Back to your roots
Evaluate rt(1 + rt(7 + rt( 1 + rt(7+......)))), where rt means square root. I have designed this so that the answer is an integer.
Labels: mathschallenge
posted by Chris at
7:50 PM
Friday, February 5, 2010
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14 Comments:
Chris...i can delete my own comments easy enough,but can't delete others(like yours in the last post-as you wanted me to)
how?
2
Knightmare. You have to do it from the comments editor; just scroll to it in that window - you should see a dustbin by every post. Sorry, I had forgotten that.
Jon. 2 it is. How did you solve it?
Knightmare. I just completely removed your deleted post. Tee hee.
Chris...i've got a grip on my super powers now.
behold and bewarned...i too,wield the forces of the invincible blog author.haahhhhahaaaaa(evil laugh)
but i still have to kick!
.. but you can delete the kicks completely and keep the page tidy. You'll also notice that the posts end up in the comment window, even if they haven't flushed trough to the main page.
so what you are saying is that with great power comes great responsibility
Oh yes! I hope you can withstand the strain of it. Pequod will be buckling (even more than usual) under the load. May the force be with her.
Chris...LOL
Without calculation i can tell it has to be "2". Analytical reasoning that is!
oookkk.... leaving aside the super powers you have got (be warned people, i have them to!! muhahahahaha)
the question's solution:
let x = rt(1 + rt(7 + rt( 1 + rt(7+......))))
then, x = rt(1 + rt(7 + x)
PS: since x is an infinite series
=> x^2 = 1 + rt(7+x)
=> x^2 - 1 = rt(7+x)
solving gives that x = 2. I don't have enough time to do it manually, so i used a program :P
btw, Chris, thanks for the prep about cube roots of unity. We did it yesterday, and i'm guessing I was the only one who understood it (cuz of your explanation, and a little of my own research after that).
PS: just a note, i won't be able to come to ToM for about... umm... till April 2nd week or so 0o. so, cya later (... a looooong while later)
Let x = rt(1 + rt(7 + rt( 1 + rt(7+......
See that x = rt(1 +rt(7 + x))
You can at this point try trial and error. In fact if you try
any initial x ≥ -7, then calculate rt(1 +rt(7 + x)), you'll get a
number that is closer to 2. Pop that back in again, over and over
again, you quickly see that you get closer and closer to 2. If you
put 2 in, you get 2 out.
x² - 1 = rt(7 + x) => (x² - 1)² = 7 + x => x^4 - 2x² - x - 6 = 0.
If you take advantage of the insight that x = 2 is a root, then
you may write (x-2)(x³ + 2x² +2x +3). However, numerical methods
are about the only practical way to solve the initial quartic.
Go to www.wolframalpha.com and enter NSolve[x^4 -2x² -x -6 == 0]
You'll get 4 roots: -1.8105, -0.09473 + (+/-) i 1.28374 and 2.
However, whichever of these values you plug in, you get an output
that is closer to 2.
Bit of a rubbish problem. Handling quartics isn't particularly
easy (without a computer). I probably had more fun choosng
a and b such that x = rt(a + rt(b + x)) = 2
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