We believe in the Fermat. And we believe that the space and the matters come into existence, when the numbers come into existence and we also believe that all cosmic materials and lives change but the number theory cannot change now and forever.
After HHS's interesting "proof" that a sum is true, and Pointless Boy's wonderings about the influence of number theory on life and the universe, I think I'll just post the solution in case we go too off topic.
\binom{2n}{n} = \sum_{k=0}{n} {\binom{n}{k}}^2
First, assume that we have 2n distinct objects and separate them into two groups containing n objects each.
Now, we calculate the number of combinations we can make taking k objects from the first pile and n-k objects from the second. It's clear that the total amount is binom(n)(k)*binom(n)(n-k), which also happens to be binom(n)(k)^2.
Summing all the possibilities from k=0 to k=n, we have taken into account all possible ways to get (n-k)+k = n objects from a group of 2n objects, which is binom(2n)(n).
The equality then follows. QED
1) Calculate the integral of f(x) dx, from 0 to 1, where f(x) is described as:
0 when x is a rational number
1 when x is an irrational number
2) Assume that a company has X working employees, excluding management. For every 10 employees, a boss is needed to supervise that group. For every 10 bosses on that level, another boss is needed.
For example, for X=100, 11 bosses are needed. For X=101, 14 bosses are needed.
What is a good function for calculating how many bosses are needed for any X number of employees?
1) The rationals form a set of measure zero, so f(x)=1 on (0,1) almost everywhere, so its Lebesgue integral is 1 (its Riemann integral is undefined; the Lebesgue integral is a proper extension of the concept of "Riemann integral" because if the latter exists, the former also does and is equal to it).
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
2) For any number X of employees, the number of bosses, B, is:
B(X) = floor(X/10) + floor(X/100)
This is the exact solution. B(X), can easily be calculated by head - the first term is the number you get when you remove the last digit of X, and the second term is the number you get when you remove the last two digits of X. Then you just add these two together.
2) Assume that a company has X working employees, excluding management. For every 10 employees, a boss is needed to supervise that group. For every 10 bosses on that level, another boss is needed.
For example, for X=100, 11 bosses are needed. For X=101, 14 bosses are needed.
What is a good function for calculating how many bosses are needed for any X number of employees?
What? You should show your derivation for that. One of us is confused because f(101)=f(100) unless f(1)=1 (which I don't think is the case, though your statement of the problem is confusing), but even if f(1)=1 then f(101)=12, not 14.
Anyway, I will first state my understanding of your question in a rigorous fashion so you at least know what question I am answering, even if I don't know what question you are asking:
A company has x level 0 employees. Small numbers of employees are able to manage themselves, however, for each 10 employees at level n, an employee of level n+1 is required to manage them. For convenience, denote any employee of level 1 or greater a boss. What is an equation for the number of bosses a company of x level 0 employees requires?
Ok. Assuming f(1)=0, ignoring your claim that f(101)=14, the following will do:
It may not look pretty but it's quite painless to calculate even for huge numbers as the number of summands grows verrrrry slowwwwwwly. Randil's solution is close but not quite right ... maybe ... depending on what you are actually asking.
For X=101, 14 bosses are needed.
What is a good function for calculating how many bosses are needed for any X number of employees?
What? You should show your derivation for that. One of us is confused because f(101)=f(100) unless f(1)=1 (which I don't think is the case, though your statement of the problem is confusing), but even if f(1)=1 then f(101)=12, not 14.
AIUI, given a particular rank N:
If there are 0 or 1 employees at rank N, we need 0 at rank N+1.
Otherwise, we need ceil(N/10) at N+1.
So the derivation is thus:
101 employees of rank 1 need ceil (101/10) = 11 managers
11 managers of rank 2 need ceil (11/10) = 2 managers
2 managers of rank 3 need ceil (2/10) = 1 managers
1 manager of rank 4 doesn't need a manager
total == 14 managers.
The total number of employees can be expressed as the infinite series:
N + floor((N+9)/10) + floor((N+9)/100) + floor((N+9)/1000)...
eventually the rounding error forces the remaining terms to 0. Without the floors, this series is equal to:
-9 + N + 9 + (N+9)/10 + (N+9)/100 + (N+9)/1000 ...
whose limit is -9 + (N+9)*10/9.
We can therefore use (N+9)*10/9-9 as an approximation to our formula, and improve this approximation by studying the systematic error of removing the floor()s. It's difficult to see how to get an exact closed-form formula because of the nonlinearity of the floor function.
1) Thank you arflech.
2) Pointless Boy, I guess I wasn't clear enough on the problem formulation for this, because f(101)=14 and f(1)=1. Randil also seems to have understood the problem in another way than I intended.
Perhaps a better explanation is: every employee needs a boss. A boss can handle at most 10 people below him/her. Every boss needs another boss, unless s/he is the head of the organization. A boss can only handle bosses below him/her of the same rank.
rhebus already solved it though.
Did I? All I did was write down the problem in maths. I thought you were after something more, such as a closed-form (non-infinite series) solution, possibly involving logs of some kind.
