Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Counting labelled graphs with k edges and n vertices Is there a way to count labelled graphs (simple graphs - without loops and without multiple edges) with k edges and n using combinatorics methods without having to draw them?
For example - How many labelled graphs are there with 3 edges over the vertices {a, b, c, d, e, f}.
Please just provide me with a way (if there's any) and I would post an answer to the example question
| HINT: An edge is between two vertices. And assuming the graph is simple, we cannot choose the same vertex pair twice. Then first of all, how many unordered vertex pairs are there when we have $6$ vertices? (Unordered means $\{a,b\} = \{b,a\}$ for all $a,b$) Secondly, how many ways are there of choosing three distinct vertex pairs among them? Note that since it's labelled graph, we don't have to consider isomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving Probability Statement via Axioms of Probability and Set Identities I am given the following problem and I am trying to figure out the last step.
Using the axioms of probability and set identities, prove that
if $(B \cap C) \subset A$, then $P(A) \geq P(B) + P(C) - 1$
The axioms:
*
*$P(A) \geq 0 \text{ for any } A$
*$P(S) = 1$
*$P(A \cup B) = P(A) + P(B) \text{ if } A \cap B = \emptyset$
So far I have:
$Z := B\cap C$
$P(A) = P(A\cap Z) + P(A - Z)$
since $Z \cup A \text{, then }A\cap Z = Z$
$P(Z)= P(Z) + P(A-Z)$
$P(A) = P(Z) + P(A \cap Z^c)$
substitue in
$P(A) = P(B \cap C) + P(A \cap (B \cap C)^c)$
$P(A) \geq P(B\cap C)$
by inclusion-exclusion principle
$ P(A) \geq P(B) + P(C) - P(B \cup C)$
Now I am trying to figure out how I get $P(B \cup C)$ to be $1$.
Thanks.
| Hint: You don't need $\mathsf P(B\cup C)=1$, just: $$-\mathsf P(B\cup C)\geq -1$$
So $\mathsf P(A)~{\geq \mathsf P(B)+\mathsf P(C)-\mathsf P(B\cup C)\\\geq \mathsf P(B)+\mathsf P(C)-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why are $-1$ and $1$ generators for the Set of integers under addition? I'm reading my textbook and I'm confused why $-1$ and $1$ are generators for the group of integers under addition.
For example for 1:
we have $1, 1+1=2, 1+1+1=3, 1+1+1+1=4,$ etc.
So shouldn't $1$ be a generator for only the group of positive integers under addition?
For $-1$: we have
$-1, -1+-1=-2, -1+-1+-1=-3$. So shouldn't $-1$ be a generator for only the group of negative integers under addition?
Or is it that both $1$ and $-1$ are generators for the set of integers under addition because in the definition of cyclic subgroup $\{a^n; n\in \Bbb Z \}$, $n$ can take on negative powers? Or am I just confused? Thanks.
| In group theory, the word "generates" has the following meaning: a set $S$ generates a group $G$ if every element of $G$ can be written as a string over $S\cup S^{-1}$ (here, if $S=\{a, b, \ldots\}$ then $S^{-1}=\{a^{-1}, b^{-1}, \ldots\}$). Equivalently, $G$ is the minimal subgroup of $G$ containing all the elements of $S$. In your example, $S=\{1\}$ so $S\cup S^{-1}=\{1, -1\}$ (there is a notation issue here...but I am sure that you can cope).
If $S=S^{-1}$ then we say that $S$ is symmetrised.
In finite groups we do not need to consider the set $S\cup S^{-1}$ as if $a\in S^{-1}$ then there exists $b\in S$ and $n\in\mathbb{N}$ such that $b^n=a$ (indeed, $b=a^{-1}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Three circles have the same radical axis?
Given three circles $\bigcirc O_1$, $\bigcirc O_2$, $\bigcirc O_3$, let $A$, $B$, $C$ be three points on $\bigcirc O_3$. If we have
$$
\frac{\operatorname{power}(A, \bigcirc O_1)}{\operatorname{power}(A, \bigcirc O_2)}=
\frac{\operatorname{power}(B, \bigcirc O_1)}{\operatorname{power}(B,\bigcirc O_2)}=
\frac{\operatorname{power}(C, \bigcirc O_1)}{\operatorname{power}(C, \bigcirc O_2 )}$$ (where $\operatorname{power}(P, \bigcirc Q)$ denotes the power of point $P$ with respect to $\bigcirc Q$), can we conclude that these circles have the same radical axis?
| I'll change your notations to make it more comfortable for me.
Let $\mathscr C_i$ be the circle centered at $O_i$ with radius $r_i$, for $1\le i\le 3$.
Denote by $P(X,\mathscr C)$ the power of point $X$ to circle $\mathscr C$.
Let $\alpha$ such that $\alpha=\frac{P(X,\mathscr C_1)}{P(X,\mathscr C_2)}$ for $X\in\{A,B,C\}$.
Claim: If $A,B,C$ are three distinct points, then for any $X\in\mathbb R^2$, we have
$P(X,\mathscr C_1)=\alpha P(X,\mathscr C_2) + (1-\alpha) P(X,\mathscr C_3)$.
In particular, if $\alpha\notin\{0;1\}$ we easily deduce that the three circles have the same radical line. (When $\alpha\in\{0;1\}$ we only have two distinct circles so the result actually still holds.)
Proof: To exploit the fact that $A,B,C$ are on $\mathscr C_3$, we re-express the power of a point:
\begin{align*}
P(X,\mathscr C_i) ~=~ \|X-O_i\|^2-r_i^2
~=~ \| X-O_3\|^2-2\langle X-O_3,\ O_i-O_3\rangle+P(O_3,\mathscr C_i)
\end{align*}
When $X\in\{A,B,C\}$ we additionally have $\|X-O_3\|^2=r_3^2$ and
$P(X,\mathscr C_1)=\alpha P(X,\mathscr C_2)$. Using the expression above we get
$$(1-\alpha)r_3^2+P(O_3,\mathscr C_1)-\alpha P(O_3,\mathscr C_2) ~=~ 2\big\langle X-O_3,\ (O_1-O_3)-\alpha (O_2-O_3)\big\rangle$$
The left hand side is independant from $X$, this implies that the right-hand side must be constant regardless of what $X\in\{A,B,C\}$ we pick.
Because $\|X-O_3\|=r_3>0$ and assuming that $A,B,C$ are distinct, this implies $$O_1-O_3 ~=~ \alpha (O_2-O_3)$$
and
$$P(O_3,\mathscr C_1) ~=~ \alpha P(O_3,\mathscr C_2) -(1-\alpha) r_3^2$$
We can then re-inject those two identities into $P(Y,\mathscr C_1)$ for an arbitrary $Y\in\mathbb R^2$:
\begin{align*}
P(Y,\mathscr C_1) &= \|Y-O_3\|^2 -2\langle Y-O_3,\ O_1-O_3\rangle +P(O_3,\mathscr C_1) \\
&= \| Y-O_3 \|^2 -2\alpha\langle Y-O_3,\ O_2-O_3\rangle +\alpha P(O_3,\mathscr C_2) -(1-\alpha)r_3^2 \\
&= \alpha P(Y,\mathscr C_2) +(1-\alpha) P(Y,\mathscr C_3)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629331",
"timestamp": "2023-03-29T00:00:00",
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Convergence of fixed-point in a gauss-seidel style I had a problem in the form
$$ \left(
\begin{array}
xx_1 \\
x_2 \\
... \\
x_n
\end{array}
\right)
=
\left(
\begin{array}
FF_1(x_2,...,x_n) \\
F_2(x_1,x_3...,x_n) \\
... \\
F_n(x_2,...,x_{n-1})
\end{array}
\right)$$
and I tried to solve this problem using fixed-point iteration as:
$$ \left(
\begin{array}
xx_1^{k+1} \\
x_2^{k+1} \\
... \\
x_n^{k+1}
\end{array}
\right)
=
\left(
\begin{array}
FF_1(x_2^{k},...,x_n^{k}) \\
F_2(x_1^{k},x_3^{k}...,x_n^{k}) \\
... \\
F_n(x_2^{k},...,x_{n-1}^{k})
\end{array}
\right)$$
The problem is that it didn't converge. So I evaluated the Jacobian and its norm was >1, this explains why the procedure does not converged. I tried to do the fixed-point iterations in a Gauss-Seidel style, re-using pre-calculated values. Fortunately, the algorithm converges.
$$ \left(
\begin{array}
xx_1^{k+1} \\
x_2^{k+1} \\
... \\
x_n^{k+1}
\end{array}
\right)
=
\left(
\begin{array}
FF_1(x_2^{k},...,x_n^{k}) \\
F_2(x_1^{k+1},x_3^{k}...,x_n^{k}) \\
... \\
F_n(x_2^{k+1},...,x_{n-1}^{k+1})
\end{array}
\right)$$
The problem is that I don't know the reason why this method converges, how do I find a condition for the convergence? I tried to evaluate the Jacobian of the system but the norm was >1, so (unless errors in the evaluation) this algorithm should not converge. To evaluate the jacobian I calculated the derivatives of functions $F$ by considering that (k+1) terms are in reality functions. For instance:
$$ J_{2,1} = \frac{\delta F_2(x_1^{k+1},x_3^{k}...,x_n^{k})}{\delta x_1^{k}} =
\frac{\delta F_2(F_1(x_2^{k},...,x_n^{k}),x_3^{k}...,x_n^{k})}{\delta x_1^{k}}
$$
and so on for all the terms. What happens is that the first row of the Jacobian is the same of previous matrix, but the other terms change. Finally the norm is still >1 so the method appear non-convergent. Something should be wrong in my evaluation, the implemented algorithm converges.
| The procedure described in this question is correct, the reason for such non convergent behavior was that the fixed point was far from the position at which the norm does not provide convergence. Indeed, even if chosen as starting point, the method may initially diverge. This cause the method to go far from the initial point but closer to the solution and/or to a point in which the Jacobian is compatible with convergence condition.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all functions $ f:\mathbb{R} \mapsto \mathbb{R} $ such that $f[x^2+f(y)] = y +[f(x)]^2$ I read this question in the book 'Problem Solving Strategies ' by Arthur Engel. This question was asked in IMO of 1992. Here's how it goes
Find all the functions $ f:\mathbb{R} \mapsto \mathbb{R} $ that satisfy
$f[x^2+f(y)] = y +[f(x)]^2$ $ \ \ \ \ x,y \in \mathbb{R}$
The book simply states that it has no solution but it is not a sufficient answer. How can I prove that no such function exists ?
| This partially answers the amended question with a different requirement: $$f(x^2+f(y)) = y+f(x)^2$$
Let $x=0$ to obtain $f(f(y)) = y + f(0)^2$. Let $y=0$ to obtain $f(x^2+f(0)) = f(x)^2$.
Suppose $f$ has a fixed point: $y=f(y)$. Then $f(x^2+y) = y+f(x)^2$, so $f(y) = y+f(0)^2$ by letting $x=0$, and hence $f(0) = 0$. Therefore $f(x^2) = f(x)^2$ for all $x$, and $f(f(y)) = y$ for all $y$; in particular, $f$ is bijective.
But then $f(-x)^2 = f((-x)^2) = f(x^2) = f(x)^2$, so $f(-x) = \pm f(x)$; since $f$ is injective, that means $f(-x) = -f(x)$, so $f$ is odd.
Now letting $y=f(u)$, obtain $f(x^2+u) = f(u)+f(x)^2$, which is to say $f(x^2+u) = f(u) + f(x^2)$; therefore $f(z+u) = f(z) + f(u)$ whenever $z$ is nonnegative. By oddness of $f$, this also holds when both $z$ and $u$ are negative; so we have shown that $f$ is self-inverse, has $f(0) = 0$, has $f(x)^2 = f(x^2)$, and distributes over $+$. (These conditions imply the original requirement; but we are still operating under the assumption that $f$ has a fixed point.)
Since $f$ is odd, that means $f(x) = \sqrt{f(x^2)}$ when $x \geq 0$, and $-\sqrt{f(x^2)}$ when $x < 0$.
Noting then that $f(1)^2 = f(1)$, we have $f(1) = \pm 1$; but $f$ is nonnegative when given nonnegative input since $f(x)^2 = f(x^2)$, so $f$ is the identity on the rationals (by following the reasoning of my answer to the other version of this question). Therefore it is also the identity on the inductively defined set $X = \cup X_n$ where $X_1 = \mathbb{Q}, X_{n+1} = X_n \cup \{\sqrt{x}: x \in X_n\} \cup \{x+q : x \in X_n\} \cup \{x^2 : x \in X_n\}$. But (for example) $\pi$ is not in this set, and I think $f$'s value is not determined on $\pi$.
If $f$ does not have a fixed point, I'm afraid I've got nowhere.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Model given for a formula The formula in question
$$ \forall x \exists y [\ \ P(x,y) \rightarrow \exists z \forall u (\lnot P(u,z))\ \ ] $$
according to my text-book, has the following model
$ \text{Domain} = \mathbb{N} $
$ \lvert P\rvert = \{ (n,m): n,m \ \in \mathbb{N},\ \ n =2m \} $
According to its description, I'd assume the set $ \lvert P\rvert $ to have the following elements:
$$ \lvert P\rvert = \{ \\(0, 0) \\ (2, 1) \\ (4,2) \\ (6,3) \\ ... \} $$
So there has to exist a $z$ , which applied to any possible $u$ is not part of set $\lvert P\rvert$ so that $\exists z \forall u (\lnot P(u,z))$ this condition can hold.
But inside $\lvert P\rvert$ every possible $z \in \mathbb{N}$ is inside the set, meaning there doesn't exist a $z$ in common to every $u$ which could satisfy the model.
Am I missing something?
| Every $z$ belongs to a pair $(2z,z)$. Thus (as you say) there is no $z$ such that $\lnot P(u,z)$ does hold for every $u$.
Thus, up to now: $∃z∀u(¬P(u,z))$ is false, and this is independent of $x$ and $y$.
Consider now $P(x,y)$; obviously, the pair $(x,x+1) \notin |P|$ and thus, we have that for every $x$ there is an $y$ such that $P(x,y)$ is false.
Now, "cook them" together: $\text { false } \to \text { false }$ is $\text { true}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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General solution to simultaneous equations - Resistors in parallel I'm designing a system that uses a grid of resistive sensors, and I'm having trouble figuring out the solution to a set of equations that is the output of this system. I haven't done serious maths for a long time, so go easy on me!
The equations are:
$$\frac{\frac{1}{\frac{1}{R2}+\frac{1}{R3}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}{R1+\frac{1}{\frac{1}{R2}+\frac{1}{R3}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}=Y_1$$
$$\frac{\frac{1}{\frac{1}{R1}+\frac{1}{R3}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}{R2+\frac{1}{\frac{1}{R1}+\frac{1}{R3}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}=Y_2$$
$$\frac{\frac{1}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}{R3+\frac{1}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}=Y_3$$
$$...$$
$$\frac{\frac{1}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}...\frac{1}{R_{x-1}}+\frac{1}{A}}}{R_x+\frac{1}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}...\frac{1}{R_{x-1}}+\frac{1}{A}}}=Y_x$$
With the number of equations equal to the number of RR terms (so for $R1−R4$, there would actually be $4$ equations). $A$ is a constant, and $Y_1,Y_2,Y_3$ etc are all known.
I have no idea really how to begin solving this. I need a general solution because the values will be constantly changing over time. Does anyone have any pointers?
edit: put back A
edit: here is a diagram of the circuit that these equations represent.
$Y1$ is the output voltage with the top switch connected to +V and the others connected to 0V. $Y2$ is the output voltage with the second switch connected to +V and the others connected to 0V. $Y3$ is the output voltage with the third switch connected to +V and the others connected to 0V, etc.
Actually, more exactly, Y is the output voltage divided by +v, to give a ratio. The circuit can be simulated using this link.
| Let $\;\dfrac{1}{R} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\ldots+ \dfrac{1}{R_x}+\dfrac{1}{A}\;$ then the equations can be written as:
$$\require{cancel}
\dfrac{\;\;\dfrac{1}{\dfrac{1}{R}-\dfrac{1}{R_k}}\;\;}{\;\;R_k+\dfrac{1}{\dfrac{1}{R}-\dfrac{1}{R_k}}\;\;} = Y_k \;\;\iff\;\; R_k\left(\dfrac{1}{R}-\cancel{\dfrac{1}{R_k}}\right)+\cancel{1}=\frac{1}{Y_k} \;\;\iff \dfrac{1}{R_k} = \dfrac{Y_k}{R} \tag{1}
$$
Summing up $\,(1)\,$ for $\,k=1,2,\ldots x\,$ gives $\,\dfrac{1}{R}-\dfrac{1}{A}= \dfrac{Y}{R}\,$ where $\displaystyle\,Y = \sum_{k=1}^x Y_k\,$, so $\,R = (1-Y)A\,$ then it follows from $\,(1)\,$ that $\,R_k=\dfrac{(1-Y)A}{Y_k}\,$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Existence of partial derivatives & Cauchy-Riemann does not imply differentiability example I learned about the Cauchy-Riemann equations today, and my instructor used the following example to show that differentiability is not guaranteed if the partial derivatives are not continuous.
Let
$$
f(z)=f(x+iy)=
\begin{cases}
\frac{xy(x+iy)}{x^2+y^2},&\quad\mbox{$z\neq 0$}\\
0+0i,&\quad\mbox{$z=0$}.
\end{cases}
$$
Similar to the example here, writing $f(z)=f(x+iy)=u(x,y)+iv(x,y)$, we can show that $u_x(0,0)=0$ and $v_y(0,0)=0$ since the partial derivatives are $0$ on the axes, but $f$ is not continuous and thus not differentiable at $(0,0)$ since the limits are different if we approach $(0,0)$ along the imaginary axis or along $x=y$.
My instructor mentioned that reason why in the above example, differentiability fails even though the partial derivatives exist and satisfy the Cauchy-Riemann equations, is that the partial derivatives are not continuous at $(0,0)$ (there is a similar comment in this document). I wrote out the partial derivatives
\begin{align*}
u_x &= \left(\frac{x^2y}{x^2+y^2}\right)'=\frac{2xy(x^2+y^2)-x^2y(2x)}{(x^2+y^2)^2}=\frac{2xy^3}{(x^2+y^2)^2}\\
v_y &= \left(\frac{xy^2}{x^2+y^2}\right)'=\frac{2yx^3}{(x^2+y^2)^2}.
\end{align*}
E.g., consider $u_x$ - if I approach $(0,0)$ from the real axis, the limit of $u_x$ is $0$; if I approach $(0,0)$ from the line $x=y$, the limit of $u_x$ would be $1/2$. I thought this is why $u_x$ is not continuous at $(0,0)$. But looking at the formulas alone, it seems that the partial derivatives are simply not defined at $(0,0)$ because the denominators would be $0$ (if so, then this would already imply the partial derivatives are not continuous at $(0,0)$ and we wouldn't even need the argument using unequal limits). But we have just shown before that the partial derivatives exist at $(0,0)$. I must be missing something. According to a post here, it may have something to do with the definition of $f$ at $(0,0)$.
Any help is much appreciated!
| You're right -- it does have to do with the definition of $f$ at $(0,0)$.
