Do you know Addition
posted by ToM at
5:48 PM
Sunday, August 16, 2009
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29 Comments:
One hundred and one dollars and 50 cents is larger.
Do you have trouble reading dollars and cents?
I think you have no clue what you are talking about. I assume these are powers
Ok, just take the take each to the 1/50 power and your asking is which is more, 199 or 101.
I've been trying pretty much but I didn't come too far with this one.
I will denote here by (n~k) "n over k" the binomial coefficient defined as: n!/(k!*(n-k)!)
We know that (a+b)^n=SUM{k=0->n}[(n~k)*a^(n-k)*b^k]
and equivalent with (a-b)^n=SUM{k=0->n} [(n~k)*(-1)^k*a^(n-k)*b^k]
We want to know if 99^50+100^50 OR 101^50 is bigger and in other words is 101^50-99^50 OR 100^50 bigger?
Why do i formulate it like this, we can now but it in the form is (a+1)^n+(a-1)^n < a^n ?
Because of b=1 our sums reduces to (a+-b)^n=Sum{k=0->n} [(n~k)*(+-1)^k*a^(n-k)]
When we subtract, every even term kills out (-1^(2,4,6...)=1) and we stay with the uneven k=>2k+1: (a+1)^n-(a-1)^n = SUM{k=0->n} [(n~(2k+1))*a^(n-2k-1)]
Now the question is reduced to: What is bigger, SUM{k=0->n} [(n~(2k+1))*a^(n-2k-1)] OR a^k ??
In numbers: SUM{k=0->50} [(50~(2k+1))*a^(49-2k)] OR 100^50 ?
I am pretty sure a^k is bigger but I can't prove it :((
Maybe that helps some guy, have luck ;)
the condition is of course (a+1)^n - (a-1)^n < a^n and not "plus"
99^50 + 100^50 ≈ 1.605*10^100
101^50 ≈ 1.645*10^100
Darned good problem Tom. Thanks.
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99^50 + 100^50 vs 101^50
[tha b...:...H was so close, if only he'd expanded a few terms at the right end.]
Consider (via binomial theorem)
(x+1)^50-(x-1)^50 = 100x^49 + (terms with +ve coeffs * x^odd power)
So LHS > 100 x^49 (if x > 0)
Subst x=100 => 101^50 - 99^50 > 100*100^49
But 100*100^49 = 100^50
Rearranging => 101^50 > 99^50 + 100^50
PS having done it the hard way, I remembered tha b...:..H's link
to wolframalpha, here's mine:
http://www.wolframalpha.com/input/?i=Expand%5B%28x+%2B+1%29%5E50+-+%28x+-+1%29%5E50%5D
you guys are making it too complicated...
i think its 101^50 because i used simpler numbers to solve:
2^2 + 3^2 < 4^2
= 4 + 9 < 16
= 13 < 16
thus the larger number (101^50) is greater. could be wrong though...
Huzzy, I had thought of trying an inductive proof that
(2n+1)^n - (2n-1)^n > (2n)^n
(this seems to be true for n > 2),
but had decided that the binomial theorem would get me there quickly enough. The induction operation looked like it was going nowhere - I gave it up very quickly.
I know the original inequality is true, you only suspect it. Just as I only suspect (but very strongly) the more general one is true - I confident that the binomial theorem approach would establish it, but I'm not going to bother.
Huzzy, I forgot to include the most important comment. That is that just because you can see a pattern in a mathematical trend, it doesn't mean that the trend is generally true. Many famous cock-ups have been made by not realising this.
Here is a genuine series, 1,2,4,8,16,31 (You'd probably think the 31 should have been 32). It comes from considering partitioning a disc with connected chords. In principle, the deviation from the expected pattern could have been at the millionth term.
I couldn't resist. Binomial (once again)=>
(x+1)^n - (x-1)^n > 2n*x^(n-1)
(x>0 n even, x<>0 n odd)
Head starting to swim, so losing confidence now.
Again, I tried induction (on n), looks equally hopeless.
Oh yeah, also checked that my previous conjecture was true.
----
My excuse for all this pontification is that I've got a broken leg and need something to pass the time.
(violins fade out...)
Ragknot, I just reread you $ and cents answer. If you'd stuck with it, yuo could have said $99.50 + $100.50 = $200.00 and is > $101.50. You would have had an nice out of the box answer.
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I believe it is the last one (101 to the power of 50). I don't know why you people have to put all these calculations in order to figure it out. All you need to know is 101 is bigger than 100 and 99.
Comment posted by: Vincent Jumper
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VJ. Your logic doesn't work. If what you say were true then
you would have to say 101 > 99 + 100, which is obviously false. If
the problem had been comparing 99^48 + 100^48 with 101^48 (or any
lesser power) you'd have got that wrong also.
It takes a little experience to be able to see that possibility
right away. You haven't got it yet, but you can learn.
The fact that you are posting all over the place here, tells me
that you do have some interest in maths. I can see that you are
not very experienced at it, you haven't got the hang of operator
precedence e.g. 2*3+4*5 = (2*3)+(4*5) = 25. I suspect that you
might calculate it as 2*(3+4)*5 = 70.
I suggest that you brush up your algebra. And read some of what's
been posted, and if you think something's wrong, check your facts
and form your own conclusions.
Especially don't be lazy, try to finish solving a problem. Don't
just accept the first vague idea that you think of. Test it if
possible, slightly change the problem and see what happens.
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