Proof by induction is a simple subset of proof by contradiction, inverted. Proof by contradiction is the almighty gift in maths and general logic. Roughly, to get induction from contradiction, assume that N is the first whole number for which your induction hypothesis H does not hold. That is, assume H(N) is false while every H(n
@@Mothuzad Not really, actually. Its the other way around; contradiction derives from Induction. Why? Simoly put, the Peano Axioms: Natural numbers are defined _by_ induction. Induction is a set property of the natural numbers (the other 2 being uniqueness and incrementing), and all other properties are derived.
Hi Steve, I sat the STEP a few weeks ago and there was a question I think you might enjoy. It contained 4 non-elementary integrals including integrating sinx / x by parts. you could check it out when the paper is released to the public
Here is a nice way to derive the sum of cubes formula that I learned from the book 'A Primer of Analytic Number Theory' by Stopple. Let s(n) = 1+2+3+...+n. We want to show that s(n)^2 = 1^3 + 2^3 + ... + n^3. This follows from two simpler identities: s(n) - s(n-1) = n s(n) + s(n-1) = n^2. This can be seen by writing (1 + 2 + ... + n) + (n-1 + n-2 + ... + 1 + 0) = n + n + ... + n (n times) Multiplying these two equations together we get s(n)^2 - s(n-1)^2 = n^3. So we have the telescoping sum 1^3 + 2^3 + ... + n^3 = (s(1)^2 - s(0)^2) + (s(2)^2 - s(1)^2) + ... + (s(n)^2 - s(n-1)^2) = s(n)^2
I think this was actually a practice problem in my number theory book, at the time I think I proved it with mathematical induction, because it was a simple algebra problem at that point.
Also, as it turns out that if you have n terms, of n, that also works. So (1)^2=1^3, (2+2)^2=2^3+2^3, (skipping 3 bc it was shown), (4+4+4+4)^2=4^3+4^3+4^3+4^3, etc
Answers include permutations of (1,2,3) and (a,-a,0) which are also solutions to x+y+z=x*y*z. But permutations of (a,-a,1), (0,1,2), (0,2,2) work, too. And of course (3,3,3). There might be rational, irrational or complex solutions, by I did not bother to find them.
Your videos are always enlightening! This is a good example of mathematical induction at work, but Nicomachus solved the same problem using geometry several hundred years ago. Now that is really impressive.
I just showed a really nice geometric proof to my 10 years old kid. All you need is to draw a square of side 1+2+3+..., draw lines from the left at distance 1, 2, 3, ..., and similar lines from the top. One 1^2, 2^2, 3^2, ... comes out naturally. The other squares of similar sizes can be found by combining rectangles from top to bottom with rectangles from right to left. All it remains is counting.
After that apply x^2 & x^3 (mod 4) LHS=perfect square=0,1(mod4) RHS=x^3+y^3+z^3 where x^3, y^3, z^3 could be 0,1,3(mod4) Now we only get 3 conditions to make RHS equal to LHS: (3,1,0), (1,0,0) & (3,1,1)
I think you can bring the x inside the log on the left as an exponent, then do something to remove the log on both sides to leave you with something easier to work with. You'd have to be careful about that though as (e.g.) what if a, b, or c are negative?
Is this only true with natural numbers? Because like (a + 2a + 3a)² = 36a² and a³ + 8a³ + 27a³ = 36a³ and a² ≠ a³ Oh wait that means that the identity is true if the "a" is equal to 1 or 0. Nice
(3 + 3 + 3)^2 was not just a pure coincidence. It was just (2 + 3 + 4)^2 which again forms a series of n(n+1)/2. Math always has a reason for everything
Bcoz sinx is equal to x for veeery smaaal x And it is actually 0.99999999..... so limit x tending to zero for [sinx/x] is equal to zero where [°]is greatest interger function or floor function... Hope it helps...
I watched your video where at the end it showed SQRT(18) - SQRT(8), and I got it; but when I plugged my own numbers of SQRT(29) - SQRT (12) I was still stumped.
That little black square is just a common textbook symbol for "end of proof" (sometimes also seen at the end of an example). It means the same thing as QED.
It feels like cheating because you can only use induction once you know the formula from another source. Induction can't find you the formula, it can only prove that the formula is true once you already know it.
Is this only applicable on numerical numbers or can we express (a+b+c)^2 as a³+b³+c³ as well because when I searched this on net I didn't get any results
You can start with any integer. It doesn't have to be 1 or even positive really. The idea is to prove that if it's true for SOME n, it must be true for n+1. So if it's true for n=-3, it must be true for -2. Now since it's true for n=-2, it has to be true for n=-1, and so on.
Interesting,could you turn this into a generic problem, eg, Given a set of integers S=a,b,c,... Find all n,m such that (sum S)^n=sum(S^m) And start with the trivial examples like {1,2}, {1,2,3},{2,3},{1,2,3,4}... And try to see if there are patterns.