By the definition of partial derivative,
$$ \begin{align} u_x(0,0) &= \lim_{h\to0}\dfrac{u(h,0)-u(0,0)}h \\ &= \lim_{h\to0} \dfrac{\frac{h^20}{h^2+0^2}-0}h \\ &= \lim_{h\to0}\dfrac0h \\ &= 0
\end{align} $$
Notice how we had to use the definition $u(x,y) = \frac{x^2y}{x^2+y^2}$ for $(x,y)=(h,0)$ but the definition $u(x,y) = 0$ for $(x,y) = (0,0)$ because of the piecewise definition of $u(x,y)$. Thus your formula for $u_x$, which (implicitly) used the definition $u(x,y) = \frac{x^2y}{x^2+y^2}$ in both places (that's how the quotient rule is derived from the definition of partial derivative), isn't valid for computing $u_x(0,0)$.
However, your formula is valid for computing $u_x(x,y)$ as long as $(x,y) \neq (0,0)$, since away from $(0,0)$ the function $u$ has a consistent definition. Thus you can use your formula to compute the limits approaching $(0,0)$ (which don't depend on the value at $(0,0)$), as you did. So your proof of discontinuity of $u_x$ is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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Then $E(X_1+X_2+X_3+X_4)^4$ equals? Let $X_1, X_2, X_3, X_4$ are i.i.d random variable taking values $1$ and $-1$ with probability $1/2$ each. Then $E(X_1+X_2+X_3+X_4)^4$ equals?
I see that each $X_i$ is standard normal and so $X_1+X_2+X_3+X_4$ is a normal variable with mean $0$ and variance $4.$ I find the answer $48$ using MGF. But the answer is $76$. Where I gone wrong? thanks.
| This seems one of those problems where just enumerating the possibilities is quick (even if dirty):
X1 X2 X3 X4 Sum Sum^4
-1 -1 -1 -1 -4 256
-1 -1 -1 1 -2 16
-1 -1 1 -1 -2 16
-1 -1 1 1 0 0
-1 1 -1 -1 -2 16
-1 1 -1 1 0 0
-1 1 1 -1 0 0
-1 1 1 1 2 16
1 -1 -1 -1 -2 16
1 -1 -1 1 0 0
1 -1 1 -1 0 0
1 -1 1 1 2 16
1 1 -1 -1 0 0
1 1 -1 1 2 16
1 1 1 -1 2 16
1 1 1 1 4 256
Now sum the last column and divide by $16$ as each row is equally probable to get an answer of $40$.
So either you haven't copied the question right or the answer your book has is wrong...
To double check, alternately,
$$(X_1+X_2+X_3+X_4)^4 = \sum X_1^4 + 4\sum X_1^3X_2 + 6\sum X_1^2X_2^2+12\sum X_1X_2^2X_3 + 24X_1X_2X_3X_4$$
(where $\sum$ is used to denote sum of similar symmetric terms - however we dont need to count most of them, as $E(X_i^n) \in \{0, 1\}$ as $n$ is odd or even, and by independence we can multiply the expectations).
$$\implies E[(X_1+X_2+X_3+X_4)^4] = 4 + 0 + 6\times \binom42+0 + 0=40$$
| {
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If an operator preserves divisibility, does that imply that it preserves multiplicability? Specifically, if it were true that
$$\int_a^b \frac{f(x)}{g(x)} \, dx = \frac{\int_a^b f(x) \, dx}{\int_a^b g(x) \, dx}\,,$$ then would that imply
$$ \int_a^b f(x) \cdot g(x) \, dx = \int_a^b f(x) \, dx \cdot \int_a^b g(x) \, dx\,. $$ Or more succinctly, is the statement
$$O\left(\frac{a}{b}\right) = \frac{O\left(a\right)}{O\left(b\right)} \implies O\left(a\cdot b\right) = O\left(a\right) \cdot O\left(b\right) $$ true for some operator $O$? How can this be proved or disproved?
| Are you sure about integration? It looks to me like you are assuming $\int{1/f}$ = $1/\int{f}$, which I don't think it true (for integrals).
As for your more general question, "preserves division" and "preserves inverses" would imply "preserves multiplication", which you should be able to prove easily, but "definite integral" is not such an operator.
| {
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Existence of a normalized vector, far from proper closed subspace. Say we have a normed vector space $(E, \|\cdot\|)$, and a proper closed subspace $F\subsetneq E$, I want to prove the following:
$$(\forall \varepsilon > 0), (\exists e \in E) : \|e\|=1 \text{ and } d(e,F) \geq 1 - \varepsilon $$
Where $d(e,F)= \inf\limits_{y \in F} \|e-y\|$.
I started proving the following fact:
Given $x \in E, \alpha \geq 1$, $\exists f \in F $ such that $\|x-f\|\leq\alpha \cdot d(x,F)$
Which is true by definition of $d$ as an infimum, because: $$ (\forall \varepsilon>0), (\exists f_{\varepsilon} \in F): d(x,F)\leq d(x,f_{\varepsilon})\leq d(x,F)+\varepsilon$$
And taking $\varepsilon = (\alpha - 1)\cdot d(x,F)$ we conclude.
So trying to use the last fact, let $\varepsilon >0,$ we define for $\bar{x} \notin F$ and $ \alpha \in \mathbb{R}_{+}$ to be determined, the $f\in F$ such that $ \|\bar{x}-f\|\leq \alpha\cdot d(\bar{x},F)$, then I don't know how to follow.
Mostly I don't know how to link $\bar{x}-f$ with the $e\in E $ that I'm trying to find, I could normalize, but I can't figure out where to get the inequality I'm looking for, my guess would be through fixing $\alpha = \frac{1}{1-\varepsilon}$ or something like that, but inequalities I manage to get are upper bounds.
Any help would be appreciated.
| I managed to follow trough with David's help.
So we choose $y=\frac{\bar{x}-f}{\|\bar{x}-f\|}$, where f is given as in the property stated in the question body, for $\alpha =\frac{1}{1-\varepsilon}$.
Clearly $\|y\|=1$, so we compute it's distance to $F$.
$$d(y,F) = \inf\limits_{g \in F} \|y-g\| = \inf\limits_{g \in F} \left\|\frac{\bar{x}-f}{\|\bar{x}-f\|}-g\right\| $$
Lets focus on a fixed $g \in F:$
$$= \left\| \frac{\bar{x}}{\|\bar{x}-f\|} -\left( \frac{f}{\|\bar{x}-f\|} + g \right) \right\| =\frac{1}{\|\bar{x} -f\|} \left\| \bar{x}-g' \right\| $$
Where $g'=f+\|\bar{x}-f\|\cdot g \in F$, because its a subspace.
$$\geq \frac{1}{\|\bar{x}-f\|}d(x,F) \geq \frac{1}{\alpha} \geq 1-\varepsilon$$
Where the last two inequalities follow from choice of $f$ and $\alpha$, since g was arbitrary, we conclude:
$$\forall \varepsilon >0 \exists y \in E : \|y\| =1 \text{ and } d(y,F)\geq 1-\varepsilon$$
| {
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Problem about continuous functions and the intermediate value theorem
Let $S^{1} := \lbrace(\cos\alpha, \sin\alpha) \subset \mathbb{R}^{2} | \alpha \in \mathbb{R}\rbrace$ be the circumference of radius $1$ and $f: S^{1} \to \mathbb{R}$ a continuous function. Prove that there exist two points
diametrically opposed at which $f$ assumes the same value.
My idea for the solution: Define $\varphi: S^{1} \to \mathbb{R}$ as$$\varphi(\cos\alpha, \sin\alpha) = f(\cos\alpha, \sin\alpha) - f(-\cos\alpha, -\sin\alpha).$$ If $f(\cos\alpha, \sin\alpha) = f(-\cos\alpha, -\sin\alpha)$ for all $\alpha \in \mathbb{R}$, the result follows. Otherwise, there exist $\alpha_{1},\alpha_{2}$ such that $$f(\cos\alpha_{1}, \sin\alpha_{1}) - f(-\cos\alpha_{1}, -\sin\alpha_{1}) > 0$$ and $$f(\cos\alpha_{2}, \sin\alpha_{2}) - f(-\cos\alpha_{2}, -\sin\alpha_{2}) < 0.$$ Applying the intermediate value theorem, we prove the result.
Is this a correct idea? I appreciate your commentd. Thanks!
| The idea is good, but the
intermediate value theorem
applies to functions mapping a closed interval $I \subset \Bbb R$ to $\Bbb R$.
It would be possible to formulate a similar statement for functions $\phi: S^1 \to \Bbb R$, but it is simpler to consider
$$
\phi: [0, \pi] \to \Bbb R, \quad
\phi(\alpha) = f(\cos\alpha, \sin\alpha) - f(-\cos\alpha, -\sin\alpha)
$$
instead, so that the IVT can be applied directly.
The remaining argument can also be simplified.
It suffices to observe that $\phi(0) = - \phi(\pi)$, so that
*
*either $\phi(0) = \phi(\pi) = 0$,
*or $\phi(0)$ and $ \phi(\pi)$ have opposite sign, and the intermediate value theorem
states that there is some $\alpha \in (0, \pi)$ with $\phi(\alpha) =0$.
| {
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Equation of Normal I'm struggling to get the same answer as the book on this one. I wonder if someone could please steer me in the right direction?
Q. Find the equation of the normal to $y=x^2 + c$ at the point where $x=\sqrt{c}$
At $x = \sqrt{c}$ then $y = 2c$, so the point given is $(\sqrt{c},2c)$
The gradient of the tangent is $2x$ so the gradient of the normal is $-\frac{1}{2}x$ and the equation of the normal will be $-\frac{1}{2}x + v$
Where $x=\sqrt{c}$
then $y = -\frac{1}{2}\sqrt{c} + v$
We know $y = 2c$ therefore $2c = -\frac{1}{2}\sqrt{c} + v$ and so
$v = 2c + \frac{1}{2}\sqrt{c}$
So the equation of the normal is $y = -\frac{1}{2}x +2c +\frac{1}{2}\sqrt{c}$
We can tidy this up to get $2y = -x + 4c + \sqrt{c}$
Unfortunately the book tells me the answer is $2y\sqrt{c} = -x+\sqrt{c}(4c+1)$
It looks like I lost a $\sqrt{c}$ somewhere? Where did I go wrong?
Thank you
Gary
| Your statement about the gradient of the normal should have been as follows:
The gradient of the tangent is $2x$ so the gradient of the normal is $ \frac {−1}{2x}$ which will be $ \frac {−1}{2\sqrt c}$
You will get the book's answer with this new slope.
| {
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Prove $\sum_{n=1}^\infty \text{Ci}(\pi n)=\frac{\ln(2)-\gamma}{2}$ I'm trying to prove that
$$\sum_{n=1}^\infty \text{Ci}(\pi n)=\frac{\ln(2)-\gamma}{2}$$
I've tried parametrizing the sum by replacing $\pi$ with $x$ and differentiating, but this creates to a divergent series (whose partial sum is too messy to integrate), so I had no luck with that.
Any hints? Emphasis on hints - please no full solutions. Just point me in the right direction.
| An overkill. Let $\mathfrak{M}\left(*,s\right) $ the Mellin transform. Using the identity $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\lambda_{k}g\left(\mu_{k}x\right),\, s\right)=\underset{k\geq1}{\sum}\frac{\lambda_{k}}{\mu_{k}^{s}}\mathfrak{M}\left(g\left(x\right),s\right) $$ we have $$\mathfrak{M}\left(\sum_{n\geq1}\mathrm{Ci}\left(nx\right),s\right)=-\zeta\left(s\right)\Gamma\left(s\right)\frac{\cos\left(\pi s/2\right)}{s}$$ for $\mathrm{Re}\left(s\right)>1$, since $$\mathfrak{M}\left(\mathrm{Ci}\left(x\right),s\right)=-\Gamma\left(s\right)\frac{\cos\left(\pi s/2\right)}{s}.$$ So, inverting, we obtain $$\sum_{n\geq1}\mathrm{Ci}\left(nx\right)=-\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}\zeta\left(s\right)\Gamma\left(s\right)\frac{\cos\left(\pi s/2\right)}{s}x^{-s}ds$$ now taking $x=\pi$ and shifting the complex line to the left we have, from the residue theorem, that $$\sum_{n\geq1}\mathrm{Ci}\left(n\pi\right)=\mathrm{Res}_{s=0}\left(-\zeta\left(s\right)\Gamma\left(s\right)\frac{\cos\left(\pi s/2\right)}{s}\pi^{-s}\right)$$ which is...
| {
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Probability that $a^2+b^2+c^2$ divisible by $7$
Three numbers $a,b,c\in\mathbb{N}$ are choosen randomly from the set of natural numbers. The probability that $a^2+b^2+c^2$ is divisible by $7$ is
Try:any natural number when divided by $7$ gives femainder $0,1,2,3,4,5,6$
So it is in the form of $7k,7k+1,7k+2\cdots ,7k+6,$ where $k\in \mathbb{W}$
Could some help me to how to solve it, thanks
| Hint
Any perfect square $x$ Is
$x\equiv 0\pmod 7$
Or
$x\equiv 1\pmod 7$
Or
$x\equiv 2\pmod 7$
Or
$x\equiv 4\pmod 7$
| {
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Commutivity of Endomorphism with Jordan Decomposition I asked this question earlier, but am still a bit stuck on the following question, and would appreciate any help.
Let $\mathbb{F}$ be algebraically closed. If $T=D+N$ is a Jordan decomposition of a linear operator, ($D$ is diagonal and $N$ is nilpotent), prove for $S\in \text{End}(V)$, that $T$ commutes with $S$ if and only if $D$ and $N$ commute with $S$. That is, show $TS=ST\iff NS=SN$ and $DS=SD$.
I got one part already, that is, suppose $N$ and $D$ commute with $S$. Then clearly:
$$TS=(N+D)S=NS+DS=SN+SD=S(N+D)=ST$$
But I am a bit stuck on the other way. I was thinking I could maybe make use of the fact that because $\mathbb{F}$ is algebraically closed, we know every polynomial $f(x)\in \mathbb{F}[x]$ factors as $f(x)=(x-c_1)^{e_1}\cdot \ldots \cdot (x-c_n)^{e_n}$, where $e_i\geq 1$ and $c_i$ are distinct roots. I know given two polynomials $f(x),g(x)$, clearly they commute, that is $f(x)g(x)=g(x)f(x)$. Thus $f(T)$ and $g(T)$ commute. I can thus represent $D$ and $N$ as polynomials in $T$, so $DN=ND$. But how can I show $DS=SD$ and $NS=SN$ given $TS=ST$? Any help would be much appreciated.
| Hint In fact, use - as you say - that $D$ and $N$ are polynomials in $T$. There is a lemma :
Lemma : The commutant of $S$,
$$
C(S):=\{S\mid ST=TS\}
$$
is an algebra (i.e. closed by products and linear combinations).
Hence, if $S$ commutes with $T$, it commutes with all powers of $T$ and polynomials $P(T)$.
| {
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Normal Coordinates on a Riemannian Manifold Consider Riemann Normal coordinates on a manifold. Consider a point other than the origin. Given that the metric has vanishing derivatives at this point, is it correct to deduce that the metric is Euclidean at this point? If the deduction is correct, how to prove/ argue this?
| Clearly: $x_k=x^ig_{ik}$. Consider differentiating on both sides: $\partial_j x_k=(\partial_j x^i)g_{ik}+x^i\partial_j g_{ik}$. Now if the metric derivatives vanish then the last term is zero and we have: $\partial_j x_k=(\partial_j x^i)g_{ik}\Rightarrow\delta_{jk}=\delta_j^ig_{ik}\Rightarrow\delta_{jk}=g_{jk}$.
| {
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Finding the minimum value of a complex number If $z$ is a complex number satisfying $|z^2+1| = 4|z|$ . Then prove that the minimum value of $|z|$ is $4$
This is how I attempted the problem ,
$\frac{|z^2+1|}{|z|} = 4$
Therefore ,
$|z + \frac{1}{z}| = 4$
How do I proceed from here ?
According to the solution of the above problem the next step would be as follows
$||z| - |\frac{1}{z}|| ≤ 4$
We can then form a quadratic in $z$ to find out the least value of $z$. However , I’m stuck at the first step of the solution . How does $|z + \frac{1}{z}| = 4$ imply $||z| - |\frac{1}{z}|| ≤ 4$ ? Please help .
| This follows from a version of the reverse triangle inequality, in your case $$\lvert\lvert z\rvert-\lvert\frac{1}{z}\rvert\rvert\leq\lvert z+\frac{1}{z}\rvert.$$
| {
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What substitution would I make to integrate this? Problem:
The integral is gonna be:
$$\int^{1}_{0}\int^{1}_{0}4xy\sqrt{x^2+y^2} dy \, dx$$
But I'm quite rusty with my calculus, and this is mainly for a statistics course. I know that $x$ and $y$ are bounded within a unit square region in the $xy$-plane, and I was considering making some sort of $\cos$ or $\sin$ substitution to get rid of that pesky square root, but then I'm not sure what limits to use, and I suspect the $dy dx$ would just become $d\theta d\theta$, which doesn't make sense.
Any help or guidance is appreciated. Please try to keep your explanation simple, and don't assume I understand things. It's been a while since my last calculus course :) Thank you!
| An alternative approach:
$$ \begin{eqnarray*}\iint_{(0,1)^2}4xy\sqrt{x^2+y^2}\,dx\,dy&=&\iint_{(0,1)^2}\sqrt{X+Y}\,dX\,dY\\&=&2\iint_{0\leq Y\leq X\leq 1}\sqrt{X+Y}\,dX\,dY\\&=&2\iint_{(0,1)^2}X\sqrt{X}\sqrt{K+1}\,dX\,dK\\&=&2\int_{0}^{1}X\sqrt{X}\,dX\int_{0}^{1}\sqrt{K+1}\,dK\end{eqnarray*}$$
by exploting the substitution $x=\sqrt{X}, y=\sqrt{Y}$, then symmetry, then the substitution $Y=KX$, then Fubini's theorem. The last integrals are elementary and they respectively equal $\frac{2}{5}$ and $\frac{2}{3}(2\sqrt{2}-1)$.
| {
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Find the limit of $\lim\limits_{x\to0^+}\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x}$ as $x$ Find the limit of $\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x}$ as $x$ approaches right of zero.
The answer is $\frac{1}{2}$ but I keep getting 1. Here's what I have:
Since $\lim_{x\to0^+} \frac{\ln (1+x)}{x}$ is of indeterminate form (0/0), we can apply L Hôpitals.
So now I have
$\lim_{x\to0^+}\frac{1}{e} \frac{e- e^ \frac{1}{1+x}}{x}$, which is also of indeterminate form (0/0).
So by L Hôpitals, $\lim_{x\to0^+}\frac{1}{e} \frac{{- e^ \frac{1}{1+x} \frac{-1}{(1+x)^2}}}{x}$ which is equal to 1.