Well, perhaps there is a better and more exact answer, but for me what you said is good enough. You gave an iterative function by which it is possible to calculate the number of bosses.
I don't have a "correct" answer if that's what you're thinking.
Perhaps a better explanation is: every employee needs a boss. A boss can handle at most 10 people below him/her. Every boss needs another boss, unless s/he is the head of the organization. A boss can only handle bosses below him/her of the same rank.
You mean to say, "A group of 2 or more employees of level n requires an employee of level n+1 to manage them, denoted their boss. A boss may manage at most 10 employees."
Anyway, a slight variation of the equation I posted above suffices for your corrected problem. I leave that as an exercise for the reader :p
Evaluate the integral of ln(1 + tan x) dx from x=0 to x=pi/4
EDIT: Confusing programming with math before, I wrote log instead of ln, not that it matters much, but...
Evaluate the integral of log(1 + tan x) dx from x=0 to x=pi/4
Let y=tan(x); then x=arctan(y) and dx=dy/(1+y2), the limits are from y=0 to y=1, and the integrand is log(1+y)/(1+y2) dy.
(Henceforth I will assume that by "log" is meant "ln" because that's easier to write; the common interpretation of "log(x)" however is ln(x)/ln(10).)
Now after setting the original integral to S, set up integration by parts, with u=1/(1+y2) and dv=log(1+y) dy; then du=-2y/(1+y2)2, v=(1+y)(log(1+y)-1), and S=ln(2)-int(2y(1+y)(1-log(1+y))/(1+y2)2,y,0,1).
Then moving that huge integral to the other side, int(2y(1+y)/(1+y2)2+(1-y2)log(1+y)/(1+y2)2,y,0,1)=ln(2)...hmmm
At this point I went to Wolfram Alpha and found that it didn't have an expression in terms of real-valued functions of a real variable.
No one? Hmm, that integral is not really hard, I think people here haven't seen the method to solve it. I'll reformulate the question pointing the right direction, but also putting one that's a little harder ;)
(a) Prove that, for every real function f, integrable in the interval [a,b], the following equality holds:
(b) Evaluate:
Come on, guys! You can do it!
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
Even I can do a)! Let's go from RHS to LHS:
I think I can do the first one in b) too. *takes a deep breath*
which is the right answer (I checked the integral numerically)
EDIT: After some work, I managed to solve the second one too:
Now let's look at 2*I:
Very good! You're correct.
More rigorously though, the second integral at (b) is an improper integral of the second kind, since lim x->0+ (ln (sin x)) = -infinity. Thus, you have to prove this integral converges before calculating it.
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
p4wn3r wrote:
More rigorously though, the second integral at (b) is an improper integral of the second kind, since lim x->0+ (ln (sin x)) = -infinity. Thus, you have to prove this integral converges before calculating it.
I think I can do this too:
ln(x) is strictly increasing on the interval 0 to pi/2, and so is sin(x). On the intervall [0,pi/2], it holds that sin(x)<x. Since ln(x) is strictly increasing we get
which shows that the integral converges.
Wrong, sorry.
ln (sin x) is non-positive in the given interval. This comparison criteria only works if the function considered is non-negative.
Try proving the convergence of \int -ln (sin x)dx in the given interval, where most methods will apply.
Joined: 8/6/2006
Posts: 784
Location: Connecticut, USA
Here's a (probably simple) question that I unfortunately can't seem to explain:
Why does something like 15/sqrt(5) equal 3*sqrt5?
Or in simpler terms, why does x/sqrt(x) equal sqrt(x)?
To clarify, it makes sense to me when you plug in numbers but I can't explain it abstractly.
Or in simpler terms, why does x/sqrt(x) equal sqrt(x)?
By the definition of square root, if x >= 0, then y = sqrt(x) <=> (y^2 = x and y >= 0).
Now if x >= 0, x/sqrt(x) = y^2 / y = y = sqrt(x).
Basically, x is square of sqrt(x), and one pair of sqrt's cancels each other out, leaving just sqrt(x).
In fact, it's not. The limit of x/sqrt(x), when x goes to 0, is 0. But 0/sqrt(0) is undefined. That's why x/sqrt(x)=sqrt(x) if and only if x>0. The function :
f : x --> sqrt(x)
is defined on [0 ; + inf[ ; the function
g : x --> x/sqrt(x)
is defined on ]0 ; + inf[.
This brings up a question which I have actually been thinking for a long time.
When simplifying equations, people will carelessly perform simplifications like x^2/x -> x. However, that equality holds only for values of x different from 0. Wouldn't that mean that from that point forward you would have to start dragging the "x != 0" condition for all the subsequent simplifications? I have seldom seen this being done anywhere.
It may not look pretty but it's quite painless to calculate even for huge numbers as the number of summands grows verrrrry slowwwwwwly. Randil's solution is close but not quite right ... maybe ... depending on what you are actually asking.
(b) Evaluate:
Come on, guys! You can do it!
