROBLEM
If I understood correctly the problem ask for the greater common odd divisor of $n$ and $n+3$. But the parity of $n$ and $n+3$ is different, because if $n$ is even and you add 3 then $n+3$ is odd, and viceversa.
So then the problem is reduced to search the GCD of $n$ and $n+3$. If GCD divides $n$ then
$$(GCD\mid n)\land (GCD\mid n+3)\implies GCD\mid 3\implies GCD(n,n+3)\in\{1,3\}$$
Then we have that the less number of squares (LNS) will be
*
*if $3\mid n$ then $\mathrm {LNS}=\frac n3*(\frac n3+1)=\frac{n^2}{9}+\frac n3$
*if $3\not\mid n$ then $\mathrm {LNS}=n*(n+3)=n^2+3n$
Cause $n>10$ we will compare the minimum n divisible by 3 ($n=12$) and the absolute minimum $n=11$:
*
*$\mathrm {LNS}(11)=11*14=154$
*$\mathrm {LNS}(12)=4*5=20$
So the minimum number of squares for a rectangle $n\times(n+3)$ where $n>10$ are $20$, and it happen when $n=12$.
NEW BEAUTIFUL AND FRESH INTERPRETATION OF THE PROBLEM ;)
Maybe the interpretation of the question wasnt correct because I see the tag "combinatorics". At the light of this information the problem may ask by the way to cut the rectangle into pieces that are squares with odd length.
If, from my previous interpretation, we know that $GCD(n,n+3)\in\{1,3\}$ then we know that
*
*we can cut the rectangle in a big square $3k\times 3k; 2\not\mid k$ plus somes $3\times 3$ or $1\times 1$ squares if $3\mid n$
*or we can cut the rectangle in a big square $n\times n$ if $2\not\mid n$ and somes $1\times 1$ squares if $3\not\mid n$
*or we can cut the rectangle in a big square $(n-1)\times(n-1)$ if $2\mid n$ and somes $1\times 1$ squares if $3\not\mid n$
It is obvious that the optimal solution is related to some n divisible by 3:
*
*for $n=12$ we can divide in a $11\times11$ square plus 4 $3\times3$ squares, plus $25$ $1\times1$ squares, being a total of 30 squares.
*for $n=13$ we can divide in a $13\times 13$ square plus 4 $3\times3$ squares plus 3 $1\times1$ squares, being a total of 8 squares
...
Analysis:
*
*if $2\mid n$ then you cant cut a square $n\times n$ because $n$ is even, so you will have a lot of tiny squares of $1\times 1$ to cover one line of, at least, length $n$, i.e, you will have at least $n+1$ squares.
*if $2\not\mid n$ and $3\not\mid n$ then $GCD(n,n+3)=1$, how I prove above, then you can cut a big square $n\times n$ because n is odd, and after cut the rest that is a rectangle $n\times4$ where, remember, $3\not\mid n$ so you will have some rest of little squares of $1\times1$, maybe 3 or 6 depending of $n \pmod 3$ plus a entire line of $n+1$. So you will had a minimum of $n+8$ squares.
*if $3\mid n$ and $2\not\mid n$ then you can cut a big square $n\times n$, because n is odd, plus somes $3\times 3$ squares because n and $n+3$ are divisible by 3. You will had here the optimal solution because there isnt squares of $1\times 1$. The minimum number of squares will be $4$ when $n=9$, but $n>10$ so the minimum will be $6$ squares.
So the solution of the problem will be the minimum n that are odd and divisible by 3. If $n>10$ this number is 15.
For $n=15$ we will have a big square of $15\times 15$ and 5 squares of $3\times 3$, being a total of 6 squares. Any other n cannot be cut in less number of squares with side of length odd.
P.S.: solutions with less or equal numbers of squares exist but for $n<10$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
}
|
What's a good motivating example for the concept of a slice category? What nice example can one give a beginner to really motivate the idea of a slice category, before they've met the more general notion of a comma category?
There's the toy example of a poset category with the slices as principal ideals -- but that doesn't exactly elicit the thought "Yep! That's pretty interesting ..."
|
If $X$ is a set, the slice category $\textsf{Set}/X$ can be thought of as the category of $X$-indexed collections of sets, where an object $f:Y\to X$ corresponds to the $X$-indexed collection of fibers $\{Y_x = f^{-1}\{x\}\mid x\in X\}$, and a morphism from $f:Y\to X$ to $g:Z \to X$ corresponds to a map of sets $Y_x\to Z_x$ for each $x\in X$.
This leads nicely into the observation that the fiber product $Y\times_X Z$ is just the product in the slice category, and in the category of sets it corresponds to taking the product on each fiber: $\{Y_x\times Z_x\mid x\in X\}$.
Moreover, this intuition about the meaning of the slice category is useful in other contexts, e.g. in algebraic geometry thinking of a map of schemes $X\to S$ as an algebraic family indexed by the base scheme $S$. Or in model theory (since you're a logician), thinking of a definable map of definable sets $X\to Y$ as a definable family of definable sets indexed by $Y$. Up to definable isomorphism, this can be arranged to be a projection map, so when we project the set $\phi(M)$ defined by $\phi(\bar{x},\bar{y})$ onto the $\bar{y}$ coordinates, we get the family $\{\phi(M,\bar{b})\mid \bar{b}\in \exists \bar{x}\,\phi(\bar{x},M)\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 0
}
|
Write $\sum_{k=1}^nk\sin(kx)^2$ in closed form $\underline{Given:}$
Write in closed form $$\sum_{k=1}^nk\sin(kx)^2$$
using the fact that $$\sum_{k=1}^nku^k=\frac u{(1-u)^2}[(n)u^{n+1}(n+1)u^n+1]$$
$\underline{My\ Work:}$
I substituted $\sin(kx)^2$ for $\frac{1-\cos(2kx)}2$ which let me rewrite the summation like this: $$\frac 12\bigg[\sum_{k=1}^nk-\sum_{k=1}^nk\cos(2kx)\bigg]$$
I could rearrange the summation again using the fact that $\cos(2kx)=\frac {e^{i2xk}+e^{-i2xk}}2$
$$\frac 14\bigg[n(n+1)-\sum_{k=1}^nk({e^{i2x}})^k-\sum_{k=1}^nk({e^{i2x}})^k\bigg]$$
Then I re-wrote the summation again using the substitution that was given to me. Using $u=e^{i2x}$ $$\frac 14\bigg[n(n+1)-\frac{e^{i2x}}{(1-e^{i2x})^2}\bigg[n(e^{i2x})^{n+1}(n+1)(e^{i2x})^n+1\bigg]-\frac{e^{-i2x}}{(1-e^{-i2x})^2}\bigg[n(e^{-i2x})^{n+1}(n+1)(e^{-i2x})^n+1\bigg]$$
Which if I multiply the $\frac{e^{-i2x}}{(1-e^{-i2x})^2}$ by $\frac {e^{4ix}}{e^{4ix}}$ it changes into $\frac{e^{i2x}}{(1-e^{i2x})^2}$ which allows me to factor it out.
Now I have: $$\frac 14\bigg[n(n+1)-\frac{e^{i2x}}{(1-e^{i2x})^2}\bigg[\big[n(e^{i2x})^{n+1}(n+1)(e^{i2x})^n+1\big]-\big[n(e^{-i2x})^{n+1}(n+1)(e^{-i2x})^n+1\big]\bigg]$$
Now this is where I'm starting to get stuck. I can't seem to organize this nicely to get back into sin's and cos's. Any help would be appreciated! Thanks!
|
We have $\dfrac{e^{2ix}}{(1-e^{2ix})^2}=\dfrac{-1}{4sin^2(x)}$ and $n(e^{2ix})^{n+1}(n+1)(e^{2ix})^n-n(e^{-2ix})^{n+1}(n+1)(e^{-2ix})^n=2in(n+1)sin(4n+2)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
finding numbers to make both sides equal Call a triple-x number an integer $k$ such that $k=x(x+1)(x+2)$ where $x \in Z$. How many triple-x numbers are there between 0 and 100,000?
I thought by doing $8!$ and $9!$ would work to see how many combinations there would be. I am not sure how to solve this problem. Can someone show me how to solve this?
|
Well, there's one with $x = 0$, one with $x = 1$, and keep going until you get something bigger than $100,000$. Equivalently, you're solving the inequality $x(x+1)(x+2) < 100,000$, which can be done in various ways (though there's no really clean way of doing it).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Eulerian circuit with no isolated vertex is connected This is my first question (ever), and I am pretty new to math. So I ask for patience and understanding in advance.
So this is the proof I came up with:
Consider $G = (V,E). $
By definition of Eulerian circuit,
$$ \exists\ W: v_0, v_1,\cdots, v_k=v_o $$
And there is no isolated vertex $$ V = \{v_0, v_1,\cdots,v_k \}$$
Take $v_i, v_j \in V $. Consider $ j = i + k $ where $k \in \mathbb{N} $.
Then, $ v_j, v_{j+1}, \cdots, v_{j+k} = v_i $ is a walk.
Hence, $\exists$ a path between $v_i$ and $v_j$. Hence connected.
|
Your proof is generally fine; I can follow your logic. If we really want to be nitpicky:
*
*You should probably mention that we are assuming that $G = (V,E)$ has an Eulerian circuit and has no isolated vertices.
*You should say: "Since $G$ has no isolated vertices, we know that $V = \{v_0, \ldots, v_k\}$."
*When you take $v_i, v_j \in V$, you should say that they are distinct and that we are assuming, without loss of generality, that $i < j$. This guarantees that $k$ really is a natural number.
*When you explicitly specify the walk between $v_i$ and $v_j$, you seem to have mixed up $i$ with $j$.
*It's usually better to use complete English sentences in proofs. Avoid unnecessary logic symbols. Finish your proof with something like: "Hence, there exists a path between $v_i$ and $v_j$ so that $G$ is connected, as desired."
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Tricky Substitution to get AM-GM inequality So, I'm reading the literature to find different proofs of the AM-GM inequality, the following proof quite hit me, and I don't seem to understand at all.
The proof is as follows:
For any positive numbers: $a_1,a_2,...a_n$. We have:
$$
\dfrac{a_1+a_2+..+a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 .... \cdot a_n}
$$
Replacing $a_k$ with $a_k \sqrt[n]{a_1 \cdot a_2 .... \cdot a_n}$ for $1\leq k \leq n$ we have $a_1 \cdot a_2 ... \cdot a_n = 1$, so that is enough to prove that $ {a_1+a_2+..+a_n} \geq n$.
My question is: why $a_1 \cdot a_2 ... \cdot a_n = 1$ ? And, in general, for any sequence of positive numbers $ a_k \neq a_k \sqrt[n]{a_1 \cdot a_2 .... \cdot a_n}$?
I would really appreciate some light in this matter.
P.s I'm annexing a photo of the section.
|
I think they mean replacing $a_k$ by $\dfrac{a_k}{\sqrt[n]{a_1a_2\cdots a_n}}$. Another way to see that is to divide both side of the original AM-GM inequality by $\sqrt[n]{a_1a_2\cdots a_n}$ to get an equivalent inequality
$$\frac1n\left(\frac{a_1}{\sqrt[n]{a_1a_2\cdots a_n}}+\frac{a_2}{\sqrt[n]{a_1a_2\cdots a_n}}+\cdots+\frac{a_n}{\sqrt[n]{a_1a_2\cdots a_n}}\right)\ge 1,$$
which is AM-GM for $\dfrac{a_k}{\sqrt[n]{a_1a_2\cdots a_n}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$ How to find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$
Let $y^2-xy-2x^2 =0...(1)$ and $y^2=x-2...(2)$
In equation (1) coefficient of $x^2 =-2; y^2=1, 2xy =\frac{-1}{2}$
We know that a second degree equation where $ab-h^2 =0 $ represent a parabola,
$ab-h^2>1$ represent an ellipse
$ab-h^2 <0$ represent a hyperbola
Here $ab-h^2 <0$ therefore equation (1) represents hyperbola and equation (2) is parabola. ( where a,b, h are coefficients of $x^2,y^2,xy$ respectively.
Now how to get the minimum distance between the two curves please suggest .. thanks.
|
$$y^2= 2x^2+xy \iff (y-2x)(x+y)=0 \iff y = -x \text{ or } y = 2x$$
so that is a couple of lines.
Hence you want to find the shortest distance from $y=-x$ or $y=2x$ to the parabola $y^2=x-2$. Let this be denoted by the line segment with end points $(a, -a)$ or $(a, 2a)$ and $(b^2+2, b)$.
Then we need the min of $(a-b^2-2)^2+(a+b)^2$ or $(a-b^2-2)^2+(2a-b)^2$.
Case 1:
We have by Cauchy Schwarz
$$((a-b^2-2)^2+(a+b)^2)((-1)^2+1^2) \ge (-a+b^2+2+a+b)^2= ((b+\tfrac12)^2+\tfrac74)^2 \ge \frac{7^2}{4^2}$$
so we get a minimum distance of $\frac{7}{4\sqrt2}$.
Case 2:
We have again by Cauchy Schwarz
$$((a-b^2-2)^2+(2a-b)^2)((-2)^2+1^2) \ge (-2a+2b^2+4+2a-b)^2= (2(b-\tfrac14)^2+\tfrac{31}8)^2 \ge \frac{31^2}{8^2}$$
so we get a minimum distance of $\frac{31}{8\sqrt5}$.
As the second case happens to be higher, the shortest distance is $\dfrac7{4\sqrt2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
}
|
Different Versions of Fourier Series? What about Uniqueness? Let $f(x)$ be a function, then for its Fourier series
$$
f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))
$$
I found two different definitions (both yielding different series).
i) I found
\begin{align*}
a_n & = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx) d x \quad n \ge 0, \\
b_n & = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(nx) d x \quad n > 0
\end{align*}
and ii)
\begin{align*}
a_n & = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(n x) d x \\
b_n & = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(n x) d x.
\end{align*}
If I evaluate $f(x) = x(1-x)$ according to ii), then I get
$$
a_n = (-1)^n \frac{4\pi}{n^2} \quad \mbox{ and } \quad b_n = (-1)^{n+1} \frac{2}{n}
$$
when I use i) I get
$$
a_n = \frac{-1}{\pi^2 n^2} \quad \mbox{ and } \quad b_n = 0.
$$
Both are different, but I guess the Fourier series is unique, or are they both "different" Fourier series, can someone explain why it is not unique and why there are both definitions around (and what is the better one?) or what is their relation?
|
The function $f$ must be $2\pi$-periodic, otherwise the whole construction becomes meaningless (since the Fourier series is $2\pi$-periodic!)
For such a function $f$ the two expressions of $a_n$ and $b_n$ do coincide.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Is my understanding right on the divisiblity rule? For a given number and a divisor. If the prime factors of the divisor can divide a number,then can I say that the divisor will divide a number.
For example - 786
divide by 21
If I break 21 in the prime factors - 3 * 7.
So, if the number is divisible by 3 as well as 7 that means the number can be divided by 21. Here 786 is not divisible by 21 because 7 cannot divide it completely.
Similarly, 42 will be broken as - 3 * 7 *2 -- this means it will not be divisible again. Since 7 couldn't divide it.
