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bogged
bogged, ppl. a. (bɒgd) [f. bog v.1 + -ed.] Plunged or entangled in a bog; bemired. Also fig.1603 B. Jonson Sejanus iv. (1692) 142 Bogg'd in his filthy Lusts. 1854 Hooker Himal. Jrnls. II. xxx. 323 My elephant got bogged in crossing a deep muddy stream. Oxford English Dictionary
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Is gmail running like bogged down trash in Chrome for ...
Is gmail running like bogged down trash in Chrome for anyone else? ... It runs perfectly and fast in Microsoft Edge, however every time in the ... www.reddit.com
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Derivative Notation Including Chain Rule While attempting this problem with derivatives, I was bogged down in the notation and was unsure as to what was being derived. The problem goes as follows: Given $f(x)=x^3$, ev...
Let $g(x)=x^3$, then $f(x^3)=f(g(x))$. Then use the chain rule $$(f(g(x)))' = f'(g(x))g'(x)$$ So we just need $f'(x)$ and $g'(x)$. $f'(x) = 3x^2$ and $g'(x) = 3x^2$ as well (because $f$ and $g$ happen to be the same function in this case). So then $$f'(g(x))g'(x) = f'(x^3)(3x^2) = (3(x^3)^2)(3x^2) =...
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If $a$, $b$ are rotations of $S^2$ through the same angle, show that they are conjugate in $SO(3)$. Basically everything to do with the question is in the title! I'm just not really sure where to begin with this quest...
Embed $\mathbf{S}^2$ in $\mathbf{R}^3$. For each rotation has a unique vector line on which it acts as the identity, and acts as a plane rotation on the orthogonal plane to this vector line, take both such vector lines for your two rotations of same angle, and take both associated orthogonal planes....
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Mitigate connection between two wireless access points I was wondering if there was a way to connect a single computer to two different access points. I'm on a network with a lot of users and the usage gets bogged dow...
Your WiFi NIC can only support being connected to 1 access point (as far as I know), irregardless of how many antennas it has connected to it. So you'd need multiple WiFi NICs. If you did have multiple NICs then you could take a look at this U&L Q&A titled: Using multiple NICs for faster internet?, ...
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Number of $x^3$ required to equal $2002^{2002}$. While working on math problems, I came upon a high-power summing problem, and got bogged down. The problem is as follows: What is the smallest positive integer $t$ such...
$2002^{2002} =(2002^{667})^3. (10^3+10^3+1^3+1^3) $ $=(2002^{667}×10)^3+(2002^{667}×10)^3+(2002^{667}×1)^3+(2002^{667}×1)^3$ It seems $t=4$ is minimum as 2002 can be expressed as a minimum of 4 cubes.
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Meaning of function mapping notation $d: X \times X \to \mathscr{R}$ I am trying to do some self-studying and am consistently getting bogged down by sophisticated mathematical notation. For example, something simple l...
The notation means that $d$ is a function that takes a pair of numbers and maps them to the reals. The pair of numbers could be thought of as an ordered pair $(x,y)$ where $x \in X$ and $y \in X$, and the tuple $(x,y)$ is in the Cartesian product of $X$ with itself: $X \times X$.
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Simple statement about the cantor set. Struggling to understand the cantor set i found this. It's really helpful but at the very start when the cantor set is defined it is stated that: "$A_k$ is a union of $2^k$ c...
You do that by induction on $k$. For $k=0$, we have $[0,1]$. Suppose that for $k$ it holds, then $A_k$ is the union of $2^k$ closed intervals, each of length $3^{-k}$. Then $A_{k+1}$ is the set generated by removing the (open) middle third of each of these intervals, the remainder is two (closed) in...
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美国从建国以来一共参与发起了多少场战争？
The U.S. then got bogged down in Asia in vicious wars in Korea (1950-1953) and wars in Vietnam (1964-1972) before venturing into the desert sands of Iraq zhihu
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If $Ax=0$ has infinite solutions, can $Ax=b\not=0$ have exactly one solution > Let $A\in\mathbb{R}^{4\times 2}$ and $b\in\mathbb{R}^4$ with $b\not=0$. > > Is it possible that: > > $Ax=b$ has exactly one solution AND...
No. Take $x_1$ to be a solution of $Ax = b$ and $x_2 \neq 0$ to be a solution of $Ax = 0$. Then, $x_3 = x_1 + x_2$ is also a solution of $Ax = b$, since $$Ax_3 = A(x_1+x_2) = Ax_1 + Ax_2 = b + 0 = b.$$
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Finding the Zeros of a Matrix-Vector Equation Here's another matrix algebra question, sorry if I'm coming at these incorrectly, but this kind of thing really isn't my forte :( Lets say we have the equation: $0 = -2...
The equation as you've written it does not have a single solution. There are $\frac{1}{2}n(n+1)$ degrees of freedom in an $n\times n$ matrix, and your equation, being a scalar equation, only places one constraint on the matrix. There is a unique solution only when $\frac{1}{2}n(n+1) = 1$; and you ca...
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How can I show that the interior of the set of interior points is an interior set? $(X^\circ)^\circ = X^\circ$ I am trying to show that the interior of a set of interior points is an interior point, that is, if $X$ is...
One way to see this for a metric space, is to use that every point of an open ball $B(x,r) = \\{y: d(x,y) 0$, and for any $z$, if $d(y,z) 0$ such that $B(x,r) \subset X$. For any $y \in B(x,r)$, there exists $s>0$ such that $B(y,s) \subset B(x,r) \subset X$, see above. But this just says that every ...
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How to prove $\ln{\frac{n+1}{n}}\le\frac{2}{n+1},\forall n\in\mathbb{N}^+$? I found an inequality: $$\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1},\forall n\in\mathbb{N}^+.$$ I tried induction. It is obvious if $k=1$...
Take $x=\frac1n$ so you will have $$\ln(1+x)\leq \frac{2x}{x+1}$$ now suppose $f(x)=\ln(1+x)- \frac{2x}{x+1}\\\x \in [0,1]$ $$f'(x)=\frac{1}{x+1}-\frac{2}{(x+1)^2}=\frac{1+x-2}{(x+1)^2} \to x=1$$ so $f'(x) \leq 0 \text { when x is } \in [0,1]$ so $f(x)$ is decreasing in this interval so $$f(x)\leq f...
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Difference of nth roots vs nth root of difference Show that for $n \ge 1, x>y \ge 0$ the following inequality holds: $x^{1\over n}-y^{\frac{1}{n}} \le (x-y)^{\frac{1}{n}}$. I have tried comparing the $n$-th powers of ...
With $t:=(y/x)^{1/n}\in[0,1)$, you can rewrite $$1-t\le(1-t^n)^{1/n}$$ or $$(1-t)^n\le1-t^n.$$ This is obvious by $$(1-t)^n\le1-t\le1-t^n.$$
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Real Analysis Beginning Proof... Alright, I've been assigned to work through a proof in my RA course and it just has me bogged down at this point. We're trying to show that If $b^2 > c$ then there exists a positive re...
After showing it is bounded above, you get a supremum, call it $b$, exists. What you want is to show neither $b^2c$ can hold, so it must be the case $b^2=c$, that is $\sqrt b=c$. One can always use Archimedianity. The claim is we may take $r=n^{-1}$, $n$ a natural number we'll specify in what follow...
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