Can you prove that without using L'Hospital's rule for counting number m and n?
Hint: I did it by using the binomial theorem.
Bonus points if you can prove it for real number m and n, because my proof can't do that and I'd be interested to see it done.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Sort by:
Top NewestUsing the binomial theorem seems overkill. Writing it like this 1−x1−xm∗1−xn1−x makes a much shorter solution apparent.
Log in to reply
How about the extension to real numbers?
It seems like L'Hopital needs to be used.
Log in to reply
This solution goes up to rational numbers,
x→1lim1−xn1−xm=x→1lim(1−xk1)(1+xk1+xk2+⋯+xkn)(1−xk1)(1+xk1+xk2+⋯+xkm)foranyk=x→1lim(1+xk1+xk2+⋯+xkn)(1+xk1+xk2+⋯+xkm)=knkm=nm
Log in to reply
let S=limit as x tends to 1, (1-x^m)/(1-x^n) =limit as p tends to zero (1-e^pm)/(1-e^pn) ,where e^p=x.
applying expansion for e^pm and e^pn in numerator and denominator, S=lim as p tends to zero, (1-(1 + pm/1 + pm^2/2x1 + .........))/ (1-(1+ pn/1 + pn/2x1 +......)) , 1- 1 =0, in numerator and denominator , then taking ' m 'and 'n' outside from numerator and denominator , and substituting p=0 for rest of values, we get m/n. :) .pls correct me frnds, if there is any mistakes in my proof.
Log in to reply
@ Archit Boobna - " Since 1/kh will tend towards infinity..." You need grouping symbols around kh when it is written out horizontally, because of the Order of Operations:
1/(kh)
"We can easily show these (1+h)^2 = 1 + 2h (1+h)^4 = 1 + 4h (1+(1/2)h)^2 = 1 + h"
The above are not true. You cannot use equalities. You can use the equivalents of "leads to."
Log in to reply
well one can easily use taylor series, but a taylor series expansion is in a way equivalent to L'hospital rule when you really think about it, so would you accept that ?
limx−−>11−(1+(x−1))n1−(1+(x−1))m≃1−(1+n(x−1))1−(1+m(x−1))=nm
Log in to reply
I suppose most of math is pretty much equivalent- All this stuff is derived from limits, right? I think that counts.
Log in to reply
Let\quad x=1+h\quad where\quad h\quad tends\quad to\quad positive\quad zero.\\ Now\quad let's\quad solve\\ \lim _{ h\rightarrow 0 }{ \frac { 1+nh }{ 1+(n-k)h } } \\ =\lim _{ h\rightarrow 0 }{ \frac { 1+nh }{ 1+nh-kh } } \\ =\lim _{ h\rightarrow 0 }{ \frac { 1+nh-kh+kh }{ 1+nh-kh } } \\ =\lim _{ h\rightarrow 0 }{ \frac { 1+nh-kh }{ 1+nh-kh } } +\frac { kh }{ 1+nh-kh } \\ =1+\lim _{ h\rightarrow 0 }{ \frac { kh }{ 1+(n-k)h } } \\ =1+\frac { 1 }{ \lim _{ h\rightarrow 0 }{ \frac { 1+(n-k)h }{ kh } } } \\ =1+\frac { 1 }{ \lim _{ h\rightarrow 0 }{ \frac { 1 }{ kh } +\frac { n-k }{ k } } } \\ Since\quad 1/kh\quad will\quad tend\quad towards\quad infinity,\quad and\quad this\quad limit\quad is\quad independent\quad of\quad other\\ "tending0s\quad or\quad tending\quad infinities",\quad \frac { n-k }{ k } \quad (a\quad finite\quad value)\quad can\quad be\quad neglected.\\ =1+\frac { 1 }{ \lim _{ h\rightarrow 0 }{ \frac { 1 }{ kh } } } \\ =1+kh\\ \\ So,\quad \frac { 1+nh }{ 1+(n-k)h } \sim 1+kh\quad as\quad h\quad tends\quad to\quad 0.\\ So,\quad \frac { 1+ah }{ 1+(a-n)h } X\frac { 1+(a-m)h }{ 1+ah } =\frac { 1+(a-m)h }{ 1+(a-n)h } \\ So,\quad \frac { 1+nh }{ 1+mh } =\frac { 1+(a-m)h }{ 1+(a-n)h } \\ Put\quad a=n,\\ \frac { 1+nh }{ 1+mh } =1+(n-m)h\\ If\quad n-m\quad is\quad constant,\quad then\quad \frac { 1+nh }{ 1+mh } \quad is\quad constant.\\ So\quad \frac { 1+(m+k)h }{ 1+mh } =\frac { 1+(m+2k)h }{ 1+(m+k)h } .\quad Put\quad m=0\\ 1+kh=\frac { 1+2kh }{ 1+kh } \\ { (1+kh) }^{ 2 }=1+2kh\\ Put\quad k=1\\ \\ We\quad can\quad easily\quad show\quad these\\ { (1+h) }^{ 2 }=1+2h\\ { (1+h) }^{ 4 }=1+4h\\ \\ { (1+(1/2)h) }^{ 2 }=1+h\\ So,\quad { (1+h) }^{ 1/2 }=1+1/2h\\ \^ \^ \quad I\quad did\quad this\quad just\quad to\quad show\quad it\quad works\quad for\quad fractions\quad also.\\ \\ So,\quad { x }^{ n }=1+nh\quad as\quad x\quad tends\quad to\quad 1\quad and\quad h\quad tends\quad to\quad 0.\\ So\quad \frac { 1-{ x }^{ m } }{ 1-{ x }^{ n } } =\frac { 1-(1+mh) }{ 1-(1+nh) } =\boxed { \frac { m }{ n } }
Log in to reply
Seems hardcore, but I'm a little confused by how you defined your variables at the beginning of the proof. Am I missing something?
Log in to reply
I am sorry, this is a bit confusing because I used "m" and "n" in the beginning also and the end also, but they have completely different roles
Log in to reply