Is it a right method?
|
Yes, that's right. It is a consequence of the fact that every integer can be written as a product of powers of primes in an essentially unique way (the only lack of uniqueness is in the order in which you write down the factors).
So in your example, $-786 = -2\cdot 3\cdot 131$ (and $131$ is prime), and $21 = 3\cdot 7$. So if $21$ divided $786$, the primes in its prime factorization would appear in the prime factorization of $786$ by uniqueness of factorization. But $7$ does not, so that $7$, and thus $21$, do not divide $-786$. And since $21\nmid 786$, it follows that $2\cdot 21 = 42$ does not divide $786$ either.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/994933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
What's the dimension of $\mathbb C$ as a vector space over $\mathbb{R}$? What's the dimension of $\mathbb C$ as a vector space over $\mathbb{R}$?
I think that the answer is $2$, because $\mathbb R^2 \cong \mathbb{C}$ (since all complex numbers can be mapped to $\mathbb R^2$ and vice versa).
Is this correct, or is it a trick question (in which case I'm wrong)?
|
Hint:
$\lbrace 1,i\rbrace$ is a basis.
As for what you have said, there is an isomorphism $$\frac{\mathbb{R}[x]}{\langle x^2+1\rangle}\simeq \mathbb{C}$$ where $\mathbb{R}[x]$ is the polynomial ring.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/995020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Entire function with vanishing derivatives? Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function.
And assume that at each point, one of it's derivatives vanishes.
What can you say about $f$?
A hint suggests that $f$ must be a polynomial.
|
we can also solve this without using baire category theorem..
we can use the fact that zeros of non zero analytic function is countable. Note that union of $A_{n}$ is $\mathbb{C}$ thus we must have some fix derivative identically 0. Now use power series and analytic continuation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/995105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
}
|
Prove statement about determinants. $A$ is a $3\times 3$ matrix over $\mathbb{R}$, I want to show that if $$\det(A + I_3)=\det(A+2I_3),$$ then $$2\det(A+I_3) + \det(A-I_3) + 6 = 3\det A.$$
Can you help me?
|
First show that you can assume $A$ diagonal. Then the question reduces to proving a basic identity between two polynomials.
More explicitely, you can write $\det (A+ x I_3) = \sigma_3 + x \sigma_2 + x^2 \sigma_1 + x^3$, where $\sigma_3=a_1 a_2 a_3$, $\sigma_2 = a_1 a_2 + a_1 a_3 + a_2 a_3$ and $\sigma_1 = a_1 + a_2 + a_3$, if the $a_i$ are the diagonal elements of $A$. The assumption gives you $\sigma_2 = -3 \sigma_1 - 7$, and the wanted equality follows readily.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/995191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
A necessary condition for a series of rational numbers to be irrational? Suppose $(q_{n})_{n\in\mathbb{Z}_{\gt 0}}$ is a decreasing sequence of positive rational numbers such that $Q:=\displaystyle{\sum_{n>0}q_{n}}$ is finite. Let's denote by $n_{i}$ and $d_{i}$ the numerator and the denominator of $q_{i}$ such that $(n_{i},d_{i})=1$. If $i\ne j\Rightarrow (n_{i}d_{i},n_{j}d_{j})=1$, is $Q$ necessarily irrational?
|
No. In fact, every positive real number $Q$ can be written this way. Here is the algorithm: pick $q_1$ to be any positive rational which lies in the interval $\left( \frac{Q}{2}, Q \right)$. Once $q_1, q_2, \dots q_n$ have been picked, pick $q_{n+1}$ to be any positive rational whose numerator and denominator are coprime to the numerators and denominators of $q_1, q_2, \dots q_n$ and which lies in the interval $\left( \frac{Q_{n+1}}{2}, Q_{n+1} \right)$, where
$$Q_{n+1} = Q - q_1 - q_2 - \dots - q_n.$$
The key point is that the set of rational numbers whose numerators and denominators do not contain a fixed finite set of primes is still dense in $\mathbb{R}$. To prove this it suffices to observe that the logarithms of the primes are linearly independent over $\mathbb{Q}$, so any two such logarithms generate a dense subgroup of $\mathbb{R}$: in fact this shows that the set of rational numbers whose numerators and denominators are divisible by two fixed (arbitrarily large) primes is still dense in $\mathbb{R}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/995372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Floating point arithmetic: $(x-2)^9$ This is taken from Trefethen and Bau, 13.3.
Why is there a difference in accuracy between evaluating near 2 the expression $(x-2)^9$ and this expression:
$$x^9 - 18x^8 + 144x^7 -672x^6 + 2016x^5 - 4032x^4 + 5376x^3 - 4608x^2 + 2304x - 512 $$
Where exactly is the problem?
Thanks.
|
$(x-2)$ is small by definition of $x$, $(x-2)^9$ is even much smaller but can be computed with small relative error.
The single terms of the expanded polynomial are much larger and therefore you will suffer from catastrophic cancellation (see e.g. Wiki or do a web search). Example:
$$x=2 + 10^{-2} \Longrightarrow (x-2)^9 = 10^{-18}.$$ This is below the machine epsilon for double! Now with your expanded polynom you have to compute
$-512 + 23.04 \pm \dots$.
And here is the actual computation with double and evaluation of the polynomial using Horner for $x=2.01:$
(x-2)^9 = 9.99999999999808E-019 poly(x) = -3.75166564481333E-012
As expected the polynomial result is completely wrong (in terms of relative error), but
note that even the relative error for the first is about $2\cdot 10^{-13},$ which is a result from the fact that $0.01$ cannot represented exactly as double.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/995489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
A unique operation on a set that makes it a group From Rotman's "Introduction to Group Theory":
Let $G$ be a group and let $X$ be a set having the same number of elements as $G$. If $f:G\rightarrow X$ is a bijection, there is a unique binary operation that can be defined on $X$ so that $X$ is a group and $f$ is an isomorphism.
Let $|G|=|X|=n$. Let the operation for $G$ be given by "$+$". Let us denote the operation for the set $X$ as "$\ast$" (we don't know if it's a group yet, though).
We can construct a bijection $f:G\rightarrow X$. Thus, $x=f(a)$ for every $x\in X$ and $a\in G$. We can see that $f(a+b)\in X\;\;\forall a,b\in G$.
My question is this: What should be my first step in approaching this problem? I was thinking of analyzing the equation:
$$f(a+b)=f(a)\ast f(b)$$
but can't think of anything useful to do with this.
I'm looking for hints on how to get started, not full answers please.
Thanks
|
this is really just an exercise in abstraction. in essence we use the bijection to pull back the group structure on $G$ into the hitherto unstructured set $X$. we may represent the image of $x$ in $G$ as $x_G$ and if the inverse of this map is $\phi$ we may define an operation $\ast$ on $X \times X$ by:
$$
x \ast y = \phi(x_G y_G) \\
x^{-1} = \phi((x_G)^{-1})
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/995580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Product of inner products Is product of innerproduct again a inner product of two vectors?
For example - Is $ (< u,v >)(< x,y >) = < m,n > $? And if yes is m and n unique and how do we calculate those?
|
Of course the product, which is after all a scalar (call it $S$) , can be written as an inner product of two vectors, e.g., $\vec{n} = (S, 0, 0, \ldots), \vec{m} = (1, 0, 0, \ldots)$. But the decomposition of that scalar is not unique, and in fact, you can find an infinite number of correct $\vec{m} $ cevtors for any specified non-zero $\vec{n}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/995662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
Holonomy representation: is it actually a class of representations? In D. Joyce's book "Riemannian Holonomy Groups and Calibrated Geometry" (2007) the author writes that if $\nabla$ is a connection on a vector bundle $E$ (over a connected base) with the fibre $\mathbb R^k$ then the holonomy group $\mathrm{Hol}(\nabla)$ comes equipped with a natural representation on $\mathbb R^k$, or equivalently, with an embedding of $\mathrm{Hol}(\nabla)$ in $\mathrm{GL}(k,\mathbb R)$. He calls this representation the holonomy representation.
But actually $\mathrm{Hol}(\nabla)$ is defined as a subgroup of $\mathrm{GL}(k,\mathbb R)$ up to a conjugation. Does it mean that actually the holonomy group comes equipped with a family of representations on $\mathbb R^k$ and "the holonomy representation" is actually a class of representations?
On the other hand, it's clear that the holonomy group $\mathrm{Hol}_x(\nabla)$ with fixed basepoint comes equipped with a natural representation on $E_x$, the fibre of $E$ over $x$.
|
Yes. The holonomy representation is an isomorphism class of representations.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/995756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Write an equation as a single power(Grade 11 Math, Function) $$\frac{10^{-4/5} \cdot 10^{1/15}}{10^{2/3}}$$
The answer is $10^{-7/5}$, which seems impossible to me. I get:
$10^{-4/5} \cdot 10^{-11/15}$. I see where the numerator $7$ comes from but the denominator is being a pest, and won't let me do anything because I have to make them equal to add them.
|
It looks like you made a sign error while combining exponents.
$$\frac1{15} - \frac23 = \frac1{15} - \frac{10}{15} = -\frac9{15}.$$
You seem to have gotten $-\frac{11}{15}$ when you should have gotten $-\frac9{15},$ perhaps by flipping the sign of $\frac1{15}.$
Of course $-\frac9{15} = -\frac35,$ and the last step is easy.
The fact that $4 - 11 = -7$ is a complete red herring.
If the problem were just a little different so the denominators happened to be equal, for example if they were both $5,$ you would have
$$ -\frac45 - \frac{11}5 = -\frac{15}5 = -3.$$
So it seems you have a tendency to make sign errors while adding or subtracting. Now that you know this, you may be able to take steps to correct it.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/995843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
A compact connected solvable Lie group is a torus I am looking for the proof of the following statement.
A compact connected solvable Lie group of dimension $n\geq 1$ is a torus, i.e., it isomorphic to the product of $n$ copies of $S^1$.
A search with Google revealed that pages 51-52 of "Lie Groups and Lie Algebras III: Structure of Lie Groups and Lie Algebras" (A.L. Onishchik) deal with this problem but the proofs are not very detailed and I'm not an expert.
Thanks,
Gis
|
@Gis: I would like to add to Tim's answer the following:
*
*The lie algebra $\mathfrak{g}$ admits a bi-invariant scalar product $<\,,>$, i.e. the adjoint action $\mathrm{ad}_x$ is antisymetric $\forall x\in\mathfrak{g}$ (since $G$ is compact, it admits a bi-invariant riemannian metric).
*$\mathfrak{g}=\mathfrak{Z}(\mathfrak{g})\oplus D(\mathfrak{g})$, where $\mathfrak{Z}(\mathfrak{g})$ is the center and $D(\mathfrak{g})=\mathfrak{Z}(\mathfrak{g})^\perp$ is the derived ideal.
*The restriction of the Killing form on $D(\mathfrak{g})$ is definite negative https://math.stackexchange.com/a/59193/14409. In particular it's semi-simple.
*Since $\mathfrak{g}$ is solvable, its semi-simple part $D(\mathfrak{g})$ must be zero. So $\mathfrak{g}$ is abelian and also $G$.
*The exponontial map $\exp : \mathfrak{g}\to G$ is surjective and since it's a local diffeomorphism its kernel is discrete (isomorphic to $\mathbb{Z}^n$) so $G\simeq \mathfrak{g}/\mathbb{Z}^n\simeq\mathbb{R}^n/\mathbb{Z}^n$.
P.S: Sorry for my bad English!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/995911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Entire functions satisfying $|f(z)|\geq |\sin z|^{10} $ for all $z\in\mathbb{C}$ How do I find all the entire functions $f(z)$ such that $|f(z)|\geq |\sin z|^{10} $ for all $z\in\mathbb{C}$
Can an entire function have essential singularity? My intuition says that "no".
But I am not sure.
Can anyone help me ?
|
If $f$ has no zeros, then $g(z)=\dfrac{(\sin z)^{10}}{f(z)}$ is entire and bounded and so is a constant.
Therefore, $f$ is a constant multiple of $(\sin z)^{10}$.
If $f$ has zeros, then they must be a subset of the zeros of $\sin z$, which are $n\pi$ for $n\in\mathbb Z$.
If $f$ has finitely many zeros, you can remove them from $\sin z$ and conclude that $g$ is constant as before.
If $f$ has infinitely many zeros, perhaps more can be said about $f$ by considering the product expansion:
$$
\begin{align}
\sin z = z \prod_{n = 1}^\infty \Bigl( 1- \frac{z^2}{n^2\pi^2} \Bigr).
\end{align}
$$
The Weierstrass factorization theorem may be useful but perhaps it's too advanced or is a sledgehammer for this problem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Proof Using cartesian products Suppose that $A$, $B$, and $C$ are sets. Prove that $(A\cap B)\times C =(A\times C)\cap(B\times C)$. Prove the statement both ways or use only if and only if statements.
|
The two sets $(A \cap B) \times C$ and $(A \times C) \cap (B \times C)$ are equal iff every element of one set is element of the other set and vice versa.
Consider $(A \times C) \cap (B \times C)$ first:
$$ a \in (A \times C) \cap (B \times C) \leftrightarrow a \in (A \times C) \wedge a \in (B \times C). $$
From the definition of the Cartesian product we have
$$ a \in (A \times C) \leftrightarrow (\exists x \in A)(\exists y \in C) \: a = (x,y). $$
Thus
$$ a \in (A \times C) \wedge a \in (B \times C) \leftrightarrow (\exists (x \in A) \wedge (x \in B) )(\exists y \in C) \: a = (x,y). $$
Where the first quantifier resolves to
$$ (\exists (x \in A) \wedge (x \in B) ) \leftrightarrow (\exists x \in A \cap B). $$
Again, from the definition of the Cartesian product,
$$ (\exists x \in A \cap B)(\exists y \in C) \: a = (x,y) \leftrightarrow a \in (A \cap B) \times C, $$
which yields
$$ (\forall a) \: a \in (A \times C) \cap (B \times C) \leftrightarrow a \in (A \cap B) \times C.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
How do you solve a system of equation with a variable as an exponent? So, I've been doing some random math on my own and I ran into a problem. I was met with the system of equations of $$x=3^y$$ $$x+1=5^y$$ How can I solve this algebraically without using the graphs?
|
Eliminating $x$ from the first equation and replacing in the second let you with the equation $$3^y+1=5^y$$ which is highly nonlinear (and you look for the intersection of these two functions).
You can make the problem nicer taking logarithms and get $$y\log(5)=\log(3^y+1)$$ which is now the intersection of a curve and a straight line.
For solving the equation $$f(y)=y\log(5)-\log(3^y+1)=0$$ let us use Newton method which, starting from a "reasonable" guess $y_0$, will update it according to $$y_{n+1}=y_n-\frac{f(y_n)}{f'(y_n)}$$ $$f'(y)=\log (5)-\frac{3^y \log (3)}{3^y+1}$$ By inspection, you know that the solution for $y$ is between $0$ and $1$ since $f(0)=-\log (2)<0$ and $f(1)=\log \left(\frac{5}{4}\right)>0$. So, let us start iterating at $y_0=1$; the successive iterates of Newton method are then $0.7159139357$, $0.7271409790$, $0.7271601514$ which is the solution for ten significant figures.