But as I said, the answer is $\frac12$. Can you tell me where did I go wrong? Thanks!
| $$\lim_{x\rightarrow0}\frac{e- e^ \frac{\ln (1+x)}{x}}{ex}=-\lim_{x\rightarrow0}\frac{e^ \frac{\ln (1+x)}{x}\left(\frac{x}{1+x}-\ln(1+x)\right)}{ex^2}=-\lim_{x\rightarrow0}\frac{x-(1+x)\ln(1+x)}{x^2}=$$
$$=-\lim_{x\rightarrow0}\frac{1-\ln(1+x)-1}{2x}=\frac{1}{2}.$$
| {
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$\pi (x) = -1 + \pi(\sqrt x) + \sum \mu (d)\lfloor \frac {x}{d}\rfloor$, Sorry about the title i didn't know what to call this other than "analytic number theory"
I am asked to show that $\pi (x) = -1 + \pi(\sqrt x) + \sum \mu (d)\lfloor \frac {x}{d}\rfloor$, where the sum is over all d for which ever prime factor is less or equal to $\sqrt x $
The furthest right hand part of the equation looks like a mobius inversion but im not sure how to do it over root x and the rest i am completely lost on.
| This should be Legendre’s formula for computing $\pi(x)$:
$$
\pi(x)=-1+\pi(\sqrt{x})+\lfloor x \rfloor-\sum_{p_i\le a}\left\lfloor \dfrac{ x }{(p_i)}\right\rfloor+\sum_{p_i<p_j\le a}\left\lfloor\dfrac{ x}{(p_ip_j)}\right\rfloor-\sum_{p_i<p_j<p_k\le a}\left\lfloor \dfrac{x}{(p_ip_jp_k)}\right\rfloor+\dots
$$
This is proved there.
| {
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6 tests in one month, each must be separated from other tests by 2 free days in between A university is determining the dates for tests in January. There can be a test on every day in January(all $31 $of them), but each two tests have to have at least $2$ free days in between them. (so if there was a test on Monday, the next one can be on Thursday or later)
How many ways are there to arrange tests in this manner?
I'm guessing inclusion-exclusion would be a way to solve this, but I'm not really sure how to put it because everything I've wanted to do seems to overcount things. Maybe it could be seen as a multiset and look for specific permutations but I'm really not sure.
I would really appreciate any hints whatsoever.
| Misunderstood Question
Before, I noticed that the title specified $6$ test, I computed all the possible test days.
Use atoms of $\left(x+x^3\right)$ representing a free day, $x$, or a test and two free days, $x^3$.
We also need to note that we can end in two or more free days, $1$, or one free day, $x^2$, or no free days, $x$, represented by $1+x+x^2$
This can be counted by the coefficient of $x^{31}$ in
$$
(1+x+x^2)\sum_{k=0}^\infty\left(x+x^3\right)^k
=\frac{1+x+x^2}{1-x-x^3}
$$
which is $183916$.
Another way to count is to add up the number of ways to have $k$ tests.
$$
\begin{align}
&\sum_{k=0}^{11}\left[\vphantom{\binom{31}{k}}\right.\overbrace{\binom{31-2k}{k}}^{\substack{\text{ending in $2$ or}\\\text{more days off}}}+\overbrace{\binom{32-2k}{k-1}}^{\substack{\text{ending in $0$}\\\text{days off}}}+\overbrace{\binom{31-2k}{k-1}}^{\substack{\text{ending in $1$}\\\text{days off}}}\left.\vphantom{\binom{31}{k}}\right]\\
&=\sum_{k=0}^{11}\binom{33-2k}{k}
\end{align}
$$
which is also $183916$.
Six Test Month
Once I notice the $6$ test days, all the question really needed was the second approach above, using only the $k=6$ term. That gives, as other answers say
$$
\binom{21}{6}=54264
$$
Another Approach
Similar to the first answer to the misunderstood question above, arrange $6$ "test and two free" days and $31-6\cdot3=13$ "free days". To handle the problem a test cannot happen on the last day or the day before the last day, we add $2$ "free days" to January and ignore them at the end. Thus, we arrange $6$ "test and two free" and $15$ "free days" giving a total number of ways of
$$
\binom{15+6}{6}=\binom{21}{6}=54264
$$
| {
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If some complex numbers lie on the one side of a line which goes through 0, then the sum of them are not 0 Let me describe the problem more specifically.
Suppose that $z_{1},z_{2},\cdots,z_{n}$ are all complex number and they all lie on one side of a straight line passing through $0$. Then $z_{1}+z_{2}+\cdots+z_{n}\neq0$.
I am asked to prove it analytically, even thought it seems quite clear geometrically. So I am thinking that I could use some inequality to prove, but don't have any idea yet..
Thank you very much for any hints or explanations!
| Well, you have to give a characterization of those numbers. Since the lines passes through $0$, let's suppose it forms with the positive $x$ semiaxis and angle $\alpha$, what can you say of the $\arg(z_i)$ for all the points on one side and what for those on the other side?
| {
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Plotting $\{z\in \mathbb{C}\mid |z| > \Re(z)-2\}$ How to plot the set of complex numbers
$$\{z\in \mathbb{C}\mid |z| > \Re(z)-2\}$$
I know that $ |z|$ should be a circle centred at $(0,0)$, but I don't know what would be its radius.
| Your condition is, with $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ :
$$\sqrt{x^2+y^2}>x-2\implies x^2+y^2>x^2-4x+4\implies y^2>-4(x-1)$$
You have a parabola there...
| {
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Main theorem of Dedekind
Main theorem of Dedekind: It claims, that for every cut $A|A'$ in the set of real numbers there exists a number $b$ that forms the cut. This number $b$ will be $1)$ the largest in the lower class or $2)$ the smallest in the higher class.
I can't get the proof. The proof in the book was given by the contradiction. It says: note by $\mathbf A$ the set of all rational number, that belong to $A$ and by $\mathbf A'$ the set of all rationals, that belong to $A'$. Therefore, $\mathbf A$ and $\mathbf A'$ form a set of all rational numbers. The cut distinguish some real number $b$. Assume, that $b$ is in the lower class $A$. We will prove that it is the maximum in $A$.
contradiction:
If it is not, we can take some number $a_0$ from $A$, that is bigger than $b$. We can also take some rational number $r$ such as $b<r<a_0$. Which means, that $r$ belongs to the $\mathbf A$ class. It is a contradiction, since the rational number $r$, which is in the lower class of the cut, is bigger than the number $b$, which is distinguhed by the lower cut.
I don't see why it can't be so. Does it want to say that the contradiction is in the fact, that if $a_0$ is bigger than $b$, it can't be in the lower class, because the cut was formed by $b$?
| This is the original proof by Dedekind (also repeated by Hardy in his A Course of Pure Mathematics) and to understand it properly you need to be very clear of the fact there are are two types of Dedekind cuts being used here. One is the cut $(A, A') $ which involves partitioning the set of reals into two sets $A, A'$. Another is a cut $(\mathbf{A}, \mathbf{A}')$ which involves partition of rationals into two sets. By convention let's assume that bold symbols deal with rationals and usual symbols deal with reals.
Once this distinction is made clear one has to show that the sets $\mathbf{A}, \mathbf{A} '$ do form a Dedekind cut and hence define a real number $b$ and this must be in exactly one of the sets $A, A' $.
Coming back to contradiction, we assume that $b\in A$ and we wish to prove that $b$ is the greatest member of $A$. If not then there is some real number $b'>b$ such that $b'\in A$. Dedekind then uses the fact that between any two distinct real numbers lies a rational number (this needs to be proved and perhaps is given earlier in your textbook). Thus there is a rational $\mathbf{r} $ such that $b<\mathbf{r} <b'$. Now this $\mathbf{r} $ must lie in either $\mathbf{A} $ or $\mathbf{A} '$. Since $b$ is the real number defined by cut $(\mathbf{A}, \mathbf{A}') $ we must have $\mathbf{r} \in \mathbf{A} '$. And since $b' \in A$ and $r<b'$ we must have $r\in A$ and hence $\mathbf{r} \in \mathbf{A} $. Note the usage of normal and bold $A$ and try to understand exactly why these statements are true. Also note the dual role played by $r$ as a rational number and as a real number (this is indicated using normal and bold symbols). And we have reached the contradiction because $\mathbf{r}$ can belong to only one of the sets $\mathbf{A}, \mathbf{A} '$.
You should also mention the conventions followed by Dedekind. Since I am familiar with his original pamphlet Stetigkeit und irrationale zahlen I could easily understand your post, but it is better to add more context for the benefit of everyone. For Dedekind a section/cut/schnitt is a procedure to partition an ordered set $F$ into two subsets $A, A'$ such that $A\neq\emptyset\neq A', A\cup A'=F, A\cap A'=\emptyset$ and every member of $A$ is less than every member of $A'$. $A$ is called the lower set and $A'$ is called the upper set.
When this procedure is carried out with $F=\mathbb{Q} $ (the set of rationals) then there are three possibilities :
*
*lower set has a greatest member
*or upper set has a least member
*neither the lower set has a greatest member nor the upper set has a least member.
The great achievement of Dedekind is to show that when the set $F=\mathbb{R} $ (the set of reals) is partitioned in the above manner only the first two possibilities listed above occur. The third possibility does not arise. This is the theorem being discussed here.
| {
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Assumption made in Proof of System of ODEs with Repeated Roots In a system of ODEs with a repeated root, what brings one to assume that the form of the equation is
\begin{align}
y(t)=v(t)e^{rt}
\end{align}
where $r$ is the repeated root. I get that multiplying $ce^{rt}$ by $t$ yields a second, linearly independent solution, but how would you arrive at the above supposition intuitively?
| In ODE and PDE research, sometimes a "guess and check" method is employed. Multiplying by $ce^{rt}$ works nicely.
Consider the ODE $y'=y$ with initial condition $y(0)=k$. Without knowing the solution is the exponential, we want to find a function (more generally, a family of functions) that is equal to its own derivative. There is only one such function that is equal to its own derivative everywhere (not just almost everywhere).
This is perhaps remarkable, because it is essentially saying out of the uncountably many functions in $C^1$, only one of them is equal to its own derivative. The proof of this fact however requires some justification, and I remember correctly is not an entirely trivial proof.
For a system of ODEs, the solution $y(t)=v(t)e^{rt}$ works. We can check to see that is unique (under some initial conditions). However, a priori, without knowing the solution to the system takes this form, we would have to use a lot of machinery to rule out other possible functions as solutions.
| {
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limit without expansion I was solving this limit instead of using sum of expansion i did this and got zero but answer is 1/5 by expansion why this is wrong
$$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots + n^4}{n^5}$$
$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n^5}$$
$$= \lim_{n\rightarrow \infty } \frac{( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n}$$
$$= \frac{(0 + 0 +0 \ldots + 1 )}{n} = 0$$ this is how i got ,
| The problem is that while each term tends to zero, the number of terms tends to infinity at the same time.
The rule “the limit of a sum is the sum of the limits” only applies if you have a finite sum with a fixed number of terms (each of which has a finite limit).
(Also, you can't really write
$$\lim_{n \to \infty} (\dots) = (\text{some expression involving $n$})$$
since there can be no $n$ left in your expression after you have let $n \to \infty$. The symbol $n$ stands for the same number everywhere, so you can't selectively let some $n$ tend to $\infty$ while leaving other $n$ unaffected.)
| {
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Find at least one solution or prove that it does not exist Please help me to find solution or prove that it does not exist.
$$
C^{2^k}_{2^n} < 2^{2^k (n - k)}, 1<k<n, k \in \mathbb{N}, n \in \mathbb{N}
$$
I tried to find solution numerically, but when $n$ > 20, the numbers grows very fast, so it seems like it has not solution (because I didn't find it when $n$ < 20). I am looking for analytical methods to prove this.
| I've found an answer.
Firstly we can introduce $N = 2^n$, $K = 2^k$. So the inequality will be rewritten as
$C_N^K < 2^{K(\log_2(N)-\log_2(K))}$
or
$\log_2(C_N^K) < K \log_2(N/K)$.
Secondly, according to Best upper and lower bound for a binomial coeficient we can use inequality
$C_N^K \ge (N / K)^K$.
It shows that there are no solutions.
| {
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Which statement is false ?(Linear algebra problem) Let $P=\dfrac{xx^{T}}{x^{T}x}$ be an a square matrix of order n where $x$ is a non zero column vector. Then which one of the following statement is False.
$(A)$ P is idempotent
$(B)$ P is orthogonal
$(C)$ P is symmetric
$(D)$ Rank of P is one
In this question i only know that rank of $xx^{T}$ is $1$. And some short notes i have in my mind are :
Eigen values of idempotent matrix are $0$ and $1$.
Eigen values of orthogonal matrix are $-1$ and $1$.
Eigen values of symmetric matrix are Real. But that is not sufficient for this problem i guess. I don't know how to deal rational functions in linear algebra. Please give me some knowledge i just started learning algebra.
| A matrix of rank $1 $ cannot be orthogonal. Orthogonal matrices have maximal rank.
| {
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A sufficient condition for $x^p-x-a$ to be primitive Let $p$ ba a prime number and $F_p$ be the finite field with $p$ elements. Characterize the set of $a\in F_p$ such that $f=x^p-x-a$ is a primitive polynomial i.e. $x$ generates the multiplicative group of $F_p[x]/(f)$.
| Further to my hint above, that
The polynomial $x^p − x − a \in \mathbb{F}_p[x]$ is primitive if and only if $a$ is primitive in $\mathbb{F}_p$.
I found the result of Cao to be of use, I think, from $2010$
On the Order of the Polynomial $x^p − x − a$
| {
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Is this $\binom{n}{p}$ for $p>n$ make a sense in mathematics or it is $0$ by convention? It is well known that gamma function is not defined at negative integers , but my question is to know how i take the value of $\binom{n}{p}$ for $p>n$ then is this make a sense or it is $0$ by convention ?
|
A common definition of the binomial coefficient with $\alpha\in\mathbb{C}$ and integer values $p$ is
\begin{align*}
\binom{\alpha}{p}=
\begin{cases}
\frac{\alpha(\alpha-1)\cdots(\alpha-p+1)}{p!}&p\geq 0\\
0&p<0
\end{cases}
\end{align*}
From this we conclude $\binom{n}{p}=0$ if $p>n \ \ (n,p\in\mathbb{N})$.
Hint: The chapter 5 Binomial coefficients by R.L. Graham, D.E. Knuth and O. Patashnik provides a thorough introduction. The formula above is stated as (5.1).
| {
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Calculate domain $f(x)=x^{\frac{x+1}{x+2}}$ I have the following function:
$$f(x)=x^{\frac{x+1}{x+2}}$$
I tried to calculate the domain, which seems easy, and my result is: $D(f)=(0,\infty)$.
When I tried to calculate it, by using Wolfram-Alpha, I obtain: $D(f)=[0,\infty)$.
Can someone explain me the reason, or if it is just a Wolfram's error?
I proceed in this way:
$$f(x)=x^{\frac{x+1}{x+2}} = e^{\frac{x+1}{x+2} \log(x)}$$
$$
\left\{
\begin{array}{c}
x+2\ne0 \ \Rightarrow\ x\ne -2 \\
x>0
\end{array}
\right.
$$
Hence: $D(f)=(0,\infty)$.
| I think Wolfram is wrong and you are right.
If we write $(f(x))^{g(x)}$ then the domain it's
$$D(g)\cap\{x|f(x)>0\},$$
where $d(g)$ it's the domain of $g$.
I think it's better to define such that even $0^{\frac{1}{2}}$ does not exist, but $\sqrt0=0.$
| {
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How many ways are there to have a collection of $10$ fruits +I am just getting started in my combinatorics class and I came across the following problem in my textbook that I am looking for some help with, thanks!
How many ways are there to have a collection of $10$ fruits from a large pile of identical oranges, apples, peaches, bananas and pears if the collection should include exactly two kinds of fruits?
Since the order is not important here we can systematically list the different combinations to see the different combinations.
How can I use this equation which I believe is the equation I need
$$C(n,r) = \frac{n!}{(n-r)!r!}$$
| Since your collection needs to include exactly $2$ kinds of fruits, this means that you have $C(5,2)$ possible choice for the fruits that will be part of your collection. Recall that $C(5,2)$ means that you pick two kind of fruits out of $5$ possible choices. But this doesn't take into account all possible collections since you don't know how many of each fruits you have. But you know that if you have $r$ fruits of kind $A$, you have $10-r$ fruits of kind $B$. Since you want two kind of fruits, it means that $r$ ranges from $1$ to $9$. Therefore, you have $C(5,2) \times 9 = 10 \times 9 = 90$ possibilities.
| {
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What is the domain of $f(x)=x^{2/4}$ What is the domain of $f(x)=x^{2/4}$
Is $f(x)= (\sqrt[4]x)^2 $ with $\operatorname{dom} (f) = [0,\infty) $
or $\sqrt[4]{x^2}$ with $\operatorname{dom} (f) = (-\infty,\infty)$
or is $f(x)=\sqrt x$ with $\operatorname{dom} (f)=[0,\infty)$
I have tried searching the internet but couldn't find anything.
| since $$x^{2/4}=x^{1/2}$$ we have $$x\geq 0$$
| {
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How many ways can $4$ couples be seated around a table if ...? I'd like to know if my solution is plausible.
4 couples are to be seated around a circular table. How many ways can they be seated if each person is sitting directly across from their spouse (i.e. there are three people between them and their spouse on either side)?
I went like this. Having to a round table gives me 7!2! to sit the first couple.
For the second couple I have 6!2! to sit them (since I have 6 spot for the first person his spouse and him can just sit across which make for 2!)
I managed to continue with that way of thinking to get in final.
7!2!6!2!4!2!2!
is this answer plausible ?
thank you and have a nice day
| Here's a different approach. There are four surnames A B C and D. If you draw a dividing line across the middle of the table, then on one side you need to arrange the letters A B C D which can be done in $4!$ ways.
For each of these surnames, there is a husband or a wife who occupies this position, with their spouse automatically sat in the place opposite. We can choose whether each surname is represented by husband or wife in $2^4$ ways.
Finally, for each arrangement, we can ask everyone to move one place to the left, but we still have the same arrangement. Likewise we can rotate each arrangement into eight new positions but each arrangement is still the same as the original. We therefore divide by eight.
Putting this all together we have a total of $$\frac{24\times2^4}{8}=48$$ arrangements.
| {
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What is $\cfrac 1n$ in this expression?
I bought a T-shirt at Kenendy Space Center a while back that had this expression on it:
$$
B > \frac1n\sum_{i=1}^n x_i
$$
And below it is captioned "Be greater than average."
I can see the "Be greater than" and the "average" in the expression, but I do not understand what the $\cfrac 1n$ is for.
So, what is the purpose of $\cfrac 1n$ in the expression above?
