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H: Cohomological definition of the Chow ring
Let $X$ be a smooth projective variety over a field $k$. One can define the Chow ring $A^\bullet(X)$ to be the free group generated by irreducible subvarieties, modulo rational equivalence. Multiplication comes from intersection. The problem is, verifying that everything is well-defined is quite a pain. My question is
Is there a definition of $A^\bullet(X)$ that is purely cohomological?
In other words, is it possible to give a definition of $A^\bullet(X)$ that works for any ringed space? Probably the fact that this definition is equivalent to the usual one will be a theorem of some substance.
Note: I am not claiming that the usual definition of the Chow ring is "bad" in any way. I just think it would be nice to know if there was a "high-level" approach.
AI: The Bloch formula (proved by Quillen, special cases by Bloch) holds for any regular scheme $X$ of finite type over a field: the group of algebraic cycles of codimension $i$ modulo rational equivalence is isomorphic to the cohomology group $H^i(X,\mathcal{K}_i)$ where $\mathcal{K_i}$ is the sheaf associated to the presheaf that sends each open subset $U$ to $K_i(U)$, where $K_i$ is the $K$-theory functor in the sense of Quillen. One can use this to define the intersection product.
Note: $K_i$ is the functor on schemes that sends $X$ to the abelian group $K_i(X)$ where $K_i$ is defined as $K_i$ of the exact category $P(X)$ of vector bundles on $X$. If $f:X\to Y$ is a morphism of schemes the inverse image functor $f^*:P(Y)\to P(X)$ is an exact functor inducing a morphism of $K$-groups $K_i(Y)\to K_i(X)$. In particular, if $U\subseteq X$ is an open subscheme then this provides the restriction, whence $U\mapsto K_i(U)$ is really a presheaf, so there is an associated sheafification.
Of course, defining $K_i(\mathcal{A})$ for an exact category $A$ requires some machinery, but the definition in Quillen's paper assumes only a basic knowledge of category theory and a bit of homological algebra. If the homotopy theory of categories is a bit fast there try Weibel's K-Book (available on his website), Ch. 4, Section 3.
The Quillen's paper "Higher Algebraic K-Theory I", Theorem 5.19 for the original proof. An excellent introduction is Gillet's paper "K-Theory and Intersection Theory" which is in the great "Handbook of K-Theory". |
H: Determinant of a nilpotent matrix
Let $A$ be a nilpotent matrix. Prove that $\det(I+A)=1$
Could someone at least give me a clue ?
AI: Since $A$ is nilpotent, we have
$A^m = 0 \tag{1}$
for some positive interger $m$. This implies every eigenvalue of $A$ vanishes, since the equation
$Av = \lambda v \tag{2}$
for non-zero $v$ (recall eigenvectors are required to be non-zero) implies
$0 = A^mv = \lambda^m v, \tag{3}$
whence
$\lambda^m = 0, \tag{4}$
since $v \ne 0$. (4) forces
$\lambda = 0 \tag{5}$
Now use the fact that for any scalars $\lambda$ and $a$, $\lambda$ is an eigenvalue of $A$ if and only if $\lambda + a$ is an eigenvalue of $A + aI$; indeed we have, from (2),
$(A + aI)v = Av + av = (\lambda + a)v. \tag{6}$
(6) allows us to conclude that every eigenvalue of $A + I$ is $1$; hence $\det (A+I)$, being the product of its eigenvalues, satisfies
$\det(A+I) = 1. \tag{7}$
QED.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!! |
H: ENS is an abbreviation of?...
In CWM Mac Lane uses the term $\mathbf {ENS}$ for a category having as objects the subsets of a given set and as morphisms the functions from these sets to these sets. What is abbreviated by the letters ENS?
AI: I believe ENS stands for ensembles, which means sets in French. |
H: Calculation of an integral via residue.
$$\int_{-\infty}^{\infty}{{\rm d}x \over 1 + x^{2n}}$$
How to calculate this integral? I guess I need to use residue. But I looked at its solution. But it seems too complicated to me. Thus, I asked here. Thank you for help.
AI: I'm pretty sure this has been done on this site, but because I cannot locate it, I will quickly work it out.
Because the integrand is even, you can write the integral as
$$2 \int_0^{\infty} \frac{dx}{1+x^{2 n}}$$
Now consider the contour integral
$$\oint_C \frac{dz}{1+z^{2 n}}$$
where $C$ is a wedge contour that goes from $[0,R]$ on the real axis, then along the arc $z=R e^{i \phi}$, where $\phi \in [0,\pi/n]$, and then along the line $z=e^{i \pi/n}t$, where $t \in [R,0]$. The contour integral is then equal to
$$\int_0^R \frac{dx}{1+x^{2 n}} + i R \int_0^{\pi/n} d\phi \, \frac{e^{i \phi}}{1+R^{2 n} e^{i 2 n \phi}} + e^{i \pi/n} \int_R^0 \frac{dt}{1+t^{2 n} e^{i 2 \pi n/n}}$$
The second integral vanishes as $1/R^{2 n-1}$the limit as $R \to \infty$. The contour integral is then equal to, in this limit:
$$\left (1-e^{i \pi/n} \right ) \int_0^{\infty} \frac{dx}{1+x^{2 n}} $$
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue at the only pole of the integrand interior to $C$, which is at $z=e^{i \pi/(2 n)}$. The residue there is equal to
$$\frac{1}{2 n e^{i (2 n-1) \pi/(2 n)}} = -\frac{1}{2 n} e^{i \pi/(2 n)} $$
Therefore the integral is given by
$$\left (1-e^{i \pi/n} \right ) \int_0^{\infty} \frac{dx}{1+x^{2 n}} = -i 2 \pi \frac{1}{2 n} e^{i \pi/(2 n)}$$
and thus
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^{2 n}} = \frac{\pi/n}{\sin{[\pi/(2 n)]}}$$ |
H: Hilbert's basis theorem for power series ring in many variables
My teacher Proved that if R is Noetherian then $R[x]$ and $R[[x]]$ are Noetherian , how can i prove that again R is Noetherian if and only if $R[[x_1,...,x_n]]$ is Noetherian ,thanks for your time and help.
AI: Hints:
A quotient ring of a Noetherian ring is Noetherian.
Secondly, notice that if you take $R[[x,y]]$ and organize monomial terms with a common maximal power of $y$ and factor this power of $y$ out, you are looking at something in $(R[[x]])[[y]]$. You can use a similar idea with induction to show it for $n$ variables. |
H: Need help finding the Laurent Series
Been struggling with this for a while now. Just looking for a nudge in the right direction :
Find the Laurent series for $f(z) = \frac{z^2-2z+3}{z-2}$ about $z=2$.
The only thing I can think to do is
$$ \frac{z^2-2z+3}{z-2} = z + \frac{3}{z-2}$$
but I don't think this is correct since the first term is not "about $z=2$," I tried using the infinite geometric series but it didn't seem to get me anywhere...
AI: If the first term is not "about $z=2$", then you can re-write your expression so that it will be : $$z+\dfrac{3}{z-2}=(z-2)+2+\dfrac{3}{z-2}.$$
More generally, looking for the Laurent series of $f$ at $z=a$ is the same as finding the Laurent series of $g:w\mapsto f(w+a)$ at $w=0$ and replace $w$ by $z-a$. |
H: Computing $\displaystyle\int \frac{1}{x^{2}\sqrt{x^{2}+7}}$
Find the primitive function of $$\frac{1}{x^{2}\sqrt{x^{2}+7}}$$
Attempt. And my answer is
$$\arcsin \left( \frac{\sqrt{\left( x^{2}+7 \right)}}{\sqrt{7}} \right)$$
Why am I wrong? See link for whole calculation.
[Wolfram](http://www.wolframalpha.com/input/?i=integrate+1/(x%5E2sqrt(x%5E2+7)).
AI: Hint: Let $x=1/u$; $dx=-du/u^2$. Then sub $v=u^2$. The result is a fairly simple integral.
EDIT
$$\begin{align}\underbrace{\int \frac{dx}{x^2 \sqrt{x^2+7}}}_{u=1/x} &= -\int \frac{du}{u^2} \frac{u^2}{\sqrt{1/u^2+7}} \\ &= -\underbrace{\int du \frac{u}{\sqrt{1+7 u^2}}}_{v=u^2} \\ &= -\frac12 \underbrace{\int \frac{dv}{\sqrt{1+7 v}}}_{y=7 v}\\ &= -\frac{1}{14} \int \frac{dy}{\sqrt{1+y}}\\ &= -\frac17 \sqrt{1+y}+C\\&= -\frac{\sqrt{x^2+7}}{7 x}+C\end{align}$$ |
H: Does the function $f: \mathbb R \to \mathbb R^2, t \mapsto (t^3, t^2)$ have a history?
I heard one mathematician briefly mentioning that the function $f: \mathbb R \to \mathbb R^2, t \mapsto (t^3, t^2)$ is very famous and has a history.
Do you know what was meant by that?
AI: “It seems that $f$ is the historic earliest example of a curved path of whom the length of compact segments of that path (similar to the length of straight path's segments) may be calculated exactly and purely algebraic (even polynomial in expressions of square root!) from the coordinates of start- and endpoint: that was not considered possible (since Aristotle) concerning curved paths.”
Roughly translated from Peter Dombrowski, Wege in euklidischen Ebenen: Springer, 1999. |
H: Axiomatic definition of complex numbers
Trying to build axiomatically the set $\mathbb C$ of complex numbers, my first attempt was to define $\mathbb C$ with three structures: addition, multiplication and conjugate: $\langle\mathbb C,+,\times,{}^*\rangle$, with $\langle\mathbb C,+,\times\rangle$ forming a commutative field and the conjugate defined as a unitary operator such that (for all $a$, $b$ in the set) $(a+b)^*=a^*+b^*$, $(ab)^*=a^*b^*$, and ${a^*}^*=a$.
Corollary $0^*=0$ and $1^*=1$ for $0$ and $1$ respectively the modules of addition and multiplication.
Of course, the minimum set over which this structure can be build is $\mathbb Z_2$ with the usual addition and multiplication, and the conjugate equivalent to the identity.
If we require the conjugate not to be trivial (not to be the identity), then, by definition, we require 4 elements (the modules of addition and multiplication whose conjugates must be themselves), another element and its conjugate. I have found the following structure over $\{0,1,A,B\}$.
$$
\begin{matrix}
+ & 0 & 1 & A & B \\
\hline
0 & 0 & 1 & A & B \\
1 & 1 & 0 & B & A \\
A & A & B & 0 & 1 \\
B & B & A & 1 & 0 \\
\end{matrix}
\qquad
\begin{matrix}
\times & 1 & A & B \\
\hline
1 & 1 & A & B \\
A & A & B & 1 \\
B & B & 1 & A \\
\end{matrix}
\qquad
\begin{matrix}
0^*=0 \\
1^*=1 \\
A^*=B \\
B^*=A \\
\end{matrix}
$$
A special case of sets with the structure $\langle C,+,\times,{}^*\rangle$ are sets derived by the Cartesian product of a field $F$ with itself, by defining the addition by ordinate and the product $(a,b)\times(c,d):=(ac-bd,ad+bc)$, and the conjugate of $(a,b)$ as $(a,-b$). For this structure to be a field it is necessary that for any $x,y\in F$, then $xx+yy\neq 0$ (unles $x=y=0$).
This means that ${\mathbb Z_3}^2$ can form a complex field, while ${\mathbb Z_5}^2$ cannot (at least not by this definition).
One way to ensure $xx+yy\ne0$ is having a total order in $F$ that is compatible with its addition and multiplication: that is that there is a positive subset $F_+\subset F$ in which addition and multiplication is closed, with $0\notin F_+$, and for any $x\ne0$ then $\{x,-x\}\cap F_+\ne\emptyset$. Given this structure it is easy to show that $xx$ is positive (and therefor non-zero) and $xx+yy$ is positive (and therefor non-zero).
The minimum field with positives are the rational numbers $\mathbb Q$, which means that we can build a complex field from $\mathbb Q^2$.
Note: in a field $F$ with positives $F_+$ the additive module $1$ must be positive and we can find an inductive subset $N$ (by having $n+1\in N$ if $n\in N$) which is analogous to the natural numbers $\mathbb N$. Completing the subfield derived by $N$ we must have a structure analogous to $\mathbb Q$ contained in $F$.
If we superimpose a convergence structure in $\mathbb Q^2$ then we will get to $\mathbb C$.
However, I want to define $\mathbb C$ axiomatically and not by construction.
Question
Besides a field structure and a non-trivial conjugate, which other structures most be imposed to a set $C, \langle C,+,\times,{}^*\rangle$ so that we end up with $\mathbb C$?
(I can define the subset $R$ of fixed points of $C$ by ${}^*$, and require positiveness in $R$ but somehow it seems like cheating.)
AI: There is no first-order theory that will do the job. For if $T$ is a first-order theory over the language you describe, or an extension of it, and $T$ has $\mathbb{C}$ as a model, then by the Upward Lowenheim-Skolem Theorem, $T$ will have models of arbitrarily large cardinality. Moreover, if the language (number of non-logical symbols) is finite or countably infinite, then $T$ will have a model that is finite or countably infinite.
The "nearest" one can get along the lines you describe is the theory of algebraically closed fields of characteristic $0$. (We need an infinite number of axioms, but that's no problem.) |
H: Countability in topology
I am looking at the following example: $\mathbb{R}$ is second countable because consider the open intervals $(a,b)$ with rational endpoints. This is countable base for the usual topology on real line $\mathbb{R}$.
How is the above base base (interval) with rational endpoints countable? Isn't it infinite as it will be a subset of a real line?
AI: Countable simply means in bijection with a subset of $\mathbb N$ (so not necessarily finite!). One can show that the rationals $\mathbb Q$ are countable and that any countable union of countable sets is also countable. How many intervals are in the set $\{ (a,b): a,b \in \mathbb Q, a<b\}$? This will be a subset of $\mathbb Q \times \mathbb Q$ which is also countable, and hence is countable itself. |
H: Completely baffled by this question involving putting matrices in matrices
This is homework, so only hints please.
Let $A\in M_{m\times m}(\mathbb{R})$
, $B\in M_{n\times n}(\mathbb{R})$
. Suppose there exist orthogonal matrices $P$
and $Q$
such that $P^{T}AP$
and $Q^{T}BQ$
are upper triangular. Let $C$
be any $m\times n$ matrix. Then show that there is an orthogonal $R$
, not depending on $C$
, such that $$R^{T} \begin{bmatrix}A & C\\
0 & B
\end{bmatrix}R$$
is upper triangular.
I haven't done any proofs before using matrices containing matrices in them. I have absolutely no idea where to begin. Any ideas?
AI: Take
$$R = \operatorname{diag}(P,Q) = \begin{bmatrix} P \\ & Q \end{bmatrix},$$
and see what happens. Ask if you get stuck. |
H: Show that f is a continuous function if and only if for every closed set C in Y, $f^{-1}(C)$ is closed in X.
Suppose X and Y are topological spaces with topology $T^x$ and $T^y$
Let $f: X \to Y$ be a function.
Show that $f$ is a continuous function if and only if for every closed set $C$ in $Y$, $f^{-1}(C)$ is closed in $X$.
I know that $f$ is continuous function if and only if for every open set $O$ in $Y$, $f^{-1}(O)$ is open in $X$.
How do I show for every open set $O$ in $Y$, $f^{-1}(O)$ is open in $X$ if and only if for every closed set $C$ in $Y$, $f^{-1}(C)$ is closed in $X$.