Now, compute $x$ from $x=3^y$ and get $x=2.223020992$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Periodicity over the prime indices of a multiplicative sequence implies periodicity? I have a real sequence $(a_p)$ indexed by the prime numbers which takes values -1, 0, or 1, having the property that $a_p=a_q$ whenever $p\equiv q \pmod m$, where $m$ is a fixed integer $>2$.
I'm wondering if by extending multiplicatively the sequence over the natural numbers,
that is, by imposing $a_{rs}=a_r a_s$ for every natural numbers $r$ and $s$, I obtain a periodic sequence, i.e., $(a_n)$ such that $a_i=a_j$ whenever $i\equiv j \pmod m$.
What kind of sequence should I start with for the extended sequence be periodic modulo $m$?
Do that conditions suffice?
Thanks in advance for any ideas or suggestions.
|
Look for sequences with values in $\{-1,0,1\}$.
Assume that $|a_p|>1$ for some integer. Then the subsequence $a_{p^n}$ tends to $\pm\infty$, which is not possible for a periodic sequence.
You may want to read about characters.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
semantics(truth) vs formal system? my first question is can we just define semantics in logic and not define a formal system ?
why do we need a formal system to prove a proposition when for example we know the proposition is true ?
e.g. ( A ^ (A->B) ) -> B if A is true and A->B is true then B is true
and this can also be shown by truth table. so why we use a formal system to show that the argument is valid when we can just use the semantics to show that . (by construction of truth tables)
since the formulas and the laws of semantics are well defined it is not possible to get to any inconsistencies (in the sense of semantics !)
i think the counterpart to my question is answered in Formal System and Formal Logical System
but what about the other way around ?
my secound question is: ShyPerson when answering the above question mentioned that we need some meaning for our propositional formulas thus we define the semantics. can not we just use the same meaning of the connectives in the human language to describe our propositions .
meaning we describe connectives just like the human language ?
|
In propositional logic we can "show" validity by truth table.
Truth table supply an algorithm to compute the truth value of every propositional formula;
in particular, truth tables can be used to tell whether a propositional expression is true for all legitimate input values, that is, logically valid [also called : a tautology].
For first-order logic, we have not a similar algorithm.
F-o logic is undecidable :
unlike propositional logic, first-order logic is undecidable (although semidecidable), provided that the language has at least one predicate of arity at least $2$ (other than equality). This means that there is no decision procedure that determines whether arbitrary formulas are logically valid.
This is the main reason why we need a deductive calculus for f-o logic :
because to show that the argument is valid when we cannot just use the semantics to show that .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
}
|
Indefinite Integral of a Function multiplied by Heaviside I want to do the following integral:
$$\int^\ x*H(x-a) dx $$
In mathematica I get that $$\int^\ x*H(x-a) dx = \frac{1}2*(x-a)*(x+a)*H(x-a)$$
where H(x-a) Is the heaviside function.
But by hand I can't find the same result.What I'm trying to do is that:
$$\int^\ x*H(x-a) dx = \frac{1}2*x^2*H(x-a)+\frac{1}2\int^\ x^2*Dirac(x-a) dx
= \frac{1}2*x^2*H(x-a)+\frac{a^2}2 $$
That is in no way near the answer the software gave me. How can I compute the initial integral by hand to match the software?
I Thank you very much for your help
|
Actually, your solution is not that far off if you keep in mind that the indefinite integral is only determined up to the addition of a constant. However, here is my way to compute the integral:
First off, we need to clarify: The Heaviside function I'll use is defined by
$$H(x) = \left\lbrace \begin{array}{cl} 1 & , ~ x \geq 0 \\ 0 & , ~ x < 0 \end{array}\right.$$
An indefinite integral is just a function that can be written as a definite integral with a variable as the upper bound:
$$\left( \int x \cdot H(x-a) ~ dx \right)(x_0) = \int\limits_a^{x_0} x \cdot H(x-a) ~ dx$$
Now we just have to divide the upper term into two cases:
If $x_0 \geq a$ then $H(x-a) = 1$ for all $x \in [a,x_0]$ and therefore the (right) intgral equals
$$\int\limits_a^{x_0} x~dx = \frac{1}{2}(x_0)^2 - \frac{1}{2} a^2 = \frac{1}{2} (x_0 - a)(x_0+a)$$
For $x_0 <a$ we have $H(x-a) = 0$ for all $x \in [x_0, a]$ and therefore the integral equals $0$.
So if we merge both cases one may write
$$\int\limits_a^{x_0} x \cdot H(x-a) ~ dx = \frac{1}{2} (x_0-a)(x_0 + a)H(x_0 - a)$$
which is the solution you obtained using Mathematica.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
How to find a bound for these (simple) integrals With help of $\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt{\pi}}{2}$ and $\int_{0}^{\infty} e^{-x} dx =1$, I would like to know how to derive the following bounds:
$$\int_{0}^m 4e^{-\frac{t^2}{8m}}dt \leq 2\sqrt{8\pi m}$$
and
$$\int_{m}^{\infty} 4 e^{-\frac{t}{8}}dt \leq 32.$$
How does one derive these?
So somehow one would do this with change of variables? As I am not very good at such stuff, I would really appreciate any help.
Thank you very much.
|
Hint
For the first, change variable $$\frac{t^2}{8m}=x^2$$ so $t=x \sqrt{8m}$, $dt=\sqrt{8m} dx$. So $$I=\int 4e^{-\frac{t^2}{8m}}dt=8 \sqrt{2m} \int e^{-x^2} dx$$
Do something similar for the second integral.
I am sure that you can take from here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Natural Logarithm - solve the equation I am having problems understanding how to solve $e^{4x}+4e^{2x}-21 = 0$.
|
Try temporarily setting $u=e^{2x}$ so it becomes $u^2+4u-21=0.$ Then solve the quadratic, discarding any negative solutions since $u=e^{2x}>0.$ After that use the $\ln$ function to get $x$ back.
For example if it came out one value of $u$ was $u=5$ then from $e^{2x}=5$ comes $2x=\ln 5$ and then $x=(1/2) \ln 5.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Why is $|x| \neq x$? I was just thinking about the absolute value function. Why does the following equality not hold?
$$|x| = \sqrt{x^2} = (x^{2})^{1/2} = x^1 = x$$
After all, there are clearly some values of $|x|$ that are not equal to $x$. Is this a domain/range issue?
|
It should be:
$$\sqrt{x^2} = (|x|^{2})^{1/2} = |x|^1 = |x|$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Determine whether $\phi$ is an automorphism Let $H$ be a proper subgroup of $G$ and let $\psi$ be an automorphism of $H$ other than the identity mapping. Define a mapping $\phi:G\rightarrow G$ by
$\phi(x)=$\begin{cases}\psi(x) \text{ if } x\in H\\x \text{ if } x\not\in H\end{cases}
Is $\phi(x)$ an automorphism?
|
Take $g \in G\backslash H$. Every element in $H$ is a product of two elements from $G \backslash H$.
$$h = g^{-1} \cdot gh$$.
If the morphism $\phi$ conincides with $\mathbb{1}_G$ on $G \backslash H$ then it is $\mathbb{1}_G$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
3D rotation around arbitrary axis I have a 3D rotation matrix, R which is a combination of rotations around x-axis , y-axis and z-axis. I know how to calculate n(the arbitrary axis around which a point rotated about theta angle and this rotation is equal to rotating that point using the 3D matrix above) i.e. by finding the eigenvector of R corresponding to the eigenvalue equal to 1. How can I calculate theta, the angle to rotate about the arbitrary axis?
|
I'll describe a computational method, assuming the rotation has been translated to be around an axis through the origin.
If you know the axis of rotation $A=(a,b,c)$, then you can find a vector orthogonal to this one. For example, if $a \ne 0$, $b \ne 0$, then $V=(b,-a,0)$ is such a vector. Or, if one component is $0$--say, for example $c=0$, then $V=(0,0,1)$ is orthogonal. And that covers all cases.
A third vector in a right-handed triple is found using the vector cross product: $W=A\times V$. Normalize $V$ and $W$ to unit vectors $\hat{V}$, $\hat{W}$.
Now, feed $\hat{V}$ into your matrix and determine the output in terms of $\hat{V}$, $\hat{W}$ using dot products
$$
R\hat{V} = \{(R\hat{V})\cdot\hat{V}\}\hat{V}+\{(R\hat{V})\cdot\hat{W}\}\hat{W}=\alpha\hat{V}+\beta\hat{W},
$$
and compare to the action of a right-handed rotation about $A$ through an angle $\theta$:
$$
\hat{V} \mapsto \cos\theta \hat{V}+\sin\theta \hat{W}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/996913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Showing a certain subspace of Hilbert space is dense Let H be the Hilbert space of square-summable sequences of reals.
A few years ago I thought I had proved that the subspace Z of real sequences with only finitely many nonzero terms, such that they sum to zero, is dense in H.
(Since then I've seen this confirmed in Rudin's text, Functional Analysis, 2nd ed., but only as a teensy subquestion in a terminally hairy exercise.)
Now I can't quite reproduce my proof, so it may have been wrong. Can someone please tell me how to prove that.
(What I tried was taking an arbitrary point c = (c1,...,ck,...) of H, and defining the point d(N) in H as follows for N $\ge$ 1:
First set AN = (c1+...+cN) / N. Then set d(N)k = ck - AN for k $\le$ N, and set d(N)k = 0 for k > N.
Clearly, the element d(N) of H lies in the subspace Z defined above. The squared Hilbert norm of its difference with c is of form N(AN)2 + T(N), where T(N) is just the squared norm of the tail of c, and so goes to 0 as N$\to\infty$.
I'm left with the expression N(AN)2, which so far I haven't been able to prove $\to$ 0 as N$\to\infty$. I suspect this is true, and it works on all the examples I've tried so far.)
NOTE: I'm only interested in a down-and-dirty proof that doesn't invoke anything but simple inequalities.
Thanks for any help you can offer.
|
Given $x \in H$ and $\epsilon > 0$, there is $y \in H$ with $\|x - y\| < \epsilon/2$ such that only finitely many elements of $y$ are nonzero. Suppose the sum of those nonzero elements of $y$ is $s$. Consider $z$ obtained from
$y$ by changing $n$ of its elements from $0$ to $-s/n$. Thus $z \in Z$,
and $\|z - y\|^2 = n (s/n)^2 = s^2/n$. If $n$ is sufficiently large, this is
less than $(\epsilon/2)^2$. So $\|x - z\| < \epsilon$. We conclude that $Z$ is dense in $H$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Algebra (group theory) Prove, without using Cauchy’s Theorem, that any finite group $G$ of even order contains an element of order two. [Hint: Let $S = \{\,g ∈ G : g \ne g^{−1}\,\}$. Show that $S$ has even number of elements. Argue that every element not in $S$ and not equal to the identity element has order two].
My Solution:
Let $S = \{\,g ∈ G : g \ne g^{−1}\,\}$. If $g ∈ S$ then $g^{-1} ∈ S$. This implies that the number of elements in $S$ must be even. Since $g\ne g^{−1}$, we can eliminate all non-identity elements. Then the number of elements of the group is $2n+1$, where $n$ is the number of pairings and $1$ is the identity element $e$. But we stated that $S$ is even. Therefore, this is a contradiction and $G$ has an element of order $2$.
|
In principle, this is fine - almost. However, you end with "Therefore, this is a contradiction" - but you didn't even start by making an assumption that can be contradicted! Presumably (but you should have explicitly said so!) you wanted to start from a finite group $G$ of even order that does not have an element of order $2$. Also, a bit more carefulness at this step would be in orde: "If $g\in S$ then $g^{-1}\in S$. This implies that the number of elements in $S$m must be even." Are you aware that there exists a subset $S'$ of $G$ that also has the property "If $g\in S'$ then $g^{-1}\in S'$", but has en odd number of elements?
By the way, you could as well keep this up as a direct proof: Once you show that $|S|$ is even, conclude that $|G\setminus S|$ is also even, contains $e$ and hence at least one second element, $a$ say, and then argue that the order of $a$ is $2$ as was to be shown.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
An equation for the third powers of the roots of a given quadradic polynomial The roots of the equation $3x^2-4x+1=0$ are $\alpha$ and $\beta$. Find the equation with integer coefficients that has roots $\alpha^3$ and $\beta^3$.
GIVEN
SR: $\alpha + \beta = \frac43$
PR: $\alpha\beta = \frac13$
REQUIRED
SR: $\alpha^3 + \beta^3 = (\alpha + \beta) (\alpha^2 + 2\alpha \beta + \beta^2)$
$\alpha^3 + 3\alpha^2 \beta + \alpha \beta^2 + \alpha^2 \beta + 3\alpha \beta^2 + \beta^3$
$\alpha^3 + 3\alpha^2 \beta + \alpha \beta^2 + \alpha^2 \beta + 3\alpha \beta^2 + \beta^3 = (\alpha + \beta)^3$
$\alpha^3+ \beta^3 = (\alpha + \beta)^3 - 3(\alpha \beta)^2 - (\alpha \beta)^2$
$\alpha^3 + \beta^3 = (\frac43)^2 - 3(\frac13)^2 - (\frac13)^2$
$\alpha^3 + \beta^3 = \frac{16}{9} - \frac13 - \frac19$
$\alpha^3 + \beta^3 = \frac43$
PR: $\alpha^3 \beta^3$
$(\alpha \beta)^3 = (\frac13)^3 = \frac{1}{27}$
EQUATION
$x^2 - \frac43 x + \frac{1}{27} = 0$
$27x^2 - 36x + 1 = 0$
I am not sure if it's suppose to be:
$(\alpha + \beta) (\alpha^2 + 2\alpha \beta + \beta^2)$
Or
$(\alpha + \beta) (\alpha^2 + 3\alpha \beta + \beta^2)$
Also sorry about the formatting, if you don't mind fixing it for me. Not sure how to do it.
Thank You.
|
Using the quadratic formula, we have $$\alpha, \beta = \frac{-(-4) \pm \sqrt{(-4)^2-4(3)(1)}}{6} \\ = \frac{4 \pm \sqrt{16-12}}{6} \\ = \frac{4 \pm 2}{6} \\ = 1, \frac{1}{3} $$ Hence a polynomial you could use with roots at $\alpha^3 = 1^3 = 1$ and $\beta^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$ would be $$f(x) = (x-1)\left(x-\frac{1}{27}\right) \\ = x^2 -\frac{28x}{27}+\frac{1}{27}$$ Or in general, for any constant $ C \neq 0$, then $C\cdot f(x)$ will have zeroes at $\alpha^3, \beta^3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
$\frac{d^2 y}{dx^2}-2y=2\tan^3\left(x\right)$ Problem:
\begin{equation}
\frac{d^2 y}{dx^2}-2y=2\tan^3\left(x\right).
\end{equation}
using the method of undetermined coefficients or variation of parameters, with $y_p\left(x\right)=\tan\left(x\right)$.