(Sorry, I have absolutely no idea how to tag this)
| The term $\dfrac 1n$ is part of the average. If we want to find the average of two numbers $a$ and $b$, the average would be $\dfrac{a + b}{2} = \dfrac 12(a + b)$. The denominator is equal to the amount of variables we have. In this case, we have two variables $a$ and $b$, so we add them together and divide it by two. This means that the average of $a, b$ and $c$ is equal to $\dfrac{a + b + c}{3} = \dfrac 13(a + b + c)$. The average is also known as the arithmetic mean, and the right hand side of the inequality is just notation for the average of $n$ given variables denoted as $x_i$.
$$\sum_{i=1}^n x_i$$ is read as the sum that goes from $x_1$ to $x_n$. In other words, it is the sum that takes the values of $x_i$ from $i = 1$ to $i = n$. We can now write this as $$\sum_{i=1}^n x_i = x_1 + x_2 +\cdots + x_n$$ and now to find the average of these $n$ different variables, we divide by the total of how many there are, which is $n$. This is the same as multiplying the entire sum by $\dfrac 1n$ which makes the right hand side of the inequality define the average of $n$ variables. Now it is clear that the symbols $$B > \frac 1n \sum_{i=1}^n x_i$$ means be greater than the average.
| {
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Let $X$ and $Y$ be i.i.d. Expo($\lambda$). Find the conditional distribution of $X$ given $X < Y$. (a) by using calculus to find the conditional PDF.
My Solution:
$$P(X \leq t \mid X < Y) = \dfrac{P(X \leq t, X < Y)}{P(X < Y)}\\
= \frac{P(X \leq t, X < Y \mid Y \leq t)P(Y \leq t) + P(X\leq t, X < Y\mid Y > t)P(Y>t)}{P(X < Y)}\\
= \frac{P(X < Y)P(Y\leq t) + P(X\leq t)P(Y > t)}{P(X < Y)} = 1 - e^{-\lambda t} + 2(1 - e^{-\lambda t})e^{-\lambda t}\\
=1 - e^{-\lambda t}-2e^{-2\lambda t}$$
I was going to take the derivative with respect to $t$ to find the conditional PDF, but realized my CDF is incorrect since the text claims
$$
P(X \leq t\mid X < Y) = P(\min(X, Y) \leq t)
$$
Which means
$$
P(X \leq t\mid X < Y) = P(\min(X, Y) \leq t) = 1- e^{2\lambda t}
$$
Any help would be greatly appreciated.
| Rather, try this:
$$\begin{align}\mathsf P(X\leqslant t\mid X\lt Y) &= \dfrac{\mathsf P(X\lt Y, X\leqslant t)}{\mathsf P(X\lt Y)}\\[1ex] &=\dfrac{\mathsf P(X\lt Y, Y\leqslant t, X\leqslant t)+\mathsf P(X\lt Y, X\leqslant t, Y>t)}{\mathsf P(X\lt Y)}\\[1ex] &=\dfrac{\mathsf P(X\lt Y\mid X\leqslant t, Y\leqslant t)\,\mathsf P(X\leqslant t,Y\leqslant t)+\mathsf P(X\leqslant t, Y>t)}{\mathsf P(X\lt Y)}\\[1ex] &=\mathsf P(X\leqslant t)\,\mathsf P(Y\leqslant t)+2\,\mathsf P(X\leqslant t)\,\mathsf P(Y>t)\\ & ~~\vdots\end{align}$$
| {
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Prove by induction that $(k + 2)^{k + 1} \leq (k+1)^{k +2}$
Prove by induction that $$ (k + 2)^{k + 1} \leq (k+1)^{k +2}$$ for $ k > 3 .$
I have been trying to solve this, but I am not getting the sufficient insight.
For example, $(k + 2)^{k + 1} = (k +2)^k (k +2) , (k+1)^{k +2}= (k+1)^k(k +1)^2.$
$(k +2) < (k +1)^2 $ but $(k+1)^k < (k +2)^k$ so what I want would clearly not be immediate from using something like If $ 0 < a < b, 0<c<d $ then $0 < ac < bd $. THe formula is valid for n = 4, So if it is valid for $n = k$ I would have to use
$ (k + 2)^{k + 1} \leq (k+1)^{k +2} $ somewhere in order to get that $ (k + 3)^{k + 2} \leq (k+2)^{k +3} $ is also valid. This seems tricky.
I also tried expanding $(k +2)^k $ using the binomial formula and multiplying this by $(k + 2)$, and I expanded $(k+1)^k$ and multiplied it by $(k + 1)^2 $ term by term. I tried to compare these sums, but it also gets tricky. I would appreaciate a hint for this problem, thanks.
| Try taking log of both sides and prove $\frac{\log x}x$ is decreasing.
Or by induction try to show $(\frac{k+1}k)^k\leq k$:
$$(1+1/k)^k\leq \sum_{i=0}^k \binom ki k^{-i}<\sum_{i=0}^k 1=k+1$$
| {
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How to prove sin(nx) has no pointwise convergent subsequence without prior knowledge of Lebesgue's Theory? In Baby Rudin 7.20 example, the author mentions to prove that the function sequence
$$f_n(x):=\sin(nx) \qquad0(\leq x\leq 2\pi)$$has no pointwise convergent subsequence would be troublesome without Lebesgue's Theorem.
Is there a proof that doesn't refer to Lebesgue's Theorem, and only requires the knowledge introduced in the first 7 chapter in Rudin?
| Here's an argument my officemate and I came up with that should work (while avoiding Lebesgue theory). Given a subsequence $\sin(n_kx)$, it constructs a sub-subsequence $\sin(n_{k_m}x)$ and a point $y$ such that $\sin(n_{k_m}y)$ fails to converge.
Suppose a subsequence $\sin(n_kx)$ is given. Then, we can find some closed interval $I_1 \subseteq [0, 2\pi]$ such that $\sin(n_1x)$ maps $I_1$ into $[1/2, 1]$. Also, set $n_{k_1} = n_1$.
Having chosen $I_1$, observe that for $k > k_1$ large enough, $\sin(n_k x)$ maps $I_1$ to $[-1,1]$ surjectively. Choose $k_2$ to be the least $k>k_1$ with this property, and let $I_2 \subset I_1$ be an interval that $\sin(n_{k_2}x)$ maps to $[-1, -1/2]$.
In general, suppose we have constructed the index $n_{k_m}$ and interval $I_m$. Then, we let $k_{m+1}$ be the least integer greater that $k_m$ such that $\sin(n_{k_{m+1}}x)$ maps $I_m$ surjectively onto $[-1, 1]$. Then, if $m$ is odd, we choose a subinterval $I_{m+1} \subset I_m$ such that $\sin(n_{k_{m+1}}x)$ maps $I_{m+1}$ into $[-1, -1/2]$, while if $m$ is even, we choose an $I_{m+1}$ which $\sin(n_{k_{m+1}}x$ maps into $[1/2,1]$.
Now, by the nested intervals theorem, $\bigcap_{m=1}^\infty I_m$ is nonempty, so it contains some point $y$. But, by construction, $\sin(n_k y)$ is both at least $1/2$ and at most $-1/2$ infinitely often, so the sequence of functions cannot converge at $y$.
| {
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Does there exist an integrable function f such that norm of f*g is equals that of g for all g? Does there exist $f \in L^{1} (\mathbb R)$ such that $||f*g||_1 =||g||_1$ for all $g \in L^{1} (\mathbb R)$? I read somewhere (long ago) that no such function exists. It is easy to see that $L^{1} (\mathbb R)$ has no unit under convolution, but this question is much harder and I still have no idea how one proves it. Thanks in advance for any hints or solution.
| No. If $\tau_x f(t)=f(t-x)$ we know that $$\lim_{x\to0}||f-\tau_xf||_1=0.$$It follows easily that if $$g_n=n(\chi_{(0,1/n)}-\chi_{(1/n.2/n)})$$then $$\lim_{n\to\infty}||f*g_n||_1=0.$$
| {
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To test whether $\sum_{n=1}^\infty\frac{n+2}{2^n+3}\sin\left[(n+\frac12)\pi\right]$ converges To determine whether the following sequence converges or divergence
$$\sum_{n=1}^\infty\frac{n+2}{2^n+3}\sin\left[(n+\frac12)\pi\right]$$
I don't know which test to use here, but my guess is it may be a comparison test but how to determine which series to use?
| Hint: I assume you mean $$\sin\left[(n+1/2)\pi\right]=(-1)^{n}$$
By using this you can see that this is an alternating series. Use the Leibniz criterion to rule out convergence.
| {
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How to prove this definite integral does not depend on the parameter? I am working on some development formulas for surfaces and as a byproduct of abstract theory i get that:
$$\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{1+\sin^2\theta}{(\cos^4\theta+(\gamma\cos^2\theta-\sin\theta)^2)^\frac{3}{4}}d\theta$$
is independent on the parameter $\gamma\in\mathbb{R}$. I thought that there was something wrong with my calculations but actually turns out that using Mathematica that the value is somewhat near $5,24412$ independently on the $\gamma$ I plug in the calculation of the integral. Is there any way to verify that actually this is a constant by direct computations, complex analysis, or at least is this kind of integrals studied?
Edit:obviously differentiating in the integral does not help much
| Put
\begin{equation*}
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{1+\sin^2\theta}{(\cos^4\theta +(\gamma\cos^2\theta-\sin \theta)^2)^{\frac{3}{4}}}\, d\theta
\end{equation*}
If $x = \dfrac{\sin\theta}{\cos^2\theta}$, $\, y = \gamma-x$ and $y = \sqrt{z}$ then
\begin{equation*}
dx = \dfrac{\cos^2\theta+2\sin^2\theta}{\cos^3\theta}\, d\theta
\end{equation*}
and
\begin{gather*}
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos^2\theta+2\sin^2\theta}{\cos^3\theta\left(1 +\left(\gamma-\frac{\sin \theta}{\cos^2\theta}\right)^2\right)^{\frac{3}{4}}}\, d\theta = \int_{-\infty}^{\infty}\dfrac{1}{\left(1 +\left(\gamma- x\right)^2\right)^{\frac{3}{4}}}\, dx = \int_{-\infty}^{\infty}\dfrac{1}{\left(1 +y^2\right)^{\frac{3}{4}}}\, dy = \\[2ex]
\int_{0}^{\infty}\dfrac{z^{\frac{1}{2}-1}}{(1+z)^{\frac{1}{2}+\frac{1}{4}}}\, dz = {\rm B}\left(\frac{1}{4},\frac{1}{2}\right) \approx 5.244115109
\end{gather*}
where ${\rm B}$ is the Beta function.
| {
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Chain rule problem: given $f(x)=\sqrt{4x+7}$ and $g(x)=e^{x+4}$, compute $f(g(x))'$. Question:
Given the functions $f(x)=\sqrt{4x+7}$ and $g(x)=e^{x+4}$, compute $f(g(x))'$.
My Approach:
I have found that found that $f(g(x))=\sqrt{4e^{x+4}+7}$. Should I now just differentiate it to get my answer or is there any simpler method to solve this problem. Any helpful suggestions or answers.
| $f(g(x))'=f'(g(x))\cdot g'(x)=\frac{4}{2\sqrt{4g(x)+7}}\cdot e^{x+4}=\frac{2e^{x+4}}{\sqrt{4e^{x+4}+7}}$
| {
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Let $f: [-1, 1] \longrightarrow [-1, 1]$ such that $f\in C^{1}$. Prove that there's exist $x_{0} \in [-1, 1]$ such that $|f'(x_{0})| \leq 1$
Let $f: [-1, 1] \longrightarrow [-1, 1]$ such that $f$ is a class $C^{1}$ function. Prove that there's exist $x_{0} \in [-1, 1]$ such that $|f'(x_{0})| \leq 1$.
I know that $f'([- 1,1])$ is compact, since $f'$ is continuous. Therefore, it is closed and limited. To prove the result, I tried to use the continuity of $f'$ in some sequence and tried to use the Weierstrass theorem, but I could not conclude anything. I would like some suggestion.
| Assume your claim is false. Your claim combined with the continuity of $f'$ implies that either $f'(x)>1$ is always true or that $f'(x)<-1$ is always true. Also, since $f'$ is continuous, we can take its integral: $$f(1)-f(-1) = \int_{-1}^1 f'(x)\,dx$$
If $f'(x)>1$ then $f(1)-f(-1)>2$. Similarly if $f'(x)<1$ then $f(1)-f(-1)<-2$. Using the triangle inequality, in both cases we have a contradiction.
| {
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Elementary question on pointwise convergence and norm continuity
Let $\;a_n \in \mathbb R\;$ be a bounded sequence, then by
Bolzano-Weierstrass Theorem it follows there exists a subsequence
$\;a_{n_k}\subset a_n\;$ such that $\;a_{n_k} \to a \in \mathbb R\;$.
If $\;f:\mathbb R \to \mathbb R^m\;$ a continuous function, then
$\;f(x-a_{n_k}) \to f(x-a)\;$ pointwise.
In addition, $\;\forall g \in C(\mathbb R;\mathbb R^m)\;$ it holds:
$\;{\vert g(x)-f(x-a_{n_k}) \vert}^2 \to {\vert g(x)-f(x-a) \vert}^2\;$ also
pointwise.
NOTE: $\; \vert \cdot \vert \;$ stands above for the Euclidean norm
As I was studying, the above part of my notes confused me a little bit.
QUESTIONS:
*
*Why is $\;f(x-a_{n_k}) \to f(x-a)\;$ pointwise? I know that uniform convergence is much stronger than pointwise but why is that the case here?
*Why does the second convergence hold? Is it because $\;\vert \cdot \vert\;$ is also continuous?
Any help would be valuable. Thanks in advance!
| Consider $y_{n_k}=x-a_{n_k}$ and $y =x-a $ for any fixed $x\in \mathbb{R}$, then $y_{n_k} \rightarrow y $ in $\mathbb{R}$. Now as both $f$ and $g$ is continuous in $\mathbb{R},$ you can have both the convergence. Also you have to use the fact $||$ is continuous and composition of two continuous maps is continuous to get the last result.
| {
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Why does a function $f(x)$ have to be bijective in order to have a $f^{-1}(x)$ (in Euclidean plane)? Yes, the title explains itself. I have no take on this one, or insights for that matter. For example, why isn't it enough for a certain function to be injective to have its $f^{-1}(x)$?
| It is the matter of the nuances of definitions.
If a function from a domain $A$ to a codomain $B$ is merely a subset of $A\times B$, as in the Wikipedia definition, then the function does not "encode" $B$. You end up with the same set of pairs regardless of whether you view $f$ as a function of $A$ to $B$, or of $A$ to a superset $B'\supset B$, or of $A$ to $f(A)=\{f(a)\mid a\in A\}$. In that sense, you are right, you can invert $f$ as a relation, and the result is a function if and only if $f$ was injective.
On the other hand, it is often argued (and this is how I was taught functions in my youth) that it is necessary for the function to "encode"/specify both its domain and its codomain too. For example (I think this probably comes from Bourbaki), a function is defined as a triplet $(f,A,B)$, where $f$ is a function in the previous sense, and $A, B$ are the domain and the codomain, respectively. Then, the inverse $f^{-1}$ exists, as a function $B\to A$ if and only if both it exists in the previous sense, and $f(A)=B$, i.e. $f$ is bijective.
As it turns out, most of the time, even intuitively, we regard the functions as defined in the second definition above (certainly not the first definition), or something similar that encodes the codomain. We implicitly assume that the function "knows" its codomain somehow, i.e. the function is not just a set of ordered pairs, but it also "knows" which (potentially larger) set the images belong to. Thus the requirement for $f^{-1}$ to exist is $f$ to be bijective, not merely injective.
| {
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Proving this limit as $x \rightarrow ∞$? I'm having trouble proving this using epsilon-delta:
$$\lim_{x\to\infty} \left|\frac{x}{x+1}\right|=1$$
I translated this into:
$$\forall \epsilon>0,∃\delta\in\mathbb R,x>\delta\implies \left|\frac{x}{x+1}-1\right|<\epsilon$$
I don't really know where to go from here. Any help would be appreciated.
| Try observing that
$$
\left|\frac{x}{x+1}-1\right|=\left|\frac{x-x-1}{x+1}\right|
=\left|\frac{1}{x+1}\right|
$$
| {
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Limit of $[\ln (3+x^2) - \ln (2+x)]$ as $x$ approaches $\infty$ I got an answer as $\infty$ but I need more clear explanation.
Please help me!
| Hint:
You can use equivalents, and the properties of log:
$$\ln(3+x^2)-\ln(1+x)=\ln \frac{x^2+3}{x+1}.$$
On the other hand,
$$\frac{x^2+3}{x+1}\sim_\infty \frac{x^2}x=\ln x\enspace\text{so}\quad\ln(3+x^2)-\ln(1+x)\sim_\infty \ln x.$$
| {
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When does the cyclist get overtaken This question is not limited to any particular method; all methods accepted. This is a homework question, stated exactly as:
A cyclist starts off in a bike at an average speed of 16 km/h. 15 minutes later, a motorcyclist sets off on the same trail with an average speed of 48 km/h. How many minutes (in total) will the cyclist have ridden when he gets overtaken by the motorcycle?
I tried to form 2 simultaneous equations for the speed and distance of both but failed. Any help??
EDIT
Thanks for the retag, and you asked to me to show my working out. Knowing that the the first guy crossed d kilos in 15 minutes, we can work that out as 4. Now for the second guy, he would cross those 4 in only 5 minutes. So I tried to plot a linear relationship for each guy and solve for an intersection point, but since they both start at the origin (do they???) The only intersection is (0, 0). And if I try to start off guy#2 at 15 minutes, the intersection point is 30. The answer at the back of the book says 22.5
| The speed of the cyclist is $3$ times slower than the speed of the motorcyclist. The motorcyclist, however goes after $15$ minutes from the same point. You can use number sense to figure this.
Or, when $x=$ amount in hours,
$16x=48(x-0.25)$, since the cyclist sets in $15$ minutes before the motorcyclist.
$0.25$ hours = $15$ minutes
$16x = 48x-12$
$-32x=-12$
$x=\cfrac {3}{8}$
The motorcyclist will overcome the cyclist after $\cfrac {3}{8}$ hours, or $22$ minutes and $30$ seconds.
| {
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Could someone check if I did this correctly? Homogeneous Differential Equation
$$y' = \frac{-y-x}{x}, y' = \frac{-y}{x} -1$$
$$F(v) = -v -1$$ Since $$y = xv$$ then $$y' = v + xv'$$
Therefore:
$$v+xv' = -v-1$$
$$xv' = -2v-1$$
$$\frac{-dv}{2v+1} = \frac{dx}{x}$$
$$-\frac{1}{2}ln\bigg |1+2v\bigg|=ln|x| + C$$
$$ln|1+2v| = -2ln|x| -2C$$
$$1+2v = \pm (x^{-2}*e^{-2C})$$
So here is where I get anxious, I'm not sure what to do with with the constants and the plus/minus from the LHS absolute value sign. But here's what I did and what we learned in class:
$$1+2v = \frac{D}{x^{2}}$$ where $D = \pm e^{-2C}$ and so $v =\frac{D}{2x^{2}} - \frac{1}{2}$
Do I substitute $v = \frac{y}{x}$ back in? So $ \frac{y}{x}= \frac{D}{2x^{2}} - \frac{1}{2}$
$$y = \frac{xD}{2x^{2}} - \frac{x}{2}$$
$$y = \frac{D}{2x} - \frac{x}{2}$$
Plugging in the initial condition where y(1) = 1:
$$1 = \frac{D}{2} - \frac{1}{2}$$
$$D = 3$$
Therefore, the general solution is...?
$$y(x) = \frac{3}{2x} - \frac{x}{2}$$
The thing I'm confused about is, are we solving for $D$ since we substituted $D$ for $e^{-2C}$?