AI: HINT: If $C$ is closed then $C = Y - O$ for an open set $O$. What is $f^{-1}(Y-O)$? That proves that the inverse of a closed set is closed. Now use the same idea to show the converse statement. |
H: Result and proof on the conditional expectation of the product of two random variables
My problem is the following:
$X$ and $Y$ are two random variables and $\mathcal{F}$ is a $\sigma$-algebra. Given that $X$ and $Y$ are independent, and that $X$ is independent of $\mathcal{F}$, can I affirm that
$$\mathbb{E}(XY \mid \mathcal{F}) = \mathbb{E}(X) \cdot \mathbb{E}(Y \mid \mathcal{F})?$$
My intuition is yes but I did not manage to find a formal proof. Any idea ?
Thanks a lot.
AI: No, in general the statement is not correct.
Example Consider sets $A_1,A_2,A_3$ such that $\mathbb{P}(A_1 \cap A_2 \cap A_3) = 0$ and $\mathbb{P}(A_i)>0$, $\mathbb{P}(A_i \cap A_j) = \mathbb{P}(A_i) \cdot \mathbb{P}(A_j)$ for any two $i,j \in \{1,2,3\}$. Set $X:=1_{A_1}$, $Y=1_{A_2}$, $\mathcal{F} := \sigma(1_{A_3})$. Then $X$ and $Y$ are independent, as well as $X$ and $\mathcal{F}$, so the assumption on the independence is satisfied. For $F:= A_3$ we have
$$\int_F X \cdot Y \, d\mathbb{P} = \int 1_{A_1} \cdot 1_{A_2} \cdot 1_{A_3} \, d\mathbb{P} = \mathbb{P}(A_1 \cap A_2 \cap A_3) = 0 \tag{1} $$
On the other hand,
$$\int_F \mathbb{E}(X) \cdot \mathbb{E}(Y \mid \mathcal{F})\, d\mathbb{P} = \mathbb{P}(A_1) \cdot \underbrace{\mathbb{E}(Y)}_{\mathbb{P}(A_2)} \cdot \mathbb{P}(A_3)>0 \tag{2}$$
where we used that $Y$ and $\mathcal{F}$ are independent (hence $\mathbb{E}(Y \mid \mathcal{F})=\mathbb{E}(Y)$). Since by the definition of conditional expectation
$$0 \stackrel{(1)}{=} \int_F X \cdot Y \, d\mathbb{P} \stackrel{!}{=} \int_F \mathbb{E}(X \cdot Y \mid \mathcal{F}) \, d\mathbb{P}$$
we conclude from $(2)$ that
$$\mathbb{E}(X \cdot Y \mid \mathcal{F}) \neq \mathbb{E}(X) \cdot \mathbb{E}(Y \mid \mathcal{F})$$
But: If $\sigma(X)$ is independent of $\sigma(Y,\mathcal{F})$ (i.e. the smallest $\sigma$-algebra containing $\mathcal{F}$ and $\sigma(Y)$), then the claim holds as the following proof shows:
Denote by $\mathcal{G} := \sigma(\mathcal{F},\sigma(Y))$ the $\sigma$-algebra generated by $\mathcal{F}$ and $Y$. Since $\sigma(X)$ and $\mathcal{G}$ are independent, we have
$$\mathbb{E}(X \mid \mathcal{G}) = \mathbb{E}(X)$$
Consequently, by the tower property
$$\mathbb{E}(X \cdot Y \mid \mathcal{F}) = \mathbb{E}(Y \cdot \underbrace{\mathbb{E}(X \mid \mathcal{G})}_{\mathbb{E}(X)} \mid \mathcal{F}) = \mathbb{E}(X) \cdot \mathbb{E}(Y \mid \mathcal{F})$$ |
H: Calculate $a^3+b^3+c^3+d^3$ for the real roots of $x^4+2x^3-3x^2-3x+2$
$a,b,c,d \in \mathbb{R}$ are the real roots of $x^4+2x^3-3x^2-3x+2$.
Calculate $a^3+b^3+c^3+d^3$.
With approximation i found out, that $a^3+b^3+c^3+d^3 = -17$, but how can I proof that without calculating the roots exactly?
Cheers
AI: Express $a^3 + b^3 + c^3 + d^3$ in terms of elementary symmetric polynomials and use Vieta's formulas.
Also, here is an alternative approach for the fans of linear algebra. Consider a matrix
$$
A = \left(\matrix{0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -2 & 3 & 3 & -2}\right).
$$
A trivial check shows that its characteristic polynomial is equal to
$$
\det (A - \lambda I) = \lambda^4 + 2\lambda^3 - 3\lambda^2 - 3\lambda + 2.
$$
So $a,b,c,d$ are the characteristic roots of $A$. Then their cubes are characteristic roots of $A^3$ (as easily follows from, say, the existence of a Jordan normal form). Then
$$
a^3 + b^3 + c^3 + d^3 = \operatorname{Tr} A^3 = -17.
$$ |
H: what is the meaning of $\mathbb{Q}(\sqrt{2},i)$
I have learnt so far that:
$\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}|a,b\in\mathbb{Q}\}$
Also
$\mathbb{Q}[i]=\{a+bi|a,b\in\mathbb{Q}\}$
What is $\mathbb{Q}(\sqrt{2},i)$ ?
Is it $\mathbb{Q}+\mathbb{Q}\sqrt{2}+\mathbb{Q}[i]$ but then we include $\mathbb{Q}$ twice as $\mathbb{Q}[i]=\mathbb{Q}+\mathbb{Q}i$
so should
$\mathbb{Q}(\sqrt{2},i)$ mean $\mathbb{Q}\sqrt{2}+\mathbb{Q}[i]$? and
$\mathbb{Q}(\sqrt{2},i)=\{a+b\sqrt{2}+ci|a,b,c\in\mathbb{Q}\}?$
Thanks in advance
AI: $\mathbb Q(\sqrt 2,i)$ is the smallest field that contains $\mathbb Q$ and $\sqrt 2 $ and $i$. The general element looks like $a+b\sqrt 2+ci+di\sqrt 2$ with $a,b,c,d\in\mathbb Q$.
Also note that in general $\mathbb Q(\xi)$ (where $\xi$ is some real or complex irrational number) need not be the set $\{\,a+b\xi\mid a,b\in\mathbb Q\,\}$. This is only true if $\xi$ is a quadratic irrational (such as $\sqrt 2$, $i$, or $\frac{1+\sqrt 5}2$), i.e. the root of a quadratic polynomial with rational coefficients.
On the other hand $\mathbb Q(\sqrt[3]2)=\{\,a+b\sqrt[3]2+c\sqrt[3]4\mid a,b,c\in\mathbb Q\,\}$ and $\mathbb Q(\pi)=\{\,\frac{a_0+a_1\pi+\ldots +a_n\pi^n}{1+b_1\pi+\ldots +b_m\pi^m}\mid n,m\in\mathbb N, a_i,b_i\in\mathbb Q\,\}$. |
H: Let $f: (a,b) \rightarrow \mathbb{R}$ a function, and $(a,b)$ contains the origin.
Let $f: (a,b) \rightarrow \mathbb{R}$ a function, and $(a,b)$ contains the origin.
Prove that if $f$ is monotone and $$\lim_{x\to 0} \frac {f(x)-f(-x)}{x}=0,$$ then f is differentiable at $0$.
I tried to add and subtract $ f (0) $ at the top, in two separate limits and use the lateral limits, but I can not guarantee that they exist.
AI: Hint: Since $f$ is monotone, we have $|f(x)-f(0)| \le |f(x)-f(-x)|$.
Therefore $0 \le \left|\frac{f(x)-f(0)}{x-0}\right| \le \frac{|f(x)-f(-x)|}{|x|} \to 0$, as $x \to 0$, i.e. $f'(0) = 0$. |
H: Sigma-finiteness of a point mass measure
Let $\mathcal{e}_{a}(x)$ denote point-mass at the point $a\in \mathbb{R}$. Let $a_{n}$ be a sequence of points in $\mathbb{R}$ and let $\nu = \sum_{n=1}^{\infty}\mathcal{e}_{a_{n}}$. What are properties of the sequence $(a_{n})_{n\geq 1}$ that render $\nu$ sigma-finite?
I would really appreciate any help you could give me!
AI: All that is needed is the sequences countability, provided all of these points have finite measure. Do you mean measure 1 for these point-masses -- I am a little unsure?
Your enumeration of the sequence should be 1-1 or you have to worry about an infinity of measure piling up at a particular point. |
H: Product of quadratic residues mod p $\equiv 1$ mod p iff $p \equiv 3$ mod 4
Let $p$ be an odd prime number. Prove that the product of the quadratic residues modulo $p$ is congruent to $1$ modulo $p$ if and only if $p \equiv 3 \pmod 4$.
I've tried using the fact that any quadratic residue modulo $p$ must be one of the numbers $1,2,\ldots,p-1$. But then I got stuck. This problem should be solvable without any Legendre symbol trickiness.
AI: Every quadratic residue is of the form $k^2$ for a $k \in \{1,\,\dotsc,\, \frac{p-1}{2}\}$. So the product of quadratic residues is
$$\prod_{k=1}^{\frac{p-1}{2}} k^2.$$
Now relate that to Wilson's theorem: Since $k^2 \equiv (-1)\cdot k \cdot (p-k) \pmod{p}$, we have
$$\prod_{k = 1}^{\frac{p-1}{2}}k^2 \equiv (-1)^{\frac{p-1}{2}}\prod_{k = 1}^{\frac{p-1}{2}} k \cdot \prod_{k = 1}^{\frac{p-1}{2}}(p-k) =(-1)^{\frac{p-1}{2}}\cdot (p-1)! \equiv (-1)^{\frac{p+1}{2}}\pmod{p}.$$
Thus the product of quadratic residues modulo $p$ is $\equiv 1 \pmod{p}$ if $p \equiv 3 \pmod{4}$ and it is $\equiv -1\pmod{p}$ if $p \equiv 1 \pmod{4}$. |
H: What approach should I take to establish this logical proof?
I need to design a logical math proof:
Write a detailed structured proof to prove that if m and n are integers, then either 4 divides mn or else 4 does not divide n.
Hint: Think about the form of the statement.
I was thinking about first proving that either 4 should divide m or n in order for the first part to be true, then use proof by cases to prove what happened when m was 4, and another case when n was 4...
But I got stuck, what do you recommend:
This is what I got so far:
Claim: (m ∈ Z ∧ n ∈ Z) => (4 | mn ˅ 4 ∤ n)
Negated claim: (m ∈ z ∧ n ∈ z) ∧ ( P(n) ∧ ¬P(mn) )
Proof by contradiction
Assume m , n ∈ z:
Assume P(n) ∧ ¬P(mn):
Then ∃ q ∈ R, n = 4q
Let q0 be such that n = 4q0
Then mn = m(4q0)
Then mn = 4(mq0)
Then ∃ q ∈ R, mn = 4q
Then 4 | mn
Then P(n) ∧ P(mn)
Then (m , n ∈ z) ∧ P(n) ∧ P(mn))
I know that is wrong, how can I fix it?
AI: Claim: $$m,n \in \mathbb{Z} \Rightarrow (4 \mid mn \vee 4 \nmid n).$$
Proof (by contradiction):
Suppose the claim is false, that is,
\begin{align*}
\neg(m,n \in \mathbb{Z} \Rightarrow (4 \mid mn \vee 4 \nmid n)) &\equiv (m,n \in \mathbb{Z})\wedge\neg(4 \mid mn \vee 4 \nmid n) \\
&\equiv (m,n \in \mathbb{Z})\wedge(4 \nmid mn \wedge 4 \mid n).
\end{align*}
All that is left to show is that any part of the negation is false, ie. $(4 \nmid mn \wedge 4 \mid n)$ is impossible. Conclude that since the negation of the claim is false, the claim itself must be true. |
H: Quotient is principal
Let $R$ be a finite commutative ring, let $J$ be a maximal ideal of $R$ and $n$ some positive integer greater or equal than $2$. Is it always true that every ideal of the quotient $R/J^{n}$ is principal?
AI: No. For instance if $R$ is local then $J^n = 0$ for sufficiently large $n$ and then you are just asking whether a finite, local commutative ring must be principal. The answer is certainly not.
Examples have come up before on this site. One natural one is $R = \mathbb{F}_p[x,y]/\langle x,y \rangle^2$. This is somehow especially instructive: the maximal ideal $\mathfrak{m} = \langle x,y \rangle$ is not monogenic since $\operatorname{dim}_{R/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2 = \operatorname{dim}_{\mathbb{F}_p} \mathfrak{m} = 2$. I felt I understood Zariski tangent spaces a bit better after thinking about this example.
Another simple example is $\mathbb{Z}[\sqrt{-3}]/\langle 2 \rangle$. More generally, if $R$ is any nonmaximal order in a ring of integers in a number field, then there will be some ideal $I$ of $R$ such that $R/I$ is finite and nonprincipal.
Added: Over the summer I had the occasion to think about finite non/principal rings. For more information about when finite local rings are principal, and especially the connection to residue rings of orders in number fields, see $\S$ 1 of this paper. |
H: Show f is continuous if and only if for any x $\in$ X and any open set O$^y$ in Y containing f(x), ...
Suppose $X$ and $Y$ are topological spaces with topology $T^x$ and $T^y$
Let $f: X \rightarrow Y$ be a function.
Show $f$ is continuous if and only if for any $x \in X$ and any open set $O^y$ in $Y$ containing $f(x)$, there is an open set $O^x$ in $X$ containing $x$, such that $f(O^x) \subseteq O^y$
I know that $f$ is continuous if and only if for every closed set $C$ in $Y$, $f^{-1}(C)$ is closed in $X$.
Also $f$ is continuous if and only if every open set O in Y, $f^{-1}(O)$ is open in $X$.
Let $x \in X$and suppose $f(x) \in O^y$ for some $O^y \in T^y$. Also there exists
an open set $O^x$ in $X$ containing $x$, such that $f(O^x) \subseteq O^y$.
$\Rightarrow x \in f^{-1}(O^y)$
AI: Hints: $x \in f^{-1}(O^y)$ and
$$f(O^x) \subseteq O^y \Longrightarrow x \in O^x \subseteq f^{-1}(O^y).$$
1) Suppose that $f$ is continuous. If $x \in X$ and $O^y \ni f(x)$, then put $O^x := f^{-1}(O^y)$.
2) Now assume that for any $x \in X$ and $O^y \ni f(x)$ there is $O^x \ni x$ such that $f(O^x) \subseteq O^y$. Let's fix some open $O^y$ and prove that $f^{-1}(O^y)$ is open. If $x \in f^{-1}(O^y)$, then there is $O^x \ni x$ such that $f(O^x) \subseteq O^y$, i.e. $x \in O^x \subseteq f^{-1}(O^y)$. Since every point of $f^{-1}(O^y)$ has some open neighborhood also belonging to that set, $f^{-1}(O^y)$ is open. |
H: Subset of A Regular Language
I need to show that a subset of a regular language is regular or not. I think it may not be regular but I could not find a counter example.
Do you have any simple example to prove that?
Thanks in advance.
AI: Let $N$ be some non-regular language over some alphabet $\Sigma$. You should know several examples of non-regular languages.
Then $$N\subset \Sigma^\ast$$
and $\Sigma^\ast$ is regular. |
H: Let $f:[0, \infty) \rightarrow \mathbb{R}$ a function of class $C^1$ in its domain, suppose $f'(x)$ is a non-decreasing function.
