3u Maths Marathon Tangent Discussion (1 Viewer)

[Pianpupodoel]

Next Question:
Change of themes here . Find ∫dx/( sqrt(x) * (1+x)).

Spoiler

Integration by Substitution:

Let u = sqrt(x)
du/dx = 1/2.x^( -1/2)
2du = dx/sqrt(x)

∫dx/[ sqrt(x) * (1+x)] = ∫[ 2/( 1+u^2)]du
= 2 [ tan^(-1)(u) + c]

= 2 tan^(-1)[ sqrt (x)] + C

I hope I'm right.

Anyways, this counting/probability problem seems rather interesting:

Assuming a 365-day year, find an expression for the probability that in a group of n people, there is at least one birthday in common. Hence find the minimum number of people required for this probability to exceed 50%.

 

[Templar]

Spoiler

23 from memory.

 

[Templar]

Spoiler

Expression for probability is:
1-Pi k=0 to n-1 (365-n)/365

 

[KeypadSDM]

Qualifying Templar's answer:

Spoiler

The probability that no person has the same birthday is the probability that they have a different birthday from all "previous" people. Thus for n people:
P_{no b-day}(n) = 365/365 * (365 - 1)/365 * (365 - 2)/365 * ... * (365 - (n-1))/365
P_{no b-day}(n) = Product[0 to n-1][(365 - k)/365]
Thus:
P_b-day(n) = 1 - P_{no b-day}(n) = 1 - Product[0 to n-1][(365 - k)/365]

 

[KeypadSDM]

Fairly easy integration question:

∫dx/(1 + e^-x)

 

[Templar]

Hint

Spoiler

Multiply by e^-x/e^-x

 

[KeypadSDM]

Hint

Spoiler

Multiply by e^-x/e^-x

Better Hint

Spoiler

Multiply by e^x/e^x

 

[KeypadSDM]

Find all real x such that:
|4x - 1| > 2 sqrt(x(1-x))

Spoiler

Firstly, we need x(1-x) >= 0

That's a quadratic with roots 0 & 1, thus:
0 <= x <= 1

|4x - 1| > 2Sqrt[x(1-x)]
is equivalent to:
(4x - 1)² > 4x(1-x), where x has the restriction above.
Rearranging gives;

20x² - 12x + 1 > 0

This has roots 1/10 & 1/2
I.e. x < 1/10, x > 1/2

Thus, within the restrictions above:

0 <= x < 1/10 & 1/2 < x <= 1

I haven't done one of these in a while. And I should have checked it.

Next Question:

Prove the Riemann Hypothesis

 

[Templar]

Next Question:

Prove the Riemann Hypothesis

Can I prove the Poincare Conjecture instead?

Anyway, stick to topic.