Or does the constant variable doesn't matter?
| $$y'x+y+x=0$$
$$(xy)'+x=0$$
Simply integrate
$$xy =-\int xdx=-\frac {x^2} 2 +K$$
$$y =-\frac {x} 2 +\frac Kx$$
then since $y(1)=1 \to K= \frac 32$
$$y =-\frac {x} 2 +\frac {3}{2x}$$
So your answer is correct..
| {
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Iterative sequence converging Consider the sequence
$$1^4,2^4,3^4,\ldots,k^4,\ldots$$
Form a new sequence, whose terms consist of the difference of the above sequence.
$$2^4-1^4,3^4-2^4,4^4-3^4,\ldots$$
Repeat the process with the terms of this new sequence. When this is done sufficiently many times, you will eventually get the sequence
$$24, 24, 24, 24, \ldots$$
Why is this the case fundamentally, from a mathematical perspective? Why 24 exactly? I suspect it might have something to do with the fact that $24 = 4!$, though this could be completely off. Will it still work if we used the same algorithm with any arbitrary exponent $n$ instead of $4$? I can not answer this question because I can't seem to formalize this process in a way that allows me to see obvious convergence to a constant sequence.
| This is the calculus of finite differences. You are starting
with a function $f(x)$ (here $f(x)=x^4$) and defining a new one by $g(x)=f(x+1)-f(x)$, then iterating the procedure.
In general if $f(x)=a_nx^n+\cdots+a_0$ is a polynomial of degree $n$,
then $g(x)=na_nx^{n-1}+\cdots$ is also a polynomial. Iterating then gives
$n(n-1)a_n x^{n-2}+\cdots$ and after $n$ stages we get the constant
$n(n-1)(n-2)\cdots 1a_n$.
| {
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Does absolute consistency imply consistency? As I understand a set of sentences $\Phi$ in first-order logic is consistent iff for all $\psi$ either $\Phi \vdash \psi$ or $\Phi \vdash \neg \psi$ is false.
On the other hand $\Phi$ is absolute consistent iff there exists a sentence $\psi$ such that $\Phi \vdash \psi$ is false.
That consistency implies absolute consistency seems self explanatory.
Since $\bot$ is usually defined by the definition schema $\forall \psi:\bot \implies \psi$ and negation is defined by $\forall \psi:\neg \psi \iff (\psi \implies \bot)$.
Question: does this mean that in first-order logic consistency and absolute consistency are equivalent?
| Yes, in first order logic these concepts are equivalent. I wull use $T$ to stand for a set of sentences, rather than $\phi$, which should stand for a single sentence.
If $T$ proves both $\psi$ and $\lnot \psi$ for some $\psi$, then $T$ proves $\rho$ for all $\rho$. So if $T$ is not consistent it is not absolutely consistent.
On the other hand, it is immediate that it $T$ is not absolutely consistent then it is not consistent.
The difference between the different forms of consistency only happens in logics other than first order logic:
*
*In complete generality, a logic might not have a negation operation. In a logic like that, we can talk about absolute consistency but not about simple consistency.
*In paraconsistent logics, it is possible that the principle of explosion does not apply. A theory can sometimes prove $\phi \land \lnot \phi$ for a sentence $\phi$, but not prove $\psi$ for some other sentence $\psi$. One motivation for studying paraconsistent logics is to handle exactly this situation, when we may be presented with contradictory assertions in one area, but we do not want to apply those to other areas.
There is also semantic consistency: a theory is semantically consistent if and only if it has a model. In first order logic, semantic consistency is equivalent to consistency. But, for example, in second order logic with standard semantics a theory can be syntactically consistent but not semantically consistent.
| {
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Proving sequence is Cauchy sequence I need to examine if this sequence is Cauchy sequence using definition.
$$a_n= 1/3 + 2^2/3^2+...+n^2/3^n$$
I start with $m,n \in N, n>m$
$$ \vert(a_n - a_m)\vert = \vert(m+1)^2/3^{(m+1)}+...+n^2/3^n\vert$$
And I don't know what is bigger than this expression and what to compare it to.
| An idea to make things simpler: take $\;n,\,\,m=n+p\;,\;\;p\in\Bbb N\;$ , so
$$\left|a_{n+p}-a_n\right|=\frac{(n+1)^2}{3^{n+1}}+\ldots+\frac{(n+p)^2}{3^{n+p}}\le p\frac{n^2}{3^n}$$
Since for any $\;p\in\Bbb N\;$ we have that $\;\lim\limits_{n\to\infty}\;p\,\cfrac {n^2}{3^n}=0\;$ , we get that for any $\;\epsilon>0\;$ and for any $\;p\in\Bbb N\;$ there exists $\;N_{p,\epsilon}\in\Bbb N\;$ s.t.
$$n>N_{p,\epsilon}\implies p\frac{n^2}{3^n}<\epsilon$$
Fill in details and complete the proof.
| {
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What is probability that a line joining two randomly selected coordinates from set form an angle $45°$ with one of axes? What is probability that a line joining two randomly selected coordinates from a set form an angle $45°$ with one of the axes?
For example, if we have set of coordinates $$\{(1, 2), (1, 3), (3, 3), (4, 4)\}$$ then, if my calculations is correct, the probability would be $0.125$
How I calculated this:
Number of ways to select coordinates is $4^2=16$ because even the same coordinates can be selected. To line form an angle $45°$ with one of axes, the slope of line have to be $1$ or $-1$. Out of these $16$ ways only $2$ have slope $1$ or $-1$ so probability is $\frac{2}{16} = 0.125$.
I am programming an algorithm that computes this probability but the algorithm have to check slope of every combination of coordinates so my question is:
Is there any other/faster way to compute this probability?
EDIT: I don't know if it helps, but coordinates will be only integers.
EDIT 2: Range of coordinates is from $-100000$ to $100000$. Number of coordinates will be no more than $10000$. Also probability will be computed to $6$ decimal places.
| For any two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ to form a $45^\circ$ angle with two axes, the equivalent condition is$$
x_1 + y_1 = x_2 + y_2 \ \text{or}\ x_1 - y_1 = x_2 - y_2. \tag{1}
$$
Note that to compute the probability, it suffices to count the number of pairs satisfying (1) and then divide it by $\binom{n}{2}$. The algorithm below counts the number of pairs $(x_1, y_1)$ and $(x_2, y_2)$ such that$$
x_1 + y_1 = x_2 + y_2,
$$
and there is an analogous algorithm that counts the number of pairs $(x_1, y_1)$ and $(x_2, y_2)$ such that$$
x_1 - y_1 = x_2 - y_2,
$$
Suppose there are $n$ distinct points (x[1], y[1]), …, (x[n], y[n]).
Step1: Compute a[i] = x[i] + y[i] for $1 \leqslant i \leqslant n$. The time complexity of this step is $O(n)$.
Step 2: Sort array a. By using merge sort, the time complexity of this step is $O(n \ln n)$.
Step 3: Count the multiplicity of each a[i]. Suppose distinct values in a are $a'_1 < \cdots < a'_t$ and their respective multiplicities are $m_1, \cdots, m_t$, then compute $s = \sum\limits_{k = 0}^t \binom{m_k}{2}$.
s = 0; m = 1;
for(i = 2; i <= n; i++) {
if(a[i] == a[i - 1]) m++;
else {
s += m * (m - 1) / 2;
m = 1;
}
}
s += m * (m - 1) / 2;
The time complexity of this step is $O(n)$.
The overall time complexity is $O(n \ln n)$.
| {
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Does $M:=\{x \in \mathbb{Q}: x^2<7\}$ $M\subseteq\mathbb{R}$ have an infimum, supremum, min, max? $M\subseteq\mathbb{R}$
$M:=\{x \in \mathbb{Q}: x^2<7\}$
Does $M$ have an infimum, supremum, min, max?
My answer would be that it doesn't because $\sqrt{7}$ and $-\sqrt{7}$ are $\not\in \mathbb{Q}$
Is that correct?
| Definition: An upper bound $u$ of $M \subset \mathbb{R}$ is an element $u \in \mathbb{R}$ such that $u \geq m$ for all $m \in M$.
Definition: A supremum of $M \subset \mathbb{R}$ is an element $x \in \mathbb{R}$ such that
*
*$x$ is an upper bound of $M$
*for each upper bound $u$ of $M$, $x \leq u$
According to the above definition, $x$ does not have to be in $M$, which is perhaps where you have gotten confused.
Example: $M = [0, 1)$ does not have a maximum, since for any $x \in [0, 1)$, we can define $y = (x + 1)/2 \in M$, noting that $y > x$.
Example: $M = [0, 1)$ has a supremum, namely $1$. Can you prove this?
Once you have tackled the above example, you should be able to answer your original question (hint: use also the fact that any real number can be defined as the limit of a sequence of rational numbers).
| {
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$a_n $ is a positive integer for any $n\in \mathbb {N} $. Let $(a_n)_{n\geq 1}$ be a sequence defined by $a_{n+1}=(2n^2+2n+1)a_n-(n^4+1 )a_{n-1} $.
$a_1=1$, $a_2=3$.
I have to show that $a_n $ is a positive integer for any $n\in \mathbb {N}, n\geq 1$.
I tried to prove it by induction but it doesn't work.
| This answer provides partial results.
For each $n$ put $b_n=\frac {a_{n+1}}{a_n}$. Then $b_1=3$ and
$$b_n=2n^2+2n+1-\frac{n^4+1}{b_{n-1}}.$$
It suffices to prove that $b_n>0$. Computer evaluation suggests that a sequence $\{c_n=b_n-n^2-2n\}$ decreases and converges to about $-5.78734$. We have $c_1=0$ and
$$c_n=n^2+1-\frac{n^4+1}{c_{n-1}+n^2-1}= \frac{c_{n-1}(n^2+1)-2}{c_{n-1}+n^2-1}.$$
It seems that we can show that $c_n<c_{n-1}$ provided the denominator $ c_{n-1}+n^2-1>0$.
| {
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Finding Parametric Equations for the right side of Hyperbola I have this equation $x^2-y^2=1$.
I can understand that it is a Hyperbola. But my question is, how can I create/find a parametric equation for the right side of this hyperbola? It seems it doesn't work, using $\sin$ or $\cos$.
Thanks
| Let's find functions $x(t), y(t)$ that satisfy $x^2 - y^2 = 1$. Perhaps, if we rewrite as $x^2 = 1 + y^2$, that'll help.
This reminds me of Pythagorean identities. In particular, it reminds me of $1 + \tan^2(t) = \sec^2(t)$.
Trying $(x(t), y(t)) = (\sec(t), \tan(t))$ has the unfortunate side effect that we may get negative values for $x$, depending on what we allow for $t$ (but it can't be fixed by choosing a small range for $t$, without losing some of the right half). Try using $|\sec(t)|$ instead, to ensure that we only get positive $x$-values.
| {
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x odd then exists y such that $x^2 = 8y+1$ How can I algebraically show that if $x$ is odd, then $x^2 = 8y + 1$?
Making sure it is a true statement, I tabulated some values of $x,y$ pairs
$(1,0)$, $(3,1)$, $(5,3)$, $(7,6)$, $(9,10)$
I let $x = 2w + 1$ (since $x$ is odd)
Then $(2w + 1)^2 = 4w^2 + 4w + 1$
I don't think I can say $4w^2 + 4w + 1 \quad ? \quad 4w + 4w + 1 = 8w + 1$, can I?
| If $x$ is odd, it is congruent to $\pm 1$ or $\pm 3$ modulo 8. In both cases, its square is congruent to $1$.
| {
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If $\lim_{x\to+\infty}f'(x)>0$, does that mean that $\lim_{x\to+\infty}f(x)=+\infty$ Let's assume some function $f: \mathbb{R}\to\mathbb{R}$, which is differentiable in some region $(a, +\infty)$, where $a\in(0,+\infty)$. It makes sense (to me) that the following holds true:
$$\lim_{x\to+\infty}f'(x)\in(0, +\infty) \Rightarrow \lim_{x\to+\infty}f(x)=+\infty$$
However, I am not quite sure how I would go about proving it, if it even is true.
| Sketch:
Assuming that $\lim_{x\rightarrow\infty}f'(x)$ exists and equals $L>0$, then there is some $M$ such that for all $x>M$, $|f'(x)-L|>\frac{L}{2}$. Therefore, $f'(x)>\frac{L}{2}$ for $x>M$.
Observe that
$$
f(x)=f(M)+\int_M^xf'(t)dt.
$$
For $x>M$, it follows that
$$
f(x)=f(M)+\int_M^xf'(t)dt\geq f(M)+\int_M^x\frac{L}{2}dt=f(M)+\frac{L}{2}(x-M).
$$
Since the RHS diverges to infinity, so does $f(x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it possible to prove associativity from $a(bc)=a(cb)$? With the given information, I'm trying to prove that an operation is both associative and commutative.
Let $(A,*)$ be a set and a binary operation such that for all $a,b,c \in A$
$$a*(b*c) = a*(c*b),$$
and
for some $e \in A$ and for any $a \in A$ $$e*a = a*e = a.$$
So far, proving that $*$ is commutative has been easy, but finding a method to prove associativity has escaped me. I don't want the solution to the problem, but could anyone help in assuring me that associativity is even possible with the given assumptions.
Thanks!
| It is not possible. Consider the operation of combining unordered binary trees, where a * b is the binary tree whose root's unordered pair of children are the roots of a and b; toss in by fiat also an identity element for this operation. This operation is clearly commutative and thus satisfies your properties, but does not satisfy associativity.
For a less abstract example, consider something like the operation $|a - b|$.
| {
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Polynomial such that $f''(x) \rightarrow2$ as $x\rightarrow\infty$ given some values what is $f(1)$? Let $f$ be a polynomial such that $f''(x) \rightarrow2$ as $x\rightarrow\infty$, the minimum of f is attained at $3$, and $f(0)=3$, Then $f(1)$ equals.
$(A) \ 1$
$(B) \ 2$
$(C) -1$
$(D) -2$
I am not sure how to deal with $f''(x) \rightarrow2$ as $x\rightarrow\infty$ . Give me a hint to try.
EDIT : Work after hint
Let $f(x)=ax^2+bx +c $
$f''(x)=2\implies 2=2a \implies a=1 $
$f(0)=3=c$
$f(x)=x^2+bx +3 $
Using the fact that minima attained at 3 we have
$f(3)=12+3b=3 \implies b=-3$
$f(x) = x^2-3x+3$
$f(1) = 1-3+3 = 1$
| Suppose $f$ has degree $n$, then we can write $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$$
We are given that $f''(x)\to 2$ as $x\to\infty$, so what could $n$ possibly be?
Suppose $n>2$, then $\deg(f''(x))=n-2>0$, so $f''(x)\to\pm\infty$ as $x\to\infty$.
Suppose $n<2$, then $f''(x)\equiv 0$ , so clearly $f''(x)\to 0\neq 2$ as $x\to\infty$.
This should get you pretty close to a solution.
| {
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if $g$ be a continuous function not differentiable at $0$ Let $g$ be a continuous function not differentiable at $0$ with $g(0)=8$. Let $f(x)=x\,g(x)$ .Find $f'(0)$
a)$0$
b)$4$
c)$2$
d)$8$
I am getting that $f'(x)=g(x)+x\,g'(x)$. But since $g'(x)$ doesn't exist for $x=0$, hence $f'(x)=8$. Please help whether it is right or wrong
| It's wrong. You should try from the definition of differentiability.
Hint: $f^{'}(0)=\lim_{h \to 0} \frac{f(0+h)-f(0)}{h}$.
| {
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Let $A,B$ be some sets such that $|A|=a, |B|=b$. Prove $\binom{a}{b}=|P_b(A)|$ is a well defined expression.
Let $A,B$ be some sets such that $|A|=a, |B|=b$. Prove $\binom{a}{b}=|P_b(A)|$(=the set of all subsets of A of cardinality $b$) is a well defined expression.
Hello all. In this question I need to show that the equality $\binom{a}{b}=|P_b(A)|$ holds for all sets $A\neq A' \land |A|=|A'|=a, B\neq B'\land |B|=|B'|=b $. I don't really know how to show it, I thought about doing some tricks with the binomial theorem but I guess it has nothing to do with that. Would love to get your help... thanks in advance :)
| Using binomial theorem could work. You could also consider the "definition" (or interpretation) of the number $\binom{a}{b}$:
$\binom{a}{b}$ is the number of ways to choose $b$ elements from a set with $a$ elements.
This is, in a very literal sense, the number of subsets of size $b$ from a set of size $a$.
This is clear if you are willing to accept this interpretation of the binomial coefficient.
On the other hand, if you want to prove it completely what you would need to show is basically that the binomial coefficient represents what it says it does: the number of subsets of size $b$ from a set of size $a$.
This is a standard reasoning when you're first introduced to the binomial coefficient, and the approach is something as follows (try to think about it first, put the pointer on the space below if you need help):
To choose $b$ elements out of a set of $a$ elements, first we determine which elements will we choose, in order (i.e. which is the first, which is the second, and so on). For this we have $n\cdot (n-1)\cdots (n-r+1)$ possibilities. Now, since the order doesn't really matter when it comes to choosing a subset, we have to divide by the number of permutations of $r$ elements, which is $r!$. The result is $\frac{n\cdot (n-1)\cdots (n-r+1)}{r!}$, which is the binomial coefficient $\binom{a}{b}$.
| {
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If $a,b,c$ are the roots of $x^3-px+q=0$, then what is the determinant of the given matrix in $a, b, c$?
If $a,b,c$ are the roots of $x^3-px+q=0$, then what is the determinant of
$$
\begin{pmatrix}
a & b & c \\
b & c & a \\
c & a & b \\
\end{pmatrix} \,\,?
$$
(A) $p^2+6q \quad$
(B) $1 \quad$
(C) $p \quad$
(D) $0 \quad$
In this equation given we have product of eigenvalues given as $-q$ and we know product of eigenvalues is determinant then why isn't the determinant is $-q$?
|
In this equation given we have product of eigenvalues given as $−q$ and we know product of eigenvalues is determinant then why isn't the determinant is $−q$?
The statement you're thinking of is that the product of eigenvalues of a matrix $A$ is equal (up to sign, depending on parity) to the constant term of its characteristic polynomial $p_A$.
In this case, however, the given matrix, though constructed from the roots $a, b, c$ of the characteristic polynomial, doesn't have the property that its eigenvalues are $a, b, c$, and therefore the above general fact does not apply.
On the other hand, using, e.g., cofactor expansion gives that the determinant of the given matrix is
$$\det \pmatrix{a&b&c\\b&c&a\\c&a&b} = 3 a b c - (a^3 + b^3 + c^3).$$
On the other hand, writing the cubic polynomial as $$x^3 - p x + q = (x - a)(x - b)(x - c),$$ expanding, and comparing like terms recovers a special case of Vieta's Formulas:
$$\begin{array}{rcl}
0 &=& - (a + b + c) \\
-p &=& bc + ca + ab \\
q &=& -a b c
\end{array} .$$
Manipulating these lets us write both terms of the expression $3 a b c - (a^3 + b^3 + c^3)$ for the determinant as polynomials in $p, q$. These manipulations give special cases of Newton's Identities.
| {
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"timestamp": "2023-03-29T00:00:00",
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Lebesgue integrable implies Riemann integrable? I'm self studying measure theory by Bartle's book and there he defined integrability for non-negative functions as follow
Definition: Let $f$ a non-negative measure function, then the integral of $f$ is
$$\int f = \sup \left\{ \int \phi : \ \phi \leq f \right\},$$
where $\phi$ is a simple function.