Let $f:[0, \infty) \rightarrow \mathbb{R}$ a function of class $C^1$ in its domain, suppose $f'(x)$ is a non-decreasing function.
Using the monotonicity of $f'$, prove that the function $g(x)= f(x)-f(c)-f'(c)(x-c)$ is nondecreasing on $[c,\infty)$ where $c> 0$.
AI: Hint: If $c \leqslant y < x$, then for some $d \in (y, x)$
$$\dfrac{f(x)-f(y)}{x-y} = f'(d) \geqslant f'(c).$$ |
H: Incongruent Solutions Modulo $p$
Let $p$ be an odd prime and $k$ a positive integer such that $\gcd(p,k)=1$. Show $x^2\equiv k \bmod p$ has zero or two incongruent solutions.
I think we are supposed to assume that $x$ is a solution, and that $y$ is also a solution where $y$ does not equal $x$. Then show $y \equiv -x \bmod p$
Any help? Thanks!
AI: If $\;a\;$ is a solution, then also $\;-a\;$ is, and
$$a=-a\pmod p\iff 2a=0\pmod p$$
But $\;p\neq 2\;$ and also $\;a\neq 0\pmod p\;$ , otherwise $\;a^2=0=k\pmod p\;$ , contradicting $\;(k,p)=1\;$ . |
H: Embarrassingly-basic fraction question
I'm trying to do a calculation at work to figure out what the average # of pages a visitor is viewing.
I am given:
47,000 visits
12% of visits do not bounce (that is, 12% navigate the site at least one link after stumbling upon it)
1.23 pages/visit is the average amount of pages a visitor sees.
My current attempt:
( 47,000 visits ) * (.12) = (1.23 pages)/(1 visit )
Solving for: the number of pages the average non-bouncing visitor is viewing, in order to bump the average to 1.23
Notice the glaring lack of variable... I have been out of hard math for maybe 4-5 years now, and I'm sure this is far below the average question you guys get, but I don't necessarily need you to tell me the solution so much as remind me of the problem-solving logic I should apply to solve this on my own... thank you very much for your time!
AI: The number of non-bouncers is $(.12)(47000)=5640$, which also means that the number who do bounce is $47000-5640 = 41360$. Let $A$ represent the average number of pages seen by a non-bouncer (what you want to solve for). Then you know that the total number of pages seen by the 47000 users is $(A)(5640)+(1)(41360) = 5640A +41360$. Therefore the average number of pages seen by all users is
$$
\frac{5640A + 41360}{47000}
$$
but you also know that this number is $1.23$. So, solve for $A$ in
$$
\frac{5640A + 41360}{47000} = 1.23
$$ |
H: Why is $\frac{d}{dx}\exp(x) = \exp(x)$?
What is the explanation for
$$(e^x)'=e^x$$
I searched the SE, 'cause this can't be the first time this has been asked. But the question seems hard to formulate and search for here and on Google. Any help?
AI: Note that for $f(x) = e^x$ you have by definition
$$
f'(x) = \lim_{h\to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h\to 0} \frac{e^{x}(e^h - 1)}{h} = e^x\lim_{h\to 0} \frac{e^h - 1}{h}.
$$
Now it all comes down to how you show that
$$
\lim_{h\to 0} \frac{e^h - 1}{h} = 1.
$$
And this will depend on how you have defined $e^x$ in the first place.
Lets say we have defined
$$
e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}.
$$
Then
$$
\frac{e^h - 1}{h} = \sum_{n=1}^{\infty} \frac{h^{n-1}}{n!} = 1 + \sum_{n = 2}^{\infty} \frac{h^{n-1}}{n!} \to 1 \text{ as } h\to 0.
$$
(You might have to think about this limit a bit.) |
H: Use a lemma to prove that $A_4$ has no subgroup of order $6$.
Use the following lemma to prove that $A_4$ has no subgroup of order $6$:
Lemma: If $H\le G$ has index $2$, i.e. $[G:H]=2$, then for any $a\in G$ we have $a^2\in H$.
The $12$ elements of $A_4$ are $(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143)$, $(234)$, and $(243)$.
AI: The order of $A_4$ is 12, so if $A_4$ has a subgroup $H$ of order $6$, then $[A_4:H]=2$ (why?). Then it follows from the lemma that $a^2\in H$ for all $a\in A_4$. List the elements of the set $\{a^2\mid a\in A_4\}$, that is, all squares in $A_4$. How many are there? |
H: Find a bijective function between two segments of R
In my homework in set theory i was asked to find a bijective function between (a,b) and (c,d) that are subsets of R. d>c and b>a.
I'm having problems thinking of a function that is surjective and injective.
i thought of the function for example f(x) = x+d-b, where x is an element in (a,b) and it is indeed injective, but not surjective...
I can use some advice please
AI: HINT:$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ |
H: Growth of exponential functions with different base and/or exponent
Assuming that the following is trivial:
$\lim_{x \to +\infty} \frac{2^x}{2.1^x}=0$
$\lim_{x \to +\infty} \frac{2^x}{2^{x^2}}=0$
What is the most simple, intuitive way to show that:
$\lim_{x \to +\infty} \frac{2^{x^2}2.1^x}{2.1^{x^2}}=0$
AI: I think I got an explanation:
$2^{x^2}=({2^2})^{\frac{1}{2}x^2}=4^{\frac{1}{2}x^2}$
when $x \to +\infty$
$\frac{1}{2}x^2>x $
$\rightarrow\lim_{x \to +\infty} \frac{2.1^x}{4^{\frac{1}{2}x^2}}=0\rightarrow 2.1^x=o(2^{x^2})$
From this point it's really trivial (using $\lim_{x \to +\infty} \frac{2^{x^2}}{2.1^{x^2}}=0$) |
H: convexity and the interior sphere condition
Consider $\Omega $ a open, convex bounded subset of $R^n$.
Let $x_0 \in \partial \Omega$. I believe that exists a open ball $B \subset \Omega$ such that $\partial B \cap \partial \Omega = \{ x_0 \}$. (this is the interior sphere condition in $x_0$) ? I have no idea how to proof ...
Someone can help me to prove or give a counter example for the question ?
thanks in advance
AI: There is an easy counter example. $\Omega = int([0,1]^n)$ and $x_0 = (1,\ldots,1)$ then there is no ball which we want. |
H: Can you split up expectation over multiplication?
I was wondering if the property exists where if you have $\ E[(Y- \mu)^3]$ you can write it as $\ E[(Y- \mu)^2] E[(Y- \mu)] $ ?
AI: Is $Y$ normally distributed by any chance? If it is then Stein's Lemma can be useful for calculating higher order moments.
Stein's Lemma:
Let $Y\sim N(\mu,\sigma^{2})$ and let $g$ be a differentiable function satisfying $\mathbb{E}|g\prime(Y)|<\infty$ then
$\mathbb{E}[g(Y)(Y-\mu)]=\sigma^{2}\mathbb{E}[g\prime(Y)]$
in your case $g(Y)=(Y-\mu)^{2}$ and so $g\prime (Y)=2(Y-\mu)$, for which $\mathbb{E}[...]=0$
and so $\mathbb{E}[(Y-\mu)^3]=0$
!but only works for normally distributed Y. |
H: α² is a cycle if and only if s is odd
let $\alpha$ be a cycle of length $s$, say $\alpha = (a_1, a_2, \ldots, a_s)$
Prove $\alpha^2$ is a cycle if and only if $s$ is odd.
Let me start off by saying I am in my 5th week of Group Theory. I often have trouble getting these problems started. This is my first proof based course.
I believe $\alpha^2 = (a_2, a_3, \ldots, a_1)$
Any tips on where to go from here would be great. Also...if there are any tips for starting proofs like these in general, I could really use them! My teacher teaches as if a proofing class was a pre-req, which it was not.
AI: HINT: Look at a couple of examples: $(1234)$ sends $1$ to $2$ and $2$ to $3$, so $(1234)^2$ sends $1$ to $3$. $(1234)$ sends $2$ to $3$ and $3$ to $4$, so $(1234)^2$ sends $2$ to $4$. $(1234)$ sends $3$ to $4$ and $4$ to $1$, so $(1234)^2$ sends $3$ to $1$. Finally, $(1234)$ sends $4$ to $1$ and $1$ to $2$, so $(1234)^2$ sends $4$ to $2$. Put all the pieces together, and you find that $(1234)^2=(13)(24)$, which is not a cycle. Do the same thing with $(12345)$, however, and you find that $(12345)^2=(13524)$, which is a cycle. Work that one through in detail to be sure, and then try to generalize these ideas to the cases $s$ even and $s$ odd. |
H: If a function $f:X \to Y$ is continuous on any compact subset $K \subseteq X$, is $f$ continuous?
I hope I phrased that right. If $X$ is not able to be covered by a family of compact spaces, how could we prove this? Am I just missing something simple?
(Let $f:X \to Y$ (EDIT: $X$ and $Y$ metric spaces) have the property such that $\forall K\subseteq X$ such that $K$ is compact, $f|_{K}:X \to Y$ is continuous. This is supposed to lead to $f$ being continuous but I am really lost...
Any help is much appreciated.
AI: For metric spaces, continuity is equivalent to sequential continuity, that is, for each convergent sequence $x_n \to x$, one has $f(x_n) \to f(x)$.
Now, if $x_n\to x$, then the set $\{x\} \cup \{ x_n : n \in \mathbb{N}\}$ is compact. |
H: Upper bound for $(1-1/x)^x$
I remember the bound $$\left(1-\frac1x\right)^x\leq e^{-1}$$ but I can't recall under which condition it holds, or how to prove it. Does it hold for all $x>0$?
AI: Starting from $e^x \geq 1+x$ for all $x \in \mathbb{R}$:
For all $x \in \mathbb{R}$
$$ e^{-x} \geq 1-x. $$
For all $x \neq 0$
$$ e^{-1/x} \geq 1-\frac{1}{x}. $$
And, since $t \mapsto t^x$ is increasing on $[0,\infty)$, for $x \geq 1$
$$ e^{-1} \geq \left(1-\frac{1}{x}\right)^x. $$ |
H: $ \lim_{x \to 0} a^x = 1 $ using the definition
Given $a > 1$, define $ f: \mathbb{Q} \to \mathbb{R}$ such that for all $ x \in \mathbb{Q}$ we have $ f(x) = a^x$ . Prove that $ \lim_{x \to 0} f(x) = 1$
I'm having issues to find my delta. I think we need to use $ \log_a$ of something in function of $ \epsilon$ but I couldn't find it.
Thanks !
AI: Pick $\varepsilon>0$ and re-write it as $\varepsilon=e^{\alpha}-1$. Want to find a $\delta$ such that $|a^{x}-1|<\varepsilon$ for $0<|x-0|<\delta$.
$|a^{x}-1|<e^{\alpha}-1$
$|a^{x}|=a^{x}<e^{\alpha}$
$x\log(a)<\alpha$
so for $|x-0|<\frac{\alpha}{\log(a)}$ it folows that $|a^{x}-1|<\varepsilon$ |
H: perfect squares possible?
If we let a, b, c, d, and x be integers is it possible that $$x^2+a^2 = (x+1)^2 + b^2 = (x+2)^2 + c^2 = (x+3)^2 + d^2$$
My initial thought is no way! I tried expanding and simplifying, getting $$a^2 = 2x+1 + b^2 = 4x+4 + c^2 = 6x+9 + d^2$$.
It seems that the difference between these perfect squares is impossible - but why? Consecutive perfect squares always differ by some $2n+1$ term, and each term must continue to increase by an odd amount...
Any ideas in the right direction are greatly appreciated!
AI: I get it. Mod 4 suffices. Well, mod 8 anyway. One of $x,x+1,x+2,x+3$ is divisible by 4. If its matching partner out of $a,b,c,d$ is even, the common sum is divisible by 4, and we actually cannot have any odd numbers involved, which cannot occur with consecutive $x,x+1,x+2,x+3.$ Required detail: the sum of two odd squares is $2 \pmod 4.$
So, the partner is odd, the sum is $1 \pmod 8.$ One of the other $x+j$ numbers is $2 \pmod 4,$ so that sum must be $5 \pmod 8.$ So, actually, impossible.
Done. |
H: Upper bound for product of exponents
From here we have the bound
$$\left(1-\frac1N\right)^N\leq e^{-1}$$
where $N$ is a positive integer.
Written another way, it is
$$\left(1-\frac1{N_1}\right)^{k_1}\left(1-\frac1{N_2}\right)^{k_2}\ldots \left(1-\frac1{N_r}\right)^{k_r}\leq e^{-1}$$ where $N_1=N_2=\ldots=N_r=N$ and $\dfrac{k_1}{N_1}+\dfrac{k_2}{N_2}+\ldots+\dfrac{k_r}{N_r}=1$.
Is it true in general that
$$\left(1-\frac1{N_1}\right)^{k_1}\left(1-\frac1{N_2}\right)^{k_2}\ldots \left(1-\frac1{N_r}\right)^{k_r}\leq e^{-1}$$
for positive integers $N_1,N_2,\ldots N_r,k_1,k_2,\ldots k_r$ such that $\dfrac{k_1}{N_1}+\dfrac{k_2}{N_2}+\ldots+\dfrac{k_r}{N_r}=1$? If not, what upper bound can we find for the left-hand expression?
AI: In full generality, for every nonnegative $n$ and every $a\lt1$,
$$\left(1-a\right)^n\leqslant\mathrm e^{-an}.
$$
Using this for $a=1/N_i$ and $n=k_i$ yields $\left(1-1/N_i\right)^{k_i}\leqslant\mathrm e^{-k_i/N_i}$ hence
$$
\prod_i\left(1-\frac1{N_i}\right)^{k_i}\leqslant\exp\left(-\sum_i\frac{k_i}{N_i}\right).
$$ |
H: If $a < b$ and $b = \infty$, then $a < \infty$?
A very simple question, but I am not sure for this moment.
I have a strict inequality $a < b$. And I prove that $b = \infty$, say $b$ is an integral. Does this prove that $a < \infty$, that is, $a$ is finite? Question seems simple but things can easily be messed up with infinity.
To be more clear, say $a = \int f dx$ and $b = \int g dx$ and I have $a < b$. If $\int g dx = \infty$, can we say that $\int f dx < \infty$?
If one can construct pathological counterexamples involve limits etc, I will be very happy.
AI: Talking about $\int fdx$ makes no sense, as that is a concrete function, and not a number.
You might want to talk about the definite integral, $\int_0^\infty fdx$ and $\int_0^\infty gdx$. You may also want to discern between showing $\int_0^\infty fdx<\int_0^\infty gdx$ (in which case your claim would be correct), and the case where $f(x)<g(x)$ for all $x$ in the domain over which you are integrating.
