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| Author |
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| 25 new of 81 responses total. |
aruba
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response 50 of 81:
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May 23 04:13 UTC 2002 |
1/cos(x) is called sec(x)
1/sin(x) is called csc(x)
1/tan(x) is called cot(x)
You can find their derivtives with the quotient formula:
(d/dt) (f/g) = (gf' - fg') / g^2
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mdw
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response 51 of 81:
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May 23 06:20 UTC 2002 |
Well, if it's completely horizontal, like a plate, then the width of the
rungs is definitely part of the equation. If it's carried such that the
rungs are horizontal, and the sides vertical, then the height of the
ceiling becomes the limiting factor on the height of the ladder. It's
only if the rungs are vertical and the sides horizontal that neither
matters. Actually, even then, unless the ladder is infinitely thin (in
which case the sharp edges will probably slice through anyone's foot)
the thickness of the ladder matters. If you could carry the ladder
diagonally, you could carry a slightly longer ladder, but if it's got
those funny rubber feet, that's going to really complicate the math,
especially if the rubber is elastic. Of course, if the ladder folds,
telescopes, or rolls up, it could also be much longer.
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remmers
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response 52 of 81:
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May 23 11:09 UTC 2002 |
All those things make it a different problem from what I had in
mind. You may assume that the "ladder" is a straight line segment
of fixed length.
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oval
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response 53 of 81:
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May 23 14:11 UTC 2002 |
is it one of those rope ladders?
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remmers
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response 54 of 81:
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May 23 15:00 UTC 2002 |
Somebody post a solution, quick! :)
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flem
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response 55 of 81:
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May 23 16:30 UTC 2002 |
I did most of this yesterday; let's see if it holds up. I wanted to go home
and do a neat writeup in LaTeX, just to hone my skills, but I watched the
Redwings instead. :)
So, L(t) = w1 sec(t) + w2 csc(t)
Using a handy table of trig derivatives,
L'(t) = w1 sec(t) tan(t) - w2 csc(t) cot(t)
Setting that to zero, we get
w1 sec(t) tan(t) = w2 csc(t) cot(t)
or
w1 sin(t) / cos^2(t) = w2 cos(t) / sin^2(t)
w1 sin^3(t) = w2 cos^3(t)
tan^3(t) = w2 / w1
t = tan^{-1} ( (w2/w1)^{1/3} )
Let's call that t0, to avoid seeing that notation again.
So,
L(t0) = w1 sec(t0) + w2 csc(t0)
is the longest ladder that will fit around under those conditions.
I fooled around trying to simplify that a bit, but didn't really get
anywhere.
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aruba
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response 56 of 81:
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May 23 20:54 UTC 2002 |
If 0 < y < pi/2 then
y = tan^{-1}(x) ==> x = tan(y) ==> x^2 + 1 = sec^2(y)
==> sec(y) = (x^2 + 1)^(1/2).
Likewise 1/x = cot(y) ==> x^-2 + 1 = csc^2(y)
==> csc(y) = (x^-2 + 1)^(1/2).
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blaise
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response 57 of 81:
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May 31 22:00 UTC 2002 |
The solution is sqrt(2*(w2+w1)^2). (The ladder forms a right triangle
touching the inside corner of the hall and touches the outside walls of the
hall at the two points w2+w1 from the outside corner of the hall. Therefore
the length of the triangle is the square root of the sum of the squares of
the sides.)
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shashee
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response 58 of 81:
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Jun 17 22:33 UTC 2002 |
x^2 = x + x + ... + x (x times)
differentiating both sides we have,
2x = 1 + 1 + ... + 1 (x times)
2x = x
2 = 1
Find the mistake ...:)
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rcurl
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response 59 of 81:
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Jun 17 23:43 UTC 2002 |
Given x a continuous variable, required for the differentiation, the
first statement is invalid for all except integer x.
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aruba
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response 60 of 81:
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Jun 18 14:07 UTC 2002 |
No one ever got the correct answer to John's last question, though Greg came
close.
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mynxcat
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response 61 of 81:
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Jun 18 14:38 UTC 2002 |
This response has been erased.
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brighn
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response 62 of 81:
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Jun 18 16:22 UTC 2002 |
#61> x * y = x + x + ... + x (y times). For instance, 3 * 4 = 3 + 3 + 3 + 3
= 12.
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mynxcat
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response 63 of 81:
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Jun 18 16:40 UTC 2002 |
This response has been erased.
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brighn
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response 64 of 81:
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Jun 18 19:21 UTC 2002 |
Like it's MY fault...
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mynxcat
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response 65 of 81:
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Jun 18 19:57 UTC 2002 |
This response has been erased.