It's clear that this definition is the same of lower integral for Riemann integral. I think a definition equivalent to upper integral to Riemann integral is
$$\int f = \inf \left\{ \int \phi : \ \phi \geq f \right\},$$
where $\phi$ is a simple function.
My doubt is if the definition of integral by Bartle's book implies the definition of Riemann integrable, i. e., is the following true?$$\sup \left\{ \int \phi : \ \phi \leq f \right\} = \inf \left\{ \int \phi : \ \phi \geq f \right\}$$
Thanks in advance!
|
It's clear that this definition is the same of lower integral for Riemann integral (...)
That is false. As a counter-example, the function $\mathbf{1}_{\mathbb{[0,1] \backslash Q}}: [0,1] \to \mathbb{R}$ has lower integral for Riemann integral equal to $0$, and "lower" integral according to the Lebesgue definition equal to $1$. The point, as mentioned by Ian at the comments, is that not every simple function is a step function: the function above being an example.
The function above is also an example of one which is Lebesgue integrable but not Riemann integrable, so the statement as it is in the title is not true.
However, the statement
$$\sup \left\{ \int \phi : \ \phi \leq f \right\} = \inf \left\{ \int \phi : \ \phi \geq f \right\}$$
is true, if $f$ is a non-negative bounded function which is not zero only on a finite measure set.* To see this, it suffices to show a sequence of simple functions $\phi_n \geq f$ such that $\lim \int \phi_n =\int f$.
Pick $M \mathbf{1}_E \geq f$. Now, we then have $M\mathbf{1}_E - f \geq 0.$ It follows that there is an increasing sequence $s_n$ of simple functions such that $s_n \to M\mathbf{1}_E-f$ and $s_n \leq M\mathbf{1}_E-f.$ By the monotone convergence theorem, $\int s_n \to \int M\mathbf{1}_E -\int f$.
We have that $f \leq M\mathbf{1}_E-s_n$, so that $\phi_n:=M\mathbf{1}_E-s_n$ is a sequence of simple functions satisfying what we want, since
$$\int \phi_n=\int M\mathbf{1}_E-\int s_n \to\int f.$$
*If $f$ doesn't satisfy those hypotheses (i.e., bounded and not zero only on a finite measure set), the right side is always infinity so the question is a little irrelevant.
| {
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Differential equations theorems(Pure mathematics) I'm currently doing some graduate work and came upon some problems. The content of the course is of a pure form with topics such as
*
*Existence and Uniqueness of solutions
*linear system of 1st order ODE
*asympotitic behaviour of soltuions and stability analysis
*boundary value problems for 2nd order ode
There are a lot of theorems and proofs in the course and I am unable to find a suitable book/books for this type of content. Some of the theorems are as follows
Theorems
Can anyone point me in the right direction as to what books would be best suitable for content such as this? Solving differential equations is irrelevant here and more emphasis is put on theory and understanding.
| This book by Walter covers the theoretical aspects of ordinary differential equations quite well and was used for my graduate course.
Also the book by Coddington is often suggested as a good source for rigorous existence and uniqueness theorems.
I personally find the book by Hale to be rigorous yet not too hard to read.
| {
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Determining Spot Rates A three-year, 4%, par-value bond with annual coupons sells for $990$, a two-year, $1000$, 3% bond with annual coupons sells for $988$, and a one-year, zero-coupon, $1000$ bond sells for $974$. Determine the spot rates $r_1$, $r_2$ and $r_3$.
This comes from Mathematical Interest Theory textbook section 8.3 #2. I understand how to compute similar problems, however I am unsure how to solve this given that the bonds have annual coupons(not zero coupon bonds). Any help would be appreciated thank you!
| For the one year bond we have:
$$
974=\frac{1000}{1+r_1}\qquad\Longrightarrow\qquad r_1=\frac{1000}{974}-1\approx 2.66940\%
$$
For the two years bond we have the coupon $3\%\times 1000=30$ and
$$
988=\frac{30}{1+r_1}+\frac{1030}{(1+r_2)^2}
$$
Observing that $\frac{1}{1+r_1}=\frac{974}{1000}=0.974$ we have
$$
988=\underbrace{30\times 0.974}_{29.22}+\frac{1030}{(1+r_2)^2}\qquad\Longrightarrow\qquad r_2=\left(\frac{1030}{958.78}\right)^{1/2}-1\approx 3.64757\%
$$
Can you now find $r_3$?
$990=\underbrace{\frac{40}{1+r_1}}_{40\times 0.974}+\underbrace{\frac{40}{(1+r_2)^2}}_{40\times\frac{958.78}{1030}}+\frac{1040}{(1+r_3)^3} \qquad\Longrightarrow\qquad r_3\approx 4.4062\%$
| {
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Show that the collection is a basis for a topology on $\mathbb{R}$ We are given the collection, $$\mathcal{B}=\{[a,b]\mid a,b\in\mathbb{R}\}.$$ We want to show that this collection is a basis for a topology on $\mathbb{R}$. To show this we show the following:
First, for all $x\in\mathbb{R}$ there exists an element in $\mathcal{B}$, namely $[x-\epsilon,x+\epsilon]$, with $\epsilon\in\mathbb{R}$ which contains $x$.
Second, let $x\in[a,b]\cap[c,d]$. We want to show there exists a third element,
say $B$ in $\mathcal{B}$, such that $x\in B\subset[a,b]\cap[c,d]$. Consider, $B=[c,b]$.
So we have shown that $\mathcal{B}$ is a basis for some topology on $\mathbb{R}$.
The question then asks me to show that the topology generated by this basis is the discrete topology. This amounts to showing that the basis elements of either topology is open in the other.
The discrete topology is generated by single element sets. So, we have trivially that all basis elements in $\mathcal{B}$ are open in the discrete topology. Similarly, we consider the basis element $[a,a]\in\mathcal{B}$. We have that for a given $\{a\}\in\mathcal{C}$, where $\mathcal{C}$ is the discrete basis, there exists a $[a,a]\in\mathcal{B}$ such that all the basis elements in $\mathcal{C}$ are open in the topology generated by $\mathcal{B}$.
So, I conclude that the topologies are equal. My question is the following. Is it okay for me to treat the element $[a,a]=\{a\}$ as an element of $\mathcal{B}$ in this proof? The question imposes no restriction on the value of $a$ or $b$, so I think it's okay. However, I'm not sure.
|
Is it okay for me to treat the element $[a,a]=\{a\}$ as an element of $\mathcal{B}$ in this proof?
Yes, that's totally fine. If you want to avoid it, you could also write
$$
\{a \} = [a-1,a] \cap [a, a+1].
$$
| {
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Is there a formula to calculate the area of a trapezoid knowing the length of all its sides? If all sides: $a, b, c, d$ are known, is there a formula that can calculate the area of a trapezoid?
I know this formula for calculating the area of a trapezoid from its two bases and its height:
$$S=\frac {a+b}{2}×h$$
And I know a well-known formula for finding the area of a triangle, called Heron's formula:
$$S=\sqrt {p(p-a)(p-b)(p-c)}$$
$$p=\frac{a+b+c}{2}$$
But I could not a formula for finding the area of a trapezoid in the books.
| This problem is more subtle than some of the other answers here let on. A great deal hinges on whether "trapezoid" is defined inclusively (i.e. as a quadrilateral with at least one pair of parallel sides) or exclusively (i.e. as a quadrilateral with exactly one pair of parallel sides). The former definition is widely considered more mathematically sophisticated, but the latter definition is more traditional, is still extensively used in K-12 education in the United States, and has some advantages.
As the other responses have pointed out, if one defines "trapezoid" inclusively, then any parallelogram is automatically a trapezoid, and as the side-lengths of a parallelogram do not determine its area, it is not possible (even conceptually) that there could be a formula for the area of a trapezoid in terms of its side lengths.
However, if "trapezoid" is defined exclusively, then things are quite different. Consider a trapezoid with parallel bases of length $a$ and $b$ with $b>a$. Let $\theta$ and $\phi$ respectively denote the angles formed by the legs $c$ and $d$ with the base $b$. Then we have the following relationships:
$$c\cos\theta + d\cos\phi = b-a$$
$$c\sin\theta = d\sin\phi$$
These conditions uniquely determine $\theta$ and $\phi$, and therefore among non-parallelogram trapezoids, choosing the lengths of the parallel sides and the lengths of the bases uniquely determines the figure. In particular we would have $$\cos\theta = \frac{(b-a)^2+c^2-d^2}{2c(b-a)}$$.
The height of the trapezoid would then be $h=c\sin\theta$ (or if you prefer $h=d\sin\phi$, which is equal to it), so the area of the trapezoid can (in principal) be computed. If you really want to carry it out, you would have
$$\sin\theta = \sqrt{1-\left( \frac{(b-a)^2+c^2-d^2}{2c(b-a)} \right)^2}$$
so the area would be
$$A=\frac{a+b}{2}c\sqrt{1-\left( \frac{(b-a)^2+c^2-d^2}{2c(b-a)} \right)^2}$$
I am not sure if there is a simpler expression, however.
| {
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Set-theoretic equality of splitting fields within a fixed algebraic closure Let $F$ be a field and let $f(x)\in F[x]$ be a polynomial. Recall the following two facts:
(1) algebraic closures are unique up to isomorphism
(2) splitting fields are unique up to isomorphism
Fix an algebraic closure $\overline F$ of $F$. Is it true that any two (necessarily isomorphic) splitting fields $E,E'\subseteq \overline F$ for $f$ are equal $as$ $sets$ in addition to being isomorphic?
My intuition tells me that this should be the case: fixing an algebraic closure allows us to fix roots of $f$ within this particular ambient field, and since any splitting field is the smallest field containing these roots, any two such fields must actually be equal as sets, in addition to being isomorphic. Does this sound about right? Is this totally trivial?
| I don't think they are equal as sets. Remember that in our way to get splitting field for $\;f(x)\in F[x]\;$ , we first get (assuming $\;f\;$ is irreducible, otherwise we take one of its irreducible factors) the quotient ring (field) $\;K:=F[x]/\langle f(x)\rangle\;$ .
Here, we already have no more $\;F\;$ but an isomorphic copy of $\;F\;$ within $\;K\;$ , and there could be several ways to obtain that copy...
| {
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Prove inequality using the Mean Value Theorem I'm trying to hone my problem-solving skills using the Mean Value Theorem and in one exercise, where $x \in (0, +\infty)$, I have to prove that:
*
*$(1+x)^a>ax+1$, if $a > 1$.
*$(1+x)^a<ax+1$, if $a \in (0, 1)$.
What I've tried:
I've tried to solve this problem using the function $f(t)=(1+t)^a$ in the closed set $[0,x]$ as follows:
*
*First, I calculated the derivative of $f$, which is $f'(t)=a(1+t)^{a-1}$.
*Then, I used the Mean Value Theorem:
$$
f'(k)={{f(x)-f(0)}\over{x-0}}={{(1+x)^a-1}\over{x}}
\Leftrightarrow
a(1+k)^{a-1}={{(1+x)^a-1}\over{x}}\\\Leftrightarrow
(1+x)^a=ax(1+k)^{a-1}+1
$$
The equation I found seems to be on the right track, so I decided based on instinct to examine the following cases:
*
*$a=1 \Rightarrow (1+x)^a=ax+1$
*$a>1 \Rightarrow (1+x)^a>ax+1$
*$a \in (0, 1) \Rightarrow (1+x)^a<ax+1$
Question:
My solution, and more specifically the part where my instinct kicks in, feels rather incomplete and rushed. Is there a better way to solve this problem using the Mean Value Theorem?
| Your “instinct” is correct, and it requires only small additions to
make it a full proof.
The mean value theorem implies that for $x > 0$
$$
(1+x)^\alpha = 1 + \alpha x (1+k)^{\alpha-1}
$$
for some $k \in (0, x)$. It is relevant that $k$ is strictly positive,
so that one can continue to argue
$$
\alpha > 1 \Longrightarrow (1+k)^{\alpha-1} > 1
\Longrightarrow (1+x)^\alpha > 1 + \alpha x \, , \\
0 < \alpha < 1 \Longrightarrow (1+k)^{\alpha-1} < 1
\Longrightarrow (1+x)^\alpha < 1 + \alpha x \, .
$$
| {
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Interpretation of Sampling Distribution and Relationship to Test Statistics Explain what a sampling distribution is and why it is important to understand the sampling distribution of a test statistic.
This is the way that I understand it:
A sampling distribution is the probability distribution of a statistic that comes from a large number of independent samples drawn from a population. The sampling distribution of a test statistic is important because this allows us to assign a probability to the occurrence of an event that we may be interested in; it allows us to know the likelihood that the collected data tells us something useful. Without knowing the sampling distribution, this probability can’t be assigned accurately.
How accurate is my description that I've given? What would you change based on the way that you understand it?
| It is often impossible to test every member of a population, but if it were, one can know for certain the characteristics of the population. Since it is not practical to test the entire population it is often the case that a randomly chosen sample, or subset, of the population is analyzed. The sample is ran through tests in hopes of finding relationships in the data which not only reflect the current state of the population via the sample but allow reliable predictions (inferences) of future states. The relation or correlation of frequencies, if one exists, is the sample distribution which models the randomly selected sample in such a way as to accurately represent the population as a whole. This is why the larger the sample, the more accurate the distribution, because the sample is approaching the population size.
| {
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Find limit of $\lim_{x\to 0}\frac{\sin^{200}(x)}{x^{199}\sin(4x)}$, if it exists I'm practising solving limits and the one I'm currently struggling with is the following: $$\ell =\lim_{x\to 0}\frac{\sin^{200}(x)}{x^{199}\sin(4x)}$$
What I've done:
*
*Since this is an obvious $0/0$ , I tried using de L'Hospital's Rule consecutively only to see both the numerator and the denominator grow so much in size that each couldn't fit in one row.
$$
\begin{align}
l
& =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}\sin(4x)}}\\
& = \lim_{x→0}{{200\sin^{199}(x)\cos(x)}\over{x^{198}\left(199\sin\left(4x\right)+4x\cos\left(4x\right)\right)}}\\
& = \lim_{x→0}{{39800\cos^2\left(x\right)\sin^{198}\left(x\right)-200\sin^{200}\left(x\right)}\over{x^{198}\left(800\cos\left(4x\right)-16x\sin\left(4x\right)\right)+198x^{197}\left(199\sin\left(4x\right)+4x\cos\left(4x\right)\right)}}
\end{align}
$$
*Another solution I tried was through manipulation and the use of trigonometric identities and formulae but to no avail. I tried substituting:
*
*$\color{red}{\sin(4x)}$ with $\color{blue}{4\sin(x)\cos(x) - 8\sin(3x)\cos(x)}$ and then
*$\color{red}{8\sin(3x)\cos(x)}$ with $\color{blue}{4\sin(4x)+4\sin(2x)}$.
$$
\begin{align}
l
& =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}\sin(4x)}}\\
& =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}(4\sin(x)\cos(x) - 8\sin(3x)\cos(x))}}\\
& =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}(4\sin(x)\cos(x) - 4\sin(4x)+4\sin(2x))}}\\
\end{align}
$$
No matter what I try, the limit remains $0/0$.
Question:
Does the above limit exist? If so, what I path should I follow to work out a solution?
| Since $\sin x= x+o(x)$ we have, $$\frac{\sin^{200}x}{x^{199}\sin(4x)}= \frac{x^{200}+o(x^{200})}{x^{199}(4x+o(x))}=\frac{x^{200}+o(x^{200})}{4x^{200}+o(x^{200})}=\frac{1+o(1)}{4+o(1)} =\to\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the value of $f(100)$?
We have $f:\Bbb R\to \Bbb R^*$, a function that admits primitives and admits the relations $$\cos \left(f(x)\right)=1,\ ∀x\in \Bbb R, \quad\text{and}\quad|f(\pi )−\pi |≤\pi .$$
What is the value of $f(100)$?
My thought. We obviously have
$$\cos (f(100)) =1\overset{?}{\implies} f(100) =\arccos (1),$$
but this seems not to make any sense at all.
How can I use the provided inequality $|f(\pi)−\pi|≤\pi$?
| Clearly $f(x)=2\pi k_x$ where $k_x\in\mathbb Z$ Besides
$$|2\pi k_x-\pi|\le\pi\iff-\pi\le2\pi k_x-\pi\le\pi\Rightarrow 0\le k_x\le1$$
Since $k_x$ is an integer the problem gives two solutions $f(x)=0$ and $f(x)=2\pi$ but because of the function is from $\Bbb R$ to $\Bbb R^*$ the only solution is
$$f(x)=2\pi$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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The sequence satisfying $a_2^{a_3^{\dots^{a_n}}} = n $ $a_2 = 2$
$a_2^{a_3 } = 3$
So $a_3= \ln(3)/\ln(2)$.
I wonder about all solutions $a_n$ such that
$a_2^{a_3^{\dots^{a_n}}} = n$
For all $n$.
How does $a_n$ behave? What are the best asymptotics?
Of course $a_n$ goes quickly towards values between $\exp(1/e)$ and $1$ that is trivial.
But I am not even sure If $a_n$ is strictly decreasing or if its limit exists.
Also If it is strictly decreasing with limit $A$ , I do not know the value of $A$.
I assume $A=1$.
I assume $A=1$ Because $a_2 = 2$.
We could consider other starting values and define
( assuming a limit ) $A(a_2)$ as a function. If that function is even continuous or $C^1$ is another matter.
| We can't have $a_n\to a<\exp(1/e)$. If we did, then there must exist some $N$ such that $a_n<\exp(1/e)$ for all $n>N$, and then we'd have
$$a_2^{a_3^{\dots^{a_n^{a_{n+1}^{\dots}}}}}<a_2^{a_3^{\dots^{a_n^{\exp(1/e)^{\dots}}}}}=a_2^{a_3^{\dots^{a_n^e}}}$$
which is bounded, while $n$ is unbounded.
We also can't have $a_n\to a>\exp(1/e)$, since then the power tower would have a lower bound that grows asymptotically faster than $n$ ($x^{x^{\dots}}$ diverges for $x>\exp(1/e)$).
Hence, we conclude that we must have $a_n\to\exp(1/e)$, if it approaches any limit at all. We also deduce that we must have $a_n\ge\exp(1/e)$ for infinitely many $n$, since $x^{x^{\dots}}$ converges for $x=\exp(1/e)$. So if the sequence is monotone, it must be decreasing.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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On tensor product $U\otimes V$, $0\otimes w = 0$ for any $w\in V$ I've tried using the definition of tensor product: $U\otimes V = \mathcal{F}_{U\times V}/\langle S\rangle$, where
$$S = \{(\lambda_1u_1+\lambda_2u_2,v)-\lambda_1(u_1,v)-\lambda_2(u_2,v); (u,\lambda_1v_1+\lambda_2v_2)-\lambda_1(u,v_1)-\lambda_2(u,v_2)\}$$
and so I got that $$0\otimes w = (0,w)+(1-\lambda_1-\lambda_2)(\beta u,\alpha v)$$ for some $\alpha, \beta\in \mathbb{F}$. But I don't know how this can give me the zero element of $U\otimes V$. Any help?
