20070807, 16:24  #1 
May 2004
New York City
3×1,409 Posts 
Range of Inequality
For what (natural number) values of n is this inequality true:
99^{n} + 100^{n} > 101^{n} 
20070807, 17:22  #2  
Nov 2003
2^{5}×233 Posts 
Quote:
We can get an immediate approximation to the answer by observing that (1  a)^n ~ 1  n*a for very small a. Whence (1a)^n + (1b)^n > 1 ==> 2  na  nb > 1 for a,b ~ .01 yields n ~ 50 Indeed, n < 49 is the answer. 

20070807, 17:35  #3 
Aug 2002
Buenos Aires, Argentina
2^{3}·3·5·11 Posts 
It appears that there is a problem with the immediate approximation.
99^n + 100^n > 101^n (99/101)^n + (100/101)^n > 1 (12/101)^n + (11/101)^n > 1 Using Bob's approximation: (1  2n/101) + (1  n/101) > 1 2  3n/101 > 1 n < 101/3 = 33.667 Ouch! It appears that the O(n^2) term is very high to discard it. Last fiddled with by alpertron on 20070807 at 17:38 
20070807, 17:42  #4  
Nov 2003
2^{5}×233 Posts 
Quote:


20070807, 20:28  #5 
∂^{2}ω=0
Sep 2002
República de California
3×3,769 Posts 
I took a slightly different tack to obtain the same firstorder estimate as Bob:
Code:
[x1]^n + x^n > [x+1]^n ==> [x+1]^n  [x1]^n < x^n [x+1]^n = x^n + n*x^(n1) + [n^2/2]*x^(n2) + [n^3/2]*x^(n3) + ... [x1]^n = x^n  n*x^(n1) + [n^2/2]*x^(n2)  [n^3/2]*x^(n3) + ... ==> [x+1]^n  [x1]^n = 2*{n*x^(n1) + [n^3/2]*x^(n3) + ... } < x^n ==> 2*{n*x^(n1) + [n^3/2]*x^(n3) + ... } < x^n Actual breakover occurs between n=48 and 49, so not bad, despite the fact that x ~= 2n means n is *not* small vs x. [That kind of "luck" is common in these kinds of regularperturbation type problems, however.] 
20070807, 21:06  #6 
Aug 2002
Buenos Aires, Argentina
528_{16} Posts 
The number 50 is actually a secondorder estimate because you consider the error in O((1/x)^3). The firstorder estimate is 33.67 as I showed above.
A small correction: [n^3/2]*x^(n3) should be replaced by [n^3/6]*x^(n3) Last fiddled with by alpertron on 20070807 at 21:06 
20070807, 21:48  #7  
∂^{2}ω=0
Sep 2002
República de California
3×3,769 Posts 
I should have said "leadingorder"  as you point out, since I am using what amounts to the analog of a centraldiffference scheme [using the terminology of finitedifference approximations], that leads to a 2ndorder scheme.
Quote:
Anyway, retaining the leading *two* LHS terms in my analysis above gives the breakover point as 2*n*x^(n1) + (n^3/3)*x^(n3) = x^n Plugging in x=100 and using a NewtonRaphson iteration [with f(n) as the function whose zeros we seek and n as the independent variable w.r.to which we take derivatives] converges to n = 48.14056..., which is quite close to the exact value of n = 48.2275... Last fiddled with by ewmayer on 20070807 at 22:03 

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