If you simplify the RHS at the very start to [n(n+1)/2]^2 which of course becomes [k(k+1)/2]^2 then it becomes slightly easier to follow maybe since you can state at the WTS part that RHS needs to be [(k+1)(k+2)/2]^2 at that point. You know what ur chasing a bit sooner. Love these videos.
Very nice mathematics teacher who will not blame or shout at students who got poor grade in Mathematics. Unlike our eastern strict Mathematics teachers.
It's interesting. It works for other multiples as well (in the video's case, it's a series of multiples of 1) but we have to multiply by an additional value (in the video's case, it's x1, which is why it's not shown): Here's a series of multiples of 7: 7^3 + 14^3 + 21^3 + 28^3 = (7+14+21+28)^2 x 7 Here's for 14: 14^3 + 28^3 + 42^3 + 56^3 + 70^3 + 84^3 = (14+28+42+56+70+84)^2 x 14 Here's for 166: 166^3 + 332^3 + 498^3 + 664^3 + 830^3 = (166+332+498+664 + 830)^2 x 166 In general it's: r^3 + (2r)^3 + (3r)^3 + (4r)^3 +...+ (nr)^3 = r(r + 2r + 3r + 4r +...+ nr)^2 where r is the common ratio and n is the number of terms. It only works if it follows specifically this general form, if it doesn't start at (r times 1)^3 then you'd have to subtract terms out from the answer.
Induction is easy. I wanna see someone proving identities like this and 1²+2²+3²+...+n²=n(n+1)(2n+1)/6 by deduction, as those expressions don't come from thin air.
They do: Let f(x)=e^x + e^2x + e^3x +...+ e^nx. Geometric sequence, so f(x)=e^x*(e^nx - 1)/(e^x - 1). f(x)=e^nx - 1 + (e^nx - 1)/(e^x - 1) What interests us is the limit of k-th derivative of f(x) as x->0. Examples: Sum of n 1's (1=1⁰=2⁰=n⁰; k=0) is lim as x->0 of e^x*(e^nx - 1)/(e^x - 1)=lim as x->0 of n*e^nx/e^x=n*e^(0*n)/e^0=n*1/1=n (way to define natural numbers?) It requires more calculations for bigger k.
@@samueldeandrade8535 Yes. k-th derivative of f(x)=e^x + e^(2x) +...+e^(nx) is e^x + (2^k)e^(2x) + (3^k)e^(3x) +...+ (n^k)e^(nx); plug in x=0 to get 1+2^k+3^k+...+n^k. f(x) can be written as (e^x)(1-e^(nx))/(1-e^x), here you cannot plug in x=0 but we can look for the limit as x->0 using d'Hospital's rule.
@@samueldeandrade8535 Yeah, quite cool. If one integrated the right side of the equation, they would get stuff like 1+1/2+1/3+...+1/n or even the Riemann Zeta function in a limit.
If you use Difference Theory these formulae are easy to prove.Note my post on this current proof. After you do the difference spadework, You arrive at a=1/3,b=1/2,c= 1/6 and d=0. S2(n)=1^2+2^2+3^2+..........+n^2= n^3/3+n^2/2+n/6+0*d =(2n^3+3n^2+n)/6 =n(2n^2+3n+1)/6 =n(n+1)(n+2)/6 which is the formula we need QED. for the S4(n) you will need to solve a 5X5 simultaneous group. The hard bit is building up and solving the equations without making mistakes.
Let's start with a little algebra fact:1 (k+1)³-k³=3k²+3k+1 We will add all versions of this equality for k=1,2,...,n sum[k from 1 to n](k+1)³-k³=sum[k from 1 to n] (3k²+3k+1) First let's work a little on the Left Hand Side sum[k from 1 to n](k+1)³-k³=(2³-1³) + (3³-2³) + .... + ((n+1)³-n³) To make the next thing clearer, I will invert the order of summation: sum[k from 1 to n](k+1)³-k³= ((n+1)³-n³) + (n³-(n-1)³) + ((n-1)³ - (n-2)³) + .... + (2³-1³) Note how the -n² can be cancelled with the +n² just after? This goes for almost every term, except (n+1)² and -1²... this is an example of a *telescopic sum*; sum[k from 1 to n](k+1)³-k³= (n+1)³ + (-n³ + n³) + (-(n-1)³ + (n-1)³) + (-(n-2)³ + (n-2)³) + .... + (-2³ + 2³) - 1³ sum[k from 1 to n](k+1)³-k³=(n+1)³-1=n³+3n²+3n Take note on that... we will come back to this expression later Now let's work on the Right Hand Side. Given that addition and multiplication have some properties(commutativity, associativity and distributive)... we can sum[k from 1 to n] (3k²+3k+1) = 3(sum[k from 1 to n] k²) + 3(sum[k from 1 to n] k) + (sum[k from 1 to n]1) sum[k from 1 to n]1 is just 1+1+1+...+1, but n times... so n sum[k from 1 to n]k is 1 + 2 + 3 + .... + n, which is n(n+1)/2... with this method you must know the sum of the first n 0,1,....(p-1)-powers before getting to the formula for the sum of p-powers sum[k from 1 to n] k² is what we want.. call it S After all our work on the sides... make them equal: n³+3n²+3n = 3S + 3n(n+1)/2 + n Then it's just solving for S. Because I did (k+1)³-k³ in the start... this will get us the sum of k²; If we did (k+1)²-k, we'd have the sum of k's; If we did for (k+1)⁴-k⁴ in the start... we would have an equality between (n+1)⁴-1⁴, the sum of k³, the sum of k², the sum of k; A bit messy calculations but you can solve for the sum of k³ after you know the sum of k² and the sum of k;
My idea before watching the video: n = 1: correct n = 2: correct Suppose that 1^3 + ... + n^3 = (1 + ... + n)^2. We need to prove that 1^3 + ... + n^3 + (n + 1)^3 = (1 + ... + n + n+1)^2 (1 + ... + n + n+1)^2 = (1 + ... + n)^2 + 2(1 + ... + n)(n + 1) + (n + 1)^2 We need to prove that (n + 1)^3 = 2(1 + ... + n)(n + 1) + (n + 1)^2 which is easy to prove if we know 1 + ... + n = n(n + 1)/2
can you please explain at which moment you verify the assomption that the proposition is true for n=k ? I mean everything is based on that. What if it is wrong for a "random" k ?
