You are not logged in.
What about 15,15 or 30,30?
Do you know why we have to choose e^(something) for these, why no other solutions are possible
and also why the general solution is c1.(a particular solution) + c2.(another particular solution)?
Happy Easter, everyone!
Let
[list=*]
[*]
and suppose
[list=*]
[*]
for some v ≠ 0. Then
[list=*]
[*]
Since the LHS is zero for some x, y not both zero, the determinant of the matrix on the LHS must be 0.
Note that eigenvalues can be complex.
The most general theorem of all:
(NB: If N contains one or more 0, we allow Nʹ to start with a 0. For example, if N = 1024, Nʹ can be 0241; in this case N is a 4-digit number whereas Nʹ = 241 is a 3-digit one.)
Proof of the most general theorem of all:
We know that a positive integer is congruent modulo 9 to the sum of its digits. For example, 12345 ≡ 1+2+3+4+5 = 15 ≡ 1+5 = 6 (mod 9) (i.e. it leaves a remainder 6 when divided by 9).
This is based on the fact that is divisible by 9 for all non-negative integers k (these are numbers of the form 0, 9, 99, 999, 9999, etc). Therefore if N is a k-digit number with digits , we have[list=*]
[*]
and adding up the i gives
So N is congruent to the sum of its digits mod 9. Similarly Nʹ is congruent to the sum of its digits mod 9. But the sum of the digits of N and the sum of the digits of Nʹ are the same since the digits of one are those of the other rearranged. This means
[list=*]
[*]
and so their difference is divisible by 9. QED.
Why did you guys remove {x}
Sorry, it was a typo. I've put it back in post #9.
Strictly, as you have one thing ≥ x and another ≥ 0, when you add them you get (one thing + another) ≥ x
No, I have the first thing > x and the other ≥ 0.
Okay, this should work. For all x,
[list=*]
[*]
Also, for all x,
[list=*]
[*]
(Fractional parts cannot be neagative.)
for all x.Is this better, Bob?
Oops, I did it wrong.
Back to the drawing board.
Here's an algebraic proof. For all x,
[list=*]
[*]
No solution.
"So, Peter, you've finally come to believe in the existence of the fourth dimension of reality."
"Yes, Paul, it's about time."
Is there a section on group theory or abstract algebra? Or algebraic topology?
San Andreas should be blamed for all the earthquake activity in California because it's his fault.
Oh! Thank you Bob, I forgot about the tan formula.
[list=*]
[*]
Implicit differentiation gives
[list=*]
[*]
Hence the angle between the two tangents is
[list=*]
[*]
PS: It's actually
[list=*]
[*]
since the gradient to the curve will always be steeper than the gradient to the circle.
This is a much more interesting puzzle than I thought. If indeed all variations end with mate in three, it must have taken a super genius brain to work them out, given there were no computers in the 19th century. I can see why it won first prize.
#26. Having a will of one's own? (7)
Oh, I didn't see that. Thanks, SteveB.
This is a chess problem from The Illustrated London News, a 19th-century publication. According to the author it is possible for white to mate in three moves but I can't find a solution.
[list=*]
[*][/*]
[/list]
The best I can do is I analysed the problem at http://www.chesslab.com/PositionSearch.html but it couldn't find a mate-in-three either; however it did find an
Can anyone find a mate in three, if there is one? Phrontister?
I think the last one should be
#23. Legal heir will cut up ginger. (4,9)
#24. Stick insect and mammal. (7, 3)
#25. Seriousness is what can pull you down. (7)