Find all the positives? Maybe you mean all positive integers, bc there are infinite many of positive (real) triples exist that satisfies (x+y+z)^2=x^3+y^3+z^3
The first 1,000 people to use the link will get a 1 month free trial of Skillshare: skl.sh/blackpenredpen07221
Thanks
48 minutes late (to this comment ?)
Good
What is/are some good mathematical journal(s) you recommend?
Can't wait to use this to prank all my imaginary friends thanks
Lol
I love my √-1 friends 😁
Imaginary friends???????????????????????????????
Your friends sound like they’re rather complex
@@peted2783 and i am one of them
One of the most suprising summation identities i would say. It still kind of blows my mind how perfect a coincidence it is.
nothing is coincidence in maths.
@@thecritiquer9407 … then how come 64/16 equals 4 when you cancel the 6 ?
@@TheOiseau that's not a coincidence 🤦♂️ that's magic 😂
@@TheOiseau Before Einstein physicists thought that gravity acting as an acceleration was just a quirky coincidence
@@TheOiseau thats like saying why is 100 10 when u divide it by 10
Other math pranks such as
log1 + log2 + log3 = log(1+2+3) = log6
3! = 3+2+1
Also....
3!=1+2+3
230-220 x 0.5 = 5!
@@kyamb3890 wowww
@@kyamb3890 bro forgor pemdas and bodmas exist
mathematical induction is literally almighty gift in mathematics
Proof by induction is a simple subset of proof by contradiction, inverted. Proof by contradiction is the almighty gift in maths and general logic.
Roughly, to get induction from contradiction, assume that N is the first whole number for which your induction hypothesis H does not hold. That is, assume H(N) is false while every H(n
@@Mothuzad You are just saying that was is not false is true here ;)
@@Mothuzad Not really, actually. Its the other way around; contradiction derives from Induction. Why?
Simoly put, the Peano Axioms: Natural numbers are defined _by_ induction. Induction is a set property of the natural numbers (the other 2 being uniqueness and incrementing), and all other properties are derived.
Hi Steve, I sat the STEP a few weeks ago and there was a question I think you might enjoy. It contained 4 non-elementary integrals including integrating sinx / x by parts. you could check it out when the paper is released to the public
Yea. I will see. Thanks for letting me know. 😃
Really you have find that ??
Was that STEP 2?
bro the infinite product one would be sick for this channel
Here is a nice way to derive the sum of cubes formula that I learned from the book 'A Primer of Analytic Number Theory' by Stopple.
Let s(n) = 1+2+3+...+n. We want to show that s(n)^2 = 1^3 + 2^3 + ... + n^3. This follows from two simpler identities:
s(n) - s(n-1) = n
s(n) + s(n-1) = n^2. This can be seen by writing (1 + 2 + ... + n) + (n-1 + n-2 + ... + 1 + 0) = n + n + ... + n (n times)
Multiplying these two equations together we get s(n)^2 - s(n-1)^2 = n^3. So we have the telescoping sum
1^3 + 2^3 + ... + n^3 = (s(1)^2 - s(0)^2) + (s(2)^2 - s(1)^2) + ... + (s(n)^2 - s(n-1)^2) = s(n)^2
I think this was actually a practice problem in my number theory book, at the time I think I proved it with mathematical induction, because it was a simple algebra problem at that point.
Hermoso
For the last question the answer can be (3,3,3) (3,2,1) (1,3,2) (1,2,3) (3,1,2) (2,3,1) (2,1,3)
Also, as it turns out that if you have n terms, of n, that also works. So (1)^2=1^3, (2+2)^2=2^3+2^3, (skipping 3 bc it was shown), (4+4+4+4)^2=4^3+4^3+4^3+4^3, etc
Nice 👍!!
@@cosmicvoidtree
In general (n+n+n+n...n times)^2 = n^3 + n^3 + n^3 ...n times
As,
LHS: (n*n)^2 = n^4 = n*n^3 = n^3 + n^3 + n^3 .....n times : RHS
But what about 0 ?
(0;0;0) is a solution
(0;1;2) works too
@@miyo.7792 but is 0 a positive number?
Answers include permutations of (1,2,3) and (a,-a,0) which are also solutions to x+y+z=x*y*z. But permutations of (a,-a,1), (0,1,2), (0,2,2) work, too. And of course (3,3,3).
There might be rational, irrational or complex solutions, by I did not bother to find them.
Your videos are always enlightening! This is a good example of mathematical induction at work, but Nicomachus solved the same problem using geometry several hundred years ago. Now that is really impressive.
Thank It is so useful
(X+Y)^2: “I think I forgot something important.”
X^2+Y^2: ”If you didn’t remember it, then it probably wasn’t important.”
