You can simplify this product further by converting sin to its complex equivalent, that is:
sin(x) = (eiπx - e-iπx)/2i
simplify?
complex numbers?
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ALRIGHT FULL PROOF BITCHES. Also, triple post because I feel like it and also to bump this awesome occasion!
Since g(1) = n, we are done. Q.E.D.
FUCK YEAH.
Without having calculated anything, I can already tell that the fields can have varying amounts of sheep. Only A = C = F = H, and B = D = E = G, and there is a linear relationship between the diagonal fields and the horizontal/vertical fields.
That would mean A+B=C+E=H+G=F+D in which case the amount of sheep would be divisible by 4 and it isn't.Without having calculated anything, I can already tell that the fields can have varying amounts of sheep. Only A = C = F = H, and B = D = E = G, and there is a linear relationship between the diagonal fields and the horizontal/vertical fields.
That would mean A+B=C+E=H+G=F+D in which case the amount of sheep would be divisible by 4 and it isn't.
Hah, that's clever. Well then, there's a fucked up amount of sheep
What if some of the sheep were cut up
there could be half a sheep on every field xD
Without having calculated anything, I can already tell that the fields can have varying amounts of sheep.
The riddle is inherently flawed in the sense that there are multiple answers.
A(3) B(6) C(8) D(6) E(6) F(8) G(6) H(3) - ABC=ADF=FGH=CEH=17 - A+B+C+D+E+F+G+H=46
A(4) B(6) C(7) D(6) E(6) F(7) G(6) H(4) - ABC=ADF=FGH=CEH=17 - A+B+C+D+E+F+G+H=46
So yeah...
Your first intuition is usually right, I've royally messed up several exams by overanalyzing the simplest of shit :3Well then I guess I was right? Now I feel kind of silly haha.
Here's a cool problem I found while browsing xkcd:
So, let's say you are prisoner of an evil dictator, which will give you a chance to give you back your freedom if you win at this little game :
He has put 100 boxes in a room, and every box has a number from 1 to 100. Additionally, each box contains a card with a number between 1 and 100. Those numbers are chosen randomly, several boxes can contain cards with the same number.
When you enter the room, you have to chose one box, and open it. If the card it contains has the same number as the box itself, then you win. If it does not, then you are allowed to open the box with the same number as the card you just found, provided it still has not been opened. You can open as many boxes as you need to, the game stops when the number on the card is the same as the one on the box that contained it (you win), or when the number on the card matches a box that has already been opened (you lose).
Short example : You chose box 10, it contains card 20. You open box 20, it contains card 20, you win.
You chose box 10, it contains card 20. You open box 20, it contains card 10, box 10 has already been opened, you lose.
So, it is obviously not a question of strategy, the question is to determine the winning probability.
Let's call the number of boxes already opened n, and the chance of winning eventually f(n).
You have a 1/100 chance of winning instantly.
You have a n/100 chance of losing instantly.
You have a (99 - n)/100 chance of continuing the game.
f(n) = 1/100 + (99 - n)/100 * f(n + 1)
f(99) = 1/100
f(98) = 1/100 + 1/10000
f(97) = 1/100 + 2f(98)/100
f(96) = 1/100 + 3f(97)/100
f(95) = 1/100 + 4f(96)/100
...
*fiddles with formulas in Excel*
f(0) = 0.122099606 (approximately)
I don't think there's a good convenient way to give an explicit direct formula for it, you just have to go through the calculations.
So, you have a slightly less than 1/8 chance of winning.
Anyone looking to prove their mettle, design a postulate for electron energy levels accounting for multiple-electron systems (such as for instance lithium)
I win at difficult tasks kthx?![]()
Stop it...the math...it hurts!
Math high?
Nice solving i understood 1% of that solution but im a maths retard so hec.
Here a problem for you guys.
There is a square house. Each side of the house has a window> And there are 9 fields surrounding the house. It is so you can see 3 fields through each window.
