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I am wondering how to prove (or disprove) that
if $A$ is a matrix $n \times n$ and
$b_1....b_k$ are $k$ vectors in $\mathbb{R}^{n}$
so that $Ab_1, ..., Ab_{k}$ is a set of generators in $\mathbb{R}^{n}$
then so is the family of vectors $b_{1},...,b_{k}$.
Thanks.
In essence, this question is asking if rank(A*B)=n implies rank(B)=n. This is a consequence of
rank(A*B) <= min( rank(A), rank(B) )
and the fact that for reasons of dimension of the spaces involved, rank(A) <= n and rank(B) <= min(k, n), so that the combined chain
n = rank(A*B) <= min( rank(A), rank(B) ) <= min(k, n)
leaves not much wiggle space.
As the question, this answer is off-topic for SO and belongs to math.SE.
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I'm doing Tim Roughgarden's Algorithms course and he has a slide with an integer multiplication algorithm.
Whats the rule that makes 10(n/2)a * 10(n/2)c become 10(n)ac ?
What do you do when multiplying fractional exponents like that?
It's based on the First Index Law, where:
am * an = am + n
in your case, the powers add to give n/2 + n/2 = 2n/2 = n
Base is the same so you just add the power of 10 i.e., (n/2) + (n/2) = n. Then it's basic multiplication 10(n)ac= 10(n)ac.
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How would you convert the constraint |x| >= 2 so that it would work in a Linear Program (in particular, solving using Simplex).
I understand how to convert |x| <= 2 as that would become x <= 2 and -x <= 2
However the same logic does not work when you have a minimum constant.
There is just no way to shoehorn an equation like |x|>=2 into a pure (continuous) LP. You need to formulate x <= -2 OR x >= 2 which is non-convex. This will require a binary variable making the problem a MIP.
One formulation can be:
x >= 2 - delta*M
x <= -2 + (1-delta)*M
delta in {0,1}
where M is judiciously chosen large number. E.g. if -100<=x<=100 then you can choose M=102.
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I cannot find a function which I can use.
It has to have diminishing returns and f(0)=1/2 and when x-->infinity f(x)-->1
Do any of you have a suggestion?
Thank you in advance!
2/pi atan(x + 1)
Simple. 2/pi for being a 1 on infinity and then solving equation: 2/pi atan(x) == 1/2 to get the offset: 1.
Wolfram:
f[x_] := 2/\[Pi] ArcTan[x + 1];
f[0] (* 1/2 *)
Limit[2/\[Pi] ArcTan[x - Tan[1/2] - 1], x -> \[Infinity]] (* 1 *)
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int length = (int) floor( log10 (float) number ) + 1;
My question is essentially a math question: WHY does taking the log10() of a number, flooring that number, adding 1, and then casting it into an int correctly calculate the length of number?
I really want to know the deep mathematical explanation please!
For an integer number that has n digits, it's value is between 10^(n - 1)(included) and 10^n, and so log10(number) is between n - 1(included) and n. Then the function floor cuts down the fractional part, leaves the result as n - 1. Finally, adding 1 to it gives the number of digits.
Consider that a four-digit number x is somewhere between 1000 <= x < 10000. Taking the log base 10 of all three components gives 3.000 <= log(x, 10) < 4.000. Taking the floor (or int) of each component and adding one gives 4 <= int(log(x, 10))+1 <= 4.
Ignoring round-off error, this gives you the number of digits in x.
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In R, I would like to create a function that returns the smallest n such that the n-th repetition of the natural logarithm gives a value smaller than one. Ex.: fun(9182) = 3 because ln(ln(ln(9182))) = 0,793 < 1.
Any suggestions will be appreciated!
logstar<-function(x){if (x<1) 0 else 1 + logstar(log(x))}
#mrip's answer works well for single values. If you'd like a function that works for vectors, you'll want to use ifelse() rather than if:
> logstar <- function(x){ifelse(x<1,0,1 + logstar(ifelse(x<1,x,log(x))))}
> x = c(0.5,1,100,10000,1E8)
> logstar(x)
[1] 0 1 3 3 4
The ifelse() in the recursive call to logstar() prevents log() from generating NaN in some cases.