In the latter case take $f(x)=1$ and $g(x)=2$. |
H: The conditions of applying L'Hospital's rule
I just finished calculus 1,2 last semester and I am learning calculus 3 now. I saw this question and I post a solution as follow:
Prove series convergent,consider the limit:
$$\lim_{n\rightarrow\infty}\frac{\tan{n}}{1.5^n} = \lim_{n\rightarrow\infty}\frac{\frac{1}{\cos^2{n}}}{1.5^n\cdot\ln1.5} =\lim_{n\rightarrow\infty}\frac{1}{\ln1.5\cdot\cos^2n\cdot1.5^n} = 0$$
Hence, by the definition of limit,
$$\forall\varepsilon\gt0,\exists N\gt0,\text{such that} ~n \gt N,~|\frac{\tan{n}}{1.5^n}| \lt \varepsilon$$
Let $\varepsilon = 1$, we have
\begin{align*}
\ &\Longrightarrow|\frac{\tan{n}}{1.5^n}|\lt 1 \Longrightarrow \tan{n} \lt 1.5^n
\\ &\Longrightarrow \frac{\tan{n}}{2^n}\lt \frac{1.5^n}{2^n} =(\frac{3}{4})^n
\\ &\Longrightarrow \sum_{n=1}^{\infty}\frac{\tan{n}}{2^n}\lt\sum_{n=1}^{\infty}(\frac{3}{4})^n ~~~~\text{for some}\space n \gt N
\end{align*}
Since the geometric series $\sum_{n=1}^{\infty}(\frac{3}{4})^n$ is convergent,by the comparison test,$\sum_{n=1}^{\infty}\frac{\tan{n}}{2^n} \text{is convergent.}$
But @i707107 told me that I cannot apply the L'Hospital's theorem for $\lim_{n\rightarrow\infty}\frac{\tan{n}}{1.5^n}$, so my whole proof is wrong. I did go back to check the scope of application of L'Hospital's theorem and I even read the proof in the appendix of our textbook, which only makes me more confused. So I am expecting somebody are able to explain why I can not apply L'Hospital's theorem here. Also, Since this question is beyond my knowledge right now, It would be perfect if you could tell me what material I should read first in order to totally understand it. Thanks very much.
AI: While $$\displaystyle\lim_{x\rightarrow \infty} 1.5^n$$ goes to $+\infty$, $$\displaystyle\lim_{x\rightarrow \infty} \tan x$$ does not go to $+\infty$ or $-\infty$ (in fact, I'm pretty sure that limit isn't defined at all), so it doesn't hit either of the cases needed in L'Hospital's Rule.
As your little snippet pointed out, we need either both limits diverging or both going to 0. We don't have that here. |
H: Hypercube and dihedral group
Let $G_n$ denote the subgroup of the orthogonal group $O_n$ of elements that send the hypercube to itself, the group of symmetries $C_n$, including the orientation-reversing symmetries.
It would like to show that $G_2$ is isomorphic to the dihedral group $D_4$ of order 8
I determined a formula for the order of $G_n$ which is $n!2^n$. Therefore, we already notice that both sets have the same order.
AI: OK, we have established that your definition of $D_4$ is
$$
D_4 = \langle a, b \mid a^4 = b^2 = 1, b^{-1} a b = a^{-1} \rangle.
$$
Now, in the group $G_2$ choose $A$ to be a rotation by 90 degrees and $B$ to be a mirror reflection. Clearly, there are such $A$ and $B$ in $G_2$. It is also clear that $G_2$ is generated by $A$ and $B$.
To prove that $G_2$ is isomorphic to $D_4$, we will build homomorphisms $\varphi: D_4 \to G_2$ and $\psi: G_2 \to D_4$ that are inverses of each other.
To build $\varphi: D_4 \to G_2$, let us note that in group $G_2$ elements $A$ and $B$ satisfy these relations: $A^4 = 1$, $B^2 = 1$, $B^{-1} A B = A^{-1}$ (check this by hand). It follows that there exists a unique homomorphism $\varphi: D_4 \to G_2$ that sends $a$ to $A$ and $b$ to $B$.
Now let us build $\psi: G_2 \to D_4$. Note that every element $C \in G_2$ can be represented as $C = A^m B^n$. Define $\psi(C) = a^m b^n$.
First of all, we need to check that this definition is correct. If we represent $C$ in two different ways, $C = A^m B^n = A^{m'} B^{n'}$, we need to check that the same value is assigned to $\psi(C)$, i.e. $a^m b^n = a^{m'} b^{n'}$. Indeed, if $A^m B^n = A^{m'} B^{n'}$, then $A^{m-m'} = B^{n'-n}$. $A$ preserves orientation and $B$ doesn't. It follows that $n-n'$ is even, so $A^{m-m'}=1$, so $m-m'$ is a multiple of $4$. But then $$a^{m'}b^{n'} = a^m a^{m'-m} b^n b^{n'-n} = a^m b^n,$$ because $a^4 = b^2 = 1$ in $D_4$. So map $\psi$ is defined correctly.
We also need to check that $\psi$ is a homomorphism. This is quite mundane and involves a little case analysis, so I will omit the check.
Now that we have built $\psi$ and $\varphi$, we see that $\psi\varphi(a) = a$ and $\psi\varphi(b) = b$, so $\psi\varphi$ is the identity map of $D_4$. In a similar way $\varphi\psi$ is the identity map of $G_2$. It follows that $\psi$ and $\varphi$ are inverses of each other, so they are isomorphisms, qed. |
H: Unique maximal ideal implies set of non-units is an ideal
This is not for homework, but I would just like a hint please. The question asks
If a commutative ring $R$ (with $1$) has a unique maximal ideal, then the set of non-units in $R$ is an ideal.
This is actually an 'if and only if', but I have shown one direction. I'm not sure how to go about proving this direction, though. I don't think contradiction or contrapositive are useful, because assuming the set of non-units is not an ideal doesn't seem to give anything useful. However, showing this directly seems difficult too, because we don't know anything about the set of non-units in an arbitrary ring (it may not even be closed under addition). I also have the following fact at my disposal:
When $R$ is a nonzero ring (with $1$), then every ideal of $R$ except $R$ itself is contained in a maximal ideal.
I would really appreciate a hint about how to go about approaching this.
AI: Every nonunit is contained in a maximal ideal. |
H: question about cluster points of a set of cluster points
Let A be a set in a metric space X. A′ is the set of cluster points of A. Is it A′′ ⊆ A′ ? I think it is not. But I can not give a counterexample or proof. Could someone give me a clue? Thank you!
AI: HINT: Yes, it is always the case that $A''\subseteq A'$. If $x\in A''$, then every open nbhd $U$ of $x$ contains some point $y\in A'$. But then $U$ is an open nbhd of $y$, so it must contain ... what? |
H: Limit point of a bounded sequence
My textbook says that if $x_n$ is a bounded sequence ($\exists m,M: m \le x_n \le M$) and a is a limit point of this sequence, then $m \le a \le M$. Can somebody explain why is that so? How can I show that it is true?
AI: HINT: Suppose that $a>M$, and let $\epsilon=M-a>0$. The sequence has a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ converging to $a$, so by definition there is some $m\in\Bbb N$ such that $|x_{n_k}-a|<\epsilon$ whenever $k\ge m$. Is that actually possible?
Now make a similar argument for $m$. |
H: What is the easy way to calculate the roots of $z^4+4z^3+6z^2+4z$?
What is the easy way to calculate the roots of $z^4+4z^3+6z^2+4z$?
I know its answer: 0, -2, -1+i, -1-i.
But I dont know how to find? Please show me this. I know this is so trivial, but important for me. Thank you.
AI: Add 1 to both side to get:
$$z^4+4z^3+6z^2+4z+1 = 1$$
i.e.
$$(z+1)^4 = 1$$
can you finish from there? |
H: Solving coupled langrangian derivatives
I have been told that the solution to
$$\frac{Du}{Dt}=2\Omega v, \frac{Dv}{Dt}=-2\Omega u$$
is
$$u(t) = u_0 \cos2\Omega t+ v_0 \sin 2\Omega t$$
$$v(t) = -u_0 \sin 2\Omega t + v_0 \cos 2\Omega t$$
But how would I arrive at this solution by analysis? Is there a method I can use, or is it simply a pattern I need to recognise?
AI: Yes, there is. Take the time derivative of the 2 equations.
\begin{align}
\frac{d^2}{dt^2} u (t) &= 2\Omega \frac{dv(t)}{dt} = -4\Omega^2 u(t)\\
\frac{d^2}{dt^2} v (t) &= -2\Omega \frac{du(t)}{dt} = -4\Omega^2 v(t)
\end{align}
i.e.
\begin{align}
\frac{d^2}{dt^2} u (t) +4\Omega^2 u(t) = 0\\
\frac{d^2}{dt^2} v (t) +4\Omega^2 v(t) =0
\end{align}
Then you solve for both, use the Euler identity to recover the cos/sin form. It can also be noted through chain derivatives that $du/dv=-v/u\implies u=-v$, which allows you to only solve for one of the above. |
H: Mistake in excercise with differentiation
I have an excercise given where $f: \mathbb{R} \rightarrow \mathbb{R}$ and $a \in \mathbb{R}$ and we have that $x(t):=\int_0^t exp(a(t-s))f(s)ds$. Now we are supposed to show that $x$ solves the differential equation $d_t(x(t))= ax(t)+f(t)$.
I would say that this is impossible, as the fact that $f$ is just a bounded function does not mean, that there is any meaning in the integral expression as this integral does not exist (in general) for functions that are just bounded and I would say that if you would for example look at Dirichlet's function, than this theorem would probably not hold, but I wanted to get your confirmation about this, before complaining about this. (Maybe there is anybody here who knows what we need to demand for f(probably continuous is sufficient, but maybe the fact that the integral exists is sufficient).
AI: Well, let's make some hypotheses about $f$ so that we know that the integral makes sense. In particular, let's assume that $f$ is continuous. The issue is to differentiate correctly the integral $$\int_0^t g(t,s)\,ds\,,$$
assuming, for example, that $g$ and $\dfrac{\partial g}{\partial t}$ are continuous.
(Otherwise, the Fundamental Theorem of Calculus will be hopeless, as will — without some care — the hypothesis for differentiating under the integral sign.) |
H: How to expand of the plane wave in Legendre polynomials
There is the expression for the plane wave:
$$
e^{i(\mathbf k \cdot \mathbf r )} = e^{ikrcos(\theta )} = e^{\frac{kr}{2}\left( ie^{i \theta} - \frac{1}{ie^{i \theta }}\right)} = e^{\frac{t}{2}\left( \omega - \frac{1}{\omega}\right)}.
$$
This is the generating function for Bessel's functions, so
$$
e^{\frac{t}{2}\left( \omega - \frac{1}{\omega}\right)} = \sum_{m = -\infty}^{\infty}J_{m}(kr)\omega^{m} = \sum_{m = -\infty}^{\infty}J_{m}(kr)i^{m}e^{im\theta}. \qquad (.1)
$$
It may be represented as
$$
\sum_{m = -\infty}^{\infty}J_{m}(kr)i^{m}e^{im\theta} = \sum_{m = -\infty}^{\infty}\sum_{n = 0}^{\infty}J_{m}(kr)i^{m}a_{mn}P_{n}(cos(\theta)). \qquad (.2)
$$
I read that full expansion of plane wave expression into special functions polynomial is
$$
e^{i(\mathbf k \cdot \mathbf r )} = \sum_{m = 0}^{\infty}(2m + 1)J_{m}(kr)i^{m}P_{m}(cos(\theta )). \qquad (.3)
$$
So how to get $(.3)$ from $(.2)$ (or from $(.1)$)? I don't know what to do with $e^{im \theta}$.
AI: Formula (.3) is not correct. Instead of $J_m(kr)$ it should be $j_m(kr)$ which correspond to Spherical Bessel functions. Work backwards from the spherical harmonic expansion
$$e^{i\mathbb{k}\cdot\mathbb{r}}=4\pi\sum_{\ell=0}^\infty i^\ell j_\ell(k r)\sum_{m=-k}^k Y^*_{\ell m}(\theta,\phi)Y_{\ell m}(\theta',\phi')$$
and use the spherical addition theorem specialized to the Legendre polynomial with $\cos(\gamma)=\cos(\theta)\cos(\theta')+\sin(\theta)\sin(\theta')\cos(\phi-\phi')$:
$$P_\ell(\cos\gamma) = \frac{4\pi}{2\ell +1}\sum_{m=-\ell}^\ell Y^*_{\ell m}(\theta,\phi)Y_{\ell m}(\theta',\phi')$$ |
H: Scalar product in vector/coordinate form
As I know, $a*b = |a|*|b|*cos(a,b)$ in vector form
And $a*b = (a_1,a_2)*(b_1,b_2) = a_1*b_1+a_2*b_2$.
1)
$$a*b=?$$
$$a=2i-3j+5k$$
$$b=i+2j+8k$$
SOLUTION: $a*b = (2,-3,5)*(1,2,8) = 2*1+(-3)*2+5*8 = 36.$
I'm unsure about the angle between them? Do I have to do only multiplication of those two?
$$a*b = (2i-3j+5k)*(i+2j+8k) = \dots$$
2)
$$a*b=?$$
$$a=(1,2,3)$$
$$b=(4,5,6)$$
Do I have to do the same?
SOLUTION: $(a*b) = (1,2,3)*(4,5,6) = 1*4+2*5+3*6 = 32$
3) Find the angle between $a$ and $b$?
$$a=2i-j+2k$$
$$b=i+j+k$$
SOLUTION: a*b = (2,-1,2)*(1,1,1) = 3
|a| = 3
|b| = sqrt(3)
cos(a,b)=?
What's the formula for calculating the angle?
AI: Hint: for 1, you are given the components, so use the formula you have after "And" but with three components and you are there. Note that $i,j,k$ are unit vectors, so $i\cdot i=j \cdot j = k \cdot k =1$ For 2, the same. For 3, get the dot product the same way, evaluate $|a|=\sqrt {a \cdot a}$ and $|b|$, then use these to get $\cos (a,b)$ by combining your first two equations. |
H: Heuristic approach to winding number
I'm working on problem 8.23 of Rudin's PMA, that is:
Let $\gamma:[a,b]\to\mathbb C$ be a closed curve, $\gamma \in C^1([a,b])$ and $\gamma(t) \neq 0 \ \forall t\in [a,b]$. Show that $$\text {Ind}(\gamma)=\frac{1}{2 i \pi} \int _a ^b \frac{\gamma '}{\gamma}$$
is an integer.
My idea was to express $\gamma$ in polar coordinates: $$\gamma(t)=r(t)e^{i\theta (t)},$$
so that $$\gamma '(t)=r'(t)e^{i\theta(t)}+r(t)e^{i\theta (t)}\cdot i\theta'(t)$$
and $$\frac{\gamma'(t)}{\gamma(t)}=\frac{r'(t)}{r(t)}+i\theta'(t).$$
Integrating over $[a,b]$ and noting that the first term vanishes:$$\frac{\theta(b)-\theta(a)}{2\pi}.$$
Since $\gamma (a)=\gamma (b)$ and $e^{i\theta}$ has period $2\pi$, this is an integer.
Now, there are some issues that I don't know how to fix. Let $x,y$ denote $\text{Re}\gamma $ and $\text {Im} \gamma$. Since $\gamma \in C^1$, follows that $x,y\in C^1$ and consequently $r=\sqrt{x^2+y^2}\in C^1$.
What about $\theta$? Suppose for example that I define $\theta (a)$ to be the smallest positive that satisfies $\frac{\gamma(a)}{r(a)}=e^{i\theta(a)}$. I know that, then, I could define $\theta$ locally via $\tan ^{-1}$ and I would get a $C^1$ function. But how can I define a function $\theta\in C^1$ that keeps count of the windings around zero?
