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Those with no math ability - stay out

DrPizza

DrPizza

Administrator Elite Member Goat Whisperer
Obviously,
sqrt(0 + sqrt(0 + sqrt(0 + sqrt(0 +.....
infinitely many nested square roots of 0 plus the next square root..
= 0


Now, this is rather non-intuitive:
replace 0 with x and take the limit as x approaches 0 instead.

Does the answer approach zero? No!
The limit is actually 1

Lim (as x -> 0) of sqrt(x + sqrt(x + sqrt(x + sqrt(x +....

= 1

Picture of the equation here:


If you didn't understand that .999... = EXACTLY 1, then this thread isn't for you 🙂
This thread is to challenge the minds of those who have taken calculus, although to demonstrate the validity of the statement takes less than calculus.
 
I figured since some of the people in here actually do attend college and have taken calculus, they might appreciate the non-intuitive nature of this problem. The rest of you high school kiddies, well, you just keep enjoying your christmas break. 😛
 
Maybe I should start a thread for the rest of you... 2 + 2 = 5 for large values of 2
(true statement, not a joke actually)
 
Originally posted by: DrPizza
I figured since some of the people in here actually do attend college and have taken calculus, they might appreciate the non-intuitive nature of this problem. The rest of you high school kiddies, well, you just keep enjoying your christmas break. 😛
Click to expand...

well it makes perfect sense - probably a non-convergant limit but I can't recall how to express your nested square roots. Some kind of a series?
 
Originally posted by: DrPizza
I figured since some of the people in here actually do attend college and have taken calculus, they might appreciate the non-intuitive nature of this problem. The rest of you high school kiddies, well, you just keep enjoying your christmas break. 😛
Click to expand...

maybe you forgot, this is ATOT
 
That's pretty interesting. And the people who posted below me have clarified it in my mind, thanks. 🙂
 
I don't get wat you want to discuss as it is rather obvious.

If you take the sqrt of 0 the answer will be 0 and not matter how many times you nest that it will always be 0.

The sqrt of a number between 0 and 1 is always larger than the number you are taking the square root of as two numbers between 0 and 1 give a smaller number when multiplied together. does not matter what you take x to be if you nest it deep enough the answer will always be 0.999...

What is so odd about this? It is just common sense.
 
Originally posted by: DrPizza
I figured since some of the people in here actually do attend college and have taken calculus, they might appreciate the non-intuitive nature of this problem. The rest of you high school kiddies, well, you just keep enjoying your christmas break. 😛
Click to expand...

What about those of us who have been out of college for years and see no use for this knowledge and do not find it all that interesting despite having taken lots of calculus?

Might as well come up with a random sorting algorithm and tell us that it is O(log(n)). I find it more interesting to discuss how uninteresting this factoid-oid is than to actually discuss the factoid-oid itself.

Edit: though in retrospect the sorting algorithm sounds kind of fun.
 
Interesting. It's "intuitively plausible" to me since for numbers less than 1, the square root is larger than the number itself (e.g. sqrt(1/100) = 1/10).

I'm too lazy to work it out on my coffee break though -- it would use too much of my precious nef time 🙂

 
Well, there are two issues here: First, Limit as x->0 is NOT the same as plugging in x=0 and evaluating, even when the latter is defined.
Secondly, the limit essentially boils down to the equation y^2 = y, which (surpise, surprise) has two solutions y=0 and y=1.
 
Originally posted by: DrPizza
Maybe I should start a thread for the rest of you... 2 + 2 = 5 for large values of 2
(true statement, not a joke actually)
Click to expand...

Soo, you saying if I order 2 small pizzas of larger size, I get more than if I order 1 large pizza?
I knew math could be used in real life somehow... I'll take mine with extra cheese pls, thanks!
 
That's because as you multiple decimals the numbers gets smaller because you have a fraction of the original number. Makes perfect sense if you think about it.
 
Umm... I've been out of college for a while now. Not a complete explanation, but still:
If we rewrite the square root series into smth like:

(x + (x + (x+....)^ 1/2) ^1/2) ^1/2

because x approaches 0, but is never equal to, you can see, the sequence essentially evaluates to 1.
 
Originally posted by: mitch2891
I don't get wat you want to discuss as it is rather obvious.

If you take the sqrt of 0 the answer will be 0 and not matter how many times you nest that it will always be 0.

The sqrt of a number between 0 and 1 is always larger than the number you are taking the square root of as two numbers between 0 and 1 give a smaller number when multiplied together. does not matter what you take x to be if you nest it deep enough the answer will always be 0.999...

What is so odd about this? It is just common sense.
Click to expand...

I was thinking the same thing
 
You try to graph it? ...it is a root function so it is going to approach a value infinately...
 
Pretty straightforward, as mentioned above. Nothing surprising here, move along. 🙂 (like you've never seen a graph with a hole in it? whoop-dee-do)
 
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