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This thread is just what the title says it is. It is all about Math. You can post math jokes(the jokes don't have to be good), math questions, what you like about Math, or even why you hate Math.
My math joke: Resistance is not futile. It is voltage divided by amps.
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who knows the most digits of Pi? the ones i know are below
I know all of them.
Hey master565,
u cant know all the digits of pi, pi goes on forever
I might have a bit of it.
If you look at the first letter of each word, it goes LLLLL!OOO!OOOOO!KKKKK!YYYYY!
I honestly am NOT sure.
Are you absolutely sure you didn't mess up anywhere in the pattern?
Anyway, my friend and I are in a disagreement.
For the past infinitely many days, I have alternatively given him a duck, and then the next day took a duck away. This is part of a agreement I made a long time ago in order to pay off a gambling debt. My friend started out with no ducks at all, which is a very sad thing because ducks make great companions.
While it is true that some days I may have messed up the order a little bit, and given him a duck when I should have taken one away (or vice versa), I always eventually corrected myself by taking a duck on a day when I normally would have given him a duck. I am 100% sure that he currently has as many ducks as he should have, even though he did not get them in the right order.
However, he currently has 57 ducks, and has accused me of giving him and excess of duck. He claims that I should take some ducks away, in order to get back on track.
However, I think I actually am back to being on track, and today is an odd day since we began, so I actually OWE him a duck.
Can you guys help me out? Am I right, or did I not manage to balance everything out?
I have alternatively given him a duck, and then the next day took a duck away.
So on Day X you would give him a duck and on day X+1 you would take a duck away? How could he accumulate ducks this way?
So on Day X you would give him a duck and on day X+1 you would take a duck away? How could he accumulate ducks this way?
Well, Ideally, that is what I should have done. But, you know, things come up and I might have given a duck two days in a row. But I swear I accounted for any mistakes by taking the appropriate actions, and after infinitely many days everything should have evened out.
On every day, I either took one duck or gave one duck. I never took or gave more than one duck on any one day. And the number of days I gave a duck is equal to the number of days I took a duck.
So, obviously, there should be 57 ducks, right? My friend doesn't seem to be able to understand this. Maybe you could explain it better than I could?
(hint: the number of ducks is equal to the following sum:
1-1+1-1+1-1+1-1+1-1+1...... infinitely many times.)
Then you have given him 57 too many ducks.
Day 1: Give duck Total: 1 duck
Day 2: Take duck Total: 0 ducks
Every odd day should be 1 duck and every even day 0 ducks.
Heres another question my math teacher gave me (her numbers were different though:P)
using 2, 6, 3, 11, 9 how can you make the answer equal 13?
heres a posssible example: (11+9)-(6+3)+2=13
Then you have given him 57 too many ducks.
That's what he said!
But, I just added all of the days together, and got that I should have given him infinitely many ducks. So, How can I have given him 57 "too many"? How could I have given him more than infinitely many ducks? Where would I even get that many ducks?
2, 6, 3, 11, 9 how can you make the answer equal 13?
Hmmm... (3^2)+9+6-11
That's what he said!
And I think he is correct.
But, I just added all of the days together, and got that I should have given him infinitely many ducks.
You must have done something wrong. The least number of ducks that would be needed is 1. Look at my previous example. The duck would be going between you and your friend every day(you,friend,you,friend...).
But in your example you said that it was possible for you to make a mistake a give him say 2 ducks in a row. The number of ducks needed is lets say X. X would be given by how many times in a row you gave him a duck. Done properly X would only ever be 1, but if you gave him 2 ducks in a row then X would be 2.
In the end you said that you equaled out the number of times you gave a duck and the number of times you took a duck. That means you would end up with X ducks and your friend would end up with 0 ducks.
You must have done something wrong
Are you sure? Because over infinitely many days I would have to have given him infinitely many ducks. Think about it: after four days, if I made no mistake I would have given him two ducks (and taken two away). So, after 8 days, I should have given him 4 ducks, and after N days I should have given him N/2 ducks. This is not counting, of course, the number of ducks took away.
After everything is counted up, I think I still gave him infinity/2 ducks and took the same number away. That is why he has 57 ducks.
I think I still gave him infinity/2 ducks and took the same number away.
Correct.
infinity/2 - infinity/2 = 0
Let's do a finite example.
A= total # of days
B= # of days giving duck
C= # of days taking duck
D= Total # of ducks friend started with(0),so it can be ignored
Let A=30 days. B would equal A/2=15. C also equals A/2=15. Therefore B and C equal each other. You could give 15 ducks in a row but then you would have to take away 15 ducks. It doesn't really matter what day you do what as long as B=C in the end
So for this problem(and any other just with different numbers):
15-15=0
infinity/2 - infinity/2 = 0
Hmmm.... Thats odd. I got Infinity/2 -infinity/2=57.
Wait! I just remembered the order that I gave the ducks in!
For the first 57 days, I gave him a duck. Then, to make up for that, I took a duck for the next 57 days. But, being me, I messed up AGAIN and gave him a duck for the next 57 days. So, I had to make for that by taking a duck for the next 57 days.
I guess I made of habit of that, because I repeated that pattern for all of the infinitely many days. Of course, since I am constantly correcting myself, everything should even out.
IE, 1-1+1-1+1-1+1-1.... surely must be equal to 57-57+57-57+57-57...., as long as the +57s are equal to the -57s, right?
And, doing the math:
57-57+57-57+57-57+57....
=57 + (-57) + 57 + (-57) +57+ (-57) +57.... since adding a negative number is the same as subtracting a positive, right?
= 57 + (-57 + 57) + (-57 +57)+ (-57 +57)....
Since by the properties of numbers, a+b+c = a+(b+c)
=57 + 0 + 0 + 0 + 0.....
= 57.
So, unless I can't add, I think he should 57 ducks. I don't see where I messed up, do you?
since adding a negative number is the same as subtracting a positive, right?
Correct.
I don't see where I messed up, do you?
I think it depends on where you are in the cycle. You can be anywhere in the range [0,57].
I think it depends on where you are in the cycle. You can be anywhere in the range [0,57].
It's not a cycle; it's a sum. Everything is added together at the same time, it doesn't really matter what order you do it in*. Or at least, that is my reasoning for why 57-57+57-57+57-57.... is just 1-1+1-1+1-1+1.... rearranged. So, to more directly answer your question, we are at day "infinity" of the sum, which is why the sum does not terminate with some final term.
Earlier you said I should be in the range [0,1]. What changed?
*Or does it?
Earlier you said I should be in the range [0,1]. What changed?
A [0,1] range would be if you never messed up. Key the word should.
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