Help Needed: Struggling to Get Started with Physics

[mynameisfunk]

https://www.physicsforums.com/attachments/28380Have really no idea where to start with this one. Anyone want to help me get started?

 

[mathwonk]

well that is about as complicated a problem i have seen, to bash out a simple idea.

basically it says to try 1/2. if that doesn't work then you try next either 1/4 or 3/4, and more precisely you try 1/4 if 1/2 was too big, and you try 3/4 if 1/2 was too small.

say 1/2 was too small and you next try 3/4. then eiother it works or it doesn't.

if 3/4 is too small you try next 7/8, and if 3/4 is too big try next 5/8.get it? eventually (after the end of the world that is) you have an infinite number of tries x all getting nearer each other and also ll values x^n getting nearer to y.
so you work to prove that the x's converge to something whose mth power must converge to y.

for this you need an axiom that tells yiou when a sequence of =reals converges.

or as this problem puts it, you need an axiom that says a shrinking sequence of nested, bounded, closed intervals contains at least one common point, and exactly one point if they shrink to zero in length.

or maybe it is rigged to use the fact that a bounded monotone sequence of reals converges.
.

to do the problem as asked just put your nose down and try to follow the steps one at a time blindly.

i myself dislike such problems. i mean what do you learn from them? maybe a little technical skill.

 

[mynameisfunk]

THANK YOU. You have made that much clearer. Tomorrow I will work on it and post my work.

 

[mynameisfunk]

OK, I am confused, the first 3 parts seem trivial and i can't imagine doing anything but going through the process a few times and showing that they are all true by induction...
Part (d) however, i do not see how this is true. How am I to know how far away a_n and b_n are from each other? I would have thought they could be anything. I specifically don't see how/why |a₁-b₁|=1/2

 

[mathwonk]

well without looking back i recall the next numbers ere the average of the previous ones so they seem to get closer by one half each time.

 

[mynameisfunk]

eventually (after the end of the world that is) you have an infinite number of tries x all getting nearer each other and also ll values x^n getting nearer to y.
so you work to prove that the x's converge to something whose mth power must converge to y.

for this you need an axiom that tells yiou when a sequence of =reals converges.

or as this problem puts it, you need an axiom that says a shrinking sequence of nested, bounded, closed intervals contains at least one common point, and exactly one point if they shrink to zero in length.

or maybe it is rigged to use the fact that a bounded monotone sequence of reals converges.

Would this theorem be the Bolzano-Weierstrass Theorem??(Every bounded infinite subset of R^k has a limit point in R^k). Specifically for part (e).

 

[Office_Shredder]

|a₁-b₁|=1/2

You start with b₀=1 and a₀=0. Then we check to see if $$\left( \frac{1}{2}^n\right)=y$$. If so, we're done. If not, there are two possibilities: we make a₁=1/2 or we make b₁=1/2, and leave the other number alone depending on whether $$\left(\frac{1}{2}\right)^n$$ is too big or too small.

Then |a₁-b₁| is either |0-1/2| or |1/2-1|

 

[mynameisfunk]

I would like to post my work on this and see if anybody has a problem. I am very unsure on how to do this...
(a)initially, a_n$$\leq$$b_n and a_n$$\leq$$c$$\leq$$b_n. If c^m>y we set a_n=a_n-1 and b_n=c so that we still have a_n-1$$\leq$$a_n$$\leq$$b_n$$\leq$$b_n-1. If c^m<y then we we set a_n=c and b_n=b_n-1 so that we still have a_n-1$$\leq$$a_n$$\leq$$b_n$$\leq$$b_n-1 and the sequence satisfies part (a).

(b)Since all b_n and a_n are reals, they are ordered. Since a₀$$\leq$$a₁$$\leq$$...$$\leq$$a_n$$\leq$$b_n$$\leq$$...$$\leq$$b₁$$\leq$$b₀, then [a_n,b_n]$$\subseteq$$[a_n-k,b_n-k] for any positive integer k$$\leq$$n.

(c) Note that a_n<c and b_n>c. Since in our process, if c^m$$\neq$$y, we shift our nested intervals(infinitely many times if necessary) so that y^1/m is contained in all of them, then the bounds of these sequences, namely a_n and b_n for all n satisfy a_n<y^1/m<b_n$$\Rightarrow$$ a^m_n<y<b^m_n for all n.

(d)Our process of setting either b_n=c or a_n=c depending on whether y^m>c or y^m<c makes our interval smaller by 1/2 each time, where initially, the interval is of length 1 ( |0-1| ). Hence |b_n-a_n|=2^-n

(e) I would like to just use the Bolzano Weierstrass Theorem here... This is as far as I have done..

 

[Office_Shredder]

There are two parts to point e: First is that the intersection is non-empty, and the second is that there is only one point. You can use a convergence argument to get that it's not empty: the sequence a_n is bounded above and increasing, so converges to something we'll call a; the sequence b_n is bounded below and decreasing, so has a limit also called b. By preservation of weak inequalities, $$a\leq b$$ (make sure this is true! otherwise you're stuck in the mud)

Prove that [a,b] is in the infinite intersection, and use that to prove that a=b also

 

[trambolin]

I think the buzzword for this kind is the bisection method.