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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
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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 always have trouble with factorials. Can someone walk me through simplifying this expression?
(x+1)! - 1 + (x+1)(x+1)!
I'm trying to get it to equal to (x+2)! - 1.
It can be solved as :-
(x+1)! - 1 + (x+1)(x+1)!
= (x+1)! + (x+1)(x+1)! - 1
= (x+1)!.{1+(x+1)} - 1
= (x+1)!.{x+2} - 1
= (x+2)! - 1. // since n!.(n+1) = (n+1)!
Hence proved.
Note that (x+1)! = (x+1)*x!
(x+1)! - 1 + (x+1)(x+1)!
= (x+1)!((x+1)+1) - 1
= (x+1)!(x+2) - 1
= (x+2)! - 1
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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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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!