This is what I have so far:
\begin{equation}
r^2-2=0\implies r=\pm\sqrt{2},\:\:\:\:\:\therefore\:\: C_1 e^{\sqrt{2}t}+C_2 e^{-\sqrt{2}t},
\end{equation}
so that is my "complimentary solution" but I do not know what to do with it.
|
I don't have the reputation to comment, but the general solution of the nonhomogeneous equation is just the fundamental set of solutions to the homogeneous equation added to the particular solution. So you have everything, just put it all together.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Find the derivative of $y = x^{\ln(x)}\sec(x)^{3x}$ What is the derivative of
$$y = x^{\ln(x)}\sec(x)^{3x}$$
I tried to find the derivative of this function but somewhere along the way I seem to have gotten lost. I started off with using the product rule and then the chain rule about 4 times and things are getting messier. Now I just don't know what to do. I would show my work here but that would be roughly 2 whole pages of calculations...
|
Try logarithmic differentiation, or somehow make use the fact that $a=e^{\ln a}$ for $a>0$. (you have to decide what $a$ should be).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Is there a formula for $k\pi ^n$, if $n$ is an odd number and $k$ is a rational number? I always find the formula for $k\pi ^n$ when $n$ is an even number and $k$ is a rational number, but I did't find for an odd number.
|
The Dirichlet beta function may satisfy you :
$$\beta(x):=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^x}$$
with the table of values :
\begin{array} {c|c}
n&\beta(n)\\
\hline
1&\frac {\pi}4\\
2&K\\
3&\frac {\pi^3}{32}\\
4&\beta(4)\\
5&\frac {5\,\pi^5}{1536}\\
\end{array}
with $K$ the Catalan constant.
Here the $n$ even cases are the difficult ones as opposed to the $\zeta$ odd cases !
A parallel with $\zeta$ is proposed in this thread (Euler numbers replacing Bernoulli numbers...).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
find common ratio of $\sum_{k=1}^\infty \frac{1}{k(k+1)}$ I have this problem, I need to find the sum.
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{k(k+1)}$$
The problem is that the ratio is not conclusive, Any idea how to find the ratio?
Thanks!
|
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots+(\frac{1}{k}-\frac{1}{k+1})$$ Do you see how to do it now?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.
Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ implies $|f(x) - f(a)| < ε.$
Then suppose $x, a$ are elements of $\Bbb R. $
Now
\begin{align}
|f(x) - f(a)|
&= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right|
\\&= \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right|
\\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}
\\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)}
\\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right]
\end{align}
I don't know how to simplify more. Can someone please help me finish? Thank very much.
|
According to the mean value theorem,
$$
\left|\frac1{1 + x^2} - \frac1{1 + a^2}\right| = f'(c)|x-a|
$$
where $f(x)=\dfrac 1 {1+x^2}$ and $c$ is somewhere between $x$ and $a$. But $|f'(c)|\le\max |f'|$, the absolute maximum value of $|f'|$. In order for this to make sense, you need to show that $|f'|$ does have an absolute maximum value. But that is not hard. So you have
$$
|f(x)-f(a)|\le M|x-a| \text{ for ALL values of $x$ and $a$},
$$
(where $M$ is the absolute maximum of $|f'|$). So $f$ is Lipschitz-continuous and therefore uniformly continuous.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 4
}
|
State space and linearization I have a question about state space representation. How can I represent an equation in which I have only the second and first derivatives? For example
where $u$ is the control input.
If I put $x_1=x$ and $x_2=\dot{x}$ I will not have $x_1$ in my state space representation, and when finding equilibrium points or Jacobian to check controllability, I will obtain zero in the partial derivatives corresponding to $x_1$. Is there a way to overcome this with a more suitable state representation?
I'm not sure how to solve this problem as I have to linearize the system around an operating point using SS. Thanks in advance.
|
By selecting $x_1=x$ and $x_2=\dot{x}$ you can write
$$\begin{align*}
\dot{x}_1 &= x_2 \\
\dot{x}_2 &= k_1 x_1 (1 - |x_1|) + k_2 u
\end{align*}$$
Now the fixed points are $(0, 0), (1, 0), (-1, 0)$. The Jacobian around $(1, 0)$ (as an example) is
$$ \begin{bmatrix} 0 & 1 \\ k_1(1 - 2 x_1) & 0 \end{bmatrix} |_{x_1=1} = \begin{bmatrix} 0 & 1 \\ -k_1 & 0 \end{bmatrix} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Finding the length of a graph I am asked to find the length of the graph of the function: $\dfrac{x^2}{8} - \ln (x)$ , $x\in[1,4]$.
I am really struggling with this problem. As far as I know, we use the formula:
$L(c)= \sqrt{1 + [f'(x)]^2}$
Either I am deriving it wrong or I am missing steps because I am not getting the answer the book gives.
|
The length is
$$\int_1^4 \sqrt{1+\Bigl(\frac x4-\frac1x\Bigr)^2}\,dx
=\int_1^4\sqrt{\Bigl(\frac x4+\frac1x\Bigr)^2}\,dx\ .$$
I'm sure you can finish it from here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Proving the existence of a homomorphism $\overline f:G/H\rightarrow G'$ such that $\overline f \pi = f.$ I'm working on a problem where I'm given that $G$ is a group, $H$ is a normal subgroup of $G,$ $f:G\rightarrow G'$ is a homomorphism, and $H\subseteq \ker(f).$ I need to show that there exists a homomorphism $\overline f: G/H\rightarrow G'$ such that $\overline f \pi = f.$
I have an answer which I believe to be too simple... Define $\overline f:G/H\rightarrow G'$ as $\overline f(gH) = f(g).$ Then $\forall g\in G,$ $\overline f(\pi(g)) = f(g).$ But I didn't even use the fact that $H\subseteq \ker(f),$ so this can't be right. Any advice?
|
Hint: You need to check your map $\overline{f}$ is well defined. Right now it seems to depend on the choice of a representative of a coset. $G/H$ consists of equivalence classes, and equivalence classes may have different representatives. It is here that you need $H \subset ker(f)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Use the techniques from integration to find the volume of a cylinder of height h and radius r I was thinking of starting with a quadrilateral or rectangle.
starting with a horizontal line a $y = r$, then set the area under the curve by integrating from $0$ to $h$.
$y=r$ then $y = r^2$
$$\pi \int_{0}^{h} r^2 $$
how's this?
|
Actually I don't have any plotting software with me but soon I'll show you Images that'll help understanding these methods
Solution 1
Assuming that You know that Area of Circle with radius R is $\pi R^2$
Now consider a disc (differential element) of Radius $R$ and Thickness $dh$
It's volume will be $$ dv=\pi R^2 dh$$
Volume of cylinder can be written as $$V=\int_0^H dv=\int_0^H\pi R^2 dh={\pi R^2 H}$$
Solution 2
Assuming that you know that cirumference of circle with radius R is $2\pi R$
Now consider a cylindrical layer ( differential element ) at distance $r$ from center with thickness $dr$
It's volume will be $$dv=2\pi r H dr$$
Volume of cylinder can be written as $$V=\int_0^R dv=\int_0^R 2\pi Hr dr=\frac{2\pi R^2 H}{2}={\pi R^2 H}$$
If possible I'll add more methods.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/997944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Probability conundrum Good morning,
wondered if you could help me please?
I would like to work out the probability of and event happening 5 times out of 6.
all 6 events have a 1 in 60 chance of a particular outcome. I would like to know what the probability is when the event happens 6 times, and 5 of which have the same outcome.
if you could help it would be appreciated.
Thanks
Neil
|
Your situation correspond to binomial distribution with parameters $n=6$ and $p=1/60$ where each of 6 trials can have 60 possible outcomes.
First, let us focus on the $i$th term of the binomial expansion that corresponds that one fixed outcome (say $1$) occurs
$$
{{n}\choose{i}}p^i(1-p)^{n-i}
$$
Specifically:
$$
{{6}\choose{5}}\left(\frac{1}{60}\right)^5\frac{59}{60}
$$
Since each of the $n=6$ can have values $1,2,\dots 60$, we have to multiply this term by 60, i.e.
$$
{{6}\choose{5}}\left(\frac{1}{60}\right)^5 59
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/998021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How would one solve this system $$12x^2=6z\\2y=-z\\6x-y=7$$
It's been many years since I've dealt with system equations, and now find myself in need to solve them. I am not quite sure what to do; I am interested in finding $x$ and $y$, so assuming that $z \neq 0$ (can I do this?), I isolate it and get
$$2x^2 = z = -2y \\ -x^2 = y \\ 6x - (-x^2) = 7 \\ 6x + x^2 = 7 $$ which gives $$x^2 + 6x - 7 = 0 \\ (x+7)(x-1)=0$$
so $$ x = -7 \\or \\x = 1$$
which gives $$y = -(-7)^2 = -49 \\or \\ y = -(1)^2= -1 $$
But the answer is supposed to be $(1,-1)$ and $(7,35)$ ?? Where's my mistake?
|
Given is $f(x,y)=4x^3+y^2$ with the constraint $y=6x-7$. The easiest way to solve this, is plugging in $y$ into the function, which gives
$$f(x)=4x^3+(6x-7)^2=4x^3+36x^2-84x+49$$
and we are asked to find the extrema. Differentiating:
$$f'(x)=12x^2+72x-84=0$$
Dividing by $12$ gives $f'(x)=x^2+6x-7=(x+7)(x-1)=0$. From this follow the solutions $x=-7$ and $x=1$.
Another method is with Lagrange multipliers. The gradient of the constraint is $\left(\begin{array}{c}6\\-1\end{array}\right)$. The gradient from the function $f(x,y)$ is $2\left(\begin{array}{c}6x^2\\y\end{array}\right)$. These two gradients must be parallel to have an extremum at a given point $(x,y)$.
From this follows that $\frac{6x^2}{y}=\frac{6}{-1}$ and $x^2=-y$. Plugging this expression for $y$ into the constraint $6x-y=7$ gives $x^2+6=-7$, which is the same quadratic equation as before.
To summarize; there is a local maximum at $f(-7,-49)=1029$ and a local minimum at $f(1,-1)=5$.
To answer the question, your approach is correct, so maybe the answer model is incorrect.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/998124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Does every algebra automorphism preserve augmentation ideal filtration? Let $A=\displaystyle\bigoplus_{n\geq0}A_n$ be a graded algebra and let the augmentation homomorphism $\varepsilon:A\to A_0$ be the projection. Define the augmentation ideal, denoted by $A_+$, to be the kernel of $\varepsilon$. Let $A_+^k$ to be the $k$th power of $A_+$ in the sense of ideal multiplication.
If $f:A\to A$ is a ring (not necessarily graded) automorphism, is it always true that $f(A_+^k)\subseteq A_+^k$?
|
$A=k[x],\epsilon(x)=0, f(x)=x+1,f^{-1}(x)=x-1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/998243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Black Scholes PDE How to show that $V_1(S,t)=S\frac{\partial V(S,t)}{\partial S} $ satisfies Black-Scholes PDE given as $\frac{\partial V}{\partial t} + \frac{\sigma^2 S^2}{2}\frac{\partial^2V}{\partial S^2} + rS\frac{\partial V}{\partial S} -rV = 0$ ?
|
Let $V(S,t)$ --- solution of the PDE.
Then
$$
S\frac{\partial}{\partial S}\left(\frac{\partial V}{\partial t} + \frac{\sigma^2 S^2}{2}\frac{\partial^2V}{\partial S^2} + rS\frac{\partial V}{\partial S} -rV\right) = 0
$$
$$
S\left(\frac{\partial^2 V}{\partial t \partial S} + \frac{\sigma^2 }{2}\left(2S\frac{\partial^2V}{\partial S^2} + S^2\frac{\partial^3V}{\partial S^3}\right) + r\frac{\partial V}{\partial S} + rS\frac{\partial^2 V}{\partial S^2} - r\frac{\partial V}{\partial S} \right) = 0
$$
$$
\frac{\partial}{\partial t}S\frac{\partial V}{\partial S} + \frac{\sigma^2 S^2 }{2}\left(2\frac{\partial^2V}{\partial S^2} + S\frac{\partial^3V}{\partial S^3}\right) + rS^2\frac{\partial^2 V}{\partial S^2} = 0.
$$
Here we note that $\frac{\partial^2}{\partial S^2}\left(S\frac{\partial V}{\partial S}\right) = 2\frac{\partial^2V}{\partial S^2} + S\frac{\partial^3V}{\partial S^3}$
and $rS\frac{\partial}{\partial S}\left(S\frac{\partial V}{\partial S} \right) =
rS\frac{\partial V}{\partial S} + rS^2\frac{\partial^2 V}{\partial S^2}$.
And finally
$$
\frac{\partial V_1}{\partial t} + \frac{\sigma^2 S^2}{2}\frac{\partial^2V_1}{\partial S^2} + rS\frac{\partial V_1}{\partial S} -rV_1 = 0.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/998341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Norm of a functional I'm facing the problem of calculating the norm of the following functional:
$\displaystyle \phi : L_p([0,1]) \rightarrow \mathbb{R}, ~~ \phi (f) = \int\limits_{0}^{1} e^x f(x) dx $
I have no idea where to begin. I need to see the supremum of $\displaystyle \phi $ on the set $ B(0,1)= \{f \in L_p([0,1]): \displaystyle\int_0^1|f(x)|^p dx \leq 1\} $
|
for every $f\in B(0,1)$ we have
$$|\phi(f)|\leq \left(\int_0^1e^{qx}dx\right)^{1/q}||f||_p$$ where $1/p+1/q=1$, so $||\phi||\leq \frac{1}{q}(e^q-1)^{\frac1q}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/998425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
How can I prove $(a+b+c)!>a!b!c!$ In fact, I couldn't prove the inequality because I don't know which method is used for this.
The condition for this inequality is $$(a+b+c)>1$$
|
An adaptation of this answer. If you have $a$ adults, $b$ boys, and $c$ girls, then $(a+b+c)!$ counts the number of ways you can line everyone up, while $a!b!c!$ counts the number of ways you can do so with the additional requirement that all adults come before all children, and all boys before all girls. Then $(a+b+c)!\geq a!b!c!$ always holds, while $(a+b+c)!>a!b!c!$ holds as soon as at least two categories are non-empty, since then there will be some line-up that violates the additional requirement.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/998571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
}
|
Relationship between divergence operators defined with respect to two different volume forms. Let us assume that you have a volume form $\mu$ defined on a manifold $\mathcal{M}$. Then you can define the divergence operator with respect to this metric, such that the following relationship holds for all $\mathcal{U}\subset \mathcal{M}$ and $v \in T\mathcal{M}$ : $$ \int_{\mathcal{U}}{\operatorname{div} v \ \mu} = \int_{\partial \mathcal{U}}{ \mathbf{i}_v \mu }$$
Now suppose that you have two distinct non degenerate volume forms $\mu_1$ and $\mu_2$. Then there exists a scalar field $\alpha$ such as $\mu_1 = \alpha \mu_2$.
How are related (in terms of $\alpha$) the corresponding divergence operators (let's call them $\operatorname{div}_1$ and $\operatorname{div}_2$) ?
|
I've actually worked out the answer to the question from the following definition of the divergence : $\operatorname{div}(v)\mu = \operatorname{d}(\mathbf{i}_v \mu)$.
Short-circuiting a few computation lines, the result is : $$ \alpha \operatorname{div}_1(v) = \operatorname{div}_2(\alpha v) $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/998655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
The second and third Chern classes of Calabi-Yau threefolds Let $X$ be a smooth projective Calabi-Yau threefold. Then the first Chern class vanishes: $$c_1(X)=c_1(T_X)=0.$$
Is anything known about $c_2(X)$ and $c_3(X)$? What about $c_2$ of a
K$3$ surface?
