Can't we argue as follows: ( f(x) -f(x+1))/f(x) = 1 - f(x+1)/f(x). So, the given integral becomes integral, from 0 to infinity, of dx minus another integral. This first integral diverges. So the given integral diverges.
a diverging integral subtracted from another diverging integral could converge, consider the integral of 1/x + 1/x^2 - the integral of 1/x, both integrals diverge, however, when subtracted it's just the integral of 1/x^2 and that converges
Nice solution sir Can you solve this (nx)ⁿ+(nx)+(n)=0 In this equation it does not ask from international mathematics Olympiad Just i created this equations. Please solve this equations sir!!
wait why can we write this inequality at 6:05 if the inequality 1-x >= e^(-2x) is applicable for x being small? at a certain point the inequality wouldn't be true or am i not getting something?
@ yeah but if x is small, wouldn't (x + n) be too big for a certain integer n to substitute in this inequality? or the only condition is x being positive?
What part of this video makes you think this is an April Fools video? Nothing in this video fools anyone else. You are such a weirdo. Keep being confused
I have an idea for a solution, but I don't know how to add rigor to it: . We know f(x) is strictly decreasing and it converges to 0. I don't know how to prove this part, but logically, at some point, the slope keeps flattening out or approaching 0. Let's call this point "N". The expression f(x)-f(x+1) is -1 times the average rate of change from x to x+1. Because we just said at some point N, the derivative keeps flattening out, we know that at any x-value greater than or equal to N, f'(x) is steeper than f(x)-f(x+1). Given that the function is strictly decreasing, we can say more specifically that -f'(x) > f(x) - f(x+1). Given this inequality, we can say that replacing the numerator of the integral with -f'(x) will yield a greater result of the integral. So let's plug it in. By doing so, you can take the antiderivative with ease and get --ln[f(x)]. Remember, we assumed the inequality with the derivative kicks in after x=N, so the integral has to be bounded below and above by N and infinity respectively. Remember, because the original function converges to 0, we get ln(0)-ln(f(N)) as our solution to the integral, and ln(0) blows up and diverges, so by comparison, so must the original integral. QED BABY!!!!!
This is absolutely RIGOROUS proof speaking to the points needed for the proof and also very sophisticated work presented. Nice work professor
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True mathematician
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My Prof Ralph Masenge wrote a book about it In Tanzania
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My favorite math professor and RUclipsr😍
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Can't we argue as follows:
( f(x) -f(x+1))/f(x) = 1 - f(x+1)/f(x). So, the given integral becomes integral, from 0 to infinity, of dx minus another integral. This first integral diverges. So the given integral diverges.
a diverging integral subtracted from another diverging integral could converge, consider the integral of 1/x + 1/x^2 - the integral of 1/x, both integrals diverge, however, when subtracted it's just the integral of 1/x^2 and that converges
@@UnofficialKoala True
Well said Koala👍👍👍
Nice solution sir
Can you solve this
(nx)ⁿ+(nx)+(n)=0
In this equation it does not ask from international mathematics Olympiad
Just i created this equations.
Please solve this equations sir!!
Wow nice question you made! For sure haha👍👍👍
Can you plz explain why limg(x)=0 as x tends to infinity?
Very nice video professor!🎉
Thanks a lot my friend haha👍👍👍
Another cool video
Thanks a lot my friend haha👍👍👍
Definitely rigorous video. No need to explain anything other than this to prove it. Well done professor
Haha thanks a lot my friend for your support👍👍👍
wait why can we write this inequality at 6:05 if the inequality 1-x >= e^(-2x) is applicable for x being small? at a certain point the inequality wouldn't be true or am i not getting something?
I already designated x to be positive and x to be small. Look at their graphs. When x is a small positive number, see which graphs y is greater
@ yeah but if x is small, wouldn't (x + n) be too big for a certain integer n to substitute in this inequality? or the only condition is x being positive?
Please watch the video.
@@datfry7791 Why x+n? He said x, why x+n to consider on your own?
@@drpkmath12345my bad! got a closer look it's g(x) for all x positive. got it sorry!
Pk running after integral
Integral video is scheduled already at 1pm today👍👍👍
Love to see u working on basics of alzebra n all 4 learning
It converses, not diverses 😂
Lol not funny😂
This looks like an April fools video, but it was made in October, so I’m confused
What part of this video makes you think this is an April Fools video? Nothing in this video fools anyone else. You are such a weirdo. Keep being confused
I have an idea for a solution, but I don't know how to add rigor to it:
.
We know f(x) is strictly decreasing and it converges to 0. I don't know how to prove this part, but logically, at some point, the slope keeps flattening out or approaching 0. Let's call this point "N". The expression f(x)-f(x+1) is -1 times the average rate of change from x to x+1. Because we just said at some point N, the derivative keeps flattening out, we know that at any x-value greater than or equal to N, f'(x) is steeper than f(x)-f(x+1). Given that the function is strictly decreasing, we can say more specifically that -f'(x) > f(x) - f(x+1).
Given this inequality, we can say that replacing the numerator of the integral with -f'(x) will yield a greater result of the integral. So let's plug it in. By doing so, you can take the antiderivative with ease and get --ln[f(x)]. Remember, we assumed the inequality with the derivative kicks in after x=N, so the integral has to be bounded below and above by N and infinity respectively. Remember, because the original function converges to 0, we get ln(0)-ln(f(N)) as our solution to the integral, and ln(0) blows up and diverges, so by comparison, so must the original integral. QED BABY!!!!!
Haha nice sharing this my friend👍👍👍