(X+Y)^2: “Yeah, you’re right.”
2XY: -_-
LoL 😂✨🤣
🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣
I just showed a really nice geometric proof to my 10 years old kid. All you need is to draw a square of side 1+2+3+..., draw lines from the left at distance 1, 2, 3, ..., and similar lines from the top. One 1^2, 2^2, 3^2, ... comes out naturally. The other squares of similar sizes can be found by combining rectangles from top to bottom with rectangles from right to left. All it remains is counting.
My favorite number prank is root(3) to the power of 3 is equal to root(3) times 3
For non-trivial solution. Let x become the highest number, we can deduce that
(x+y+z)^2
After that apply x^2 & x^3 (mod 4)
LHS=perfect square=0,1(mod4)
RHS=x^3+y^3+z^3 where x^3, y^3, z^3 could be 0,1,3(mod4)
Now we only get 3 conditions to make RHS equal to LHS:
(3,1,0), (1,0,0) & (3,1,1)
With this method, I only have to solve 42 possible solutions in the end
the best thing is that you can change the awkward moment of changing 2 to 3 by saying it's 1¹+2¹+3¹ and 1+2=3 in the power
Very nice I have never noticed such a great relation
Wow i never know how to prove the summation of n well now i know thanks to you
I remember proving this by mathematical induction back in high school.
Now, I can't even do simple math lol
Maybe because you started watching anime, anime ruins everything and the community is toxic
No U
One of my favorites mathematicians ❤️❤️❤️
This is really amazing 🤘🤘
bro I'm already watching this while I was drunk you're just making it more confusing
Please make a full detailed video on Math Induction topic
I have a shirt from Dr. Mike Penn with that equation on it. How come I can't buy a black 2.718 shirt? I only see blue on your site.
For some reasons it was gone. But now I just re-added it. Thanks for letting me know.
@@blackpenredpen thanks. just ordered one
Please help me solve this.
Is there a way to isolate x??
xlog(b/a)=log(c*(a^-x)-1)
I think you can bring the x inside the log on the left as an exponent, then do something to remove the log on both sides to leave you with something easier to work with. You'd have to be careful about that though as (e.g.) what if a, b, or c are negative?
@@SimonClarkstone
Thanks
I'm looking at equations in the form
a^x +b^x =c
No x cannot be made subject of the formula unless the base is the same
Best calculus teacher ever🌷🌹
computationally (1+2+3+...+n)^2 is more efficient
computationally 1/4 * n * (n+1) is most efficient lol
definitely
@@Liam-pf7ih actually n^2*(n+1)^2/4
That's a cool proof
Proving it with induction feels like taking the easy way out.
Is this only true with natural numbers?
Because like
(a + 2a + 3a)² = 36a²
and
a³ + 8a³ + 27a³ = 36a³
and a² ≠ a³
Oh wait that means that the identity is true if the "a" is equal to 1 or 0. Nice
👍🏻very nice !!
(3 + 3 + 3)^2 was not just a pure coincidence. It was just (2 + 3 + 4)^2 which again forms a series of n(n+1)/2. Math always has a reason for everything
so does that mean that no matter what our first term is, as long as it is less than n, the equation will be satisfied?
This is why I love proofs
I found visual proofs of this on the main math sub-Reddit.
This was so cool
What I really learned is how to write using 2 different coloured pens. Thx👍
The Algebraic Power Rule
GOAT
Can you explain why the limit when x approaches 0 of sin(x)/x is 1
That is one of the most covered questions in mathematics on the internet....
Bcoz sinx is equal to x for veeery smaaal x
And it is actually 0.99999999..... so
limit x tending to zero for [sinx/x] is equal to zero where [°]is greatest interger function or floor function... Hope it helps...
It involves the squeeze theorem and some geometry, theres alot of video about it
Also i think bprp already made a video about the proof a few years ago
Yes.
Do you have a video about / proof for the pascals identity?
sum: 1^q + 2^q + … + n^q = ?
n and q are positive integers
Explanation in brief: sum of n^3 =
(n(n+1)/2) ^2
But 1+2+3..=n(n+1)/2
Therefore, (1+2+3..) ^2= sum of n^3
What is the time period of zero function??pls explain by proof!!
Nice one
I was just watching this on my instagram from another math guy. Love it.
Are there any other integer values of m != 2 and n != 3 such that (1+2+3)^m = 1^n + 2^n + 3^n ?
Only 1,1 and 2,3 work
Sure: m=n=1 will do.
@@blackpenredpen And this is really proven? Or is it a statement similar to Last Fermat Theorem whose proof was a mystery for 400 years?
I watched your video where at the end it showed SQRT(18) - SQRT(8), and I got it; but when I plugged my own numbers of SQRT(29) - SQRT (12) I was still stumped.
