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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!
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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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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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My professor gave an example located on slide 3 of this pdf: can anybody explain to me how he ended up with m_n = 2^(n) - 1. Thanks!
The step is from
mn =2nā1 +2nā2 +...+22 +2+1.
to
mn = 2n ā 1
There are two ways to make the step. One is to recognize this as a geometric series, and know the rule:
sum=(1-rn)/(1-r)
The other is to have played around enough with powers of two to know that if you add up a bunch of them starting from 1, you get the next one, minus one.
There is a formula for the sum of the first n terms of a geometric series.
1 + 2 + 2^2 + 2^3 + ... + 2^{n-1}
= (1 - 2^n) / (1 - 2)
= (1 - 2^n) / (-1)
= 1/(-1) - 2^n/(-1)
= 2^n - 1
It's just one of the relations of series that people have figured out over the years:
2^(n-1) + 2^(n-2) + ... + 2 + 1 == 2^n - 1
You can think of it a lot like the sum of binary numbers:
000001
000010
000100
001000
+ 010000
------
011111 == 1000000 - 1
Actually,
Mn=2^0+2^1+.........+2^(n-1)+2^(n-2)
is the Nth term of the sequence Mk=.....
And this Nth term itself is a sum of a geometrical progression whose 1st term is 1(2^0) and common ratio=2.
And this sum(Mn) is
=a[(r^n)-1]/[r-1]
where a is 1st term and r common ratio
=1*[(2^n)-1]/[2-1]
Mn=2^n - 1
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I have a test coming up, and I need some help with a practise question... Need to prove this by induction:
Reccurence relation: m(i) = m(i-1) + m(i - 3) + 1, i >= 3
Initial conditions: m(0) = 1, m(1) = 2, m(2) = 3
Prove m(i) >= 2^(i/3)
Here is what I have been able to do so far:
Base case: m(3) >= 2 -----> 5 >= 2. Therefore it holds for the base case.
Induction Hypothesis Assume there is a k such that m(k) >= 2^(k/3) holds.
Now I must prove that it holds for k+1.
So we have: m(k+1) >= 2^((k+1)/3)
which equals (by substituting hypothesis):
m(k) + m(k-2) + 1 >= 2^((k+1)/3)
This is where I am stuck. I'm not sure where to go from here. Any help will be appreciated. Thanks guys!
Hints:
Prove that m(k) >= m(k-2). (This is trivial.)
Since m(k+1) = m(k) + m(k-2) + 1, you can replace = with >= to get m(k+1) >= m(k) + m(k-2) + 1.
You can make substitutions on the right-hand side of >=, as long as what you put in is less than or equal to what you take out. Start by using #1 to make a substitution in #2.
Consider your base case: you show that for 3 prior consecutive given values for m(0), m(1), and m(2), that the formula holds for m(4). Then show that m(k+1) formula works if you assume that it's true for 3 prior values m(k), m(k-1), and m(k-2) [this is valid for induction].
By initial condition
m(k+1) = m(k) + m(k-2) + 1
Substitution:
m(k+1) >= 2^(k/3) + 2^((k-2)/3) + 1
Factor the right hand side in terms of 2^((k+1)/3) [HINT: leave the +1 alone] and it should fall out from there.