(I am sorry if this is very well-known. This question is just a cultural curiosity: I started asking myself such questions while doing a Chern class computation on a threefold, and I realized I could not replace $c_i(X)$ by anything I knew...)
Thanks!
|
In general the top Chern class is the Euler class of the real bundle underlying the holomorphic tangent bundle, so its degree is the Euler characteristic of the manifold.
So for $X$ a $K3$ surface, we always have $c_2(X)=24$.
Unfortunately for Calabi–Yau threefolds the Euler characteristic can vary wildly, and in fact it's not even known whether the Euler characteristics are bounded. See this MO question for some good information. According to one answer there, the smallest known value for $c_3$ of a CY threefold is -960.
As for $c_2$ of a Calabi–Yau threefold, I find I have even less to say. Of course now it is a class in $H^4(X)$, rather than an integer; by cup-product, one can view it as a linear form on $H^2(X)$. This paper of Kanazawa and Wilson has some information on the properties of this linear form (and other things relevant to your question). For example, they quote the interesting fact (attributed to Miyaoka) that this form is nonnegative on nef divisor classes in $H^2(X)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/998797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
$a_n=2^n+3^n+6^n-1$. Find all positive integers that are primes to all terms of the sequence. Let the sequence $a_n=2^n+3^n+6^n-1, n\in\mathbb N_{> 0}$. Find all positive integers that are prime to all terms of this sequence. I have no idea how to approach this, but I know that I CAN'T use Euler's theorem or any generalization/extension of fermat's little theorem as it's on a later chapter in the textbook.
If possible, fully explain all steps.
|
Thought it could be useful to cite the solution from The IMO Compendium A Collection of Problems Suggested for The International Mathematical Olympiads: 1959-2009:
We will show that $1$ is the only such number. It is sufficient to prove that for every prime number $p$ there exists some $a_m$ such that $p \mid a_m$. For $p=2,3$ we have $p\mid a_2=48$. Assume now that $p>3$. Applying Fermat's theorem, we have
$$
6a_{p-2}=3\cdot 2^{p-1}+2\cdot 3^{p-1}+6^{p-1}-6\equiv 3+2+1-6=0 \pmod{p}.
$$
Hence $p \mid a_{p-2}$, i.e. $\gcd(p,a_{p-2})=p>1$. This completes the proof.
Notice this is very similar to @Exodd's answer, although it avoids fractions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/998897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
How to calculate this $\sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{4\pi}{9}$? I'm stuck with the expression
$$\sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{4\pi}{9}.$$
I have no idea how to begin, please give me a hint!
(The answer should be $\sqrt3/8$.)
|
Remember that the polynomial $P(x)=3x-4x^3$ has the property $P(\sin t)=\sin 3t$. Therefore $x_1=\sin \pi/9$,
$x_2=\sin 2\pi/9$ and $x_3=-\sin4\pi/9$ are all solutions of the equation
$P(x_i)=\sin \pi/3=\sqrt3/2$. In other words they are zeros of $P(x)-\sqrt3/2$. The leading coefficient of $P$ is $-4$, so
$$
-4x^3+3x-\sqrt3/2=P(x)-\sqrt3/2=-4(x-x_1)(x-x_2)(x-x_3).
$$
Expand the product on the right hand side, and compare the constant terms.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/999034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
What to do if the critical point is not a real number? I have a function and I already differentiate it, but when I put it equals to zero I don't get a real number. What am I doing wrong?
$f(x) = x\sqrt{x^2+1}$
$f'(x) = \sqrt{x^2+1}+\frac{x^2}{\sqrt{x^2+1})}$
$f'(x) = \frac{2x^2+1}{\sqrt{x^2+1}}$
Then I equalizes it to 0
$\frac{2x^2+1}{\sqrt{x^2+1}}=0$
$2x^2+1=0$
$2x^2=-1$
$x^2=-1/2$
So I don't know what to do next.
When reach the second derivative I get the same problem. What it means?
|
This function does not have a critical point. Notice that $f'(x)=\frac{2x^{2}+1}{\sqrt{x^{2}+1}}>0$ for all $x\in\mathbb{R}$. So the function $f$ is a monotonic increasing function.
If you are doing this for finding maxima of minima, then the answer is maxima or minima do not exist on $\mathbb{R}$. However if you restrict your attention on a compact interval (like $[a,b]$), then the absolute maxima and absolute minima of $f$ on $[a,b]$ are respectively $f(b)$ and $f(a)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/999146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
$X=C[0,1]$ is a Banach space, $M=\{f\in X: f(0)=0\}$, prove $M$ is closed, find explicit formula for the quotient norm, and find an isomorphism. Here is my question:
Let $X$ be a Banach space $C[0,1]$ with the supremum norm. Let $M=\{f\in X: f(0)=0\}$. Show that $M$ is closed. Find an explicit formula for the quotient norm $\|[f]\|$ for $[f]\in X/M$. Find an isometric isomorphism from $\mathbb{R}$ to $X/M$.
Here is what I have:
$M$ is closed: Let $f_n\in M$ such that $f_n\to f$. Let us assume that $M$ is open, that is to say that $f(0)=y\neq 0$. Then, given $\epsilon >0$ we know there exists some $N>$ such that for a fixed $x\in [0,1]$, $\|f_n-f\|<\epsilon$ for all $n\geq N$. Set $\epsilon < y$ and choose $x=0$. Then:
$$\|f_n(0)-f(0)\|=\|0-y\|=sup\{|-y|\}=y\gt\epsilon$$
So we have a contradiction, therefore $f(0)=0$ and $M$ is closed.
Explicit formula for the quotient norm:
$$\|[f]\|=\|f+m\|=inf\{\|f+m\|_\infty:m\in M\}=inf\{sup\{|f+m|\}:m\in M\}$$
Isomorphism:
This one I am having some trouble with.
|
Continuing your solution: First I will show that $\|[f]\|=|f(0)|$. Notice that for a fixed $f\in C[0,1]$ and for any $m\in M$, we have $\|f+m\|_{\infty}=\sup_{x\in[0,1]}|f(x)+m(x)|\geq |f(0)+m(0)|=|f(0)|$. Therefore $\|[f]\|=\inf\{\|f+m\|_{\infty}:m\in M\}\geq |f(0)|$. Conversely, define the function $m(x)=f(0)-f(x)$. Clearly $m\in M$, and $\|f+m\|_{\infty}=|f(0)|$. Therefore $\|[f]\|=|f(0)|$.
Now define a function $F:X\to\mathbb{R}$ as $F(f)=f(0)$. Obviously $\ker(F)=M$. So $X/M$ is isomorphic to $\mathbb{R}$. This isomorphism is also an isometry, which follows from the fact that $\|[f]\|=|f(0)|$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/999224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
$X$ is the vector space $C[0,1]$ with the norm $\|f\|_1=\int_0^1|f(t)|dt$, and $M=\{f\in X:f(0)=0\}$, show that $M$ is not closed. Here is my question:
Let $X$ be the vector space $C[0,1]$ with the norm $\|f\|_1=\int_0^1|f(t)|dt$. Let
$$M=\{f\in X:f(0)=0\}$$
Show that $M$ is not closed. Show that the “quotient norm" $inf\{\|f-m\|_1:m\in M\}$ is not a norm on $X/M$.
Here is what I have:
For the closure, I am trying to find a function $f$ such that a sequence $\{f_n\}\in M$ which converges to $f$, but $f\notin M$. This would mean that for any $n$, $f_n(0)=0$ but $f(0)\neq 0$. I cannot seem to find such an $f_n$ and $f$.
As for the quotient norm not being a norm on $X/M$, I believe this follows from $M$ not being closed, as with quotient norms, $\|x-M\|=0$ if and only if $x\in M$.
So any suggestions on finding the $f$? Thanks.
|
As for the "quotient norm", have a look at mookid's counter-example and see if you can modify it to force a violation of one of the norm axioms. I'd aim to show that $$\| [f] \|_{\hbox{am I a norm?}} = 0 \ \ \ \Longrightarrow \ \ \ f = 0$$
is violated.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/999293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Embedding of a ring into a ring with unity I was reading the theorem on Embedding of a ring into a ring with unity which is as follows:
Let R be ring and $R\times \mathbb Z=\{(r,n)|r\in R,n\in \mathbb Z\}$.
This is a ring with addition defined as $(r,n)+(s,m)=(r+s,n+m).$ and multiplication defined as :$(r,n).(s,m)=(rs+ns+mr,nm)$ .this ring has unity as $(0,1)$.
Now we can easily show that
If we define an homomorphism from $R\to R_1$ as $f(r)=(r,0)$ $\forall r\in R$ then $R \cong f(R)\subseteq R_1$.
Hence $R$ is embedable in $R_1$ which has unity $(0,1)$.
I can't understand why in the ring $R\times \mathbb Z=\{(r,n)|r\in R,n\in \mathbb Z\}$ we had to define multiplication as $(r,n).(s,m)=(rs+ns+mr,nm)$, why can't we define it as $(r,n).(s,m)=(rs,nm)$ ?
|
With your multiplication, what is your proposed unity?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/999370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
How to show this Legendre symbol problem Let $n \in \mathbb{N}$ and $p$ an odd prime number such as $p \nmid n$.
Prove that:
$\exists x, y \in \mathbb{Z};\,\, \gcd(x, y) = 1$ such as $x^{2} + ny^{2} \equiv 0\, (\mod p) \iff \Bigg(\displaystyle \frac{-n}{p}\Bigg) = 1$, where $\Bigg(\displaystyle \frac{-n}{p}\Bigg)$ represent the Legendre Symbol.
I've been trying but I have not idea where to start. Thanks in advance.
|
Here is an outline, see if you can fill in more details.
First part: if $x^2+ny^2\equiv0\pmod p$ then $y$ is not a multiple of $p$. (If so then $x$ is a multiple of $p$, but this is impossible since $\gcd(x,y)=1$.) So $y$ has a multiplicative inverse $z$ modulo $p$, so $-n\equiv(xz)^2\pmod p$.
Converse: if $\bigl(\frac{-n}p\bigr)=1$ then for some $x$ we have $-n\equiv x^2\pmod p$, that is, $x^2+n\equiv0\pmod p$, and now you should have no trouble finding a suitable value of $y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/999472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $x>y$, then $\lfloor x\rfloor\ge \lfloor y\rfloor$, formal proof For x ∈ ℝ, define by: ⌊x⌋ ∈ ℤ ∧ ⌊x⌋ ≤ x ∧ (∀z ∈ ℤ, z ≤ x ⇒ z ≤ ⌊x⌋).
Claim 1.1: ∀x ∈ ℝ, ∀y ∈ ℝ, x > y ⇒ ⌊x⌋ ≥ ⌊y⌋.
Assume, x, y ∈ ℝ # Domain assumption
Assume x > y # Antecedent assumption
Then x ≥ ⌊x⌋ # By definition
Then x ≥ z # By definition
Then ⌊x⌋≥ z # Also by definition
Let z = ⌊y⌋ # By assumption & definition since ∀z ∈ ℤ, y ∈ ℝ and ℤ ∈ ℝ
Then ⌊x⌋ ≥ ⌊y⌋ # Substitution
Then x > y ⇒ ⌊x⌋ ≥ ⌊y⌋ # Introduce implication
Then ∀x ∈ ℝ, ∀y ∈ ℝ, x > y ⇒ ⌊x⌋ ≥ ⌊y⌋ # Introduce universal
|
$x\geq y$ combined with $\lfloor y\rfloor\leq y$ gives $\lfloor y\rfloor\leq x$.
Here $\lfloor y\rfloor\in\mathbb Z$ so we conclude that $\lfloor y\rfloor\leq \lfloor x\rfloor$.
Proved is now: $$x\geq y\Rightarrow \lfloor x\rfloor\geq \lfloor y\rfloor$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/999551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Solving for N in a binomial distribution Mid-term study...
Two dice are rolled. How many times must the dice be rolled so that the probability of getting a sum of 10 or greater on at least one roll is larger than 0.9?
So am I correct in thinking that it would be:
$$\sum_{j=1}^{n}\sum_{k=1}^{j}{n \choose k}P^k(1-P)^{j-k} \geq .9$$
Given that the probability of getting a sum of 10+ is $\frac{6}{36}$
I get:
$$\sum_{j=1}^{n}\sum_{k=1}^{j}{n \choose k} \left( \frac{6}{36}\right)^k \left( \frac{30}{36}\right)^{j-k} \geq .9$$
Am I right in my formula? If not, what should it be? And then how do I solve for N?
Thanks.
EDIT / RESPONSE:
What I wrote above is the same as saying
$${n \choose 0}P^0(1-P)^n < .1$$
The first parts are both just 1, which plugging in numbers leaves me with
$$\left(\frac{30}{36} \right)^n < .1$$
To solve, I just need to do the following
$$\log_{\frac{30}{36}}{\left(\frac{30}{36} \right)^n} < log_{\frac{30}{36}}{(.1)}$$
Which according to my calculations should be 12.6, which means 13.
|
So am I correct in thinking that it would be
$$\sum_{j=1}^{n}\sum_{k=1}^{j}{n \choose k}P^k(1-P)^{j-k} \geq .9$$
No, not quite.
Rolling two die such that their sum is less than 10 is a trial with probability of success $p$.
The number of successes, $N_n$, in a series of $n$ trials has a binomial probability distribution. $N_n\sim \mathcal{Bin}(n, p)$.
You simply want to find the number of trials, $n$, such that $\mathsf P(N_n\geq 1)\geq 0.9$. That is:
$$ \sum_{k=1}^n {n\choose k}p^k(1-p)^{n-k}\geq 0.9$$
However the probability of the complement event is easier to much calculate, so find $n$ such that: $\mathsf P(N_n=0) \leq 0.1$. That is:
$$(1-p)^n \leq 0.1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/999650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
how to integrate $\mathrm{arcsin}\left(x^{15}\right)$? Integral by parts:
$$
I = x\sin^{-1}\left(x^{15}\right) - \int\frac{15x^{15}}{\sqrt{1-x^{30}}}dx
$$
then what?
The answer by wolfram gives an answer contains hypergeometric ${}_2F_1$ function,because it has no elementary answer. The question I want to know is, how can we find the integral of
$$
\frac{15x^{15}}{\sqrt{1-x^{30}}}
$$
in terms of hypergeometric function?
|
$\int\dfrac{15x^{15}}{\sqrt{1-x^{30}}}dx$
$=\int_0^x15t^{15}(1-t^{30})^{-\frac{1}{2}}~dt+C$
$=\int_0^{x^{30}}15t^\frac{1}{2}(1-t)^{-\frac{1}{2}}~d(t^\frac{1}{30})+C$
$=\dfrac{1}{2}\int_0^{x^{30}}t^{-\frac{7}{15}}(1-t)^{-\frac{1}{2}}~dt+C$
$=\dfrac{1}{2}\int_0^1(x^{30}t)^{-\frac{7}{15}}(1-x^{30}t)^{-\frac{1}{2}}~d(x^{30}t)+C$
$=\dfrac{x^{16}}{2}\int_0^1t^{-\frac{7}{15}}(1-x^{30}t)^{-\frac{1}{2}}~dt+C$
$=\dfrac{15x^{16}}{16}~_2F_1\left(\dfrac{1}{2},\dfrac{8}{15};\dfrac{23}{15};x^{30}\right)+C$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/999744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Why is Q dense in R? Consider a topological space $S$ and an arbitrary subset $E$. Then if the closure of $E$ given by $\bar{E} = S$, we say that $E$ is dense in $S$.