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sholmes
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response 66 of 81:
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Jun 19 04:03 UTC 2002 |
Somewhere I have read .. any single digit positive number when raised to the
5 leaves the same remainder as that number itself when divided by 5
e.g 2^5 = 32 leaves the remainder 2 when divided by 10 .
there was a proof too ..but either I forgot or I dint understand it in the
first place.
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mcnally
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response 67 of 81:
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Jun 19 05:18 UTC 2002 |
Seems like the sort of thing that's extremely easy to prove by
demonstration -- I'm not sure why you'd want a more complicated proof.
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sholmes
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response 68 of 81:
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Jun 19 05:27 UTC 2002 |
its easy to prove by demonstration , but I wanted to know if a more elegant
proof exists.
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brighn
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response 69 of 81:
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Jun 19 14:05 UTC 2002 |
actually, since 10^5 will always end in 0, the "any single digit positive
number" is unnecessarily specific. If it works for 1..9 (which it does), it
also works for *any* integer (positive or negative, or 0).
E.g. 562 ^ 5 = 56063676972832; 6791 ^ 5 = 14443373841053414951
Also, the pattern is that the last digit is the same; the "remainder from 5"
condition is also unneeded. Perhaps the way it was presented was, the last
digit of x^5 = x when x is a single positive number. What I just described
is:
mod(x, 10) = mod(x^5, 10) for all integers x.
I can prove that's true for all integers if it's true for all integers 0..9,
but I'm not sure how to prove it's true for all integers 0..9 (except by
demonstration: 1, 32, 243, 1024, 3125, 7776, 16807, 32768, 59049).
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sholmes
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response 70 of 81:
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Jun 19 14:16 UTC 2002 |
there _is_ a proof , only that earl;ier I dint understand it and now I dont
have the book .
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jp2
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response 71 of 81:
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Jun 19 14:54 UTC 2002 |
This response has been erased.
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brighn
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response 72 of 81:
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Jun 19 15:33 UTC 2002 |
I can also prove:
mod(x, 10) = mod(x^(4*y+1), 10) for all integers x and all positive integers
y.
This is done by taking the final digit of x*x as a pattern, and involves a
similar proof-by-exhaustion. For instance, mod(7*7, 10) = 9; mod(9*7, 10) =
3, mod(3*7, 10) = 1, mod(1*7, 10) = 7. The patterns for each digit are:
0,0,0,0,0
1,1,1,1,1
2,4,8,6,2
3,9,7,1,3
4,6,4,6,4
5,5,5,5,5
6,6,6,6,6
7,9,3,1,7
8,4,2,6,8
9,1,9,1,9
Think of the house party ramifications... won't it impress your friends when
you accurately report that the final digit of 327^43 = 3 with only a moment
or two of thought? (While the final digit of 327^48 is, of course, 1.)
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aruba
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response 73 of 81:
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Jun 19 18:03 UTC 2002 |
The theorem that Abhijit is looking for is called "Fermat's Litle Theorem",
to distinguish it from Fermat's Last Theorem (which wasn't actually a
theorem until a few years ago). Fermat's Little Theorem says:
If p is prime and a is any number, then p divides (a^p - a).
Proof: If p divides a, then the theorem is clearly true. So suppose p
does not divide a. Then if -p < n < p, p divides neither n or a, so p
does not divide na.
It follows that the numbers a, 2a, 3a, ..., (p-1)a all have distinct,
nonzero remainders when divided by p, since the difference between any two
of them is of the form na described above. But there *are* only p-1
distinct nonzero remainders mod p; namely, 1, 2, ..., p-1. So
a, 2a, ..., (p-1)a must be 1, 2, ..., p-1 in some order, by the pigeon
hole principle. Therefore
a * 2a *** (p-1)a is congruent to 1 * 2 *** (p-1) (mod p)
==> (p-1)! * a^(p-1) is congruent to (p-1)! (mod p)
==> p divides (p-1)! * (a^(p-1) - 1)
==> p divides (a^(p-1) - 1) since p doesn't divide (p-1)!
==> p divides a^p - a.
So, in particular, if a is any number, a^5 has the same remainder as a
when you divide by 5.
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aruba
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response 74 of 81:
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Jun 19 18:11 UTC 2002 |
If you learn a little group theory, you realize that Fermat's Little Theorem
is just a special case of a very elementary property of finite groups, which
says that the order of an element of a group divides the order of the group.
Since the numbers 1, 2, ... , p-1 form a group under multiplication (mod p),
the order of each element divides p-1. Which means that a^(p-1) gives 1
when you do multiplication mod p.
But Fermat didn't know about group theory, since it hadn't been invented
yet.
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