I'm so glad this channel exists. Despite my enormous passion for the subject, my maths skills mysteriously vanished during high school and I blame much to the overcomplicated way people teach in my country. This explanation is so easy that even a 6 years old kid could understand it. Thank you so much for allowing me to enjoy maths again! ❤❤❤
So what he's basically doing is showing that, (1 + a)^k * (1 + a)^1 >= 1 + (k + 1)a + ka^2 >= 1 + (k + 1)a He never actually drops the ka^2, what he's really doing is saying that since we know ka^2 will be at least zero, we know that 1 + (k + 1)a will always be less than or equal to itself plus that ka^2. Since the we know that (1 + a)^k * (1 + a)^1 >= 1 + (k + 1)a + ka^2 (base on the induction hyp.). And we know that 1 + (k + 1)a + ka^2 >= 1 + (k + 1)a, we can then conclude that (1 + a)^k * (1 + a)^1 >= 1 + (k + 1)a, and that is what needed to be shown.
If the LHS is greater than or equal to the RHS when ka^2 is part of the RHS, then the LHS will STILL be greater than or equal to the RHS when we make the RHS a bit smaller (since ka^2 is either 0, or it's a positive number). E.g. if LHS = 20, RHS = 10. We take away ka^2 = 2, so then RHS becomes 8. 20 is still greater than or equal to 8 so the statement is still true. I guess he didn't really 'take away' ka^2 since he didn't subtract it from both sides, essentially he just removed it from the picture since it didn't affect the inequality.
Nice. From this you can prove a series of inequalities that lead to the famous Gibbs inequality that is important in machine learning (which is why I was here).
Thank you for the video!!... Math: formulas written by people who are lazy, so they short handed everything into hard-to-understand formatting that requires formal education. Sometimes without knowing what level or type of math that's applied, it becomes hard to provide the answer. Suppose x + y = (x+y) is true. prove it. AHH!! Which way?!?! Elementary school style (count those apple, oranges, which are all fruits) or advanced math using proofs that take 10x longer? (Edit for slight grammar issue)
gets a little murky the logic dealing with the sign of (1+a ) at 3:45 , kind of reverse logic. see math.stackexchange.com/questions/181702/proof-by-induction-of-bernoullis-inequality-1xn-ge-1nx Nice video, i am hooked to math (and your videos).
I'm so glad this channel exists. Despite my enormous passion for the subject, my maths skills mysteriously vanished during high school and I blame much to the overcomplicated way people teach in my country. This explanation is so easy that even a 6 years old kid could understand it. Thank you so much for allowing me to enjoy maths again! ❤❤❤
I have been trying to find out why a>-1 and you are the only one who mentions it so thanks a lot.
literally the only video with an unknown im so grateful for you thankuuuuu
😀
@@TheMathSorcerer :D
Can you tell me why ka^2 being greater than 0 enables us to drop the term?
So what he's basically doing is showing that,
(1 + a)^k * (1 + a)^1 >= 1 + (k + 1)a + ka^2 >= 1 + (k + 1)a
He never actually drops the ka^2, what he's really doing is saying that since we know ka^2 will be at least zero, we know that 1 + (k + 1)a will always be less than or equal to itself plus that ka^2. Since the we know that (1 + a)^k * (1 + a)^1 >= 1 + (k + 1)a + ka^2 (base on the induction hyp.). And we know that 1 + (k + 1)a + ka^2 >= 1 + (k + 1)a, we can then conclude that (1 + a)^k * (1 + a)^1 >= 1 + (k + 1)a, and that is what needed to be shown.
Awesome, thanks.
@@patricksalmas1877 Godbless
If the LHS is greater than or equal to the RHS when ka^2 is part of the RHS, then the LHS will STILL be greater than or equal to the RHS when we make the RHS a bit smaller (since ka^2 is either 0, or it's a positive number).
E.g. if LHS = 20, RHS = 10. We take away ka^2 = 2, so then RHS becomes 8. 20 is still greater than or equal to 8 so the statement is still true.
I guess he didn't really 'take away' ka^2 since he didn't subtract it from both sides, essentially he just removed it from the picture since it didn't affect the inequality.
as an example 3+5>3 because 5 is positive we didnt really drop the 5 here
Thanks for the great video!
Dude this is great. Love this vido
Thanks man!
Thank you very much for this video. It was really useful to me. 👍🏼
Nice. From this you can prove a series of inequalities that lead to the famous Gibbs inequality that is important in machine learning (which is why I was here).
Thank you so much
I see you used the theorem to prove it, I really do not get it
Thank you so much.
thank you
Thank you!
np
Thank you for the video!!... Math: formulas written by people who are lazy, so they short handed everything into hard-to-understand formatting that requires formal education. Sometimes without knowing what level or type of math that's applied, it becomes hard to provide the answer. Suppose x + y = (x+y) is true. prove it. AHH!! Which way?!?! Elementary school style (count those apple, oranges, which are all fruits) or advanced math using proofs that take 10x longer? (Edit for slight grammar issue)
Thanks!!
a is greater than -1 .why used ≥ this sign
absolute chad
gotta love this shit
Not so clear but thanks
why it is not a>-1?
Some time ago I saw an article where a>-1 is taken instead of a=>-1. So I guess this guy had a mistake. (sorry for my English)
❤️❤️❤️
well i understand Bernoulli's Inequality but i still hate pure math
where numbers
gets a little murky the logic dealing with the sign of (1+a ) at 3:45 , kind of reverse logic.
see math.stackexchange.com/questions/181702/proof-by-induction-of-bernoullis-inequality-1xn-ge-1nx
Nice video, i am hooked to math (and your videos).
Thank you so much