How can we prove that $\mathbb{Q}^{n}$ is dense in the vector space $\mathbb{R}^{n}$ with the standard metric topology?
I intended to prove that $\mathbb{Q}^{n}$ is dense in the vector space $\mathbb{R}^{n}$ in the topological sense, not the usual "between two $\mathbb{R}$ there is a $\mathbb{Q}$". Of course, both might be equivalent, but that remains to be proved.
|
Hint:
*
*$\mathbb{Q}$ is dense in $\mathbb{R}$.
*It suffices to show $\forall x=(x_i)\in \mathbb{R}^n$, and $x\in I=(a_1,b_1)\times (a_2,b_2)\cdots \times (a_n,b_n)\cap \mathbb{Q}^n\neq \emptyset$. That is $\exists q=(q_i)\in \mathbb{Q}^n$, s.t. $q\in I$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/999872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
}
|
Homology and Fundamental Group (Algebraic Topology - Allen Hatcher) I have some questions regarding 2 parts of the theorem of this section:
1) Having $f = \sum_{i,j}(-1)^jn_i\tau_{ij}$, it pairs $\tau_{ij}$ with opposite signs (in any way I assume), and says that the non paired $\tau_{ij}$'s are $f$.
I don't get why necessarily only some of the $\tau_{ij}$'s are $f$, and why it's ok to just pair any $\tau_{ij}$'s with opposite sign, because identifying the edges of the paired $\tau_{ij}$'s it then forms a $\Delta$-complex $K$, and then extends the maps $\sigma_i$ to a map $\sigma : K \to X$, but for $\sigma$ to be well defined, shouldn't the paired $\tau_{ij}$'s be exactly the same?
2)In the last paragraph of the proof it writes $[f] = \sum_{i,j}(-1)^jn_i[\tau_{ij}] \in \pi_1(X)_{ab}$ in additive notation, so the sums mean composition, but that means that $f$ is (or is homotopic to) $\prod_{i,j}\tau_{ij}^{(-1)^jn_i}$ as a loop, and I don't know where do you get that information, the information you have is $f = \sum_{i,j}(-1)^jn_i\tau_{ij}$, but as I see it that doesn't mean $f = \prod_{i,j}\tau_{ij}^{(-1)^jn_i}$ as a loop, since we have $f\simeq g \Rightarrow f \sim g$ but not $f\simeq g \Leftrightarrow f \sim g$.
Sorry for the extended "question" and thank you in advance.
|
1) This information is not directly for gluing, rather it is information on the orientation of the simplex. The goal is to obtain a coherently oriented two-dimensional $\Delta$ complex with boundary equal to $[f]$. Look back at the beginning of the chapter when Hatcher writes down the formula for the boundary of a 3-simplex. Notice that each of the 1-simplices appears twice, each time in an opposite direction. This means that they don't contribute to the boundary of the 2-complex.
2) The misconception here is that the sum represents loop composition, which it doesn't. Rather, the sum is to be interpreted in the abelianization of the fundamental group. As an example of what I mean here, let $X = S^1 \vee S^1$, which has a fundamental group which is free on two generators we will call $a$ and $b$. In particular, $f = aba^{-1}b^{-1}$ is not trivial. However, $[f] = [ aba^{-1}b^{-1}] = [a] + [b] - [a] -[b] = 0$, since homology is abelian. What the theorem says is that anything in the kernel of the homomorphism $\pi_1 \to H_1$ can be decomposed into loops which can be reordered into a null-homotopic loop, just as $f = aba^{-1}b^{-1}$ can be reordered to get $g = aa^{-1}bb^{-1}$. The claim isn't that $f$ and $g$ are homotopic, but that $[f] = [g]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/999956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Is $I=\langle7, 3+\sqrt{19}\rangle$ a principal ideal of $\Bbb Z[\sqrt{19}]$?
Is $I=\langle7, 3+\sqrt{19}\rangle$ a principal ideal of $\Bbb Z[\sqrt{19}]$?
I defined the norm :
$$N(a+b\sqrt{19})=(a+b\sqrt{19})(a-b\sqrt{19})=a^2-19b^2$$
Then we can see the multiplicative property :
$$N((a+b\sqrt{19})(c+d\sqrt{19}))=N(a+b\sqrt{19})N(c+d\sqrt{19})$$
Since $N(7)=49$, $N(3+\sqrt{19})=-10$, if $z | 7$ and $z | 3+\sqrt{19}$ then $N(z)=1$. But it means that
$$u^2-19v^2=1$$
I know that this is called Pell's equation and it has a solution. In fact, $u=170, v=39$ satisfies equation.
I need your help. Thanks in advance.
|
You know that any common divisor $z$ of $7$ and $3+\sqrt{19}$ has norm $1$, so it is a unit. Then any principal ideal $J\subseteq\mathbb{Z}[\sqrt{19}]$ containing $7$ and $3+\sqrt{19}$ also contains $1$ and is therefore equal to $\mathbb{Z}[\sqrt{19}]$. You now have to check whether $I=\mathbb{Z}[\sqrt{19}]$: If yes, then $I=\langle 1\rangle$, else $I$ is not a principal ideal in $\mathbb{Z}[\sqrt{19}]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1000046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
derivative of $y=\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$ $y=\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$
The answer is $\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5} \left(\dfrac{2}{x}+\dfrac{1}{2(x+1)}-\dfrac{1}{x+2}-\dfrac{5}{x-3}\right)$
I know that the quotient rule is used but I don't know how to do this problem. Would you multiply together all the terms and then differentiate?
|
It's a long derivative. First use the quotient rule:
$$\frac{\frac{d}{dx}\left(x^2\sqrt{x+1}\right)(x+2)(x-3)^5- x^2\sqrt{x+1}\frac{d}{dx}\left((x+2)(x-3)^5\right)}{((x+2)(x-3)^5)^2}.$$
Then do the derivatives.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1000136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
α and ω possible limit sets of points
*
*What are all the $α$ and $ω$ possible limit sets of points for:
$$A=\begin{pmatrix}-4&-2\\3&-11 \end{pmatrix}.$$
I am not really sure what to do..
$$\dot{x}=\begin{pmatrix}-4&-2\\3&-11 \end{pmatrix}x$$
|
The eigenvalues of $A$ are $-5$ and $-10$, both negative, hence:
*
*For every "initial" condition $x(0)$, $x(t)\to0$ when $t\to+\infty$
*For every "initial" condition $x(0)\ne0$, $\|x(t)\|\to+\infty$ when $t\to-\infty$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1000217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to prove $\int_{0}^{-1} \frac{\operatorname{Li}_2(x)}{(1-x)^2} dx=\frac{\pi^2}{24}-\frac{\ln^2(2)}{2} $ $\def\Li{\operatorname{Li}}$
I wonder how to prove:
$$
\int_{0}^{-1} \frac{\Li_2(x)}{(1-x)^2} dx=\frac{\pi^2}{24}-\frac{\ln^2(2)}{2}
$$
I'm not used to polylogarithm, so I don't know how to tackle it. So any help is highly appreciated.
|
$\def\Li{{\rm{Li}}_2}$Set $x\mapsto -x$ followed by integration by parts, we have
\begin{align}
\int_0^{-1}\frac{\Li(x)}{(1-x)^2}\,dx&=-\int_0^{1}\frac{\Li(-x)}{(1+x)^2}\,dx\qquad\Rightarrow\qquad u=\Li(-x)\,\,\mbox{and}\,\,dv=\frac{1}{(1+x)^2}\\
&=\left.\frac{\Li(-x)}{1+x}\right|_0^{1}+\int_0^{1}\frac{\ln(1+x)}{x(1+x)}\,dx\\
&=-\frac{\pi^2}{24}+\int_0^{1}\frac{\ln(1+x)}{x}\,dx-\int_0^{1}\frac{\ln(1+x)}{1+x}\,dx\\
&=-\frac{\pi^2}{24}-\Li(-x)\bigg|_0^{1}-\frac{1}{2}\ln^2(1+x)\bigg|_0^{1}\\
&=\frac{\pi^2}{24}-\frac{\ln^22}{2}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1000349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
}
|
Limit of $\frac{2^n+2.71^n+\frac{4^n}{n^4}}{\frac{4^n}{n^4}+n^33^n}$ - what is wrong with my proof? Here's quite a hairy sequence, the limit of which I need to find as $n\rightarrow\infty$:
$$\frac{2^n+2.71^n+\frac{4^n}{n^4}}{\frac{4^n}{n^4}+n^33^n}$$
The Squeeze Theorem seemed like a good idea so here's what I've done:
$\frac{2^n+2^n+2^n}{2^n+2^n}<=\frac{2^nn^4+2.71^nn^4+4^n}{4^n+n^73^n}=\frac{2^n+2.71^n+\frac{4^n}{n^4}}{\frac{4^n}{n^4}+n^33^n}<=\frac{n^n+n^n+n^n}{n^n+n^n}$
The left and right sequences have the $\lim_{n\rightarrow\infty}=\frac{3}{2}$. So theoretically the sequence for which $\lim$ is being searched should also converge to $\frac{3}{2}$. But WolframAlpha tells me that the correct answer is $1$. So where is an error in my logic?
One more question, in my "proof" I couldn't show that for sufficiently large $n$: $n^33^n<=n^n$ and this is important in order for my transformation to be considered as correct. Is there any way to show it without derivatives?
|
The problem in your approach is this.
$a,b,c,d>0$, if $a>b$ and $c>d$ you cannot conclude $\frac{a}{c}>\frac{b}{d}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1000434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Throw a dice 4 times. What is the probability `6` be up at-least one time? First time I approach a probability question (:
Throw a dice 4 times. What is the probability 6 be up at-least one time?
Intuitively, I would say: $\frac{1}{6}\times4$.
I would explain as: If you throw one time, probability is $\frac{1}{6}$.
If you do it 4 times, then multiply by 4.
According to the answers I'm wrong.
Can you explain please? thanks in advance.
|
Throw a dice six times - are you certain to get at least one six?
The mean number of sixes in four throws is indeed $\frac 23$. But there are combinations with $2, 3 \text{ or } 4$ sixes, and these reduce the number with just one six (apply the same argument to six throws, where it is more intuitive).
If you get no sixes in four throws, there are $5$ possibilities for each throw, and therefore $5^4=625$ possibilities with no six out of the $6^4=1296$ possibilities in total. So you get the probability by taking the $1296-625 = 671$ possibilities which must include at least one six, out of the $1296$ possibilities altogether.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1000488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
}
|
Set theory by Julia Robinson I used to have a set theory textbook downloaded free from the internet. I lost my laptop in the airport of a city in Eastern Europe, and it was not found (or perhaps “not found”) by the airport security. I now try to rediscover the file. I remember the author was a female mathematician from Berkeley. From an internet search I believe that it might have been Julia Robinson, but I still can’t find the book.
It had a fair number of exercises and I believe it covered Gödel’s incompleteness theorem.
I would be grateful for help to find the book.
|
if you google the title of your questions, one of the links shown is to the collected works of Julia Robinson, and google books does indeed show you a discussion of Godel's incompleteness theorem on one of the pages
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1000551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Fibers of $\operatorname{Spec}(R)\to\operatorname{Spec}(S):\mathfrak{q}\mapsto \mathfrak{q}\cap S$ are discrete? Suppose $S$ is a subring of a commutative ring $R$, such that $R$ is finitely generated as an $S$-module. I"m curious about a property of the map $\operatorname{Spec}(R)\to\operatorname{Spec}(S):\mathfrak{q}\mapsto\mathfrak{q}\cap S$, which shows up frequently in the discussion around the Going Up Theorem.
The fibers of the map are precisely the sets of primes lying above a prime in $S$. I know that these fibers are finite. However, can anybody explain why are these fibers also discrete? Since $\operatorname{Spec}(R)$ is generally not Hausdorff, I'm having trouble separating each prime from the others, so that each point would be open in the fiber in the induced topology.
I know $\operatorname{Spec}(R)$ is compact, but I don't think it's necessarily true that a finite subset of a compact space need be discrete in general. Thanks!
|
Since $R$ is finitely generated as an $S$-module, it has relative dimension zero. That implies if $\mathfrak{q}_1$ and $\mathfrak{q}_2$ are primes of $R$ lying over $\mathfrak{p}\in\operatorname{Spec} S$, then neither is contained in the other.
Suppose $\mathfrak{q}$ is some prime over $\mathfrak{p}$ and $\mathfrak{q}_1,...,\mathfrak{q}_n$ are all the other ones. Pick $f_i \in \mathfrak{q}_i - \mathfrak{q}$ and let $f$ be their product. Then $D(f)$ separates $\mathfrak{q}$ from $\mathfrak{q}_i$ in $\operatorname{Spec} R$, therefore its restriction to the fibre over $\mathfrak{p}$ does the same in the induced topology.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1000667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
If $2^{\aleph_{\beta}}\geq \aleph_{\alpha}$, then $\aleph_{\alpha}^{\aleph_{\beta}}=2^{\aleph_{\beta}}$. If $2^{\aleph_{\beta}}\geq \aleph_{\alpha}$, then $\aleph_{\alpha}^{\aleph_{\beta}}=2^{\aleph_{\beta}}$.
Proof: Note that if $\beta \geq \alpha$, then we have $\aleph_{\alpha}^{\aleph_{\beta}}=2^{\aleph_{\beta}}$. I don't know whether the condition given $2^{\aleph_{\beta}}\geq \aleph_{\alpha}$ will give $\beta \geq \alpha$ or not. It it does, then we are done. If it is not, then we still have to consider other cases.
UPDATE: Clearly $2 ^ {\aleph_{\beta}} \leq \aleph_{\alpha}^{\aleph_{\beta}}$. Note that $\aleph_{\alpha}^{\aleph_{\beta}} \leq (2^{\aleph_{\beta}})^{\aleph_{\beta}}=2^{\aleph_{\beta}}$. By Cantor-Bernstein Theorem, we have the desired result.
Can anyone help me to check whether my update is correct or not.
|
You don't necessarily have that $2^{\aleph_\beta}\geq\aleph_\alpha$ implies $\beta\geq\alpha$; consider that $2^{\aleph_0}\geq\aleph_1$, but $0$ certainly isn't $\geq 1$. Instead, try proving that you have both $\aleph_\alpha^{\aleph_\beta}\geq 2^{\aleph_\beta}$ and $\aleph_\alpha^{\aleph_\beta}\leq2^{\aleph_\beta}$ and then using Cantor-Bernstein. One of these should be trivial; for the other you'll have to do a little cardinal arithmetic using your hypothesis.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1000757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to solve $a^x=a+x$ Given the equation $a^x = a + x$, how do you express $x$ in terms of $a$?
|
This equation has a solution expressed in terms of Lambert function and the solution is $$x=-a-\frac{W\left(-a^{-a} \log (a)\right)}{\log (a)}$$ In fact, any equation which can be written as $$A+Bx+C\log(D+Ex)=0$$ has explicit solution(s) in terms of Lambert function.
