/sci/ - Science & Math

bumping cause I dont post often

>>10081313
Well the average value you'll be adding by is (1+2+5+5+7+10+12+12+15))/9=7.6

The average value you will be multiplying by (1+1+2+2+2+3+3+4+5)/9=2.5

Since you have 6 variables to be filled and getting either is an equal probability you can just do each one 3 times (7.6+7.6+7.6)+(2.5)^3=38.4

>>10081373
This is surely a too simple answer
I think you have misunderstood
the order of the sequence matters
the averages are actually 23/3 and 23/9
and in your case it would be
(3*(23/3))*(23/9)^3

I think you meant
(7.6+7.6+7.6)*(2.5)^3=356.25
not
(7.6+7.6+7.6)+(2.5)^3=38.4

>>10081373
you can't do that lol
You know when you have an idea you should really test it on a smaller example to see if it's true.

So let's say the adders are: {3, 7}, the multipliers are: {2, 6}, and the sequence length is 2.
The real expected value would be: ((3+3) + (3+7) + (3*2) + (3*6) + (7+3) + (7+7) + (7*2) + (7*6)) / 8 = 15
While the expected value according to your method would be: 5 + 4 = 9


>>10081313
Are you allowed to use programming OP?

strawberry = 1
orange = 5
perry = 139
grapes = infinite
lemon = 1
cot = wtf?

>>10081313
how do you calculate cot^-1 please?

>>10081411
I am pretty sure I could do this with some simple program, there a lot of possibilities but not so much that its impossible for a computer.
I am however trying to find an analytical solution.

>>10081413
4*(4/tan^-1(5) - 1/tan^-1(139) = 9.09
eggplant = 9.09
cherry & watermelon ---> do not exist

>>10081313
>>10081431
Let's call the expected addition [math] a [/math] and the expected multiplication [math] m [/math].

Let's say [math] f(n) [/math] is the expected value after [math] n [/math] steps (the first number doesn't count as a step).

Then we can express [math] f [/math] recursively as:
[eqn] f(0) = a \\
f(n) = (f(n-1) + a + f(n-1) \cdot m) / 2 [/eqn]
This results in:
(I don't know how to do this by hand, I used Wolfram Alpha: http://www.wolframalpha.com/input/?i=f(n)+%3D+(f(n-1)+%2B+a+%2B++f(n-1)+*+m)+%2F+2,+f(0)+%3D+a

[eqn] f(n) = \frac{a(2^{-n}m(m+1)^n-1)}{m-1} [/eqn]
Filling in [math]a = \frac{69}{9}[/math], [math]m = \frac{23}{9}[/math] and [math]n = 5[/math] gives the solution:

[eqn] f(5) = \frac{\frac{69}{9}(2^{-5}\frac{23}{9}(\frac{23}{9}+1)^5-1)}{\frac{23}{9}-1} = \frac{77496223}{354294} \approx 218.734 [/eqn]

This answer is consistent with a small program I made to doublecheck :)

>>10081313
>>10081431
Let's call the expected addition [math] a [/math] and the expected multiplication [math] m [/math].

Let's say [math] f(n) [/math] is the expected value after [math] n [/math] steps (the first number doesn't count as a step).

Then we can express [math] f [/math] recursively as:
[eqn] f(0) = a \\
f(n) = \frac{f(n-1) + a + f(n-1) \cdot m}{2} [/eqn]
This results in:
(I don't know how to do this by hand, I used Wolfram Alpha: http://www.wolframalpha.com/input/?i=f(n)+%3D+(f(n-1)+%2B+a+%2B++f(n-1)+*+m)+%2F+2,+f(0)+%3D+a)

[eqn] f(n) = \frac{a(2^{-n}m(m+1)^n-1)}{m-1} [/eqn]
Filling in [math]a = \frac{69}{9}[/math], [math]m = \frac{23}{9}[/math] and [math]n = 5[/math] gives the solution:

[eqn] f(5) = \frac{\frac{69}{9}(2^{-5}\frac{23}{9}(\frac{23}{9}+1)^5-1)}{\frac{23}{9}-1} = \frac{77496223}{354294} \approx 218.734 [/eqn]

This answer is consistent with a small program I made to doublecheck :)

>>10081313
>>10081431
Let's call the expected addition [math] a [/math] and the expected multiplication [math] m [/math].

Let's say [math] f(n) [/math] is the expected value after [math] n [/math] steps (the first number doesn't count as a step).

Then we can express [math] f [/math] recursively as:
[eqn] f(0) = a \\
f(n) = \frac{f(n-1) + a + f(n-1) \cdot m}{2} [/eqn]
This results in:
(I don't know how to do this by hand, I used Wolfram Alpha: http://www.wolframalpha.com/input/?i=f(n)+%3D+(f(n-1)+%2B+a+%2B++f(n-1)+*+m)+%2F+2,+f(0)+%3D+a

[eqn] f(n) = \frac{a(2^{-n}m(m+1)^n-1)}{m-1} [/eqn]
Filling in [math]a = \frac{69}{9}[/math], [math]m = \frac{23}{9}[/math] and [math]n = 5[/math] gives the solution:

[eqn] f(5) = \frac{\frac{69}{9}(2^{-5}\frac{23}{9}(\frac{23}{9}+1)^5-1)}{\frac{23}{9}-1} = \frac{77496223}{354294} \approx 218.734 [/eqn]

This answer is consistent with a small program I made to doublecheck :)

>>10081313
Eggplant is pi, lemon is e. No rational solution.

>>10081313
Strawb=1
Orange=5
Pear=114
Grape=infinity
Lemon=e
Eggplant=16*Tan(5)-4*Tan(e)
Cherry and watermelon have infinite solutions.

>>10081713
Thank you very much for making the effort.
I have however one quarrel with your solution and that is that you have implied that the expected addition (69/9) and expected multiplication (23/9) is applied at each step n?
but at each step it should be one or the other, so is it not appropriate to half the average expected values?

>>10081713
>>10081830
>>10081830
I spoke too soon. I didn't want to lose you.
After actually working through the problem myself, I am happy to accept this as a suitable solution.
I hope you enjoyed working on it.

>>10081790
>infinite
Prove there's one.

>>10081845
ok. Watermelon over cherry is equal to lemon plus eggplant. Lemon = e, which is a finite real number, and eggplant equals 16*tan(5)-4*tan(114). The tangent function is real valued, except for divergent points at multiples of pi. Neither 4, nor 114 is a multiple of pi, therefore eggplant is a real number.

Lemon plus eggplant is a pretty exotic transcendental number, yes, but watermelon can be made equal to cherry*(lemon + eggplant), at which point cherry can take on any value, thus generating infinite solutions.