I know that the converse holds, that is, if $u\otimes v = 0$, then $u=0$ or $v=0$.
| Note that $0\otimes w = (0+0)\otimes w = 0\otimes w + 0 \otimes w$ by bilinearity,
so you get that $0\otimes w$ is the zero element of $V\otimes W$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The system of exponential equations How do you solve:
$$\begin{cases}
x\cdot2^{x-y}+3y\cdot2^{2x+y-1}=1 \hspace{0.1cm},\\ x\cdot2^{2x+y+1}+3y\cdot8^{x+y}=1\\
\end{cases}$$
I subtracted the equations, factorized by grouping, and got two terms equal zero. When I tried to use one of the terms, by expressing one variable through another and put it back in the first equation, it became complicated, so I gave up.
Result is: $ (x=1, \hspace{0.1cm}y=-1) $
| We obtain $$x(2^{x-y}-2^{2x+y+1})+3y(2^{2x+y-1}-2^{3x+3y})=0$$ or
$$x2^{x-y}(1-2^{x+2y+1})+3y(1-2^{x+2y+1})=0,$$ which gives
$$x2^{x-y}+3y=0$$ or
$$x+2y+1=0.$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to take the conjugate of a number with more than 2 square roots I was doing some abstract algebra and I came across the problem of figuring out if $\mathbb{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} : a,b,c,d \in \mathbb{Q}\}$ is a ring and further if it is a field. Part of this is proving that every element is invertible for a field which I believe to be the case in this example.
My question is, is there a general way to take a conjugate of a number of this form and more generally numbers of the form,
$$x=a_{0}+ \sum_{n \neq k^2} a_{i}\sqrt{n}$$
Where each $a_i \in \mathbb{Q}$?
In other words a number $y$ such that $\frac{1}{x} \times \frac{y}{y}$ involves no square roots in the denominator.
| The background story here is that your field is a splitting field of the polynomial $(x^2-2)(x^2-3)$, and as such the Galois group $\operatorname{Gal}(\mathbb Q(\sqrt 2, \sqrt 3):\mathbb Q)$ acts on it. It turns out that this group has four elements:
*
*$I$: Identity,
*$\Phi_2$: Automorphism that maps $\sqrt 2$ to $-\sqrt 2$ but leaves $\sqrt 3$ in place,
*$\Phi_3$: Automorphism that maps $\sqrt 3$ to $-\sqrt 3$ but leaves $\sqrt 2$ in place,
*$\Phi_6$: Automorphism that maps $\sqrt 2$ to $-\sqrt 2$ and $\sqrt 3$ to $-\sqrt 3$.
Now, for an element $x=a+b\sqrt 2+c\sqrt 3+d\sqrt 6$, its conjugates are defined as the maps of $x$ using all those automorphisms. In our case, what you will get is:
*
*$I(x)=x=a+b\sqrt 2+c\sqrt 3+d\sqrt 6$
*$\Phi_2(x)=a-b\sqrt 2+c\sqrt 3-d\sqrt 6$
*$\Phi_3(x)=a+b\sqrt 2-c\sqrt 3-d\sqrt 6$
*$\Phi_6(x)=a-b\sqrt 2-c\sqrt 3+d\sqrt 6$
The product of all those: $N(x)=I(x)\Phi_2(x)\Phi_3(x)\Phi_6(x)$ must be mapped into itself by all those automorphisms, because those automorphisms make up a group. For example,
$$\Phi_2(N(x))=\Phi_2(I(x))\Phi_2(\Phi_2(x))\Phi_2(\Phi_3(x))\Phi_2(\Phi_6(x))=\Phi_2(x)I(x)\Phi_6(x)\Phi_3(x)=N(x)$$
And similar for $\Phi_3$ and $\Phi_6$.
Thus, as the Galois theory teaches us, $N(x)$ belongs to the field fixed by all the automorphisms, which coincides with $\mathbb Q$. In other words, $N(x)$ is always rational.
Moral: if $x=a+b\sqrt 2+c\sqrt 3+d\sqrt 6$, with $a,b,c,d$ rational: to find $\frac{1}{x}$, multiply both the numerator and denominator by $\Phi_2(x)\Phi_3(x)\Phi_6(x)=(a-b\sqrt 2+c\sqrt 3-d\sqrt 6)(a+b\sqrt 2-c\sqrt 3-d\sqrt 6)(a-b\sqrt 2-c\sqrt 3+d\sqrt 6)$.
Edited to add: Wolfram Alpha has calculated for me that you will end up with the following in the denominator:
$$a^4 - 4 a^2 b^2 - 6 a^2 c^2 - 12 a^2 d^2 + 48 a b c d + 4 b^4 - 12 b^2 c^2 - 24 b^2 d^2 + 9 c^4 - 36 c^2 d^2 + 36 d^4$$
(yep, I know, it's horrid, but it is rational!)
| {
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"url": "https://math.stackexchange.com/questions/2639342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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One-to-One function? In text it states that a one to one function is “A function f is one to one if for any two range values f(u) and f(v), f(u)=f(v) implies that u=v. What exactly does this mean? I thought if there are two equal y values it is NOT a one to one function?
| One to one means: For every output there is exactly one input.
Or in other words, If two times you got the same out put, then you must have had the same input.
Or in other words. If you got $f(u) = f(v)$ that means $u = v$.
=====
Or... one to one means. If $u \ne v$ then $f(u) \ne f(v)$.... so if you ever DO find yourself with $f(u) = f(v)$ the only way that can possibly happen is if $u = v$.
Basically the two statements:
i) If $u \ne v \implies f(u) \ne f(v)$ and
ii) If $f(u) = f(v) \implies u = v$
are both equivalent. ii) is just the contrapositive of i).
"Cheap food is not good" is the contrapositive of "Good food is not cheap". THey both mean exactly the same thing.
======
To do an example.
Suppose $f(x) = x^3$ we'll take it for granted that we know $f$ is one-to-one.
Suppose $f(u) = f(v) = 8$. Then what is $u$ and what is $v$?
Well $v^3 = 8$ so $v =\sqrt[3]8 = 2$. And $u^3 = 8$ so $u = \sqrt[3]8 = 2$ and .... what do you know! $u = v$! What were the odds!
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that completely regular is a topological property Show that completely regular is a topological property.
Let $ X $ be a completely regular space and let $ h:X \rightarrow Y $ a homeomorphism. We will prove that $ Y $ is completely regular.
My Attempt
Suppose that $ C $ is a closed subset of $ Y $ and that $ y \in Y \setminus C $ is an arbitrary point. Then, there exists $ x \in X $ such that $ h(x) = y $ and $ x \not\in h^{-1}(C) $. Since $ h $ is continuous, $ h^{-1}(C) $ is closed in $ X $.
Since $ X $ is completely regular, there is a continuous function $ f $ such that $ f(x) = f(h^{-1}(y)) = 0 $ and $ f(h^{-1}(C)) = 1 $. Thus, $ Y $ is completely regular.
I'm not sure of the proof, mainly of the last two lines.
| To prove $Y$ is completely regular, you need to prove there exists a continuous function $g:Y\to [0,1]$ such such that $g(y)=0$ and $g(c)=1$ for all $c\in C$. You haven't actually exhibited such a function, so you haven't proved that $Y$ is completely regular. However, you've done most of the work needed to do so. Based on what you have, what could you define $g$ as?
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the derivative of $(2x)^{4x}$? This is quite simple. I know. I am having a problem when comparing my answer to online calculators like Symbolab and such.
$$\begin{array}{rll} y &= **(2x)^{4x}** & \text{equation}\\
\ln(y) &= \ln((2x)^{4x}) &\text{ take ln of both sides to bring 4x out front}\\
\ln(y) &= 4x \ln(2x) & \text{ use log property}\\
1/y * y' &= 4\ln(2x) + 4x (1/2x) 2 & \text{ use product and chain rule}\\
y' &= y ( 4\ln(2x) + 4x (1/2x) 2 ) &\text{ multiply both sides by y}\\
y' &= 2x^{4x}( 4\ln(2x) + 4) & \text{simplify }4x*2 = 8x / 2x = 4 \\
y' &= 8x^{4x}\ln(2x) + 8x^{4x} &\text{further simplify}\\
y' &= 8x^{4x}(\ln(2x) + 1).&
\end{array}$$
However, the answer on symbolab gives $8x^{4x}(\ln(x) + 1)$. <--- Disregard that. That was based on my incorrect input of the first line.
Am I wrong? If so, how?
| Given,
$$ f(x) = y = (2x)^{4x}$$
Taking natural log on both sides,
$$ \ln y = 4x \ln 2x$$
Now differentiating w.r.t. x,
$$ \frac{1}{y}\frac{dy}{dx} = 4x.\frac{1}{2x}.2 + ln 2x . 4$$
$$ \frac{dy}{dx} = y \,(4+4\ln 2x)$$
$$ = 4(2x)^{4x}(1+\ln 2x)$$
$$ = 2^{(4x+2)} . x^{4x} (1+ \ln 2x) $$
That's even the answer given in symbolab. :)
| {
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"timestamp": "2023-03-29T00:00:00",
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How to study the strong and weak convergence of this sequence of functions? Discuss the strong and weak convergence of the sequence
$$u_n(x)=\sin(x)+\frac{1}{n}\sin^2(nx)$$
in the Sobolev space $W^{1,2}(0,1)$.
I know that a function $u(x)$ belong to $W^{1,p}(a,b)$ iff the $L^p$ norms of $u(x)$ and $u'(x)$ are finite. In particular we can express the norm in a Sobolev space in this way
$$||{u}||_{W^{1,p}}=||u||_{L^p}+||u'||_{L^p}$$
where
$$||u||_{L^p}^p=\int_\Omega|u|^pd\omega$$
Since the following integrals are hard to compute i think there is another way to discuss this problem.
$$||u_n||_{L^2}^2=\int_0^1|\sin(x)+\frac{1}{n}\sin^2(nx)|^2dx$$
$$||u'_n||_{L^2}^2=\int_0^1|\cos(x)+2\cos(nx)\sin(nx)|^2dx$$
Maybe could be helpfull knowing that
$$\lim_{n\rightarrow\infty}u_n(x)=\sin(x)$$
Could someone help me?
| The $L^2$ norm of $u_n(x)$ is not so hard to deal with. Remember, we just have to show that it is finite:
\begin{align}
\int_0 ^1\left|\sin(x)+\frac{1}{n}\sin^2(nx)\right|^2\;dx &\leq \int_0^1\left(|\sin(x)| + \frac{1}{n}\sin^2(nx)\right)^2\;dx\\
&=\int_0^1\sin^2(x)+\frac{2}{n}|\sin^3(x)|+\frac{1}{n^2}\sin^4(x)\;dx \\
&\leq \int_0^1 1+\frac{2}{n}+\frac{1}{n^2}\;dx \\
&=1+\frac{2}{n}+\frac{1}{n^2} \\
&< \infty
\end{align}
So $u_n(x)\in L^2((0,1))$. You can do similarly for $u_n'(x)$. Now we look at weak convergence in $L^2$. You made the guess $u_n\rightharpoonup \sin(x)$. We have to show that:
$$\lim_{n\rightarrow\infty}\int_0^1 [u_n(x) - u(x)]\phi(x)\;dx=0,\qquad\forall\phi\in L^2$$
We can show this by finding an upper bound on the magnitude of this integral:
\begin{align}
\left|\int_0^1 [u_n(x) - u(x)]\phi(x)\;dx\right| &=\left|\int_0^1 \frac{1}{n}\sin^2(nx)\phi(x)\;dx\right|\\
&\leq\frac{1}{n}\int_0^1\sin^2(x)\left|\phi(x)\right|\;dx\\
&\leq\frac{1}{n}\int_0^1\left|\phi(x)\right|\;dx \\
&= \frac{M}{n}
\end{align}
where $M = \int_0^1|\phi(x)|\;dx <\infty$ because $\phi\in L^2$. Taking the limit, we obtain:
$$\lim_{n\rightarrow\infty}\left|\int_0^1 [u_n(x) - u(x)]\phi(x)\;dx\right|= 0$$
For weak convergence of $u_n'$, we make use of the fact that step functions are dense in $L^2$, so that for any $\phi(x)\in L^2$, there exists a sequence of step functions $\phi_\nu(x)$ that converges strongly to $\phi(x)$ in $L^2$. This allows us to just consider the integral of $[u_n'-u']\phi_\nu$ on each interval $[c_i,c_{i+1})$ on which $\phi_\nu$ is constant:
\begin{align}
\int_{c_i}^{c_i+1}[u_n'(x) - u'(x)]\phi(x)\;dx &= \int_{c_i}^{c_i+1} \sin(2nx)\phi(x)\;dx\\
&=\int_{c_i}^{c_i+1} \sin(2nx)\phi_\nu(x)\;dx + \int_{c_i}^{c_i+1}\sin(2nx)(\phi(x) - \phi_\nu(x))\;dx \\
&=\phi_\nu(c_i)\int_{c_i}^{c_i+1}\sin(2nx)\;dx+\int_{c_i}^{c_i+1}\sin(2nx)(\phi(x) - \phi_\nu(x))\;dx \\
&= \phi_\nu(c_i)\frac{\cos(2nc_{i}) - \cos(2nc_{i+1})}{2n} +\int_{c_i}^{c_i+1}\sin(2nx)(\phi(x) - \phi_\nu(x))\;dx
\end{align}
Strong convergence of $\phi_\nu$ to $\phi$ means we can make the second integral arbitrarily close to zero. Taking the limit as $n\rightarrow \infty$:
$$\lim_{n\rightarrow \infty}\int_{c_i}^{c_i+1}[u_n'(x) - u'(x)]\phi(x)\;dx =0$$
So $u_n\rightharpoonup \sin(x)$ in $W^{1,2}$. Can you deal with strong convergence?
| {
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How many numbers from $1$ to $99999$ have the sum of the digits $= 15$? The problem:
How many numbers from $1$ to $99999$ have the sum of the digits $= 15$?
I thought of using the bitstring method and $x_1 + x_2 + x_3 + x_4 + x_5$ will be the boxes therefor we have $4$ zeros and $15$ balls. I'd say the answer would have to be $\binom{19}{15}$ is it right?
| Your answer is incorrect since you have not considered the restriction that a digit in the decimal system cannot exceed $9$.
We want to find the number of positive integers between $1$ and $99~999$ inclusive that have digit sum $15$. Since $0$ does not have digit sum $15$, we get the same answer by considering nonnegative numbers less than or equal to $99~999$ with digit sum $15$.
A nonnegative number with fewer than five digits such as $437$ can be viewed as a string of five digits by appending leading zeros. In this case, $437$ can be represented as the string $00437$. Hence, we can view the problem as finding the number of five-digit decimal strings with digit sum $15$. Hence, we seek the number of solutions in the nonnegative integers of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 15 \tag{1}$$
subject to the restrictions that $x_j \leq 9$ for $1 \leq j \leq 5$.
As you determined, the number of solutions of equation 1 is
$$\binom{15 + 5 - 1}{5 - 1} = \binom{19}{4} = \binom{19}{15}$$
From these, we must subtract those solutions in which one or more of the variables exceeds $9$. Since $2 \cdot 10 = 20 > 15$, at most one of the variables can exceed $9$.
Suppose $x_1 > 9$. Then $x_1' = x_1 - 10$ is a nonnegative integer. Substituting $x_1' + 10$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 10 + x_2 + x_3 + x_4 + x_5 & = 15\\
x_1' + x_2 + x_3 + x_4 + x_5 & = 5 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{5 + 5 - 1}{5 - 1} = \binom{9}{4}$$
solution. By symmetry, there are $\binom{9}{4}$ solutions that violate the restrictions for each of the five variables that could exceed $9$. Hence, the number of positive integers less than or equal to $99~999$ with digit sum $15$ is
$$\binom{19}{4} - \binom{5}{1}\binom{9}{4}$$
| {
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Polynomials: gaining irreducibility by adding a constant EDITED:
Let $f\in\mathbb{Q}[X]$. I'm interested in the factoring properties of the family $F:=\{f+c\}_{c\in\mathbb{Q}}$ of rational polynomials that differ from $f$ only by a constant. Specifically:
1) I think there must be at least one irreducible polynomial in $F$. How can we prove it?
2) If $f$ is reducible, can we effectively determine an specific $c$ such that $f+c$ is irreducible?
3) How are the asymptotic growth and the density of those $c$ such that $f+c$ is irreducible? Are there at least an infinite number of them? How does this depend on the coefficients of the polynomial?
E.g, if $f(X):=X^2$ then any $c>0$ makes $f(X)+c$ irreducible, but it looks like $X^3+3X$ can be made reducible by infinite (ever sparser) values of $c$.
4) If $f\in\mathbb{Z}[X]$, can something be said about the residue distribution of the $c\in\mathbb{Z}$ modulo any prime?
5) How do the results generalize to other fields? What happens with degree 2 polynomials in real closed fields? What happens exactly in fields of characteristic $p$?
| The case of degree $2$ polynomials in $\mathbb{R}$ or $\mathbb{Q}$ can be handled. Let $f(x) = ax^2 + bx$. By the quadratic formula, if $b^2 - 4ac < 0$, then we have $f(x) + c$ irreducible.
| {
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How to Find the number of tangents to the curve y=$f\left(x\right)$ parallel to line $x+y=0$
Question For $x$$>0,$ let
$f\left(x\right)=\int_{1}^{x}\left(\sqrt{\log t}-\frac{1}{2}\log\sqrt{t}\right)dt$
The number of tangents to the curve y=$f\left(x\right)$parallel to
line x+y=0 is _________________
MY approach $x+y=0\Longrightarrow\frac{dy}{dx}=-1$
$f\left(x\right)=\int_{1}^{x}\left(\sqrt{\log t}-\frac{1}{2}\log\sqrt{t}\right)dt$$\Longrightarrow f'\left(x\right)=$$\sqrt{\log x}-\frac{1}{2}\log\sqrt{x}$
I know that parallel lines have same slope,but it would not give me
number of tangents.
| Hint: Finding the number of tangents of $f(x)$ parallel to line $y=-x$ is equivalent to finding the number of roots of the equation $f'(x)=-1$.
| {
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} |
Is the set of terms of a sequence countable? According to Rosen, an infinite set A is countable if $|A|= |\mathbb{Z}^+|$ which in turn can be established by finding a bijection from A to $\mathbb{Z}^+$.
Also, a sequence is defined as a function from $\mathbb{Z}^+$ (or $\{0\} \cup \mathbb{Z}^+$) to some set.
With the above, a sequence is certainly enumerable. However, it need not be a bijection, e.g. Fibonacci(1) = Fibonacci(2) = 1.
This implies that not every sequence is countable which seems counterintuitive. Are there any results in this regard? Is there a mistake in the reasoning above?
| Every sequence has a countable or a finite set of values.
