Limit of a sequence (Calculus 1)

Why is the following statement True ?

limn→+∞ an = L then limn→+∞ 1/n Σk=1n ak = L

If you're using epsilon-delta definitions, here's a proof:

For e>0 there's an N such that for k>N, |ak - L|<e/2.

Take M = |(a1-L) + (a2-L) + .... + (aN-L)|, then for k>N

|a1+...+ak - k*L| <= M + (k-N)e/2

dividing by k

|(a1+....+ak)/k - L| <= M/k + ((k-N)/k)(e/2)

and in the limit k->inf that's e/2, so there's some N' where k>N' implies that's <e.

That means for the given e>0, that N' satifies the limit definition for

lim (a1+...+ak)/k = L

Integer of (1+x²)/abs(1-x²)

If 1 - x² => 0 :

int ( 1 + x² ) / | 1 - x² | dx

= int ( 1 + x² ) / ( 1 - x² ) dx

= int ( x² - 1 + 2 ) / ( 1 - x² ) dx

= int -1 + 2 / ( 1 - x² ) dx

= int -1 + 2 / [(1 - x)(1 + x)] dx

2 / [(1 - x)(1 + x)] = a / (1 - x) + b / (1 + x)

=> 2 = a(1 + x) + b(1 - x) = a + ax + b - bx = (a - b)x + a + b

=> a = b and a + b = 2

=> a = 1 and b = 1

So :

int ( 1 + x² ) / | 1 - x² | dx

= int -1 + 1 / (1 - x) + 1 / (1 + x) dx

= int -1 - (-1) / (1 - x) + 1 / (1 + x) dx

= [-x - ln(|1 - x|) + ln(|1 + x|)]

Can you do 1 - x² < 0 ?

There are 10 spots in a parking lot arranged in a single row. Three cars are parked randomly but in such a way that no two cars are adjacent to each other. What is the probability that there are at least two empty spots between any two cars?

You can view this as counting integer solutions. x1,x2,x3,x4 representing the spacing. x1 is the spots before the first car, x2 between the first and the second, x3 between the second and third, x4 after the third.

For the no two cars adjacent the constraints are: x1,x4>=0, x2,x3>=1, x1+x2+x3+x4=7. For at least two spaces, x1,x4>=0, x2,x3>=2, x1+x2+x3+x4=7.

Then you can use stars-and-bars to count the solutions.

http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29

 

Subtracting 1 from x2,x3 turns x1,x4>=0, x2,x3>=1, x1+x2+x3+x4=7 into x1,x2,x3,x4>=0, x1+x2+x3+x4=5, and there are C(5+3,3) ways.

Subtracting 2 from x2,x3 turns x1,x4>=0, x2,x3>=2, x1+x2+x3+x4=7 into x1,x2,x3,x4>=0,x1+x2+x3+x4=3, and there are C(3+3,3) ways.

C(3+3,3)/C(5+3,3) = 5/14

For the subject Becquerel(Bq) to disintegerate from 11.5Mbq to 4.6 MBq takes 8 hours. How much percent does it disintegerate per hours?And how big is it after a 24 hours if it starts from 11.5 MBq?

Final = Initial*e^rt

4.6 = 11.5e^8r

4.6/11.5 = 2/5 = e^8r

ln(2/5) = ln(2) - ln(5) = 8r

(ln(2) - ln(5))/8 = r ~ -11%/hour.

To find how much is left after 24 hours:

Final = 11.5e^{[24(ln(2)-ln(5))/8]} = 11.5e^{[3(ln(2)-ln(5))]} = 11.5(2/5)³

In the following pseudocode, the statement print questionCounter will be executed ____ times.

num PARTS = 5 num QUESTIONS = 3 partCounter = 1 while partCounter <= PARTS questionCounter = 1 while questionCounter <= QUESTIONS print questionCounter questionCounter = questionCounter + 1 endwhile partCounter = partCounter + 1 endwhile Answer: 15

What is the value of names[4] in the following array? String[] names = {"Jeff", "Dan", "Sally", "Jill", "Allie"};

Allie

What is the value of times[2][1] in the array that follows? double[][] times = { {23.0, 3.5}, {22.4, 3.6}, {21.3, 3.7} };

3.7