In the real domain, $W(t)$ only exists if $t \geq -\frac{1}{e}$. For your problem, the only restriction seems to be $a>0$.
In order to approach it, you can write $$a^x=\frac{a^{x+a}}{a^a}$$ and so the equation becomes $$a^{x+a}=e^{(x+a)\log(a)}=(x+a)\log(a)\frac{a^a}{\log(a)}$$ and define $$y=(x+a)\log(a)$$ which make the equation $$e^y=y\frac{a^a}{\log(a)}$$ or still better $$ze^{z}=-\frac{log(a)}{a^a}$$ where $z=-y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1000904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Is $\sqrt{m}$ irrational iff at least one prime occurs with an odd exponent in the factorisation of $m$? Thinking about it, I think I found the following criterion for irrationality of $\sqrt{m}$ if $m$ is a positive integer.
Let $p_1^{a_1}\cdots p_k^{a_k}$ be the prime factorization of $m$. Then $\sqrt{m}$ is irrational if and only if $a_i$ is odd for at least one $1\le i\le k$.
Anyone knows if this is correct and, if so, a proof thereof?
|
Yes this is true. A nice way to think about this, is that if you extend the notion of prime factorisation to allow for negative exponents, you can show that all rational numbers have a unique prime factorisation as well. Then if you square a rational number of the form $r=p_1^{a_1}\cdots p_k^{a_k}$, where $a_i\in\mathbb{Z}$, then the square of this has the unique prime factorisation
$$r^2=p_1^{2a_1}\cdots p_k^{2a_k}$$
So the square of any rational number has only even exponents in its prime factorisation. The converse is pretty obvious.
Note that from this line of thinking its easy to see that if $m$ is an integer, then $\sqrt{m}$ is either irrational or an integer too.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1000998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How many even numbers will $99$ dice show if we roll them forever under a certain condition? Consider a six-sided die with numbers from $1$ to $6$. Imagine you have a jar with $99$ of such dice. You throw all dice on the floor randomly. You look at one of the dice on the floor at a time. For each die, you do the following:
*
*If it landed at an even number $(2,4,6)$, you turn the die so that it lands on the number $1$.
*If the die landed on an odd number $(1,3,5)$, you throw the die up in the air, so it can land on any number.
After you finish doing the above for all dice on the floor, you come back to the first die and repeat the entire process again. You keep on doing this until eternity (for a billion years, let’s say). If I come into the room after a billion years, how many dice on the floor will have even numbers up?
|
OK, its not about Mathematica, but let's make it about using Mathematica to visualize the result, just for fun.
Analytic
The first run produces some number of nE even numbers with probability
pe0 = PDF[BinomialDistribution[99, 1/2], nE]
Any number from 0 to 99 is possible, justifying the last rule.
Irrespective of the initial distribution, on equilibrium the probability of changing an even number for an odd number must be equal to changing from an odd to an even
Solve[pe == (1 - pe)/2, pe]
{{pe -> 1/3}}
That means in average 1/3 even, 2/3 odd.
Simulation
start = RandomChoice[Range[6], 99];
BarChart[Apply[Labeled, Reverse[Sort@Tally[start], 2], {1}]]
f = Block[
{rndIndx = RandomInteger[{1, 99}], new},
new = If[EvenQ@Part[#, rndIndx], 1, RandomInteger[{1, 6}]];
ReplacePart[#, rndIndx -> new]
] &
reslist = NestList[f, start, 100000];
evol = Transpose[(#/Total[#]) &[Part[Sort@Tally[EvenQ[#]], All, 2]] & /@ reslist];
ListLogLinearPlot[evol, Joined -> True, Epilog -> {Line[{{0, 1/3}, {100000, 1/3}}], Line[{{0, 2/3}, {100000, 2/3}}]},
PlotRange -> {{1, 100000}, {0, 1}},
Frame -> True
]
Analysis
As we iterate there will be random fluctuations continuously changing the number of odd and even numbers. The probability of creating a 1 is bigger than any other number. After a couple thousand iterations most of the numbers are 1
Sort@Tally[Last@reslist]
{{1, 37}, {2, 9}, {3, 10}, {4, 17}, {5, 15}, {6, 11}}
BarChart[Apply[Labeled, Reverse[%, 2], {1}]]
And the distribution of even and odd numbers is centred around
N[Mean /@ evol]
{0.666522, 0.333478}
or 66.65% Odd, and 33.35% even, very close to 2/3 and 1/3 predicted.
The variation about that average is considerable:
N[StandardDeviation /@ evol]
{0.048476, 0.048476}
so the answer can not be given as a specific number, but as a distribution.
SmoothHistogram[evol, 0.01, Frame -> True,
Epilog -> {Line[{{1/3, 0}, {1/3, 9}}], Line[{{2/3, 0}, {2/3, 9}}]},
PlotLabel -> "Probability Distribution",
FrameLabel -> {"Proportion", "Probability density"}]
Conclusion
After a not so long time, provided a couple of thousand iterations have pass, with more than 95% probability you will find a pile with roughly between 57% to 76% odd numbers.
Only the average number of odds tends to 2/3 in the limit where the samples goes to infinity. The exact number of a particular instance can not be determined, but the most likely outcome will be 33 even, 66 odds of which a one (1) will be chosen to be rolled again.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1001079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Question about recurrence relation problem.
solve the following recurrence relation, subject to given initial conditions.
$a_{n+1} = 6a_n -9,$
$a_0 = 0,$
$a_1 = 3.$
Here is what I have done.
$a_{n+1} - 6a_n +9 = 0$
$a_n = r^n$
$r^{n+1} - 6r^n + 9 = 0$
$r + 6 = 0$
$r = -6$
$a_n$ = $X(-6)^n$
$a_1 = 3 = X(-6)^1$
$X = -1/2$
Therefore
$a_n = -1/2(-6)^n$
Am I on the right track or are there any mistakes I made somewhere?
Edited: added the capture of the problem
|
$$\begin{align}
a_{n+1}&=6a_n-9\\
a_{n+1}-\frac95&=6a_n-\frac{54}5\\
&=6\left(a_n-\frac95\right)\\
u_{n+1}&=6u_n\\
u_n&=6u_{n-1}=6^2u_{n-2}=\cdots=6^{n-1}u_1\\
&=6^{n-1}\left(a_1-\frac95\right)\\
&=6^{n-1}\left(3-\frac95\right)\\
a_n-\frac95&=\frac65(6^{n-1})\\
&=\frac{6^n}5\\
a_n&=\frac15\left(9+6^n\right)\qquad\blacksquare
\end{align}$$
Check:
$$\begin{align}
6a_n-9&=\frac65(9+6^n)-9\\
&=\frac95+6(6^n)\\
&=\frac15(9+6^{n+1})\\
&=a_{n+1}
\end{align} $$
Hence formula is correct.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1001193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Prove $\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4}$ How does one prove the following integral
\begin{equation}
\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4}
\end{equation}
Wolfram Alpha and Mathematica can easily evaluate this integral. This integral came up in the process of finding the solution this question: Evaluating $\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$. There are some good answers there but avoiding this approach. I have been waiting for a day hoping an answers would be posted using this approach, but nothing shows up. The integral cannot be evaluated separately since each terms doesn't converge. I tried integration by parts but the problem arises when substituting the bounds of integration.
I would appreciate if anyone here could provide an answer where its approach using integral only preferably with elementary ways.
|
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{\ln\pars{2} - \ln\pars{1 + x^{2}} \over
1 - x}\,\dd x}
\\[5mm] = &\
-\int_{x\ =\ 0}^{x\ =\ 1}\bracks{\ln\pars{2} -
\ln\pars{1 + x^{2}}}\,\dd\ln\pars{1 - x}
\\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\, &
\int_{0}^{1}\ln\pars{1 - x}\pars{-\,{2x \over 1 + x^{2}}}\,\dd x =
-2\int_{0}^{1}{x\ln\pars{1 - x} \over \pars{x + \ic}\pars{x - \ic}}\,\dd x
\\[5mm] = &\
-\int_{0}^{1}\ln\pars{1 - x}\pars{{1 \over x + \ic} +
{1 \over x - \ic}}\,\dd x =
-\,\Re\int_{0}^{1}{\ln\pars{1 - x} \over x + \ic}\,\dd x
\\[5mm] = &\
-\,\Re\int_{0}^{1}{\ln\pars{x} \over 1 + \ic - x}\,\dd x =
-\,\Re\int_{0}^{1}{\ln\pars{x} \over 1 - x/\pars{1 + \ic}}
\,{\dd x \over 1 + \ic}
\\[5mm] = &\
-\,\Re\int_{0}^{\pars{1\ -\ \ic}/2}{\ln\pars{\bracks{1 + \ic}x} \over
1 - x}\,\dd x
\\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,&
-2\,\Re\int_{0}^{\pars{1\ -\ \ic}/2}{\ln\pars{1 - x} \over x}\,\dd x
\\[5mm] = &\
2\,\Re\int_{0}^{\pars{1\ -\ \ic}/2}\mrm{Li}_{2}'\pars{x}\,\dd x =
2\,\Re\,\mrm{Li}_{2}\pars{1 - \ic \over 2}
\\[5mm] = &\
\mrm{Li}_{2}\pars{{1 \over 2} - {1 \over 2}\,\ic} +
\mrm{Li}_{2}\pars{{1 \over 2} + {1 \over 2}\,\ic}
\\[5mm] = &
{\pi^{2} \over 6} - \ln\pars{{1 \over 2} - {1 \over 2}\,\ic}
\ln\pars{{1 \over 2} + {1 \over 2}\,\ic}
\label{1}\tag{1}
\end{align}
In (\ref{1}) I used the
Euler Reflection Formula.
Then,
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{\ln\pars{2} - \ln\pars{1 + x^{2}} \over
1 - x}\,\dd x} =
{\pi^{2} \over 6} -
\verts{-\,{1 \over 2}\,\ln\pars{2} - {1 \over 4}\,\pi\ic}^{\, 2}
\\[5mm] = &\
\bbx{{5\pi^{2} \over 48} - {1 \over 4}\,\ln^{2}\pars{2}}
\approx 0.9080
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1001269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
}
|
Algebras: Finite Additivity $\implies$ Countable Additivity (Idea?) Given an algebra $\mathcal{A}$.
Consider a set function $\mu:\mathcal{A}\to\mathbb{R}_+$:
Then countable additivity follows from finite additivity:
$$\mu(A+B)=\mu(A)+\mu(B)\implies\mu(A)=\sum_{n\in\mathbb{N}}\mu(A_n)$$
Can you explain the idea behind the proof?
(I always forget the precise steps probably because I haven't understand it yet.)
|
I think there is something missing in yiur question.
To see this, consider the algebra(!) of all subsets $A \subset \Bbb{N}$ which are finite or for which $A^c$ is finite.
Define $\mu(A)=0$ if $A$ is finite and $\mu(A)=1$ otherwise. It is easy to verify that $\mu$ is finitely additive, but not countably additive.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1001457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Proof the concave transformation of the tail distribution is always above the tail distribution I need to prove that for a given continuous non-decreasing distribution $F_X(x)$, and a concave non-decreasing distortion function $g(.)$ defined on $[0,1]$, the following holds: $$g(1-F_X(x)) \ge 1-F_X(x)$$
We know that $g(0) = 0$ and $g(1) = 1$.
I was thinking to use the fact that if $g$ is concave then for $t$ in $[0,1]: g(tx + (1-t)y) \ge tg(x) + (1-t)g(y)$ , for any $x,y$ in the domain of the distribution but can't seem to find a solution. It's probably pretty obvious, but I'm stuck so any help would be truly appreciated.
Best,
tb
|
If you're not familiar with it already, Jensen's Inequality is applicable here:
$E[g(1-\mathbf{1}_{\leq x}(X))]\leq g(1-E[\mathbf{1}_{\leq x}(X)])=g(1-F_X(x))$ [for $g$ concave].
Since $g(0)=0$ and $g(1)=1$ then
$E[g(1-\mathbf{1}_{\leq x}(X))]=E[\mathbf{1}_{\geq x}(X)]=1-F_X(x)\implies g(1-F_x)\geq 1-F_X$
$\square$...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1001591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Limits of Exponent Laws I have recently learned (discovered) that the exponent law $b^{mn} = {(b^m)}^n$ is not universally applicable. To demonstrate, if it were we could conclude that $(-1)^{\frac{3}{2}}$ (or by extension -1 to any power) is equal to 1.
$(-1)^{\frac{3}{2}} = (-1)^{2*\frac{3}{4}}$
= $((-1)^2)^{\frac{3}{4}}$
= $1^{\frac{3}{4}}$
= $1$
Under exactly what circumstances (natural/integer/rational/real bases or exponents) are the various exponent laws applicable? For example, can I use the above law ( $b^{mn} = {(b^m)}^n$ ) with a negative base when the exponents are required to be integers?
|
Basically, the one thing you need to remember is : $b^m$ does not have a signification when the two following condition are simultaneously met :
-b is negative
-m is not an integer
So in your example, $(-1)^{\frac{3}{2}}$ does not have any signification. Hence obviously, you should not write it.
Just remember that and you can only write correct equalities.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1001665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Writing iterated integral of a function Write an iterated integral of a function f for the region given by a triangle with vertices at point (1,1), (1,2), (3,0).
I figured that I'm supposed to first find the equations of the three lines representing the sides of the triangle.
(1,2) and (1,1) make $x=1$
(1,1) and (3,0) make $y= -(1/2)x+(3/2)$
(1,2) and (3,0) make $y= -x+3$
Is that all I need to do? In other words, is my iterated integral the following?
$$ \int_{1}^3 \int_{\frac{1}{2}x+\frac{3}2}^{-x+3} f(x,y) dydx $$
It seems too simple and I'm afraid I'm missing something.
|
Yep, it's as easy as that! Note that you got the slope of your second line wrong: it should be:
$$
y = \frac{-1}{2}x + \frac{3}{2}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1001760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solve $\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$ for $y$ I have the equation:
$\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$,
where $k$ and $c$ are arbitrary constants.
