they say this is a strong example but he just pulls that formula at 3:40 outta nowhere, i did the math and idk where the 40 and 80 is coming from wtfffff
You may not need this anymore since you asked 2 months ago, but just in case, here's how: Expanding the three terms individually: (k + 1)^3 = k^3 + 3k^2 + 3k + 1 -(3k + 1)^2 = -3k^2 - 6k -3 2(k + 1) = 2k + 2 So: k^3 + 3k^2 + 3k + 1 -3k^2 - 6k -3 + 2k + 2 Combine like terms: = (k^3 - k) then just add the (k - 2)! back in for the answer of (k^3 - k)(k - 2)!
In the last example, can I write (k+1) as (K-4)+5 and continue the proof that way? Also in that case, I would make sure that my base cases include 8, 9, 10, 11 and 12. Is it possible?
Thank you so much I know you have 69 comments and that number is perfect but I really want to tell you thank you because up until this video I had no idea wth I was doing with strong induction. Thank you Dr. Barrus
This is so far in my research the best discourse about strong induction. Thank you. Still I don't see the usefulness of assuming that g_i =i! for all 3=
Very helpful to a cs student taking fundamental structures and learning the induction method. But please write out your methods. It helps alot looking at your example I kept up with you until you started your expansions then you lost me completely.
I don't understand how you knew [(k+1)k(k-1))](k-2)! is equal to (k+1)! can you explain this? I would never be able to think that this is equal to (k+1)! at the top of my head.
(K+1)! is what we were looking for, so it sorta pops up in the mind as the goal. (K+1)! = (k+1)(k+0)(k-1)(k-2)! (Although the 0 doesn't need to be written).
No, in your induction hypothesis you need to allow the possibility that i equals 1 or 2 or 3 in order to have your proof hold for n=4, n=5, and n=6. In general, you'll typically want to keep your attention on the theorem statement (that g_n = n! for all natural numbers...meaning for all integers n from 1 on) and use that first possible value (1, here) in saying what i should be allowed to start from. (Be careful not to be distracted by the fact that the recurrence applies when n is at least 4--we're not proving the recurrence (we're just using it), so the important thing is to focus on the statement you're trying to prove.)
Thank you! Your explanation made much sense. I was trying to form a pattern for the inequality of i's from the examples so I got slightly confused. I understand that strong induction revolves around the idea of the numbers preceding k to be true but it's hard to establish the grounds for the method sometimes. In time I'll get the hang of it
This was super helpful! I didn't understand how to decide how many base cases to do before watching this.
This is the best explanation of Strong Induction I have found. Thank you!
they say this is a strong example but he just pulls that formula at 3:40 outta nowhere, i did the math and idk where the 40 and 80 is coming from wtfffff
Best introduction on Strong Induction I have seen on UA-cam. Elegantly presented sir.👍
who else is here cuz of the discrete final exam
Speaker of Memes porn hub got boring.. and my proofs finals coming up.
i just came here to cram for my final and honestly im feeling so attacked right now
I am here cuz of my First quiz in Algorithm Design and Analysis Course!
I am here cuz can't solve a problem
*algebra I final
That was wonderful. So clear and to the point.
I have a higher math midterm on Tuesday. You’re a lifesaver!
Excellent tutorial. Thanks!
who else is here cuz of the discrete mid term
lol same
ayy lmao
1 year later ... me too lol
Me too, 1 year later... FML
same
Scroll back up and pay attention! Thank me later
thank you
This was so helpful as an extension to my proofs class, thank you!
Thank you Mr. Barrus from a UCSC CS major!
11:10 How does that go from line two to line three? I have a polynomial in my line three.
You may not need this anymore since you asked 2 months ago, but just in case, here's how:
Expanding the three terms individually:
(k + 1)^3 = k^3 + 3k^2 + 3k + 1
-(3k + 1)^2 = -3k^2 - 6k -3
2(k + 1) = 2k + 2
So:
k^3 + 3k^2 + 3k + 1 -3k^2 - 6k -3 + 2k + 2
Combine like terms:
= (k^3 - k) then just add the (k - 2)! back in for the answer of (k^3 - k)(k - 2)!
Thank you Michael. Great explanation!
In the last example, can I write (k+1) as (K-4)+5 and continue the proof that way? Also in that case, I would make sure that my base cases include 8, 9, 10, 11 and 12. Is it possible?
