I just cranked through it by moving sqrt(n-1) to the other side, squaring both sides, isolating the sqrt(n-1) and then both squaring both sides again. But it's definitely cool to think about solving these problems different ways 👍
Me too! 👍 I did that without really knowing what to expect, I feel lucky that it worked out. What's nice is that after moving sqrt(n-1) to the right side and squaring both sides the first time, you wind up with n on both sides that cancel out. After that, isolating sqrt(n-1) and squaring both sides again makes it really easy to solve for n (assuming for a minute that n is real, then set it to an appropriate integer that satisfies the inequality).
You are correct. The difference of larger numbers square roots become easier to figure out by reciprocals of 1/100 and the reciprocal of the difference. The √n - √(n - 1) reciprocal being √n + √(n - 1) and when it passes 100 makes the solution easier! 👍
We square the equation and isolate the remanining root on one side, then square again. The n² terms cancel out so we get n > 100020001 / 40000 = 2500.500025 so n = 2501.
I was thinking of a derivative. A bound of 1/100 in the difference means that we're looking for a place on f(x) = sqrt x where f(x) climbs less than 1/100 when x is incremented by 1. But that's basically the slope. So for a starting point, we ought to look for where the derivative of sqrt x is 1/100. Surprise, it's exactly 2500. So we check it, increment it to 2501, bam.
@@maxhagenauer24 that actually ends up being more or less the same. Instead of Lim {h->0} sqrt x - sqrt x-h / h, simply evaluate it when h = 1. You end up with a derivative approximate for x = x0, which you can reverse as an approximate for sqrt x0 - sqrt x0 - 1 since the denominator is just 1. It's more of using the derivative to REALLY hone down the upper and lower bounds of the estimate that you did in the beginning. Since f' is monotonically decreasing on x > 0, solving for where the derivative is precisely 1/100 will serve at the lower bound, then it's just a matter of checking sqrt 2500 - sqrt 2499 vs sqrt 2501 - sqrt 2500
This is a very simple problem: it boils down to solving (sqrt(N)-1/100)^250*(1+1/10000). Therefore, N>2500+0.5+1/100000000. So N = 2500+1 = 2501, since N is an integer. Done
I went about it less creatively, but still efficiently I think. Cross multiplication and radical separation gives 100sqrt(n)-1(10^8+2×10^4+1)/(4×10^4). The right side can be rearranged to (1/4)×(10^4+2+1/10^4), or 2500 + (1/4)×(2+1/10^4). The second term of this sum is clearly positive but less than 1, hence the sum is somewhere between 2500 and 2501, which brings to the conclusion
Nice solution. That's a handy trick I didn't know that could be useful in other types of problem. I looked at sqrt(n) - sqrt(n-1) = 1/100 and figured it was a close approximation to d(sqrt(n))/d(n) = 1/100. This gives 1/(2sqrt(n)) = 100, which gives n=2500. At this point n=2499 and n=2501 haven't been ruled out (because of the approximation), but this is possible with a little more calculus.
Set n=m^2 and substitute. You get sqrt(m^2-1)>100-m. Square both sides you get m^2-1>10,000-200m+m^2 ==> 200m>10001 ==> m>50 ==> n > 2500. So n = 2501
I just cranked through it by moving sqrt(n-1) to the other side, squaring both sides, isolating the sqrt(n-1) and then both squaring both sides again. But it's definitely cool to think about solving these problems different ways 👍
Me too! 👍 I did that without really knowing what to expect, I feel lucky that it worked out. What's nice is that after moving sqrt(n-1) to the right side and squaring both sides the first time, you wind up with n on both sides that cancel out. After that, isolating sqrt(n-1) and squaring both sides again makes it really easy to solve for n (assuming for a minute that n is real, then set it to an appropriate integer that satisfies the inequality).
You are correct. The difference of larger numbers square roots become easier to figure out by reciprocals of 1/100 and the reciprocal of the difference. The √n - √(n - 1) reciprocal being √n + √(n - 1) and when it passes 100 makes the solution easier! 👍
We square the equation and isolate the remanining root on one side, then square again. The n² terms cancel out so we get n > 100020001 / 40000 = 2500.500025 so n = 2501.
Wow!
Just arranging LHS and RHS, and then squaring both sides yield the same result without all the fuss.
Thanks as always ;)
I was thinking of a derivative. A bound of 1/100 in the difference means that we're looking for a place on f(x) = sqrt x where f(x) climbs less than 1/100 when x is incremented by 1. But that's basically the slope. So for a starting point, we ought to look for where the derivative of sqrt x is 1/100. Surprise, it's exactly 2500. So we check it, increment it to 2501, bam.
But that's for minimum of sqrt(n), not of sqrt(n) - sqrt(n - 1).
@@maxhagenauer24 that actually ends up being more or less the same. Instead of Lim {h->0} sqrt x - sqrt x-h / h, simply evaluate it when h = 1.
You end up with a derivative approximate for x = x0, which you can reverse as an approximate for sqrt x0 - sqrt x0 - 1 since the denominator is just 1.
It's more of using the derivative to REALLY hone down the upper and lower bounds of the estimate that you did in the beginning. Since f' is monotonically decreasing on x > 0, solving for where the derivative is precisely 1/100 will serve at the lower bound, then it's just a matter of checking sqrt 2500 - sqrt 2499 vs sqrt 2501 - sqrt 2500
👍If we make the inequality (from the beginning equal to 1/100, and solve it for the value of (n)
This is a very simple problem: it boils down to solving (sqrt(N)-1/100)^250*(1+1/10000). Therefore, N>2500+0.5+1/100000000.
So N = 2500+1 = 2501, since N is an integer. Done
I went about it less creatively, but still efficiently I think. Cross multiplication and radical separation gives 100sqrt(n)-1(10^8+2×10^4+1)/(4×10^4). The right side can be rearranged to (1/4)×(10^4+2+1/10^4), or 2500 + (1/4)×(2+1/10^4). The second term of this sum is clearly positive but less than 1, hence the sum is somewhere between 2500 and 2501, which brings to the conclusion
clear answer !
Ｉcompletely understood.
Glad to hear that!
For real values instead of integers exact solution is: x > 2500 + 20001/40000
There can be no "exact" solution to this problem.
Nice!
Thanks!
Nice solution. That's a handy trick I didn't know that could be useful in other types of problem.
I looked at sqrt(n) - sqrt(n-1) = 1/100 and figured it was a close approximation to d(sqrt(n))/d(n) = 1/100. This gives 1/(2sqrt(n)) = 100, which gives n=2500.
At this point n=2499 and n=2501 haven't been ruled out (because of the approximation), but this is possible with a little more calculus.
Noice