magguu
  • Member for 8 years, 10 months
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$xy=1 \implies $minimum $x+y=$?
4 votes

The above answer does it totally. However you can visualize this problem graphically. Imagine the three axes $x,y,z$ and the first quadrant part of the curve $xy=1$ drawn on the $xy$ plane. The ...

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composition of continuous functions
3 votes

Yes it is continuous. Compositions of two continuous functions is always continuous. In this case you can see it by the sequential definition of continuity. $$x_n\rightarrow x \Rightarrow f(x_n)\...

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Is this a new expression for $e$?
2 votes

You are right. Use the fact $$\lim_{x\rightarrow\infty}\bigg(1+\frac{1}{x}\bigg)^x = e$$

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A set containing one element is an open set. Why?
1 votes

I am not sure whether you mean a) and b) in general or just in a discrete metric space so let me answer for both cases A) for the discrete metric space it is always true by definition. for a general ...

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Function analytic on a disk maps linear segments to line segments
1 votes

since it is analytic so the complex derivative computed along x-direction and y-direction is the same and is the same as the derivative. you compute the derivative in the x-direction. it is real. so ...

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How to find the smallest enclosing ellipse around two circles?
Accepted answer
1 votes

Join the two centers and extend the line. wherever it cuts the two circles at their outer extremes are the points which are crucial say $A$ and $B$. take the mid point of these two points. for ...

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Tilting a line and a cloud of 3D points around the line
Accepted answer
1 votes

You are looking for a linear (or affine) isometry which maps $q_1$ to $t_1$ and $q_2$ to $t_2$. First map $q_1$ to $t_1$. This gives a translation $x\mapsto x + t_1 - q_1$. call this map $T$. Then ...

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How to approximate using the differential?
0 votes

Go by the meaning of the derivative. It means when you change the parameters by a small amount, the value changes linearly as a function of this change. ie, $$f(x+\Delta x) \approx f(x) + df(x).x$$ ...

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Finding a connecting line between two lines in 3D space, with specific requirements for its length.
Accepted answer
0 votes

PART - 2 I keep the notation as above. To wit, $P,Q$ are the fixed points on $L_1,L_2$. The vectors $v,w$ are the direction vector along $L_1,L_2$ and $P',Q'$ are the points to be found. Define two ...

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Finding a connecting line between two lines in 3D space, with specific requirements for its length.
0 votes

PART 1 You probably want an algorithm to find the answer. I am ignoring the trivial case of parallel lines. Consider the points $P'$ and $Q'$ as moving points along their respective lines. Now ...

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Show that a surjective function from $X$ to $J$ does not exist (with a twist!!)
0 votes

Note that you can write $\mathcal{P}(X)$ as set of all functions from $X$ to $\{0,1\}$ which I use as a standard 2 element set. So let me write $\mathcal{P}(X)$ as $\textrm{Map}(X,\{0,1\})$ If $Y$ ...

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