We love you Mr.Barrus !!!
Thank you so much I know you have 69 comments and that number is perfect but I really want to tell you thank you because up until this video I had no idea wth I was doing with strong induction. Thank you Dr. Barrus
This is so far in my research the best discourse about strong induction.
Thank you.
Still I don't see the usefulness of assuming that
g_i =i! for all 3=
I'm sorry but at video section 1:12 what do mean 10 does go into zero? That doesn't make sense.
meaning 0 can be divided by 10. (0/10).
Ok now that is the correct way to express that 0 divides by 10 not goes into 10. It confuses people.
You can actually say 10 goes into 0, 0 times.
10 divides 0 if and only if there exists an integer q such that 0 = q * 10.
Very helpful to a cs student taking fundamental structures and learning the induction method. But please write out your methods. It helps alot looking at your example I kept up with you until you started your expansions then you lost me completely.
+Kaze No Hito FYI for the first example he used binomial expansion, if knowing that is of any help.
it should say "by the binomial expansion" in a proof if it does that.
@@hathawayamato OH man thank you that makes so much more sense. this was prime 'rest of the fucking owl' material until i saw your comment.
thx for showing the pattern for example 3
I would like to know, if possible, where the example 2 was taken. Is is part of a book ? Which one ?
Thank you
Just so I'm understanding correctly at 6:05 , if K-1
Yes, since it means (less than) OR (equal to), it only needs to meet one of these conditions for it to be true.
You are awesome. Thank you!
Best explanation ! 🔥🔥🔥 Thank you ❤️
I don't understand how you knew [(k+1)k(k-1))](k-2)! is equal to (k+1)!
can you explain this? I would never be able to think that this is equal to (k+1)! at the top of my head.
[(k+1)k(k-1)](k-2)! = (k+1)k(k-1)(k-2)(k-3)...1. So, all natural numbers less than or equal to k+1 multiplied together, so equal to (k+1)!
(K+1)! is what we were looking for, so it sorta pops up in the mind as the goal.
(K+1)! = (k+1)(k+0)(k-1)(k-2)! (Although the 0 doesn't need to be written).
this vido is very helpful at all !!
very lovely explained
great video thank you very much
Very good explanation
Amazing explanation
Great video! Thank you. I just have one question: on the first example, how did you get the '2' in [(k-1) + 2]^5?
Jerry Zamora (k + 1) = (k - 1 + 2)
@@joshuastander2697 i was also confusedan you helped me
In the first example, one does not need strong induction. It can be done with the first principle too.
I whish I could press the like button twice..
Very helpful! Thank you!
Sorry but isnt 80 in the 80(k-1)² supposed to be 120?
Why did k-2 have to be greater than the base case 10?
man wtf is this topic watched the whole thing and understood absolutely nothing
It was such a Great teaching I've ever seen !
Are Strong induction and Structural induction identical?
On example 2, shouldn't i be between 4 and k instead of 1 and k?
No, in your induction hypothesis you need to allow the possibility that i equals 1 or 2 or 3 in order to have your proof hold for n=4, n=5, and n=6. In general, you'll typically want to keep your attention on the theorem statement (that g_n = n! for all natural numbers...meaning for all integers n from 1 on) and use that first possible value (1, here) in saying what i should be allowed to start from. (Be careful not to be distracted by the fact that the recurrence applies when n is at least 4--we're not proving the recurrence (we're just using it), so the important thing is to focus on the statement you're trying to prove.)
Thank you! Your explanation made much sense. I was trying to form a pattern for the inequality of i's from the examples so I got slightly confused. I understand that strong induction revolves around the idea of the numbers preceding k to be true but it's hard to establish the grounds for the method sometimes. In time I'll get the hang of it
Good stuff!
wooow be my teacher
thank you a lot
now i can solve strong induction problems.
This guy sounds Canadian. Are you Canadian?
Thank you!
Uconn represent!
nice
Thank you =)
who else is here for IB discrete math exam this monday
to anyone reading this, please help me with a strong induction proof problem:( i'm stuck
Adamsn ab 12:06
Test Neprošel
who else is here to just learn strong induction for high school olympiads?
Shree Ganesh kindergarten Olympiad’s actually bud
Would be great video if you didnt keep making that weird sounds out of your mouth every 5 seconds.
very lovely explained
Thank you!