How do I go about simplifying this and solving for $y$ in terms of $x$, excluding the obvious solution $y=-x$
|
Treat $x$ as a constant as well. We could use the standard quadratic formula, but since you noticed that $y = -x$ is a solution, let's try factoring instead. We obtain:
\begin{align*}
0
&= \tfrac{1}{2}ky^{2} - \tfrac{1}{2}kx^{2} + cy + cx \\
&= \tfrac{1}{2}k(y^{2} - x^{2}) + c(y + x) \\
&= \tfrac{1}{2}k(y - x)(y + x) + c(y + x) \\
&= (\tfrac{1}{2}k(y - x) + c)(y + x) \\
\end{align*}
Thus, the other solution can be obtained by setting the first factor equal to zero, yielding:
\begin{align*}
\tfrac{1}{2}k(y - x) + c &= 0 \\
\tfrac{1}{2}k(y - x) &= -c \\
y - x &= \tfrac{-2c}{k} \\
y &= x - \tfrac{2c}{k}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1001876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
How to determine if set of vectors is a basis for W Consider the subspace $$
W =\left\{ \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3}\end{bmatrix} \in \mathbb{R}^3 \,| x_{1}+x_{2}+x_{3} = 0 \right\}
$$
Is the set S a basis for W? $$
S= \left\{ \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix} , \begin{bmatrix} -3 \\ 2 \\ 1\end{bmatrix} \right\}
$$
I'm not too sure how to go about this question but this is what I've tried. Let $x_{2} = s$ and $x_{3}=t$. Then I have $$
\begin{bmatrix} -s-t \\ s \\ t\end{bmatrix} =s \begin{bmatrix} -1 \\ 1 \\ 0\end{bmatrix} + t \begin{bmatrix} -1 \\ 0 \\ 1\end{bmatrix}
$$
Therefore, the basis for $W$ is $$
A= \begin{bmatrix} -1 \\ 1 \\ 0\end{bmatrix} , \begin{bmatrix} -1 \\ 0 \\ 1\end{bmatrix}
$$
Since the vectors in $S$ is not a scalar multiple of the vectors in $A$, $S$ is not a basis for $W$.
Could anyone confirm if I am correct? Thanks
|
$S$ is a basis for $W$ since: (2) span $W$; (1) is a linearly independent set.
To prove (1) you just have to solve: $\alpha (-1,-1,2)+\beta(-3,2,1)=0$ for $\alpha$ and $\beta$ to get $\alpha=0=\beta$.
To prove (2): Let $(x_1,x_2,x_3)\in W$ (i.e. $x_1+x_2+x_3=0$) you have to find $\alpha$ and $\beta$ such that $\alpha (-1,-1,2)+\beta(-3,2,1)=(x_1,x_2,x_3)$ but $\beta=-\frac{x_1+x_2}{5}$ and $\alpha=\frac{x_3-\frac{x_1+x_2}{5}}{2}$ works. Then we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1001960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
taking the inverse of power series I am working with solution to near regular singular points.
I started with:
$$y_1(x)=x^\frac{1}{2}\left[1-\frac{3}{4}x+\frac{9}{64}x^2-\frac{3}{256}x^3+\cdots\right] $$
Then I squared it:
$$y_1^2(x) = x\left(1-\frac{3}{2}x+\frac{27}{32}x^2-\frac{15}{64}x^3+\cdots\right)$$
Why is the inverse:
$$\frac{1}{x}\left[1+\frac{3}{2}x+\frac{45}{32}x^2+\frac{69}{64}x^3+\cdots\right] \text{ ?}$$
I cannot seem to see how this works out. Any pointers
|
If $$y=\sqrt{x} \left(1-\frac{3 x}{4}+\frac{9 x^2}{64}-\frac{3 x^3}{256}+\cdots\right)$$ then effectively $$y^2=x \left(1-\frac{3 x}{2}+\frac{27 x^2}{32}-\frac{15 x^3}{64} +\cdots\right)$$ Now you want to compute $\frac{1}{y^2 }$. You can write $$\frac{x}{y^2 }=\frac{1}{1-\frac{3 x}{2}+\frac{27 x^2}{32}-\frac{15 x^3}{64} +\cdots}$$ and perform the long division. Another way is to consider the rhs as $\frac{1}{1-z}$ and expand it as $$\frac{1}{1-z}=1+z+z^2+z^3+\cdots$$ and reuse $$z=\frac{3 x}{2}-\frac{27 x^2}{32}+\frac{15 x^3}{64}+\cdots$$ to affectively arrive, after substitution and simplifications, to $$\frac{x}{y^2 }=1+\frac{3 x}{2}+\frac{45 x^2}{32}+\frac{69 x^3}{64}+\cdots$$ and finally $$\frac{1}{y^2 }=\frac{1}{x}\Big(1+\frac{3}{2}x+\frac{45}{32}x^2+\frac{69}{64}x^3+\cdots\Big)$$
From a semantic point of view, I shall not call that operation the inverse of a power series (which is something different).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1002059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Free Group Norms
Hello everyone, I'm trying to solve this problem, but I'm stuck... i don't quite understand the definition of the norm, If you guys can give me a better explanation, I would appreciate it, Thanks
|
You have to calculate how many groups elements are represented by words at length at most $N$ in the group generators.
Let $G$ be free abelian with generators $a,b$. How many elements have shortest representative of length $n$? For $n=0$ there is just $1$. For $n>0$, we have $a^n,b^n,a^{-n},b^{-n}$, which is $4$. We also have $a^kb^{n-k}$ for $k=1,2,\ldots,n-1$, which makes $n-1$, and we get three more sets of size $n-1$ by replacing $a$ by $a^{-1}$ and/or $b$ by $b^{-1}$. So that makes $4+4(n-1) = 4n$ in total of length $n$. Summing these from $n=1$ to $N$ and adding on $1$ for length $0$, gives the required answer.
For the free group, the first generator in a reduced word can be any one of $a^{\pm 1}$ or $b^{\pm 1}$, making four possibilities, but there are only $3$ for subsequent generators in the word, since they must be the inverse of the preceding genertor. So, for $n >0$, there are $4.3^{n-1}$ group elements whose shortst representatives have length $n$. Again you need to some from $1$ to $N$ and add on $1$ for length $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1002165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
After removing any part the rest can be split evenly. Consequences?
Let $S$ be a finite collection of real numbers (not necessarily distinct). If any element of $S$ is removed then the remaining real numbers can be divided into two collections with same size and same sum ; then is it true that all elements of $S$ are equal ?
(I know the result is true if the "reals" are replaced by "integers".)
|
Yes, all elements of $S$ must be equal.
Clearly, $S$ has an odd number of elements; say $S=\{x_1,\dots,x_{2n+1}\}$. The condition "if any element of $S$ is removed then the remaining real numbers can be divided into two collections with same size and same sum" means that the column vector $\mathbf X=[x_1,\dots,x_{2n+1}]^T$ satisfies a matrix equation of the form $A\mathbf X=\mathbf0$ for some $2n+1\times 2n+1$ matrix $A$ with all diagonal entries $0$, all off-diagonal entries $\pm1$, and equal numbers of $+1$s and $-1$s on each row. All we have to do is show that such a matrix must have rank $2n$; its right null space must then be $1$-dimensional, so it contains only the constant vectors $[x,x,\dots,x]^T$. In fact, working in$\!\mod2\,$ arithmetic for simplicity, it's easy to see that the matrix can be reduced by elementary row operations to an upper triangular form with $2n$ ones and a zero on the diagonal. (Namely, swap the first two rows, and subtract them from all rows below them; swap the next two rows, and subtract them from all rows below them; and so on.)
To put it more generally, if $A$ is a square matrix of odd order $2n+1$, and if all diagonal entries are even integers, and all off-diagonal entries are odd integers, then the rank of $A$ is at least $2n$.
Alternatively, if the equation $A\mathbf X=\mathbf0$ had a nonconstant solution, it would also have a nonconstant integer solution; but the OP has already proved (see also JiK's comment on the question) the result for integers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1002267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
}
|
Number of positive integral solutions to $x+y+z+w=20$ with $xWhat is the number of positive unequal integral solution of the equation $x+y+z+w=20$, if $\,x<y<z<w\,$ and $\,x,y,z,w\ge1\;?$
How to solve this question?
|
For a count of the solutions without restriction on the order by inclusion-exclusion, see this answer. The result is $552$; then the number of ordered solutions is obtained as $\frac{552}{4!}=23$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1002344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Integral equation and metric spaces Let $C([0,\frac{\pi }{2}])$ be the set off all continuous functions defined on $[0,\frac{\pi }{2}]$ . Prove that this integral equation $$
f(t) = \int\limits_0^{\frac{\pi }{2}} {\arctan } (\frac{{f(s)}}{2} + t)\,ds
$$ has an unique solution on $C([0,\frac{\pi }{2}])$.
Any ideas ? I just started studying metric spaces and an older peer suggested this problem, but I don't even know where to start. I'd be very grateful to anyone who could show me how such a pretty problem can be solved .
|
Daniel is right, Banach's fixed point theorem is the way to go. Defining $A$ as he did, we see that $A: C([0,\pi/2]) \rightarrow C([0,\pi/2])$. In addition,
$||Af-Ag||_{C([0,\pi/2])} = \sup_{t \in [0,\pi/2]} \left|\int_0^{\pi/2} \arctan\left(\frac{f(s)}{2} +t\right) - \arctan\left(\frac{g(s)}{2} + t\right) ds\right|$
Since arctan is Lipschitz continuous with Lipschitz constant 1 (this follows from the fact that its derivative is uniformly bounded), we have
$||Af-Ag||_{C([0,\pi/2])} \leq \left|\int_0^{\pi/2} \frac{f(s)}{2} - \frac{g(s)}{2} ds\right| \leq \int_0^{\pi/2} \frac{1}{2} ||f - g||_{C([0,\pi/2])} ds = \frac{\pi}{4} ||f-g||_{C([0,\pi/2])}$
This shows that $A$ is a strict contraction, and thus there exists a unique fixed point, i.e. a solution to the given integral equation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1002447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$ Question:
let $x,y,z>0$ and such $xyz=1$, show that
$$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$
My idea: use AM-GM inequality
$$x^3+x^3+1\ge 3x^2$$
$$y^3+y^3+1\ge 3y^2$$
$$z^3+z^3+1\ge 3z^2$$
so
$$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$
But this is not my inequality,so How prove it? I know this condition is very important.but how use this condition? and this inequality is stronger
|
Here is a possible solution: (although it is not the most elegant one)
I will employ Mixing Variables technique here. Since the inequality is symmetric, WLOG let $x=\min(x,y,z)$. Therefore $t^2:=yz \ge 1$.
Let
$$f(x,y,z)=x^3+y^3+z^3-2(x^2+y^2+z^2)$$
I wish to show
$$f(x,y,z)\ge f(x,\sqrt{yz},\sqrt{yz}) = f(\frac1{t^2},t,t) \ge -3$$
Let us put $p^2=x, q^2=y, r^2=z$. The first inequality in the above chain is equivalent to
$$q^6+r^6-2q^3r^3 \ge 2(q^4+r^4-2q^2r^2)$$
$$\iff (q^3-r^3)^2 \ge 2(q^2-r^2)^2$$
$$\iff (q^2+qr+r^2)^2 \ge 2(q+r)^2$$
This is true since
$$(q^2+qr+r^2)^2 \ge q^4+r^4+2q^2r^2+2qr(q^2+r^2) \ge 4q^2r^2+2(q^2+r^2) \ge 2(q+r)^2$$
Therefore it enough to prove $f(\frac1{t^2},t,t)\ge -3$ for $t>0$ which is equivalent to
$$(t-1)^2((t^7-2t^5+t^3)+(t^7+t-2t^4)+(t^4-t^3+t^2)+t+1)\ge 0$$
Each term in the brackets of the large factor is greater than zero by AM-GM.
The last part is little tedious to do by hand. But you always know that $(t-1)$ has to be factor (possibly with multiplicity $2$) of that thing. That helps in simplification.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1002529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
}
|
2(n-1) induction There are $n$ cities and every pair of cities is connected by exactly one direct one-way road. Now more one-way roads have been added between some cities so that between some pairs of cities there may be two direct roads between them, for example, there may now be a road going directly from some city A to some city B and a road going directly from city B to city A. Note that now between every pair of cities there is at least one direct road.
The goal is to find a dead-end city, if it exists, i.e., a city D to which there is a direct one-way road from every other city, but there is no direct road going from D to any other city. You are allowed to make only one type of query – “Is there a direct road going from city A to city B?” The answer to this query will be a “Yes” or a “No”. Use mathematical induction to show that if there are $n$ cities then one can find a dead-end city, if there is one, using at most $2(n − 1)$ queries.
So far I've gotten:
We will prove this claim using induction on n.
IH: Assume that the claim is true when $n = k$, for some $k > 1.
$2(k-1)$
BC: $k = 2$
$2(2-1) = 2$
IS: We want to prove that the claim is true when $n = k + 1$
Am I on the right path or?
|
Yes, you're on the right track.
Hint: Your first question will rule out one city as a potential dead end. What would happen if you never asked about that city again, and simply pretended it didn't exist?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1002645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Mathematical trivia (i.e. collections of anecdotes and miscellaneous (recreational) mathematics) Can you suggest some books on mathematical trivia?
I use the word "trivia" with a double meaning in this case:
*
*curious anecdotes that enlighten what the real life of mathematicians is like (like the ones in Mathematical Apocrypha Redux);
*curious miscellaneous (also mainly recreational) mathematical ideas and topics with only occasional anecdotes and jokes (like the ones in Professor Stewart's Cabinet of Mathematical Curiosities and Professor Stewart's Hoard of Mathematical Tresures).
|
My suggestions:
1."The Penguin book of curious and interesting mathematics" by David G. Wells;
2."The Colossal Book of Mathematics" by Martin Gardner;
3."Maths Facts, Fun, Tricks and Trivia" by Paul Swan;
4."Math hysteria" by Ian Stewart;
5."In Mathematical Circles: A Selection of Mathematical Stories and Anecdotes: Quadrants I, II, III, and IV (Mathematical Association of America)" by Howard W. Eves.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1002768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 0
}
|
$\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ I'm trying to determine $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ using L'Hopital's Rule.
I can clearly see that $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}} = \frac{\infty}{\infty},$ so we can use L'Hoptial's Rule. I'm having trouble differentiating $f(x) = (\frac{x+1}{x-1})^{\sqrt{x^2-1}}$. I've used Mathematica, but I won't understand it unless I see the step-by-step process.
|
Use that if $\displaystyle\lim_{x\to+\infty}f(x)=L$ then $\displaystyle\lim_{x\to+\infty}\ln(f(x))=\ln(L)$, for $f(x),L>0$.
And that $\ln(a^b)=b\ln(a)$ for $a,b>0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1002826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
Show that if $f$ is complex differentiable and in some region $U$ it is $f'=0$, then $f$ is constant I have to show show that if $f$ is complex differentiable and in some region $U$ it is $f'=0$, then $f$ is constant. How can one prove it?
|
Let $a\in U$ and $V=\{z\in U / f(z) =f(a)\}$. Show that $V$ is open and closed in $U$.
Let $b $ in $ V$ hence $ b$ in $ U$. There is $ r>0$ st the ball $B(b,r)$ is in $U$.Fix $z$ in this ball. The map $g(t)=f(tb+(1-t)z)$ is dif and $g'=0$ then $g(0)=g(1)$. This show the opnes.
$V$ is closed in $U$ because $f$ is continuous.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1002946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
quick way to prove $\mathbb{Z}_2 \times \mathbb{Z}_3$ is isomorphic to $\mathbb{Z}_6$ I want to show that $\mathbb{Z}_2 \times \mathbb{Z}_3$ is isomorphic to $\mathbb{Z}_6$. The number of elements and their orders are equal but I don't see a way to prove that the groups are isomorphic except for writing down the Cayley table of the group operation. Is there a better way?
|
Hint:
What is the subgroup generated by $ (1_{\Bbb Z_2},1_{\Bbb Z_3}){}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$ in $\Bbb Z_2\times \Bbb Z_3$?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1003041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.