AI: You can locally, in a neighbourhood of each $t\in [a,b]$ define a family $\theta_{t,k}$ of differentiable argument functions via all possible branches of $\tan^{-1}$ or $\cot^{-1}$. Since $[a,b]$ is compact, $\gamma$ is uniformly continuous, hence there is a $\delta > 0$ such that for all $t\in [a,b]$, you can define a differentiable argument of $\gamma(t)$ at least on the neighbourhood $[t-\delta,t+\delta] \cap [a,b]$ of $t$. Then you can cover $[a,b]$ with finitely many such intervals, and glue together a family of local arguments to obtain a differentiable global argument. Starting with an arbitrary choice of $\theta_{a,0}$, extend it by $\theta_{a+\delta, k_1}$ such that $\theta_{a,0}(a+\delta/2) = \theta_{a+\delta,k_1}(a+\delta/2)$ to obtain a differentiable argument on $[a,a+2\delta]$. Since the two glued-together argument functions agree in one point of the overlap and are continuous, and their difference can take only integral multiples of $2\pi$ as values, they agree on the entire overlap of their domains. Continue until you have reached $b$. |
H: proof by negation - there is no supremum
A = { x + 1/x : x > 0 }
How do you prove that there is no supremum for this set?
I think this is the inequality needed:
M - ε > x + 1/x
M - the supremum
How do you keep from this step?
thanks
AI: Suppose the set
$$A=\{x+1/x:x>0\}$$
has a supremum. Call it $M$. Then,
$$x+\frac{1}{x}\leq M,\quad\forall x>0$$
In particular, take $x=1/n$ for $n\in\mathbb{N}$. Then, for all $n\in\mathbb{N}$,
$$\frac{1}{n}+\frac{1}{1/n}=\frac{1}{n}+n\leq M$$
$$\Rightarrow n\leq \frac{1}{n}+n\leq M,\quad\forall n\in\mathbb{N}$$
That is, $\mathbb{N}$ is bounded above by $M$. Contradiction. |
H: Max value of Anti-symmetric Relation
Here is the question: Let A be a set with |A| = n and R be a relation on A that is anti-symmetric. What is the max value for |R|? How many antisymmetric relations can have this size?
So I'm not sure how to answer this. My first inclination is to just use the formula that tells you how to many antisymmetric relations there can be : $(2^n)(3^{n^2-n}/2)$. But then I figured that R can have more than just antisymmetric relations. My next thought was to count up equivalence relations and subtract the number of symmetric relations. I'm not sure if this will work, but this just seems more complicated than it should.
What should I be doing?
No clue on how to do the second part.
Note: Not homework. This is just me trying my hand at some practice problems for a test.
Edit: I'm saying the max value of |R| is .5(n^2 +n). I got this by adding n + (n^2-n)/2. I arrived at this by considering an upper triangular matrix. If this is correct, I'm still not sure how to the second part.
AI: $\newcommand{\bb}{\color{blue}{\bullet}}\newcommand{\br}{\color{brown}{\bullet}}\newcommand{\bu}{\bullet}$Look instead at where the formula for the number of antisymmetric relations on $A$ comes from. Here’s a diagram of $A\times A$ for a set $A$ of $9$ elements. A relation on $A$ is a subset of this diagram. If that relation is antisymmetric, can it include all of the brown dots? If it includes a blue dot, can it include the brown dot symmetrically opposite it on the other side of the brown diagonal or vice versa (like the orange/cyan pair)?
$$\begin{array}{c|cc}
&a_1&a_2&a_3&a_4&a_5&a_6&a_7&a_8&a_9\\ \hline
a_1&\br&\bb&\bb&\bb&\bb&\bb&\bb&\bb&\bb\\
a_2&\bu&\br&\bb&\bb&\bb&\bb&\color{orange}\bullet&\bb&\bb\\
a_3&\bu&\bu&\br&\bb&\bb&\bb&\bb&\bb&\bb\\
a_4&\bu&\bu&\bu&\br&\bb&\bb&\bb&\bb&\bb\\
a_5&\bu&\bu&\bu&\bu&\br&\bb&\bb&\bb&\bb\\
a_6&\bu&\bu&\bu&\bu&\bu&\br&\bb&\bb&\bb\\
a_7&\bu&\color{cyan}\bullet&\bu&\bu&\bu&\bu&\br&\bb&\bb\\
a_8&\bu&\bu&\bu&\bu&\bu&\bu&\bu&\br&\bb\\
a_9&\bu&\bu&\bu&\bu&\bu&\bu&\bu&\bu&\br
\end{array}$$
Added: Every one of these largest antisymmetric relations on $A$ must include all of the diagonal pairs. It must include exactly one of each symmetric pair, like $\langle a_2,a_7\rangle$ and $\langle a_7,a_2\rangle$ in the diagram. In the comments you found that there are $\frac12n(n-1)$ symmetric pairs. From each of those pairs you make a $2$-way choice; how many ways are there to make $\frac12n(n-1)$ $2$-way choices? |
H: Grandi's Series; tends to $1/2$, but why is this considered a valid sum?
Grandi's series,
$$1+1-1+1-1+1-1+...$$
can be expressed as the below:
$$\sum_{n=0}^\infty(-1)^n$$
Two valid sums that make sense to me are $1$, and $0$, depending on how you approach the series. $(1+1)-(1+1)-(1+1)-...=0$, and $1+(1-1)+(1-1)+(1-1)+...=1$.
There is consensus, however, that the actual sum is $\frac{1}{2}$. Why? I understand the approach of finding partial means of the series, and they do indeed tend to $\frac{1}{2}$, but it seems unintuitive to assert that the sum is neither $1$ or $0$.
A more convincing method I found was assuming the series is $S$, then shifting it such that $S-1 = S$, then through algebra finding $S = \frac{1}{2}$, but again, it seems more intuitive answer is either $0$ or $1$. I say this strictly because adding and subtracting integers should equal an integer, never a fraction.
Is this a characteristic of infinite series, which is not specific to Grandi's series?
AI: Consider power series
$$
S(x) = \sum_{n=0}^{\infty} (-1)^n x^n, \qquad x\in [0;1).
$$
It is geometric series:
$$
\sum_{n=0}^{\infty} (-x)^n = \frac{1}{1-(-x)} = \frac{1}{1+x}.
$$
So,
$$
S(x)=\frac{1}{1+x}, \qquad x\in[0,1).
$$
$S(x)$ is continuous and bounded on $[0;1)$. So, we can find limit:
$$
S = \lim_{x\to 1} S(x) = \frac{1}{2}.
$$
See Abel summation for better understanding. |
H: Assignment: Find the number of parameters in the general solution to a system of linear equations
This is a question given in an assignment I'm working on:
If the coefficient matrix $A$ in a homogeneous system of 33 equations with 28 unknowns is known to have rank 12, how many parameters are there in the general solution?
I've deduced that, since this is a homogeneous system with fewer variables than equations, the only solution is the trivial solution; I'm unsure, however, how to find the number of parameters in this solution.
How would I go about finding the number of parameters in this kind of abstract situation?
AI: You add a parameter for every column without a pivot in the REF form of the coefficient matrix. That is, there are the same number of parameters as the dimension of the null space of $A$. What is the relationship between $\dim NS(A)$ and $\text{rk}(A)$?
Mouse over for relationship...
$$\dim NS(A) = \text{columns} - \text{rk}(A)$$ |
H: How can we derive the pseudo inverse of a matrix from its Singular value decomposition?
For a matrix $M$ with its singular value decomposition $UΣV^T$, the pseudo inverse of $M$, i.e., $M^+$ is $VΣ^+U^T$.
How can I derive the pseudo inverse(Moore–Penrose) $M^+$ from the singular value decomposition of a matrix $M$?
From SVD, we know that $Σ$ is a diagonal matrix which contains the square roots of the eigen values of both $MM^T$ and $M^TM$ whereas $Σ^+$ is formed by replacing the non-zero diagonal elements of $Σ$ by its reciprocal. The diagonal matrix Σ is not always full rank so I assume that $ΣΣ^+$ cannot always be an Identity Matrix. How is it possible to prove that the pseudo inverse of $M$, i.e., $M^+$=$VΣ^+U^T$ holds when $ΣΣ^+$ cannot be reduced to Identity Matrix? Is there any other approach?
AI: $M^+=V\Sigma^+U^\ast$ (i.e. $M^+=V\Sigma^+U^\top$ in the real case), where $\Sigma^+$ is a rectangular diagonal matrix whose size is identical to the size of $\Sigma^\top$. The $i$-th main diagonal entry of $\Sigma^+$ is $\sigma_i^{-1}$ if the $i$-th singular value $\sigma_i$ of $M$ is positive, otherwise the diagonal entry is zero.
You may simply prove that $M^+$ is indeed the Moore-Penrose pseudoinverse of $M$ by showing that it satisfies the four defining properties of Moore-Penrose pseudoinverse, namely, both $MM^+$ and $M^+M$ are Hermitian (or real symmetric in your case), $MM^+M=M$ and $M^+MM^+=M^+$. |
H: Why nonlinear programming problem (NLO) called "nonlinear"? What does "nonlinearity" actually mean? Is it "not linear" or something different?
My teacher in the course Mat-2.3139 presented the same definition as in Wikipedia for the nonlinear programming problem here
but he did not specify what the nonlinearity actually means or what it actually infers. It is a bit hard to make a linear problem into nonlinear without a clear definition for it. So
I. Does the nonlinearity mean nonlinear objective function?
II. For one variable $\bf x$, does the nonlinearity mean that the function
is not a linear map? Definition here.
III. or does the NLO mean a problem with nonlinear constraints?
IV. or does the NLO mean a problem with nonlinear constraints and
nonlinear objective function?
V. or instead of linear mapping is a NLO problem with an objective
function with certain smoothness?
VI. and what kind of requirements a NLO problem require: what kind of
convexity assumptions for domain and codomain? Does the function
itself need to be convex or concave? More detailed question related to convex optimisation here.
AI: Any of I, III or IV would make a problem non-linear.
The answers to VI will probably covered during your course and different convexity conditions may lead to different algorithms. |
H: Group algebras, Maschke's lemma and direct sums of matrix algebras
Let $G=\{g_1,g_2,\dots,g_n\}$ be an arbitrary finite group. We consider its representations over $\mathbb{C}$. There is Maschke's theorem which states that each representation of $G$ is a direct sum of irreducible ones. I'm trying to link this result with some properties of the group algebra $\mathbb{C}[G]$: it should be the sum of matrix algebras.
So, the regular representation, induced by the action of $G$ on $\{e_{g_1},e_{g_2},\ldots,e_{g_n}\}$ (each $g$ sends $e_h$ to $e_{gh}$) is a representation of $G$ in the group algebra, i.e. the homomorphism $G\rightarrow \mathrm{GL}(\mathbb{C}[G])$. Maschke's theorem allows us to decompose $\mathbb{C}[G]$ in the sum of irreducible representations: $V_1^{\bigoplus m_1}\bigoplus\cdots\bigoplus V_k^{\bigoplus m_k}$ (each $V_i$ is irreducible).
Well, where are matrix algebras...?
AI: You will need Schur's lemma to prove this. A proof reference can be found here, Lemma 9, Theorem 10. |
H: Computation of $n$-th order difference of falling factorial
I was reading a difference equation textbook and came across a problem. The question asks to compute ${\Delta}^nt^{\underline3}$ for $n=1,2,3,...$, where $t^{\underline3}$ is the falling factorial "$t$ to the $3$ falling" and ${\Delta}^n$ is the $n$-th order difference operator.
However, there is a theorem that says ${\Delta}^nt^{\underline{r}}=rt^\underline{r-1}$. Therefore If I am right, had if been $n$ is known (say 2), the solution would have been as easy as
${\Delta}^2t^{\underline3}={\Delta}3t^{\underline2}=6t^{\underline1}=6t$
How will the solution be for $n=1,2,3,...$ please?I really wish someone will help.
AI: since $t^{\underline3}=t(t-1)(t-2)$, it's aisy to see that ${\Delta}^nt^{\underline3}=0$ if $n \geq 4$.
for $n=3$, we have: ${\Delta}^3t^{\underline3}=6$ |
H: Why are two directions enough for the Cauchy-Riemann equations to imply differentiability?
If the complex function $f(z)$ is complex differentiable $\Rightarrow$ the Cauchy Riemann equations hold. $($This is because if $f'(z)$ is the same no matter in what direction $\delta z\rightarrow 0$. Choosing the special case of $\delta z\rightarrow 0$ along the real, then the imaginary, line yields the Cauchy Riemann equations, so it is obvious that any differentiable function will satisfy them$)$.
However, how is demonstrating that $f'(z)$ is the same for two perpendicular paths enough to show that $f'(z)$ is the same for all paths (i.e. how does one go from $\Rightarrow$ to $\Leftrightarrow$)? I have tried explicitly calculating the directional derivatives, but the mess that ensued did not enlighten whatsoever.
I do not know much about linear algebra if that helps in writing answers. Any geometric intuition would be greatly appreciated!
AI: The directional derivative in any direction in $\mathbb{R}^n$ is completely determined by the directional derivative in $n$ linearly independent directions. This somewhat counterintuitive fact follows from linearity of the directional derivative: any vector $v$ can be expressed as $\sum \alpha_i b_i$ where $b_i$ are the linearly independent directions, and then
$$Df(v) = \sum \alpha_i Df(b_i).$$
In the case of $\mathbb{R}^2$, if the directional derivative of two functions agree in two directions, they must agree in all directions.
This same phenomenon also explains why the curvature of a surface in $\mathbb{R}^3$ is completely determined by the two principle curvatures, why the bending stiffness of a rod with non-circular cross section is completely determined by the bending stiffness in two directions, etc.
EDIT: You asked me for geometry and I gave you a bunch of linear algebra, so let me say a bit more from a geometric perspective. A function $f:\mathbb{R}^2\to \mathbb{R}$ has its first derivatives at a point determined by its tangent plane at that point. If you know two lines contained in that plane (which is what the two directional derivatives tell you) then you know the entire plane, since it passes through the point in question, and has as its normal the cross product of those two lines.
If you think of a complex function as $f:\mathbb{R}^2\to \mathbb{R}^2$, nothing much changes, since the tangent-plane interpretation applies to each component of $f$. |
H: prove that there is no supremum for mn/1+m+n
D = { mn/(1+m+n) } for m,n natural numbers.
To simplify the expression, I presumed m=n, which means:
D = { n^2 / (2n + 1) }
Now, I know by intuition there is no supremum, for this series convergencing to infinity, but I couldn't find a formal way to prove it.
I thought about proving by contradiction:
- assume S is the supremum
- ϵ > 0
S - ϵ > n^2 / (2n + 1)
But then I got stuck..
Any guideness will be very appreciated.
Thanks!
AI: Hint: we can set $n = k$, $m = k-1$ to get
$$
\frac{mn}{1+m+n}=
\frac{k(k-1)}{1 + k + k-1} = \frac{k(k-1)}{2k} = \frac{k-1}{2}
$$
Thus, $\frac{k-1}{2}\in D$ for every $k\geq 2$.
So, if $D$ has a supremum, then the supremum of $D$ is bigger than $\frac{k-1}{2}$ for every integer $k$. Why does this result in a contradiction? |
H: Integral $\int_0^\infty\frac{1}{\sqrt[3]{x}}\left(1+\log\frac{1+e^{x-1}}{1+e^x}\right)dx$
Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{1}{\sqrt[3]{x}}\left(1+\log\frac{1+e^{x-1}}{1+e^x}\right)dx$$
AI: We first remark the following identity:
\begin{align*}
\int_{0}^{\infty} \frac{x^{s-1}}{ze^{x} - 1} \, dx
&= \int_{0}^{\infty} \frac{z^{-1}x^{s-1}e^{-x}}{1 - z^{-1}e^{-x}} \, dx \\
&= \sum_{n=1}^{\infty} z^{-n} \int_{0}^{\infty} x^{s-1} e^{-nx} \, dx \\
&= \Gamma(s) \sum_{n=1}^{\infty} \frac{z^{-n}}{n^{s}}
= \Gamma(s)\mathrm{Li}_{s}(z^{-1})
\end{align*}
which initially holds for $|z| > 1$, and then extends holomorphically for $z^{-1} \notin (1, \infty]$ since both sides define holomorphic functions on this region. Now, by integrating by parts,
\begin{align*}
\int_{0}^{\infty} \frac{1}{\sqrt[3]{x}} \left(1+\log\frac{1+e^{x-1}}{1+e^{x}}\right) \, dx
&= \frac{3}{2} \int_{0}^{\infty} x^{2/3} \left( \frac{1}{(-1)e^{x} - 1} - \frac{1}{(-e^{-1})e^{x} - 1} \right) \, dx \\
&= \frac{3}{2} \Gamma\left(\frac{5}{3} \right) \left\{ \mathrm{Li}_{5/3}(-1) - \mathrm{Li}_{5/3}(-e) \right\} \\
&= -\frac{3}{2} \Gamma\left(\frac{5}{3} \right) \left\{ (1 - 2^{-2/3})\zeta(5/3) + \mathrm{Li}_{5/3}(-e) \right\}.
