Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 6 years ago.
Improve this question
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.
Related
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 6 years ago.
Improve this question
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.
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 8 years ago.
Improve this question
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 *)
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 9 years ago.
Improve this question
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.
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 9 years ago.
Improve this question
limit x-->3 2x^2 + 7x-15/x-3
What i Simplified
Step 1 : 2x^2 + 10x -3x -15 / x-3
Step 2 : 2x( x + 5)-3( x + 5)/x-3
Step 3 : (2x - 3)(x + 5)/x-3
but unable to move further.. im thinking either the question ix wrong or there ix some trick which im unable to understand
thanx in advance
as x -> 3, the numerator goes to 3 * 8 = 24 and the denominator goes to 0, so the limit goes to +infinity if you approach 3 from the right, and -infinity if you approach 3 from the left
since you didn't specify which direction, the limit does not exist.
try graphing it: https://www.desmos.com/calculator
Closed. This question is off-topic. It is not currently accepting answers.
Want to improve this question? Update the question so it's on-topic for Stack Overflow.
Closed 9 years ago.
Improve this question
http://doyourmath.com/web-algebrator/#c=expand_algexpand&v5=2%5E(k%2B1)
Anyone can explain why does expand 2^(k + 1) equal to (2^k) + 1?
That's not actually possible. 2^(k+1) is always going to be an even number. 2^k + 1 is always going to be an odd number.
I think you mean
2^(k+1) = 2^k * 2^1 = 2^k * 2.
One way of looking at it is the associative property of multiplication:
(2 X 3) X 4 = 2 X (3 X 4)
No matter how you group the numbers, the outcome will always be equal. In this case we're dealing with exponents, which is a shorthand notation for multiplying a number by itself.
It is not!!!
2^(k+1) = 2^k * 2 which is greater than 2^k + 1
Instead (k+1)^2 expands to (k^2)+2k+1
http://doyourmath.com/web-algebrator/#c=expand_algexpand&v5=2%5E(k%2B1) has ERRORS!