TT's Maths Thread

[Mao]

I'm fairly sure it will. The Jacobian is just a function of u and v, so it shouldn't affect anything.

[TrueTears]

Thanks Mao  :smitten:

[TrueTears]

Um is the answer wrong for this question? I can't seem to get the right answer... it is off by a factor of 2 grrr anyways:

Evaluate where is bounded by the xz-plane and the hemispheres and .

So my solution is this:

Using spherical coordinates, we have

Thus

But the answer is hmmm how did they get this?

[Cthulhu]

Um is the answer wrong for this question? I can't seem to get the right answer... it is off by a factor of 2 grrr anyways:

Evaluate where is bounded by the xz-plane and the hemispheres and .

So my solution is this:

Using spherical coordinates, we have

Thus

But the answer is hmmm how did they get this?

I remember doing this(pretty sure it was) question last semester... I got it wrong too. My tutor showed me what I did wrong and its in a notebook somewhere  but I can't remember... sorry.

[TrueTears]

haha, damnnnn i swear i'm checking every minuscule detail and i can't find anything wrong!

NOOOOOOOOO omg what a stupid mistake... sketched it wrong lmao, should range from 0 to , because y is the axis of symmetry! NOT Z! SIGH I'M TOO USED TO SEEING z = f(x,y) instead of y = f(z,x)!

[TrueTears]

Suppose a gambler with $n plans to bet $1 a time on the toss of a coin, until he reaches $100 or $0. What is the probability of going broke starting with $n? First make a recurrence relation to represent the situation and then solve it.

Hmmm how would I start this question?

[kamil9876]

Let be the answer. Obviously otherwise this problem is no fun.

In order for him to go broke, he must eventually be at at some point. At that point his probability of going broke is . So now using this you can easily show that where is the probability of him eventually being at n-1. Now you must find . I guess the best way of doing this is again be a recursion, since is easy to find. And finding in terms of is also easy (ie it is like a "downwards" recursion).

[Mao]

Suppose a gambler with $n plans to bet $1 a time on the toss of a coin, until he reaches $100 or $0. What is the probability of going broke starting with $n? First make a recurrence relation to represent the situation and then solve it.

Hmmm how would I start this question?

Not helping, but I remember doing this when I was trying to get a strategy for winning roulette. Not sure about doing it analytically, but I ran a simulation 1,000,000 times and approximated the long-term probability

[Cthulhu]

Suppose a gambler with $n plans to bet $1 a time on the toss of a coin, until he reaches $100 or $0. What is the probability of going broke starting with $n? First make a recurrence relation to represent the situation and then solve it.

Hmmm how would I start this question?

Not helping, but I remember doing this when I was trying to get a strategy for winning roulette. Not sure about doing it analytically, but I ran a simulation 1,000,000 times and approximated the long-term probability

Link to arxiv or published paper or it didn't happen.

[TrueTears]

The Fibonnaci numbers are as F(k+1) = F(k)+F(k-1) with initial values F(0) = F(1) = 1

Show that if k is even

How do I use induction to show this? If we use strong induction on k and assume is true for 0,2,4,...,k then we must show

But F(k+2) = F(k)+F(k+1)

We have information from our inductive hypothesis about F(k) but know nothing about F(k+1)... this is where I'm stuck.

[/0]

Let

The statement is equal to:

for .

There is another fibonacci identity that may be useful: (source: wikipedia lol)

[TrueTears]

Ahh I see thanks for that, is there another way of doing it without knowing the identity... cause this is from an exam

If we were to show for

then we'd need to show F(2(n+1)) = F(2n+2) = F(2n+1)+F(2n) to be true, but we don't have any information on F(2n+1)...

[/0]

Hmmm...  oh right this might work...



I wanna get rid of that odd f(2n+1) so I keep expanding that:



Actually I can just stop at the first step: . Then I have:

no need for obscure identities xD

[TrueTears]

oh em gee that is smart!!!!

thanks bro

[kamil9876]

is an easy induction step.