\end{align*}
I can hardly believe that $\mathrm{Li}_{5/3}(-e)$ can be simplified further. |
H: One question about topology.
Why is the set $ A=\{ (x ,x^{-1}):0<x\leqslant 1\}$ is closed in $\Bbb R^2$ but is not bounded?
Why is the set $ S=\{(x,\sin(x^{-1})) :0<x\leqslant 1\}$ is bounded in $\Bbb R^2$ but is not closed?
I am just a beginner in topology, so I hope someone can answer it in detail. Thanks.
AI: I guess, you mean $A=\{(x,1/x)\,\mid\,0<x\le1\}$ and $S=\{(x,\sin(1/x))\,\mid\,0<x\le 1\}$/
Well, $A$ is not bounded because $1/x$ can be arbitrary big as $x$ is close to $0$, so actually it is not bounded from upside. $S$ is bounded because $\sin($anything$)$ is always between $-1$ and $1$, and now the first coordinate $x$ is also bounded. So, $S$ is contained in a rectangle, hence it is also contained in a disc, so it is bounded.
$A$ is closed because it contains all limit points. If $x_n$ is a sequence in $(0,1]$, and $x_n\to x$ for some $x\in (0,1]$ then we have $1/x_n\to 1/x$, so the sequence $(x_n,1/x_n)$ in $\Bbb R^2$ converges to $(x,1/x)$.
$S$ is not closed, because there are limit points at $x=0$ which are not contained in $S$, e.g. if we take $x_n:=\displaystyle\frac1{2n\pi+\pi/2}$, then $(x_n,\sin(1/x_n))\to (0,1)$ which is $\notin S$. |
H: Prove that the set $H$ is a subspace of $\mathbf{P}_3$
The question is:
Consider the vector space $\mathbf{P}_3$ of all polynomials of degree at most 3 with real coefficients.
Prove that the set $H$ of all polynomials $p$ in $\mathbf{P}_3$ which vanish at $t = 2$ (meaning $p(2) = 0$) is a subspace of $\mathbf{P}_3$ (using the definition of a subspace).
Find three polynomials $\,f, g, h$ so that $H$ from the previous problem equals $\text{span}\{\,f, g, h\}$.
I know that I need to prove $H$ is closed under addition and scalar multiplication, and it contains the zero vector. I need someone to help me get started and help explain how "which vanish at $t = 2$ (meaning $p(2) = 0$)" affects the problem.
AI: The specific value $t=2$ does not affect the problem, but restricts the polynomials to a subspace.
To start, assume that $f,g\in H$. Show that $\lambda\,f\in H$ and $f+g\in H$, using the definition of $H$.
For the second part, take $\big((t-2),\ (t-2)t,\ (t-2)t^2\big)$. |
H: Problem of understanding instructions on practice sheet
Can someone help me understand the scalar product of problem 23 in the attached practice sheet? I should show that this defines a nondegenerated symmetric scalar product on the vector space of trivariate polynomials.
My problem is that I don't understand the argument of $P$. An example would be great!
Thank you.
The sheet is in german, but all you need to know is above :)
Practice Sheet
AI: This is just a notation/formal calculation. Just plug in the differential operators into the polynomial $P$ and apply the resulting operator to $Q$ and evaluate at $(0,0,0)$. |
H: Proof that $\sin(x)$ don't have limit to infinity
I just used the Heine's definition.
Let $\alpha,\delta \in \mathbb{R}$ such that $\sin(\alpha)=a$ and $\sin(\delta)=b$. Let $(u_{n})=\alpha+2\pi n$ and $(v_{n})=\delta+2\pi n$ and $f(x)=\sin(x)$.
So one have,
$$\lim\limits_{n\rightarrow \infty} u_{n}=+\infty$$
$$\lim\limits_{n\rightarrow \infty} v_{n}=+\infty$$
But,
$\lim\limits_{n\rightarrow \infty} f(u_{n})=a$
$\lim\limits_{n\rightarrow \infty} f(v_{n})=b$
Because these last limits aren't equal, the sine function don't have limit to infinity. Is this proof correct? Thanks.
AI: Another proof would be to note that $\lim_{x \to \infty} \left( \sin(\frac{\pi}{2}+2\lfloor x \rfloor \pi) - \sin ( 2\lfloor x \rfloor \pi) \right) = 1$, hence $\sin$ has no limit as $x \to \infty$. (If $\lim_{x \to \infty} f(x) = L$, then
$\lim_{x \to \infty} \left( f(x+y)-f(x) \right) = 0$ for all $y$). |
H: Combination with quantity
We need to ship 100 kits, each containing three beads of different colors. There are four colors: purple, blue, green, silver. We have the following quantities for each respective color: 53, 53, 85, 53. What are the fewest beads we will need to order, and what colors should we order to ship 100 kits with three beads in each kit and no kit having two or more beads of the same color? Also, how did you solve?
My guess was 56 total beads, using two colors: 41 of one, 15 of another, but I'm not sure if I solved it correctly:
make 53 boxes, using purple, blue, green, leaving you with 32 green and 53 silver.
order 15 green and 41 of either purple or blue, allowing you to make 100 kits using the three extra silver you have.
Now, that I think about it. I think this way has 6 boxes with duplicate silvers. :/
AI: Hint: how many total beads do you have? How many do you need to ship? Is there any reason you can't order just enough beads, like having too many of one color?
Added: You are correct that you need to order $56$ more beads. There is no best way. I would order $5$ green and $17$ of each other color, giving me $70,70,90,70$ just because I like round numbers. Now $30$ kits of each of $pbg, pgs, bgs$ and $40$ kits of $pbs$ get you there.
Added2: Any collection of $300$ beads can fill your order as long as you don't have more than $100$ of any one color. Add white beads to each bin to make a total of $100$ beads per bin. Now pack kits, taking one bead from each bin, with exactly one white bead in each kit. Remove the white beads and ship the kits. |
H: Find this limit without L'hopital Rule : $\lim_{x\rightarrow +\infty}\frac{x(1+\sin(x))}{x-\sqrt{(1+x^2)}}$
Find this limit without l'Hopital rule : $$\lim_{x\rightarrow +\infty}\frac{x(1+ \sin x)}{x-\sqrt{1+x^2}}$$
I tried much but can't get any progress!
AI: The limit does not exist. Multiply top and bottom by $x+\sqrt{1+x^2}$. The bottom becomes $-1$. As to the new top, it is very big if $\sin x$ is not close to $-1$. However, there are arbitrarily large $x$ such that $\sin x=-1$. |
H: Limit Comparison Test - ex. prob wrong?
I'm doing some practice problems on the Limit Comparison test from this site:
http://archives.math.utk.edu/visual.calculus/6/series.14/index.html
But I'm a bit confused on this problem: $\sum_{n=1}^\infty \sin (\frac 1n)$ .
The solution says to use the harmonic series for comparison ( $\sum_{n=1}^\infty \frac 1n $ ),
which, when set up for the limit test gives (this is the solution from the webpage): $$\lim_{n\to \infty} \frac{\sin (\frac 1n)}{\frac 1n} = 1$$
What I don't understand is how they solved the limit. The $\sin (\frac 1\infty$) makes sense, since it goes to $\sin(0) = 0$, but the bottom half of the fraction is $\lim_{n\to \infty} \frac 1n = \frac 1\infty = 0$, which would result in $\frac {\sin(0)}0 $, which by my understanding would be a divide by zero error.
Changing the form to $\lim_{n\to\infty} n\sin(\frac 1n)$ also appears to be a dead end since I just get $\infty \sin(\frac 1\infty)$.
What am I missing here? How can the above limit evaluate to 1? I know that $\sum_{n=1}^\infty \sin (\frac 1n) $ is divergent, as is $\sum_{n=1}^\infty \frac 1n $, and so I would expect the LCT to give a finite, positive value confirming this, but I guess I don't understand how they did their limit evaluation. Can anyone enlighten me?
AI: You make the transformation $ \ u \ = \frac{1}{n} \ $ , so that you can write the limit as $ \ \lim_{u \rightarrow 0} \ \frac{\sin u}{u} \ $ , which has a (supposedly) familiar result. |
H: Determine if projection of 3D point onto plane is within a triangle
In 3D, given three points $P_1$, $P_2$, and $P_3$ spanning a non-degenerate triangle $T$. How to determine if the projection of a point $P$ onto the plane of $T$ lies within $T$?
AI: The question is a slight extension of the question given here: Check whether a point is within a 3D Triangle
There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.
Let $\vec{u}=P_2-P_1$, $\vec{v}=P_3-P_1$, $\vec{n}=\vec{u}\times\vec{v}$,
$\vec{w}=P-P_1$. We then have directly the barycentric coordinates of the projection $P'$ of $P$ onto $T$ as
$\gamma=\left[(\vec{u}\times\vec{w})\cdot\vec{n}\right]/\vec{n}^2$
$\beta=\left[(\vec{w}\times\vec{v})\cdot\vec{n}\right]/\vec{n}^2$
$\alpha=1-\gamma-\beta$
The coordinates of the projected point is
$P'=\alpha P_1+\beta P_2 +\gamma P_3$
The point $P'$ lies inside $T$ if
$0\leq\alpha\leq 1$,
$0\leq\beta\leq 1$, and
$0\leq\gamma\leq 1$. |
H: Example involving Uniform Continuity
Question: Could someone give an example of a sequence of uniformly continuous real-valued functions on the reals such that they converge point-wise to a function that is continuous but not uniform continuous.
My attempt so far: I managed to prove this is true in the case of uniform convergence, so I'm convinced there is an example. I considered triangles such that they were symmetrical on the y-axis such that they got taller and closer however this converges point-wise to 0 which is uniform continuous
AI: I'm not convinced this is the simplest example (I certainly wouldn't want to make it explicit), but it was fun :)
Lines are uniformly continuous and quadratics are not. Also, continuous implies is uniformly continuous on a compact domain. In addition, it's not too hard to show that if you glue two uniformly continuous functions together, the result is uniformly continuous.
Now create a function that approximates a quadratic by being identical near the vertex and at some point on each side breaking off into the tangent lines. Therefore, considering a sequence of such functions whose "breakoff points" have $x$-values that go to $\pm\infty$; these satisfy the premises and get the conclusion, so you're good. |
H: Prove f(x) <= x for all x>=0 if f ' (x) <= -2 for all x and f(0) = 0
The title basically states the whole question..I was trying to invoke the Mean Value Theorem on it but it hasn't worked..I was wondering if I'm supposed to solve it some other way. I just need hints, please. Thank you.
AI: If $f$ is continuously differentable,
$$ f(x) = f(0) + \int_0^x f'(t)\, dt \le 0 + \int_0^x -2\, dt = -2x, \qquad x \ge 0.$$
Choose $x > 0$.
Suppose $f$ is differentiable on $(0,x)$ and continuous on $[0,x]$. Then there is some point $c\in(0,x)$ so that
$$f'(c) = {f(x)-f(0)\over x} ={f(x)\over x}$$
Since $f'(c) \le -2$,
${f(x)\over x}\le -2$
so $f(x) \le -2x$. |
H: Can 'a family of sets' be an empty set?
1.Let $X$ be a set.
Let $\mathscr{A}$ be a family of subsets of $X$.
Here, what does 'a family' means precisely?
Does this mean $\{f(\alpha)\}_{\alpha\in A}$ for some $A\subset P(X)$ and $f:A\rightarrow P(X)$?
AI: Yes; the empty set is indeed a family of sets. |
H: Let $W$ be a vector space and let $U$ and $V$ be finite-dimensional subspaces.
a) Show that $U ∩ V$ is a subspace of $W$.
b) Show that $U + V = \{u + v : u \in U, v \in V\}$ is a subspace of $W$.
c) Show that $\dim(U+V) = \dim(U) + \dim(V) - \dim(U ∩ V)$.
I have no idea where to start. Any help?
AI: For part a) Use the definition of what a subspace means. It is first closed under vector addition, has an identity, and is closed under scalars. First show closure under vector addition which is basically proving this: If $x,y\in U\cap V\implies x+y\in U+V$. Now the identity is trivial to show. Since $U$ and $V$ are finite subspaces it follows by definition that... And for scalar multiplication. Let $r\in\mathbb{R}$ where $r$ will be a scalar. Let $x\in U\cap V\implies x\in U$ and $x\in V$. It follows that since $U$ and $V$ are subspaces then $rx\in U$ and $rx\in V \implies rx\in U\cap V$. Hence we conclude...
For part b) Same thing. Let $x\in U+V$. First show closure. That is if $x,y\in U+V$ where $x=a+b$ and $y=c+d$ where $a,c\in U$ and $b,d\in V$ then we see $x+y=(a+b)+(c+d)$ which is still an element of the set $U+V$. For the identity its easy. I will denote $0$ to be the identity both in $U$ and $V$ since it follows by definition of a subspace. Thus $0=0+0\in U+V$ And lastly scalar multiplication which I will leave up to you.
For part c) Recall what dimension means and remember $U+V$ is a set.
I forgot to put arrows on top of the elements but they are all vectors. So you can place arrows for clarity. |
H: Whats the difference between axiom and primitive concept?
I've read the definitions, but they are not very clear to me.
Looks like both are a premisse so evident to be accepted as true without controversy.
But, what about the axioms on the set theory?? Many of them are not evident, which contradicts the definition i've read about axiom.
Thank you!
AI: An axiom is expressible as a complete sentence, with a subject and a predicate.
A concept, on the other hand, is not expressible as a sentence, but as a word or a phrase, usually a noun or a noun phrase. A primitive concept is one that is not defined in terms of other concepts.
Thus "point" and "line" are primitive concepts in some axiomatizations of geometry, and "For any two points, there is exactly one line incident to both" is an axiom in some such systems. |
H: Calculating primitive roots
Wikipedia cleanly demonstrates that $3$ is a primitive root modulo $7$. Here
is the table, and my question is how do they calculate the 4th column? It appears that they take the exponent from the previous column and multiply it by $3$ but this pattern fails for the 4th row.
Is there some number theoretic result that allows this to be calculated so seamlessly? If not what am I not seeing?
AI: In the fourth column, they reduce $3^n$ modulo $7$ to make computations easier. As an example, $3^3=27$ which is $=6\mod 7$ for $27-6=21=3\times 7$.
