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[Vector]

YOU DID IT AGAIN, YOU MIGHT KILL MY PROF NEXT, SO PLEASE JUST STOP, THINK OF THE POOR MATHEMATICIANS.
I will point out the most obvious error:
∞/∞ =/= 1, it is an indeterminate form.

"Warning - while you were typing 2 new replies have been posted. You may wish to review your post."

Ninja'd by Vector, who explained things much better than I.

Dear KaminaSquirtle: don't be an asshole while demonstrating mathematics.

Love and bites on the face, Vector.

[KaminaSquirtle]

Once I saw your post, I knew that this was going to happen.  Now I wish I had just edited my post when I saw that, I had hoped it would have been fine as the continuation of joke, but it appears I was mistaken.  In fact that should have been obvious to me at the outset.

I didn't mean any offense qwertyuiopas, ok?

[qwertyuiopas]

I haven't really used much math (outside of programming stuff) for a year or so until a few months ago, anyway, so I should have expected to make a mistake like that. Plus, I rewrote a_whatever as a function to make writing it easier, so it was an even easier mistake to make... Well, thanks for reminding me how easy it is to make such a mistake, I hope I don't do it again any time soon.

Edit: Further issue: I started writing it in notepad...

[Soadreqm]

I tried to take the easy way out and provide a physicist's answer.

Spoiler (click to show/hide)

Of course, I don't even have that. The stupid thing approaches infinity, and since my brilliant proof begins with "Assume that the series converges because otherwise what would be the point", I'm guessing I fucked up the java syntax. It's supposed to converge, isn't it? Oh well, no big loss. It wouldn't have been real proof, anyway.

[KaminaSquirtle]

I tried to take the easy way out and provide a physicist's answer.

Spoiler (click to show/hide)

Of course, I don't even have that. The stupid thing approaches infinity, and since my brilliant proof begins with "Assume that the series converges because otherwise what would be the point", I'm guessing I fucked up the java syntax. It's supposed to converge, isn't it? Oh well, no big loss. It wouldn't have been real proof, anyway.

I'm fairly sure it converges, as, AFAIK, proving that things diverge is way easier than proving they converge.  Can Java even take limits as n approaches infinity?  I suppose it could just approximate it as n gets really, really big, right?  Nice try though, for a physicist.

[Vector]

I tried to take the easy way out and provide a physicist's answer.

Spoiler (click to show/hide)

Of course, I don't even have that. The stupid thing approaches infinity, and since my brilliant proof begins with "Assume that the series converges because otherwise what would be the point", I'm guessing I fucked up the java syntax. It's supposed to converge, isn't it? Oh well, no big loss. It wouldn't have been real proof, anyway.

Yeah, you can't assume that the series converges >_>  That is not the true way of the mathematician.

Not enough paranoia.

[Karlito]

Oooh, recursive sequence. Let me see if I can remember how to do these.

[Soadreqm]

Yeah, you can't assume that the series converges >_>  That is not the true way of the mathematician.

Not enough paranoia.

I can't do that? Really? Because I think I just did.

Yeah, what I would have done is basically just iterate some numbers out of the series. It's as simple as:

Spoiler (click to show/hide)

        double a = 2;
        for (int i=1; i<1000; i++){
                a = ((i+3/(2*i))*a+i/((i+1)*a));
                }

This proves nothing, but it gives the answer. Or would if it worked.

[Heron TSG]

Mmkay, math help.

Finding the derivative of Y = ((x^.5) / (x+10)^.5)

So far, my best answer is as follows.

Y^I = (5x²)(x+10)

Can anyone verify this? It's the one problem on this worksheet that doesn't have an answer and has two radicals divided by each other in the original equation, so I want to make sure of this one.

[Vector]

Er... didn't get anything like that.  Try again?

Rather, can you write down your steps?

EDIT: Checked, definitely wrong.

[Jim Groovester]

The chain rule followed immediately by the quotient rule? Blech. Differentiation can be a real pain sometimes.

But this one isn't so bad, now that I look at it. This'll be easiest for me if I just make some dummy functions and manipulate those around, and then just substitute them back in.

