Solve 4 Hard Math Problems Creatively

[sitedesigner]

http://img325.imageshack.us/img325/1876/mathproblem2gt.jpg

Need help solving these in the MOST CREATIVE way possible.

Thanks

 

[Euclid]

I have a formula for 1. But since this is homework, I can't just blurt out the answer. If you want, I'll tell you my "creative" method:
1) determine the first few a_n
2) make a guess for the general result
3) prove your guess by induction

Good luck.

 

[Gale]

"most creative" is very subjective. Anyway, you need to post your thoughts first on any of these.

i'd say for 3, use the fact that we're in base 10, so the one's place is 1*10^0 son on.. sum each place separately. its straightforward, so i guess that isn't very creative.

for 4, i'd start with the planes next to each other on one side of the circle, and work towards the other side. seems like they'll all end up together that way.

 

[0rthodontist]

Well, for 1. I don't think there is a single formula. It looks like you get a new sequence for each choice of a1 so long as a1 >= 0. The formula for a1 = 1, at least, is not hard though.

 

[Curious3141]

Well, for 1. I don't think there is a single formula. It looks like you get a new sequence for each choice of a1 so long as a1 >= 0. The formula for a1 = 1, at least, is not hard though.

In fact, a_1 has to be one. This is by virtue of the fact that S_1 = a_1 so by the series definition,

a_1 + (1/a_1) = 2a_1

and the only real pos soln of that is a_1 = 1.

From that we can get a_2 = \frac{\sqrt{8} - 2}{2} and so forth.

The general term can be defined recursively as

a_n = \frac{\sqrt{a_{n-1}^2 + \frac{1}{a_{n-1}^2} + 6} - a_{n-1} - \frac{1}{a_{n-1}}}{2}

and that's a fairly ugly expression and I can't put that in a direct closed form.

 

[Euclid]

From that we can get a_2 = \frac{\sqrt{8} - 2}{2} and so forth.

a_2 = \frac{\sqrt{8} - 2}{2}= \sqrt{2} - 1

and notice a_1 = \sqrt{1} - \sqrt{0}

 

[Curious3141]

a_2 = \frac{\sqrt{8} - 2}{2}= \sqrt{2} - 1

and notice a_1 = \sqrt{1} - \sqrt{0}

Ah yes, I see now thanks !

The induction proof of the general form is really very easy. This is a cool problem.

 

[VietDao29]

Need help solving these in the MOST CREATIVE way possible.

Thanks

Please don't demand / tell us to do your problems for you. You should show your thoughts first, and what have you done to tackle these problems.
It's your problems, not ours.
If you don't know how to start off these problems, there are always people out there to help you, if and only if you ask nicely. Don't be rude to people.

 

[0rthodontist]

Oh, you're right, VietDao, about the first element of the sequence.

Besides the closed form, I have a simpler recursive formula, based on the sum and not directly on previous terms:
an = -S(n-1) + sqrt(S(n-1)^2 + 1)

 

[Curious3141]

Oh, you're right, VietDao, about the first element of the sequence.

Besides the closed form, I have a simpler recursive formula, based on the sum and not directly on previous terms:
an = -S(n-1) + sqrt(S(n-1)^2 + 1)

Vietdao ? Do you mean me instead ?

 

[0rthodontist]

Vietdao ? Do you mean me instead ?

Oh, yeah. I meant you.

 

[Curious3141]

Oh, yeah. I meant you.

Cool, that's a nice recursion, just thought the one between individual terms makes calculating terms easier. But Euclid's insight is the winner.

Now, we have to figure out if there's a way to see/prove the expression of a_n directly from the series definition, rather than guessing the pattern then proving by induction. Thoughts ?

 

[0rthodontist]

Well, one can minimize the amount of the proof that depends on induction from the recursion I have. Sn = S(n-1) + an = sqrt(S(n-1)^2 + 1), and from this it is clear by inspection and induction that Sn = sqrt(n). Then an = Sn - S(n - 1).

 

[eigenglue]

For #1 just subsitute S(n) = S(n-1) + a(n) into the equation above and solve the quadratic for a(n). That will give you a(n) in terms of s(n-1). Then substitute back in get S(n) in terms of S(n-1) and the rest will be obvious. Enough of that one.

For #2, consider the two triangles APY and APX. You can use the law of sines on those two triangles to get an expression for 1/AX + 1/AY.

For #3, Write out a difference equation: y(n) = 10*y(n-1) + 1 with initial condition y(0) = 1. That will give you an expression for the '111111' parts. Then write that out as a sum, and you should see how to do the rest.

For #4, the planes can only move one airport at a time, so they can only go around the circle one way or another. You must have two flights per day, so you must make two moves at a time, and thus will always have made an even number of moves. There are two slightly different situations, one with an even number of airports on the circle and the other with an odd number of aiports on the circle. Take a look at each of those.

 

[the_tiger]

let S= 1 + 11 + 111+... => 9*S=9+99+999+...=>
9*S=(10-1)+(100-1)+(1000-1)+...+[(10^n)-1]=(10+100+...+10^n)-n=
10(1+10+...+10^(n-1))-n=10[((10^n)-1)/10-1]-n=10[((10^n)-1)/9]-n =>
S=10[((10^n)-1)/81]-n/9