Besides, you are mixing two ideas : a sequence $(u_n)_n$ is a function $n\mapsto u_n\in F$ ($F$ being any possible set) and almost never a bijection, but the set of all its values are finite or countable.
| {
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What is the automorphism group $Aut_{\mathbb{Q}} \mathbb{Q}(\sqrt{3}+\sqrt{2})$? I am trying to understand the automorphism group $Aut_{\mathbb{Q}} \mathbb{Q}(\sqrt{3}+\sqrt{2})$, aka the the automorphism group of $\mathbb{Q}(\sqrt{3}+\sqrt{2})$ over the rationals $\mathbb{Q}$. I know the following:
*
*$\mathbb{Q}(\sqrt{3}+\sqrt{2}) =\mathbb{Q}(\sqrt{3},\sqrt{2})$
*the minimal Polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $x^2-2$ and the minimal Polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})$ is $x^2-3$, hence the degree of $\mathbb{Q}(\sqrt{3}+\sqrt{2}) =\mathbb{Q}(\sqrt{3},\sqrt{2})$ over the rationals is 4.
*the minimal polynomial of $\mathbb{Q}(\sqrt{3}+\sqrt{2})$ is $x^4-10x^2+1$.
*The automorphism group of $\mathbb{Q}(\sqrt{3},\sqrt{2})$ has $4$ elements, namely the obvious ones. (positive root to negative root, all $4$ possibilities.)
Now my questions: Does $Aut_{\mathbb{Q}} \mathbb{Q}(\sqrt{3}+\sqrt{2})$ also have four elements? It must have, right? But what are these elements? how do I find four of them? I either have all 4 roots in $\mathbb{Q}(\sqrt{3}+\sqrt{2})$in which case I'd have $4!$ automorphism, or $3$ roots giving me $3!$ automorphisms, and so on. What am I missing here?
| What you are missing is the fact that a field automorphism for an algebraic extension always permutes the zeros of the minimal polynomial - but the permutations allowed are not arbitrary, they are those which always yield consistent results when applied to ANY sum or product of field elements. For example, in $\mathbb{Q}(\sqrt{3},\sqrt{2})$, consider the product expression (1 + $\sqrt{2}$) * $\sqrt{2}$. If your permutation maps from $\sqrt{2}$ to -$\sqrt{2}$, then you may apply this map to both sides of the product before multiplying, OR multiply and apply the map to the product, and the result will be the same. This is definitely not true if your permutation maps from $\sqrt{2}$ to $\sqrt{3}$, for example; therefore that permutation does NOT extend to a field automorphism.
For your extension the four roots of the minimal polynomial are $\sqrt{3}+\sqrt{2}$, $\sqrt{3}-\sqrt{2}$, $-\sqrt{3}+\sqrt{2}$, and $-\sqrt{3}-\sqrt{2}$. What permutations of these four values induce field automorphisms? Once you know that, there's your group.
| {
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Determine all pairs of positive integers $(m,n)$ that satisfies $m!+n!=m^n$ Determine all pairs of positive integers $(m,n)$ that satisfies $m!+n!=m^n$
I found easily the pairs $(2,2)$ and $(2,3)$ but i can't prove that these are the only pairs possibles.
Any hints?
| This is not quite a full answer but it has the main ideas you'll need.
First, let us get the ability to assert a lower bound on $m$ after checking only finitely many pairs. This can be done by noting that $m^n>n!$, so by Stirling's approximation, $n<em$. Thus we can divide the problem between $m<m^*$ and $m \geq m^*$ by directly checking the cases $m=1,2,\dots,m^*-1$ and $n=1,2,\dots,\lfloor em \rfloor$.
Second, let us see what happens if $m$ is large. Similar to the above, if we assume $m>e$, then Stirling's approximation applied to $m^n>m!$ implies
$$m \leq \frac{n}{1-1/\ln(m)}.$$
So it suffices to provide an upper bound on $n$ in order to finish this case.
Third, let $r=\min \{ m,n \}$. Then $r!$ divides $m^n$ and so $r\#$ divides $m$, where $\#$ is the primorial function. Thus in particular $r\# \leq m$.
So there are two cases: $r=m$ and $r=n$. Consider $r=m$. By Bertand's postulate, if $m>2$ then there is a prime $p$ between $m/2$ and $m$, and so $m\# \geq 2p>m$. So $r=m$ can only occur for $m \leq 2$.
Now consider $r=n$. In this case $n\# \leq m$. If we assume $m \geq 8$, then $1-1/\ln(m) \leq 2$, so for $m \geq 8$ the inequality above gives
$$n\# \leq 2n.$$
Applying Bertrand's postulate twice in the same manner as the previous case implies that $n\# \geq n^2/8$. Thus $n \leq 16$.
Putting the pieces together, we divide the problem between $m<8$ and $m \geq 8$. In the case $m<8$ we have $n \leq \lfloor em \rfloor$, which is on the order of 100 cases to check. In the case $m \geq 8$ we have $n \leq 16$ and $m \leq 2n$, which is again on the order of 100 cases to check.
| {
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Are imaginary numbers really incomparable? If we really don't know which is bigger if $ i $ is greater or $ 2i $ or so on then why do we plot $ i $ first then $ 2i $ and so on, on the imaginary axis of the Argand plane? My teacher said that imaginary numbers are just points and all are dimensionless so they are incomparable and the distance really doesn't matter. I want to get this more clear
| I've come to think of i as a rotation operator, so ordering may not make any sense. But these are the thoughts of an old man, and I'm not sure there is any basis to this? This probably ought to be a comment, but you can't comment on a question until you have a reputation level of 50+.
| {
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All prime ideals of $\mathbf{R}[x,y]$ and $\mathbf{C}[x, y]$. What are the all the prime ideals of $\mathbf{R}[x,y]$ and $\mathbf{C}[x, y]$? (Also, how do you prove that you've found all of them?) I'm trying to understand what the $\mathbf{R}$-algebraic vs $\mathbf{C}$-algebraic subsets of $\mathbf{C}^2$ are, defined as the zeros (in $\mathbf{C}^2$) of an ideal in $\mathbf{R}[x, y]$ and $\mathbf{C}[x,y]$, respectively.
Edit: I'm not that familiar with dimension theory, so I'd prefer to see an argument that does not rely on it.
| Here's an answer for $\mathbf C[x,y]$: Hilbert's Nullstellensatz, asserts the maximal ideals have the form $(x-\alpha,y-\beta)$ for some $\;\alpha, \beta\in \mathbf C$.
On the other hand, $\mathbf C[x,y]$ is a U.F.D. of (Krull) dimension $2$. So the prime ideals of height $1$ are principal, generated by irreducible polynomials.
Add to this list the only ideal of height $0$, and you've finished.
| {
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$s + \frac{1}{n}$ is an upper bound for $A$ and $s - \frac{1}{n}$ is not an upper bound for $A$. Show $s = \sup A.$
Let $A \subset \mathbb R$ be non empty and bounded above, and let $s \in \mathbb R$ have the property that for all $n \in \mathbb N$, $s + \frac{1}{n}$ is an upper bound for $A$ and $s - \frac{1}{n}$ is not an upper bound for $A$. Show $s = \sup A.$
I am having difficulty proving this statement. It is intuitively clear to me that it holds true but have no idea where to begin proving this. Thanks in advance for any help.
| Pure definitions.
The sup must exist as the reals have the least upper bound property.
$s - \frac 1n$ is not an upper bound for all $n$ so $\sup A > s + \frac 1n$ or all natural $n$.
All $s + \frac 1n$ is an upper bound so $\sup A \le s+ \frac 1n$ for all natural $n$.
So $s-\frac 1n < \sup A \le s+ \frac 1n$ for all natural $n$.
Now either $\sup A < s$ or $\sup A = s$ or $\sup A > s$.
If $\sup A < s$ then $s - \sup A > 0$ and there is an $n$ in $\mathbb N$ so that $\frac 1n < s - \sup A$ an $s < s - \frac 1n$ which is a contradiction.
If $\sup A > s$ then $\sup A - s > 0$ and there is an $n$ so that $\frac 1n < \sup A -s$ and $s > \sup a+\frac 1n$ which is a contradiction.
So $\sup A = s$.
Of course you may need to review why it is true that for all $x > 0$ theren is a natural $n$ so that $0 < \frac 1n < x$....
| {
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What does it mean for a function to be continuous on its domain? I never understood the phrase "continuous on its domain."
Isn't everything continuous on its own domain, since the domain are all the $x$ values that we can plug into $f(x)$ and get a defined $y$ value back? i.e. doesn't the domain by definition tell you where the function is continuous? Why would the domain ever include something not continuous / not defined?
| A function is said to be continuous if it continues at each point. This means that over the domain. Functions that are not continuous do not exist for every x value over the domain. For example if a function is defined near an open interval (the circle that is not shaded on a graph) then the function is discontinuous. However, if a function is defined near a closed circle (the shaded circle of a graph) then that function is continuous. This problem usually occurs when trying to find the Domain and Range of a function or if a problem ask to graph the Domain and Range of a function.
| {
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The number of continuous functions $f:[0,1]\to\mathbb R$ that satisfy $\int_0^1xf(x)\,dx=\frac13+\frac14\int_0^1(f(x))^2\,dx$
90) The number of continuous functions $f:[0,1]\to\mathbb R$ that satisfy
$$\int_0^1xf(x)\,dx=\frac13+\frac14\int_0^1(f(x))^2\,dx$$
is
A) 0
B) 1
C) 2
D) $\infty$
How to approach this sum? I thought of using Newton-Leibniz but the limits are constants, so that approach failed.
| Another way:
$$\int^{1}_{0}4xf(x)dx-\int^{1}_{0}(f(x))^2dx=\frac{4}{3}$$
$$\int^{1}_{0}f(x)\bigg(4x-f(x)\bigg)dx\leq \frac{1}{4}\int^{1}_{0}\bigg[f(x)+4x-f(x)\bigg]^2dx=\frac{4}{3}.$$
Equality hold when $f(x)=2x$
In $2$ line earlilier i have used the inequality $$ab\leq \frac{(a+b)^2}{4}$$ equality hold when $a=b.$
| {
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permutation of 5 digit numbers divisible by 3 "The total number of possible combination of 5 digits numbers formed from the digits(0,1,2,3,4,5,6,7,8,9) which are divisible by 3?"
This was the question given to me by my mathematics teacher during out permutation and combination lessons;I was able to solve this with ease but later I was thought of modifing the problem a little bit to
"The total number of possible combination of 5 digits numbers formed from the digits(0,1,2,3,4,5,6,7,8,9) which are divisible by 3 without repetation of any digits? eg 12345 not 33120"
I asked my teacher about the solution of the problem because i was unable to solve it with certain accuracy, he was unable to give me a satisfactory answer.
Can anyone help me in solving this problem
thank you.
| Here is a correct solution along the lines of Manthanein's answer:
There are $9\cdot8\cdot7\cdot 6 \cdot5=15\,120$ strings of length $5$ not containing a repeat or zero. Adding $1$ mod $9$ to each digit in such a string changes its sum by $2$ mod $3$. From this we can conclude that exactly one third of these strings have a sum which is divisible by $3$. It follows that there are 5040 valid strings of this kind.
Similarly, there are $9\cdot8\cdot7\cdot 6=3024$ strings of length $4$ not containing a repeat or zero, and exactly one third of these strings have a sum which is divisible by $3$. Given such a string we can insert a $0$ at four different places in order to obtain a valid string of length $5$. It follows that there are ${3024\cdot 4\over3}=4032$ valid strings of this kind.
In all there are $5040+4032=9072$ valid strings of length $5$.
| {
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Kids in wagons probability $11$ kids get on a train with $3$ wagons. What's the probability that in the $1$st wagon there are exactly $3$ kids? Isn't this the same with saying $x_1+x_2+x_3=11$ and $x_1=3$? If yes, how could one solve this?
| Well, yes, it's the same, but neither question has enough information to answer it. You don't give the underlying probability distribution, so we can't evaluate the probability.
The "natural" choice of underlying distribution might be different depending on your two presentations of the question. In the $x_1 + x_2 + x_3 =11$ presentation, it might be natural to stipulate that all valid choices of $(x_1,x_2,x_3)$ are equally likely. In the children presentation, it might be natural to stipulate that for each child, getting onto any of the three trains is equally likely -- that's a different underlying distribution.
Assuming that that is the distribution in question, the children problem can be solved by noting that the probability of any specific setup in which 3 children are on the first train and 8 are on one of the other two is $\left(\frac13\right)^3\left(\frac23\right)^8$, and then multiplying this by the number of ways we can have 3 children on the first train, which is equivalent to the number of ways to choose 3 from 11, $\binom{11}3$.
| {
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Is it possible to "re-normalize" a Dirac delta function? The delta function in spherical coordinates is given by:
$$\delta(\vec{r}_0-\vec{r})=\frac{1}{r^2}\delta(r_0-r)\delta(\cos\theta_0-\cos\theta)\delta(\phi_0-\phi),$$
(The ordering of the coordinates inside the $\delta$'s isn't important). If I have a particular location in mind, say $(r_0,\theta_0,\phi_0)=(r_0,0,0)$, is there a neat way to "re-normalize" the following:
$$\cot\theta_0\left(\frac{1}{r^2}\delta(r_0-r)\delta(\cos\theta_0-\cos\theta)\delta(\phi_0-\phi)\right),$$
since $\cot\theta_0$ is also singular at $\theta_0\in \pi \mathbb{Z}$.
To be clear, I'm trying to see if there is a way to cleverly absorb this cotangent function into the delta function so that it remains relatively unchanged. It's also possible this question is entirely unfounded, and I apologize if so.
| It seems OP wants to avoid having poles inside & outside the Dirac delta distribution. Using a hopefully obvious notation
$$s~\equiv~ \sin\theta,\qquad c~\equiv~ \cos\theta,$$
one possibility is to replace
$$ \frac{c}{s} \delta(c-c_0)~=~ \frac{sc}{|f^{\prime}(c)|} \delta(c-c_0)~=~sc~\delta(f(c)-f(c_0)), $$
where $$f(c)~:=~c(1-\frac{c^2}{3}),\qquad f^{\prime}(c) ~=~1-c^2, \qquad c~\in~[-1,1].$$
Of course such rewritings are essentially just cosmetics. The singular nature of the construction does not disappear but manifests itself in other ways.
| {
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Why the hexadecimal numbers can be converted directly into binary numbers so cleanly? Suppose we have F9 hex. If we want to convert it into binary, we just replace the hexadecimal numbers with their corresponding binary numbers. Like 9 has1001 in binary and F has 1111 in binary. By combining it becomes 1111 1001. But why does it give the right answer when we convert both into decimal number. Why we can't convert any base system into another base system which does not come in its power? As we form a base-3 number system and a base-9 number system. As 9=3^2 so two digits of base-3 can be represented by a single base-9 digit. But why this happens?
| Suppose we have a number $X$ which is written as $X_2X_1X_0$ in hexidecimal. Then we have $X = 16^0X_0 + 16^1X_1 + 16^2X_2$.
$16 = 2^4$ so we rewrite that as $X = 2^0X_0 + 2^4X_1 + 2^8X_2$. Now let's suppose $X_2$ can be written $x_3x_2x_1x_0$ in binary, then $X_2 = 2^3x_3 + 2^2x_2 +2^1x_1 +2^0x_0$. Substituting we get
$$
X = 2^0X_0 + 2^4X_1 + 2^8(2^3x_3 + 2^2x_2 +2^1x_1 +2^0x_0)
$$
$$
X = 2^0X_0 + 2^4X_1 + 2^{11}x_3 + 2^10x_2 +2^9x_1 +2^8x_0
$$
If you repeat that process for $X_0$ and $X_1$ you'll get an expression for $X$ entirely in terms of powers of two, just by substituting in the binary expansions of its digits in hexidecimal.
The special property of base 16 that makes this possible is that 16 is a power of 2, which I used to rewrite the powers of 16 as powers of 2. This is equally possible for any other pair of bases where one is a power of the other.
| {
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Basic counting trouble Suppose you have $8$ people. How many possible ways can you seat these people if
a) two persons, A and B, must sit together?
b) $4$ men and $4$ women, but no $2$ men and no $2$ women can sit together
c) there are $5$ men who must all sit together
d) there are $4$ married couples that must sit together
I suppose I'm confused as to what my error is on some of these by doing:
a) This makes it so that we are dividing the seating into a group of $2$ and $6$ groups of $1$, so we have $$\frac{8!}{2!}$$ ways to do this. But also we could view this as only having $7$ degrees of freedom, twice, so we'd have $$2 \cdot 7!$$. Which one is correct and why is the other wrong??
b) This one I do not know how to do.
c) Again, is this not just a group of five and three groups of 1? to get
$$\frac{8!}{5!}?$$ Or is it $4! \cdot 5!$ since we have 4 degrees of freedom and $5!$ distinguishable items?
d) Again, is this
$$\frac{8!}{2^4}$$ or is it $4$ groups of $2$ giving $4!2^4$?? Why do both answers here seem right??
| Are these circular tables or tables with a distinct orientation.
Based on the way you answered the first question, I am going to guess that these tables have an orientation.
a) 1 group of 2 and 6 groups of 1 is 7 groups total.
$2\cdot 7!$
b) You can but a man at the head or you can put a woman a the head. Once you have made this decision there are $4!$ to arrange the men and $4!$ ways to arrange the women.
c) there is a group of 5 and 3 groups of 1. $5!\cdot 4!$
d) 4 groups of 2. $(2!)^44!$
If you descide that this is a circular table. Then for any of the configurations above, you can always rotate the table such that Mr. Jones is at the head.
Divide the results by $8.$
| {
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Piecewise functions in MATLAB. Help! I am trying to plot a pretty repetitive function in MATLAB. The values of $f$ are the same each $y$ for a designated season in the year. We have a function $f(t)$ and:
$$ f(t) = \begin{cases}
500 & \text{ if } 0 \le x < 91 \\
1500 & \text{ if } 91 \le x < 182 \\
500 & \text{ if } 182 \le x < 273 \\
0 & \text{ if } 273 \le x < 365
\end{cases} $$
I am able to plot this for one year using a piecewise function. But I am wondering how to extend this further. What if I want to plot this for a period of ten years, without having to manually input the values of $t$? Is there something that I can use, like an elseif function? I am quite confused on how to do this.
So far, I have been using a piecewise function and manually inputting the values of $t$ is getting very tedious.
For example, I have:
$f = {\tt piecewise}(0 \le t<91,500,91 \le t<182,1500,182 \le t<273,500,273 \le t<365,0,365 \le t<456,500,456 \le t<547,1500,547 \le t<638,500,638 \le t<730,0,730 \le t<821,500,821 \le t<912,1500,912 \le t<1003,500,1003 \le t<1095,0); $
Just wondering if there is an easier way to do this.
Thank you so much!
| It looks like you have 12 segments, so I'd make an input num_segments=12; and an input t_length=92; as finally an array y_result=[500,1500,500,0]; (and so on). Then, in your function, you have, say, a for loop, with i=0:num_segments-1 and then on each iteration, programmatically define t_lengthi<=t(i+1) and extract y_result[i]. Does that help?
I apologize for the bad formatting and if my syntax is wrong on defining arrays. I work in multiple languages and always forget which one uses brackets vs. parentheses for different things.
| {
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