Remember that by construction of $\equiv$ in $\Bbb Z/n\Bbb Z$, if $a\equiv b\mod n$ and $a'\equiv b'\mod n$ then $a+a'\equiv b+b'\mod n$ and $aa'\equiv bb'\mod n$. |
H: Fermat's Little Theorem and Prime Moduli
I am given two distinct primes $p$ and $q$, where $$m = p*q$$ Also,
$$ \begin{cases} r\equiv 1\mod p-1\\ r\equiv 1\mod q-1 \end{cases} $$
I have to show that given an integer a, show that $$a^r \equiv a \mod (m)$$
I'm not sure how to get started. I know I can tie this in to Fermat's Little Theorem and I've found something here CRT + Fermat's Little Theorem that was somewhat related to my question but I had a hard time seeing where to go. Just need a little hint to get me in the right direction! Thanks.
AI: Hint #1. It's enough to show that $a^r\equiv a\pmod p$ and $a^r\equiv a\pmod q$; since $p$ and $q$ are relatively prime, it will follow from that that $a^r\equiv a\pmod{pq}$. By symmetry, it's enough to show that $a^r\equiv a\pmod p$.
Hint #2. What does $\ r\equiv1\pmod{p-1}\ \ $ mean? It means that $r-1$ is divisible by $p-1$; in other words, that $\ r=1+(p-1)m\ \ $ for some integer $m$. |
H: Starting Bisection Proof of Extreme Value Theorem
I am having difficulties beginning a proof for the following statement:
Use a proof strategy of bisection to prove that every function $f:[a,b] \to \mathbb{R}$ that is not bounded above is discontinuous at some point $c \in [a,b]$ (and discontinuous from the right or left if $c=a$ or $b$, respectively.
Although the strategy is given, I am having trouble getting started. Furthermore, how would I use this bisection argument to prove that if $f:[a,b] \to \mathbb{R}$ is continuous and $\sup{\{f(x):a \leq x \leq b\}}=M$, then $f(c)=M$ for some $c \in [a,b]$. I believe that this is a reformulation of the Extreme Value Theorem.
Many thanks in advance. I am using the textbook Introduction to Analysis by Arthur Mattuck.
AI: If $f:[a,b] \to \mathbb{R}$
is not bounded above,
there is a point $x_1 \in [a, b]$
such that
$f(x_1) > 1$.
Divide $[a, b]$
into two parts
$[a, x_1]$ and $[x_1, b]$.
In one of these parts,
there is an $x_2$
such that
$f(x_2) > 2$
(by the unboundedness of $f$).
The width of this interval
is $\le \frac{b-a}{2}$.
Again,
divide that part
at $x_2$ into two parts.
The width of each of these parts
is $\le \frac{b-a}{4}$,
and there must be an
$x_3$ in one of these parts
such that
$f(x_3) > 3$
(or $f(x_3) > $ some large value if you want).
Repeating this,
after the $n$-th division,
there is an interval
or width $\le \frac{b-a}{2^{n-1}}$
with a point $x_n$
such that
$f(x_n) > n$.
This sequence of points
$(x_n)$
converges to a limit
and $f(x_n)$
is unbounded,
so $f$ can not be continuous
at the limit. |
H: Big-O evaluation:
I have the expression:
$$f_{k}(n,m) = (n - k)(m - k) + f_{k+1}(n,m)$$
which runs until k = n or m.
What is the big theta of this function in terms of n,m?
A naive approach is to assume that m does not vary at all since:
$f_k(n,m) < (n - k)m + f_{k+1}(n,m) $
Which gives the expression:
$$nm + (n-1)m + (n-2)m. .. = m\frac{n^2 + n}{2} = O(mn^2)$$
But I want a tighter bound given that we are working with:
$$nm + (n-1)(m-1) + (n-2)(m-2) ... $$
AI: Assume $n>m$. Then the expression you are currently working with has m terms. Expanding each expression of the form $(n-a)(m-a)=nm-an-am+a^2$ and regrouping like terms and taking their sum, we get:
$nm^2-n(m(m+1)/2)-m(m(m+1)/2)+m(m+1)(2m+1)/6$
After simplifying the expression, you should get the tight bound. |
H: Assignment: determining sets are bases of $\mathbb{R}^3$
This is question from an assignment I'm working on:
Which two of the following three sets in $\mathbb{R}^3$ is a basis of $\mathbb{R}^3$?
\begin{align*}
B_1&=\{(1,0,1),(6,4,5),(-4,-4,7)\}\\
B_2&=\{(2,1,3),(3,1,-3),(1,1,9)\}\\
B_3&=\{(3,-1,2),(5,1,1),(1,1,1)\}
\end{align*}
Thanks to software, I know that the answer is $B_1$ and $B_3$, as $B_2$ the only linearly dependent set of the three - and in this case, an LD set can't be a basis for $\mathbb{R}^3$.
Manually, I've put all three sets in RREF, and all three can be reduced to
$$
\left[\begin{array}{ccc|c}
{1}&{0}&{0}&{0}\\
{0}&{1}&{0}&{0}\\
{0}&{0}&{1}&{0}
\end{array}\right]
$$
This also checks out when computed by software.
Since all three sets reduce to the same RREF, how can I prove that these sets are linearly (in)dependent?
AI: For $B_{2}$, this is linearly independent if the only solution to
$$a(2,1,3)^{T} + b (3,1,-3)^{T} +c(1,1,9)^{T} = 0$$
has the trivial solution $a=b=c=0$
or
$$
\begin{bmatrix}
2&3&1\\
1&1&1\\
3&-3&9
\end{bmatrix}\begin{bmatrix}
a\\
b\\
c
\end{bmatrix} = \begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$$
Take $-2R_{2} +R_{1}\rightarrow R_{1}$, $-3R_{2} + R_{3}\rightarrow R_{3}$ $-1/6 R_{3}\rightarrow R_{3}$
To get
$$
\begin{bmatrix}
0&1&-1\\
1&1&1\\
0&1&-1
\end{bmatrix}\begin{bmatrix}
a\\
b\\
c
\end{bmatrix} = \begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$$
So the rank is clearly less than 3 hence there are infinitely many solutions. If we continue by taking $-1R_{1} + R_{3}\rightarrow R_{3}$ and perform a row swap, we get
$$
\begin{bmatrix}
1&1&1\\
0&1&-1\\
0&0&0
\end{bmatrix}\begin{bmatrix}
a\\
b\\
c
\end{bmatrix} = \begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$$
At this point, you can either continue taking the system to RREF, note that since the system has a row of zeros and is homogeneous, it must be infinitely many solutions, or perform back-substitution beginning with $c=s$ where $s\in \mathbb{R}$.
The choice of row operations are more or less arbitrary, I mainly did them because of the $1$ in row 2 for ease of calculations. It isn't precisely Gaussian Elimination because I did not start with a row exchange so that the leading 1 is in the first pivot column, but that does not matter. As soon as we got a row of zeros, that showed that Row 3 was a linear combination of Row 1 and Row 2, hence the dimension that these three vectors spans is $2$ so it cannot span $\mathbb{R}^{3}$. |
H: If a matrix of the form I + B is singular, then ||B|| ≥ 1 for every subordinate norm.
I need some guidance showing that:
If a matrix of the form I+B is singular, where I is the identity matrix, then for any subordinate norm $\|\cdot\|$, $\|B\|\geq1$.
AI: If the matrix is singular, then for some $v$ with $\|v\| = 1$, we have $(I+B)v = 0$, that is, $Bv = -v$. Hence $\|B\| = \sup_{\|y\|\le 1} \|By\| \ge \|Bv\| = \|-v\|=1$. |
H: Sum of squares of the quadratic nonresidues modulo $p$ is divisible by $p$
Let $p$ be a prime number with $p > 5$. Prove that the sum of the squares of the quadratic nonresidues modulo $p$ is divisible by $p$.
My idea is to use the fact that any quadratic residue is congruent modulo $p$ to one integer in the set $\{1^2,2^2,\ldots,\left( \frac{p-1}{2} \right)^2 \}$. And none of the quadratic residues are congruent to each other. So the quadratic nonresidues must all be in the set $\{ \left( \frac{p+1}{2} \right)^2 , \left( \frac{p+3}{2}\right)^2 \ldots, (p-1)^2\}$. So the sum of the quadratic nonresidues must be given by $$\sum_{k=1}^{p-1} k^2 - \sum_{k=1}^{\frac{p-1}{2}}k^2 = \frac{(p-1)p(2p-1)}{6} - \frac{\left( \frac{p-1}{2} \right) \left( \frac{p+1}{2} \right) p}{6} = \frac{p}{6} \left( (p-1)(2p-1) - \frac{1}{4}(p-1)(p+1) \right).$$
Is this the correct approach? Prove that the term in parenthesis is an integer divisible by $6$?
AI: Using primitive roots finishes the problem quickly: Let $g$ be a primitive root, so that the sum of the squares of the quadratic non-residues is just
$$\sum_{i=1}^{\frac{p-1}{2}}{(g^{2i-1})^2} \equiv g^2\sum_{i=0}^{\frac{p-3}{2}}{g^{4i}} \equiv g^2\frac{1-(g^4)^{\frac{p-1}{2}}}{1-g^4} \equiv \frac{g^2(1-(g^{p-1})^2)}{1-g^4} \equiv 0 \pmod{p}$$
where since $p>5$, $p \nmid 1-g^4$ so the manipulations above are valid.
If you don't want to appeal to primitive roots, then something similar to what you tried can also be done. What we want is to take the sum of the squares of all non-zero elements and subtract the squares of the quadratic residues. So we get
\begin{align}
&\sum_{k=1}^{p-1}{k^2}-\sum_{k=1}^{\frac{p-1}{2}}{(k^2)^2} \\
&\equiv \frac{(p-1)p(2p-1)}{6}-\frac{(\frac{p-1}{2})(\frac{p+1}{2})(p)(3(\frac{p-1}{2})^2+3(\frac{p-1}{2})-1)}{30} \pmod{p}\\
& \equiv p[\frac{(p-1)(2p-1)}{6}-\frac{(\frac{p-1}{2})(\frac{p+1}{2})(3(\frac{p-1}{2})^2+3(\frac{p-1}{2})-1)}{30}] \pmod{p}\\
& \equiv 0 \pmod{p}
\end{align}
where we may treat $\frac{1}{30}=30^{-1}$ as an element in $\mathbb{Z}_p$ since $p>5$. |
H: Pigeonhole question and generalization
Let H be a regular hexagon with side length 1 unit.
(a) Show that if more than 6 points are specied inside H then the points of at least one pair of them are at most 1 unit apart.
(b) State and prove a generalization of the result in (a) to the situation where there are in excess of $2^{2n+1}3$ points inside H.
I did part (a), I solved it using the Pigeonhole Principle. On the other side, I don't understand what part (b) is about. Any suggestions is more that appreciate it!!
AI: Hint: An equilateral triangle with sides of unit length can be split into $2^{2n}$ triangles of side length $\frac{1}{2^n}$.
As you noticed before, a regular hexagon with side lengths 1 can be split into 6 equilateral triangles of unit length. Thus, a regular hexagon with side lengths 1 can be split into $2^{2n+1}3$ equilateral triangles of side length $\frac{1}{2^n}$.
Now, how can you use the pigeonhole principle to prove a more general result for your part (b)? |
H: How many solution with the equation $f_{2013}(x)=\frac{x}{2013}$
let $f(x)=f_{1}(x)=\mid \cos{(2\pi x)}\mid,f_{2}(x)=f(f_{1}(x))=\mid \cos{(2\pi (\mid\cos{(2\pi x)}\mid)}\mid$
$f_{n}(x)=f(f_{n-1}(x))$,
Question:
How many solution with the equation
$$f_{2013}(x)=\dfrac{x}{2013}$$
This problem is from china students ask me,But I consider sometimes,and I can't it,I hope someone can help.
My try: I find $$\cos{(2\pi x)}=\dfrac{x}{2013}$$
AI: The map
$$
\varphi : [0,1] \to [0,1], \qquad \varphi(t) = |\cos(2\pi t)|
$$
is surjective and continuous. During the interval $[0,1/4]$, $\varphi$ goes from $1$ to $0$. Then during the interval $[1/4,1/2]$, $\varphi$ goes from $0$ to $1$. Again during $[1/2,3/4]$, $\varphi$ goes from $1$ to $0$, and finally during $[3/4,1]$, $\varphi$ goes back from $0$ to $1$.
The line $y = x/2013$ must therefore cross the graph of $\varphi$ 4 times, namely the number of times the curve $(t,\varphi(t))$ goes from $\varphi(x_1) = 0$ to $\varphi(x_2) = 1$ or vice-versa.
If we iterate $\varphi$ to get $\varphi^2$, over each of those intervals of size $1/4$, $\varphi$ will go back and forth as explained above $4$ times, for a total of $16 = 4^2$ times. For $\varphi^3$, it will go back and forth $64 = 4^3$ times.
The number of such $x$ is therefore $4^{2013}$.
EDIT! Perhaps I should add : in the intervals where $\varphi$ goes from $0$ to $1$ (resp. $1$ to $0$), $\varphi$ is increasing (resp. decreasing), hence will intersect the line $y = x/2013$ precisely once.
Hope that helps, |
H: If a sequence converges, then every subsequence converges to the same limit -- but how do I know a subsequence exists?
I have been reading the following post:
Prove: If a sequence converges, then every subsequence converges to the same limit.
I understand the idea, but I wonder, does this proof imply that such a subsequence actually exists?
That is, suppose a sequence $s_n$ converges. Then every subsequence $s_{n_k}$ of $s_n$ converges to the same limit. But my question is: does there necessarily exist such a subsequence $s_{n_k}$?
AI: Take your sequence $s_n$. It is nonempty. Therefore it has a subsequence. |
H: In how many ways can one divide 10 people into 4 unequally sized groups?
Many questions on this site involve counting the number of ways one can divide a set of n people into equally-sized groups, but how would one do so for unequally-sized groups?
The answers for this question don't provide an explanation of how to do this.
More specifically, what are the number of ways I can divide a set of 10 people in the following groups:
Two groups will contain 3 people.
One group will contain 2 people.
One group will contain 1 person.
For this specific question, I was thinking
$$
\frac{\dbinom{10}{3} \dbinom{7}{3} \dbinom{4}{2} \dbinom{2}{1}}{4!}
$$
The numerator represents the grouping and the denominator represents the fact that the order of the grouping does not matter.
However, the answers to the referenced question show that this is incorrect.
AI: Because the groups of $2$ and $1$ are distinct sizes, you don't have to consider reordering those groups. The denominator should be $2!=2$ because there are $2$ ways to order the two groups of three. The numerator is correct. |
H: How do I prove the definition of a homomorphism?
The question is asking me to prove that $f(a \circ b) = f(a) \circ f(b)$.
This I believe is referencing our previous proof which tells us:
Assume $g:x \rightarrow y $ is a bijection and for an $a \in S(X)$ set $f(a)= g \circ a \circ g^{-1}$ and then I proved $f$ is a bijective function from $S(X)$ to $S(Y)$.
We haven't been given any background on homomorphisms so I'm a little unsure on how to proceed.