Spoiler: The solution (click to show/hide)

f(x) = x^1/2
g(x) = x
h(x) = x + 10

f'(x) = 1/2 x^-1/2
g'(x) = 1
h'(x) = 1

d/dx f(g/h) = f'(g/h) * (g/h)' = f'(g/h) * ( (gh' - hg')/h²)
d/dx f(g/h) = 1/2 (x/(x+10))^-1/2 * ( (x - x - 10)/(x + 10)² )
d/dx f(g/h) = -1/2 ((x+10)/x)^1/2 * (10/(x + 10)² )

This can be reduced further to:

d/dx f(g/h) = -5/(x^1/2 * (x + 10)^3/2)

So there you go.

But yeah, you should tell us what your steps are, so we can correct whatever error you made that led you astray.

[Vector]

h(x) = (x+10)^.5, Groovester...

[Jim Groovester]

x^1/2/(x+10)^1/2 = (x/(x+10))^1/2

The outer function is x^1/2.

_{Vector correction combo x3!}

[Virex]

I tried to take the easy way out and provide a physicist's answer.

Spoiler (click to show/hide)

Of course, I don't even have that. The stupid thing approaches infinity, and since my brilliant proof begins with "Assume that the series converges because otherwise what would be the point", I'm guessing I fucked up the java syntax. It's supposed to converge, isn't it? Oh well, no big loss. It wouldn't have been real proof, anyway.

Yeah, you can't assume that the series converges >_>  That is not the true way of the mathematician.

Not enough paranoia.

You can disprove something by assuming it's true though


Anyway, I think I got the solution, I just need to prove 1 thing for it to work and I know it's true. I just get a bit stuck on it and I'm tired Got it! scratch that, now I got it:

Spoiler (click to show/hide)

a₀=2
a_n=((n+3)/(2n))*a_n-1+n/((n+1)*a_n-1), for all integers n>=1

lim_n->inf a_n = a_n-1/2 + 1/a_n-1
for a₀ > 0, a_n > 0, since no term in the function can become negative for positive n and a_n-1

1.) If the series converges, it converges to 2^1/2:
x = x/2 + 1/x
x² = x²/2 + 1
x²/2 = 1
x² = 2
x = 2^1/2


2.) The series converges from above:
a_n = a_n-1/2 + 1/a_n-1
For values larger then 2^1/2 we can state that
a_n-1/2 > 1/a_n-1, so a_n-1/2 + 1/a_n-1 < a_n-1
This means that for any value a_n-1 that is larger then 2^1/2, a_n < a_n-1
since a_n > 0 for any value of n, we can state that:a_n-2 > a_n-1 > a_n > 0
If the series converges, then a_n-2 - a_n-1 > a_n-1 - a_n. If that is true, a_n-2 - a_n-1 - (a_n-1 - a_n) > 0
This means that a_n-2 - 2*a_n-1 + a_n > 0
a_n-2 - 2*a_n-1 + a_n-1/2 + 1/a_n-1 >0
a_n-2 - 1.5 a_n-1 + 1/a_n-1 > 0
a_n-2 - 1.5*(a_n-2/2 + 1/a_n-2) + 1/a_n-1 > 0
1/4 * a_n-2 - 1/a_n-2+ 1/a_n-1 > 0
a_n-2 > a_n-1, so 1/a_n-2< 1/a_n-1 and thus 1/4 * a_n-2 - 1/a_n-2+ 1/a_n-1 > 0 is true, Q.E.D.


From this we can conclude that since the difference between 2 consecutive terms decreases as n increases, yet stays positive, the difference must eventually decrease to infinitesimal values for n -> inf and a_n -> 2^1/2, in other words, the function converges from above.

3.) The series converges from below:

Following the same reasoning as at 2, it can be shown that for values below 2^1/2, a_n-1/2 < 1/a_n-1, so a_n-1/2 + 1/a_n-1 > a_n-1. With that proven, the same proof as with 2 can be used to prove that the series will converge for values of a_n-1 lower then 2^1/2

[KaminaSquirtle]

Well, since you were able to find it convergent, it must be bounded.  So I should be able to prove that it is bounded.
Is that the right sort of thing to prove at all?  Am I on the right track?
I don't want to try to prove that and up wasting a bunch of time.

And congratulations on solving the problem!  It always feels good to solve a more difficult problem, at least for me.
Since you were able to solve it in one day, I now have more confidence in my ability to solve it in a greater span of time.