Bro summer holidays really got you mad bored huh 😂
At 7:25 you wrote a red "+1" which should have been a blue one. Terrible misteak!!! :-o
lol
What a terrible beefstake 😱
(n(n+1))/2 is also the formula for triangular numbers, how interesting.
hold on, why can you just draw a square and fill it at the end of a proof? I've just now seen that done in my new school books too, what does it mean?
That little black square is just a common textbook symbol for "end of proof" (sometimes also seen at the end of an example). It means the same thing as QED.
Does the pattern continues such as (1 + 2 + 3)^3 = 1^4 + 2^4 + 3^4 + 4^4?
does this work if you start with a number higher than 1?
Yep. Works for all positive integers. That's the point of an induction proof
@@williamwilliam4944 No it does not work. 5 squared is not equal to 2 cubed plus 3 cubed
@@bscutajar that isn't what the proof is saying, now is it?
I haven't seen that proof for the triangular number formula before.
8!(36+8):2......8!(36-8):2......9!
Where is your Pokemon
Wow that's a cool formula!
Proving something using Math. Induction sometimes feels like cheating
It feels like cheating because you can only use induction once you know the formula from another source. Induction can't find you the formula, it can only prove that the formula is true once you already know it.
Is this only applicable on numerical numbers or can we express (a+b+c)^2 as a³+b³+c³ as well because when I searched this on net I didn't get any results
The sum of powers shown in the video is Faulhaber’s formula for p = 1 and p = 3. The formula only works for positive integers.
7:10 this formula can work with complex num like (I+(i+1)+(i+2)...(I+k) )² equal to i³+(I+1)³....+(I+k)³ ???
I don't think so
Let k=0
So (i)^2 = i^3
Which is false
you can see output of the formula on channel Boris Trushin, its easy
#blackpenredpen There should be a whole square instead whole cube in your description
Fixed. Thank you for pointing that out
Awesome
I assumed it relied on the classic 1+2+3=1*2*3. Would be a better intro with another term, to get out of the misleading special case.
Because int x^3 dx = (int x dx)^2.
For anyone taking Calc AB/BC or Stats, what did you get on the AP Exam? I got a 5! (a 5 not 120 😂)
a 5 in stats
5 in Stats
5 in bc
I got 5 in bc, looks like watching bprp helps with math! (or maybe people who didn't get a 5 didn't comment)
When he said that the summ is n*(n+1)/2 i stopped, though about it, figured it out, continued watching *and then he explained the whole thing*
Only one confusion...in pmi base value starts for n=1 then let n=m then to prove true for n=m+1.But here why n=0...Can anyone explain??
You can start with any integer. It doesn't have to be 1 or even positive really. The idea is to prove that if it's true for SOME n, it must be true for n+1. So if it's true for n=-3, it must be true for -2. Now since it's true for n=-2, it has to be true for n=-1, and so on.
Couldn’t this also be explained as ((1^1)+(2^1)+(3^1))^2. You could add the sum of the powers
That's what I was thinking. You have ^1 on the inside and ^2 on the outside so clearly when you combine them you get ^3.
That's not what is happening. Firstly it is a coincidence, secondly powers would multiply not add up.
One of my friends depises this prank. He was irate. He is passionate, but hateful.
if i had friends they would find this funny.
ua-cam.com/users/shortsO_5CnZh-G7Q?si=xtT9TqqUFjUi-69u
Geometric demonstration for n = 4 base case. Can you prove the induction step geometrically?
Is this for real ?
this is very interesting but it took me a while to follow the calculations.
I haven't done this in a while and the trick was surprising
230 - 220 x 0.5 = 5!
This is definitely my favorite one
Multiply both sides by 2: 230 - 220 = 10. Math xheck out ;)
You probably won't believe it, but the answer is 5!
Retro 14's 🔥
Math
🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
It's my humble request to you
Can you pls do a series on partial differential equations ?
Edit : I forgot the " ? " in my comment .
Why is this a prank? What's the prank? This is just proving an identity
Interesting,could you turn this into a generic problem, eg,
Given a set of integers S=a,b,c,...
Find all n,m such that
(sum S)^n=sum(S^m)
And start with the trivial examples like {1,2}, {1,2,3},{2,3},{1,2,3,4}...
And try to see if there are patterns.
Find all the positives? Maybe you mean all positive integers, bc there are infinite many of positive (real) triples exist that satisfies (x+y+z)^2=x^3+y^3+z^3
Not surprising to me😅
you kidding me right I discovered it when I am 14 years in a math competition
😎
Of course it is. Quit being astonished by the mundane.
I dont know anyone that is even interested in math
1🥰
Funny
very old...
Hello
just my opinion, i dont like facial expressions on thumbnails, they make me cringe
i dont understanda why is this a prank. is it supposed to be funny?😂😂😂
Hi