AI: Note that $f(ab)=gabg^{-1}=ga1bg^{-1}=gag^{-1}gbg^{-1}=\cdots$? |
H: Proving an Combination formula $ \binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1}$
Proving an Combination formula $\displaystyle \binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1}$
$\bf{My Try}$::$\displaystyle{\binom{n-1}{k}+\binom{n-1}{k-1}=}$
$\displaystyle{\frac{\left(n-1\right)!}{k!\left(n-k-1\right)!}+\frac{\left(n-1\right)!}{\left(k-1\right)!\left(n-k\right)!}=}$
$\displaystyle{\frac{\left(n-1\right)!}{k\,\left(k-1\right)!\left(n-k-1\right)!}+\frac{\left(n-1\right)!}{\left(k-1\right)!\left(n-k-1\right)!\,\left(n-k\right)}}=$
$\displaystyle \frac{(n-1)!}{(k-1)!\cdot (n-k-1)!}\left(\frac{1}{k}+\frac{1}{n-k}\right) = \binom{n}{k}$
My Question is How can i prove using combinational argument
Help Required
Thanks
AI: Given $n$ people we can form a committee of size $k$ in ${n\choose k}$ ways. We can count the same thing by counting the number of ways in which person $x$ is in the committee and person $x$ is not in the committee. The number of ways person $x$ is not in the committee is ${n-1\choose k}$. We have $n-1$ people to work with because we are excluding the possibility of person $x$ being in the committee. The number of ways person $x$ is in the committee is ${n-1\choose k-1}$. We have $n-1$ people to work with since person $x$ is in the committee by default and we choose $k-1$ people because person $x$ is in the committee. Thus ${n\choose k}={n-1\choose k}+{n-1\choose k-1}$. |
H: Sheaf of differetials on $\mathbb{P}^n_A$
I am not clear how to display the sheaf of differetials $\Omega_{X/A}$ on $X=\mathbb{P}^n_A$ explicitly,
What is its gobal section $\Omega_{X/A}(X)$ and section on the complement of the hyperplane $T_0=0$, $\Omega_{X/A}(D_+(T_0))$, and its stalk ${\Omega_{X/A}}_x$?
I am reading a proof that the sheaf of differetials $\Omega_{X/A}$ on $X=\mathbb{P}^1_A$ isomorphic to $\mathcal{O}_X(-2)$.(Q.Liu, Algebraic geometry and arithematical curves, P217, Example 1.22)
It is argued that on $D_+(T_0)\cap D_+(T_1)$, we have $T_0^2d(T_1/T_0)=T_1^2d(T_0/T_1)$. This induces a global section of $\mathcal{O}_X(2)\otimes\Omega_{X/A}(X)$. It can be checked isomorphic to $\mathcal{O}_X$. So $\Omega_{X/A}$ is isomorphic to $\mathcal{O}_X(-2)$.
I am not clear how to check $\mathcal{O}_X(2)\otimes\Omega_{X/A}(X) \cong \mathcal{O}_X$
, on the stalk?
And does it hold or not for higher dimension projective spaces?
AI: I'm going to first answer your second question which is establishing the isomorphism
$$\mathcal{O}_X(2)\otimes\Omega_{X}(X) \cong \mathcal{O}_X$$
for $X=\mathbb{P}^1.$
For constructing the cotangent sheaf, we start with the module of relative differentials $\Omega_A$ and then by definition, $\Omega_X$ will be the coherent sheaf associated to $\Omega_A.$ Let's find out what $\Omega_A$ is?
If you review the construction of the module of differentials, you'll see that we start with a free module generated by $dT_0, dT_1$ i.e. $AdT_0 \oplus AdT_1,$ then we mode out by relations, which in our case, there is only one relation, $T_0^2d(T_1/T_0)=T_1^2d(T_0/T_1)$ in the overlap $D_+(T_0) \cap D_+(T_1)$ which after differentiation leads to $T_0dT_1=T_1dT_0.$
Therefore,
$$\Omega_A=AdT_0 \oplus AdT_1/(T_0dT_1-T_1dT_0).$$
It's enough to establish the isomorphism on stalks, so let $p \in D_+(T_0)$ and $\mathfrak{p}$ be the prime ideal of our point. We claim that
$$A_\mathfrak{p}dT_0 \oplus A_\mathfrak{p}dT_1/(T_0dT_1-T_1dT_0) \cong A_\mathfrak{p}$$
Proof: Take an element $fdT_0+gdT_1 \in A_\mathfrak{p}dT_0 \oplus A_\mathfrak{p}dT_1/(T_0dT_1-T_1dT_0)$ and send it to $fT_0+gT_1 \in A_{\mathfrak{p}}.$ Conversely, take an element $h \in A_{\mathfrak{p}}$ and send it to $(h/T_0)dT_0 \in A_\mathfrak{p}dT_0 \oplus A_\mathfrak{p}dT_1/(T_0dT_1-T_1dT_0).$ It's easy to show that it's in fact an isomorphism.
The case for localizing at a point $q \in D_+(T_1)$ is quite similar. I'll leave it to you to fill the gaps. Also, note that we're using the relation $dT_0/T_0=dT_1/T_1$ on the overlap $D_+(T_0) \cap D_+(T_1)$ to have a consistent morphism.
You can easily generalize this isomorphism and show that $\Lambda^n \Omega_{\mathbb{P}^n} \cong \mathcal{O}_{\mathbb{P}^n}(-n-1).$
By now, you should be able to somehow answer your first question. |
H: Proof by Induction (discrete series)
Let P(n) denote the statement that n^p = n + kp
Base case - P(n) is true for n=1
Inductive step - Assume P(n) is true
- Show that P(n+1) is true
Show that (n+1)^p = (n+1) + cp
= RHS of P(n) + (c-k)p + 1
= LHS of P(n) + (c-k)p + 1
= n^p + (c-k)p + 1
I can't close the gap between that last step and the LHS of P(n+1), i.e. (n+1)^p
Any advice would be greatly appreciated!
AI: HINT:
Using Binomial Expansion,
$$(m+1)^p-(m+1)-\{m^p-m\}=\sum_{1\le r\le p-1}\binom prm^{p-r}$$
Now, prove that $p$ divides $\binom pr$ for $1\le r\le p-1$
$\implies (m+1)^p-(m+1)- (m^p-m)$ is divisible by $p$
But by inductive hypothesis, $m^p-m$ is divisible by $p$ |
H: Different real roots polynomial, roots of $P'+aP$
Let $P$ be a polynomial of degree $n$ with real roots $t_1<t_2\ldots<t_n$. Show that $P' + aP$, with $a\in\mathbb{R}$, has only real roots.
Is easy to conclude if $a=0$, by Rolle's theorem. But I can't see what to do if $a\neq0$. Any hint?
AI: Hint: Use the function $f(x) = e^{ax}P(x)$ and see that its derivative is $f'(x) = e^{ax}\{P'(x) + aP(x)\}$.
Since $f(x)$ has $n$ distinct real roots we get $(n - 1)$ distinct roots of $f'(x)$ and so of $P' + aP$ and since this is a degree $n$ polynomial there is only one more root and it has to be real as complex roots appear in pairs. |
H: The Derivative of $\cos(x-2)$
I would think that the solution would be $-\sin(x-2)$, but when i use WolframAlpha it says that the answer is $\sin(2-x)$. Are these $2$ answers equivalent or I am missing some fact here? Thanks in advance.
AI: Since $\sin(x)$ is an odd function, $\sin(-x)=-\sin(x)$.
In particular, $\frac{d}{dx}[\cos(x-2)]=-\sin(x-2)=\sin(2-x)$.
Likewise, $\cos(x)$ is an even function, so $\cos(-x)=\cos(x)$. |
H: How to efficiently compute $17^{23} (\mod 31)$ by hand?
I could use that $17^{2} \equiv 10 (\mod 31)$ and express $17^{23}$ as $17^{16}.17^{4}.17^{3} = (((17^2)^2)^2)^2.(17^2)^2.17^2.17$ and take advantage of the fact that I can more easily work with powers of ten ($17^2 \equiv 10 (\mod 31), (17^2)^2 \equiv 100 (\mod 31) \equiv 7 (\mod 31)$, etc.). While this makes the computation easier, I'm thinking there might be a better way to do this.
AI: $17^2\equiv10\pmod{31}$
$17^4 \equiv 7\pmod{31}$
$17^3 \equiv 170\pmod{31}\equiv15\pmod{31}$
$17^7=17^3\times17^4\equiv15\times7\pmod{31}\equiv12\pmod{31}$
${(17^7)}^3=17^{21}\equiv23\pmod{31}$
$17^{23}=17^{21}\times17^2\equiv230\pmod{31}\equiv13\pmod{31}$ |
H: Prove: Sum and Difference of two distinct positive integers are both perfect squares.
I'm trying to prove that there exists two distinct positive integers whose sum and difference are both perfect squares. I cannot find any pattern or characteristic between the pairs of numbers that work i.e.
4, 5
6, 10
8, 17
10, 6
10, 26
Any help will be greatly appreciated!
Thanks!
AI: Let those positive integers be $a$ and $b$.Then,
$a+b=x^2$
$a-b=y^2$
Adding the two we get $a=\frac{x^2+y^2}{2}$ and $b=\frac{x^2-y^2}{2}$
Since, a and b are integers we must have $x^2+y^2$ and $x^2-y^2$ must be even, for that we must have x and y both even or both odd.
Now for finding such pairs take any even $x,y$ for example let x=8 and y=4
which gives $a=40$ and $b=24$, we have $a+b=64=8^2$ and $a-b=16=4^2$ |
H: $A_5$ has no subgroup of order 15 and 20
Show that $A_5$ has no subgroup of order 15 and 20.
I have been thinking about this problem for so much time but I'm still clueless. Can anyone tell me how to do this problem? Thanks.
I looked up online and saw some proofs with simple groups or Sylow Theorem. Can somebody solve this problem without using simple groups or Sylow Theorem? Thanks.
AI: Hint:
Prove the following:
Claim: If a group $\;G\;$ has a subgroup $\;H\;$ of index $\;n\;$ then it has a normal subgroup contained in $\;H\;$ and of index at most $\;n!\;$
Proof: Hint: make $\;G\;$ act on the left cosets of $\;H\;$, look at the induced homomorphism $\;G\to S_n\;$ and to its kernel...
From the above follows that a simple group cannot have subgroups of index $\;n\;$ if the order of the group doesn't divide $\;n!\;$ . |
H: Symmetric matrices are a subspace of the space of $n\times n$ matrices
Hey, I'm trying to learn how to properly prove this problem. Any advice on how to go about this problem?
AI: It is easy to check that the sum of symmetric matrices is symmetric, and that any multiple of a symmetric matrix is symmetric. Hence it is a subspace.
To see that ${\cal B}$ is a basis, you must show that (1) all symmetric matrices can be created using the linear span of ${\cal B}$, and (2) that the elements of ${\cal B}$ are linearly independent.
Note that a symmetric $A$ is completely defined by its entries $[A]_{ij}$ where $1\le i\le j\le n$. Hence we can write $A= \sum_{1\le i\le j\le n} [A]_{ij} S_{(i,j)}$, so it follows that all symmetric matrices lie in the linear span of ${\cal B}$.
Suppose $\sum_{1\le i\le j\le n} \alpha_{ij} S_{(i,j)} = 0$. If $e_k$ is the vector of zeroes with a one in the $k$th position, we have
$e_a^T(\sum_{1\le i\le j\le n} \alpha_{ij} S_{(i,j)}) e_b = \sum_{1\le i\le j\le n} \alpha_{ij} e_a^T S_{(i,j)} e_b = \alpha_{ab} e_a^T S_{(a,b)} e_b = n^2 \alpha_{ab} = 0$. It follows that $\alpha_{ab} = 0$, and so ${\cal B}$ is linearly independent.
The dimension of a subspace is the cardinality of a basis. Hence $\dim S = |{\cal B}|$. We see from the above that $|{\cal B}| = |\{ (i,j) | 1\le i\le j\le n \} | = n+ \frac{1}{2}(n^2-n) = \frac{1}{2}n (n+1)$. |
H: Continuity implies Borel-measurability?
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function. Is it necessary that $f$ is Borel-measurable?
I'm considering $A=f^{-1}((a,\infty))$ where $a\in\mathbb{R}$. Is $A$ necessarily a Borel set? It looks like it should be, but I'm not sure.
AI: By definition of continuity, and since $(a, \infty)$ is open,
$$f^{-1}((a, \infty))$$ is an open subset of $\mathbb{R}$. Every open set is Borel. |
H: how to prove mean value property for harmonic functions?
For a harmonic function $u(x)$, on domain $\Omega$ where $x \in \Omega \subset \Bbb R^n $, how to show that
$$ u(x) = \frac{1}{\omega_n R^{n-1}}\int_{\partial B_R(x)} u(\sigma) d\sigma$$
where $\omega_n$ is the area of the unit sphere $\partial B_1(x)$.
I am looking for simple case $n=2$ centered at $0$. An argument in my note uses the definition
$\displaystyle g(r) \overset{(def)}{=} \frac{1}{2\pi r }\int_{\partial B_r(0)} u(\sigma) d\sigma$ and shows that $g'(r) = 0$. I don't see how does it prove above? I think it proves $g(r) = \text{constant}$.
AI: Hint: consider $\displaystyle\lim_{r\to 0}g(r)$. |
H: About assumptions in the monotone convergence theorem
Why is the hypothesis that $\left\{f_n \right\}$ be an increasing sequence essential to the monotone convergence theorem? Could someone provide a nice, easy to understand counterexample if I were to assume otherwise? Thank you.
AI: Take for instance $f_n = \frac{1}{n}$, $f_n$ are deacreasing.
Then $\lim _{ n \rightarrow \infty} \int_{-\infty}^{\infty } f_n(x) \rm{d}x = \infty$
But $ \int_{-\infty}^{\infty }\lim _{ n \rightarrow \infty} f_n(x) \rm{d}x =0$ |
H: Easy question about set notation
I have 2 sets and am wondering what B actually is.
A = {a, b} and B = {A}
Does B = {a, b} or B = {{a, b}}
AI: $B$ is a set that consists of another set $A$ therefore $B = \{A\} = \{\{a,b\}\}$ hence why there are two brackets instead of just one. |
H: integration using substitution: symbols $\frac{dy}{dx}$ are used as variables?
I'm learning integration using substitution and the symbols $\frac{dy}{dx}$ are used as variables which is confusing me as I thought they weren't normal variables. So if I'm integrating something and have to substitute $x^2$ so that $u = x^2$ then $\frac{du}{dx} = 2x$ and now I'm at the point that's confusing to me, I multiply both sides of $\frac{du}{dx} = 2x$ with $dx$ so I get $du = 2x dx$ and finish the substitution. So is $dx$ just a normal variable here?
AI: This is one of those frustrating things about the notation used in introductory calculus. Technically $dy$ and $dx$ are not variables, but thanks to some weird property, you can use them just as if they were variables.
See this answer for more detail.
EDIT: I should note that while you can treat $dy$ and $dx$ just as if they are variables, it is extremely important that you do NOT think of them like $dy=d\cdot y$. Doing that would land you in a world of trouble. |
H: How would I differentiate $\sin{x}^{\cos{x}}?$
How can I differentiate $\displaystyle \sin{x}^{\cos{x}}$? I know the power rule will not work in this case, but logarithmic differentiation would. I'm not sure how to start the problem though and I'm not too comfortable with logarithmic differentiation.
AI: Let $y = \sin x ^{\cos x } $ , then
$$ \ln y = \cos x \ln (\sin x ) \implies \frac{y'}{y} = -\sin x \ln(\sin x) + \cos x\frac{\cos x}{\sin x}$$
$$ \therefore y' = \sin x ^{\cos x } \left( \frac{\cos^2 x}{\sin x} - \sin x \ln(\sin x)\right) $$
In general,
Let $f, g$ be any functions. Let $y = f^g \implies \ln y = g \ln f $
$$ \therefore \frac{y'}{y} = g' \ln f + g\frac{f'}{f} \implies \frac{ df^g}{dx}= y' = f^g \left( f' \ln f + \frac{g f'}{f} \right)$$ |
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