Through each window you can see the same amount of sheep.
Yet there is a total of 46 Sheep.
How many sheep are on each field?
FIELD A FIELD B FIELD C
FIELD D HOUSE! FIELD E
FIELD F FIELD G FIELD H
fladdermasken said:A(3) B(6) C(8) D(6) E(6) F(8) G(6) H(3) - ABC=ADF=FGH=CEH=17 - A+B+C+D+E+F+G+H=46
That's because Americans don't put an S on the end. The S makes it soft.
Mathsssss...
<----Canadian.
f(n)=(n-1)^2-n^2
Where N=3
Then after every full calculation N becomes the output.
First find out what happens in the output. And 2ndly explain why it does that![]()
f(n)=(n-1)^2-n^2
Basically its like this
(1-Ans)^2-ANs^2=Y
So
Here is an example
(1-3)^2-Ans^2=-5
(1--5)^2--5^2=11
(1-11)^2-11^2=-21
Etc
The following is a list of the sequence (read downwards)
3 -1365 699051 -357914000 -1.466x10^12
-5 2731 -1398101 715828000 2.93x10^12
11 -5461 2796203 -1.14532x10^10 5.86x10^12
-21 10923 -5592405 -2.2907x10^10 1.17x10^13
43 -21845 11184811 4.5814x10^10 0
-85 43691 -22369621 -9.163x10^10 1
171 -87381 44738243 1.8326x10^10 -1
-341 174763 -89478480 -3.666x10^10 3
683 -349525 178956960 7.33x10^11 End of loop
So why does it loop back to 3 again?
r
Okay obvs you dont understand me i have made a vbs programn inputting the equasion for you and outputting numbers up to a point. As i did not bother looping it.
To quit off of it just go CTRL Shift Delete then close the process wscript
Ok guys, I wrote that problem a month ago, so I forgot the actual answer
Here's a new one:
f(11) = 56
f(9) = 22
f(8) = 14
f(2) = 2
Can you find f(12)? ^^
Let's just say that this problem is best solved using recursion in programming and it causes an exponential growth in the number of function calls ^^
WolframAlpha, Google, Maple, matlab, anything that doesn't have terrible precision. Also, what are you talking about -1 and 1?Brambleclaw said::L But that doesnt explain -1 and 1 :L Also if its about precision what do you suggest i use to calculate? And C++ is not an option due to faulty Windows
Dude.Brambleclaw said:Also 2.33 * 8 = 18.66664 not -15 so i wonder when u got that from :L
Answer is 594/7. Using the same technique as here. Pretty straightforward.
int fibonacci(int i) {
if (i==0)
return 0;
else if (i==1)
return 1;
else
return fibonacci(i-1) + fibonacci(i-2);
}
// This is the actual function
int f(int i) {
return 1 + fibonacci(i);
}
Nope.
Math isn't about calculations and shit, it's about Trial and Error, Dumb Luck, etc..
That's what makes it fun sometimes
In that sequence, I used the fibonacci series and just added one xD
The function cant be represented mathematically, but here's what it looks like in C++
Code:int fibonacci(int i) { if (i==0) return 0; else if (i==1) return 1; else return fibonacci(i-1) + fibonacci(i-2); } // This is the actual function int f(int i) { return 1 + fibonacci(i); }
lol I'm so used to programming xD
f(0) is the first number in the fibonacci series + 1
f(1) is the second number in the fibonacci series + 1
etc..
f(12) = 90
Ok how abouot this.
"Find a power sereis which describes the indefinite antiderivate of the function f(x) = x^x."
I know this really silly and pointless, since this is that kind of function can not be represented trough any other kind of special or elementary function. But when i get bored i work out those kind of things.
Where Q is the regularized gamma function of the lower limit which is usually defined as![]()
Also i worked out a n-ary sum if x is a natural number![]()
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This thread just leveled up.