Fir more calrification , when we get tge base case n=1 is true And assume that for n=k>=1 is true and found that it is true k+1. We done and some might say why. I tell them that if we assume that it is true for k and find out that it is true for k+1 Then we go back to base case k=1 If it is true for n=1 , it is true fir n=2" And if true for n=2, it true for n=3 And we go on
no, we can't . we just have to trust the so called 'farmer's logica'. the matter is : first: you show that the thesis is true for let's say n = 1 , or like this video, for n = 3 . second : you suppose, of course based on some expected outcome, but you just suppose that the thesis would be true for some value n = k . third : you try to prove mathematically , based on that supposition, that in that case the thesis will also be true for the next value n = (k + 1) . ((if not, then not, of course .)) conclusion: after the proof succeeded, and convinced , you can decide that based on the shown validity of the first value of n, i.e. n = 3, the thesis is also true for n = 4, and therefore for n = 5, because of this also for 6 and for each following natural number . all this based on the logical steps done from "second" to "third" , which provides a proof *Only* for the follower of the already *proven number* . in europe we call this kind of proof a proof by complete, or full, induction, inductio plena, vollständige Induktion, volledige inductie. good luck .
There's kinda of a proof using the well ordering principle: any non-empty subset of ℕ has a least number; Given a statement P(n) over natural numbers n... also given P(0) and P(k) -> P(k+1)... Consider the set X={n ∊ ℕ | P(n) is false}... this is the set of all natural numbers n for which P(n) is false; We can prove that X=∅ by contradiction: assume X≠∅and consider p the least number of X(which it must exist under the well ordering principle). We know 0 ∉ X because it is given that P(0) is true... so 0
I doubt anyone at this level still needs to be explained why 1+2+3...+n = n(n+1)/2 but just in case. First we write the series: 1+2+3...+n Call this series p Then beneath it we write the same series but backwards n+n-1+n-2...1 Call this series q Now we know both of these series have the same amount of terms, n of them. So we can add these series term by term, and they will be equal to p+q. Take the first two terms and add them, that's 1 + n The second two terms added is 2 + n-1 = 1 + n The third two terms added is 3 + n-2 = 1+n And this pattern continues, all the way to the last terms that also add to 1+n so you have n groups of terms that are all equal to (n+1), leading to a total addition of n(n+1) You wrote the series twice, once forward and once backwards, and you added them. Therefore, the sum of one of these series is n(n+1)/2
The teacher first shows that the equation is true for n=1,n=2,n=3 Then he shows that it is always true for n+1/the next integer. So n=4 is true, n=5 is true, etc. equation is true by induction.
think, suppose that when she was young like we were once, she tried on a piece of paper the first cubes 1 _ 8 _ 27 _ 64 _ 125 _ 216 .. then maybe tried their row of differences , 7 _ 19 _ 37 _ 61 _ 91 .. weird somehow , but yet isn't there a kind of regularity between these , 12 _ 18 _ 24 _ 30 .. well that is , there could be more than one interesting thing about or between or around them . then , on another day , trying their sums like in the video 1+8 , which produces 9 . and 1+8+27 , which produces 36 , hey nice. and 1+8+27+64 , producing 100 . uff, too nice, what's up here . what could be the buzz. all of them seem to be squares . squares of what ? what's between them ? etcetera .. triangle numbers ? how ? why ? and so on . all hypothetical , out of childish curiosity , nothing more .
You can use difference theory to prove this. (1) S1(n)= 1^1+2^1+3^1+..........+n^1= n(n+1)/2 (n is a member of natural numbers) (2) S2(n)= 1^2+2^2+3^2+..........+n^2=n(2n+1)(2n+2)/6 (3) S3(n)= 1^3+2^3+3^3+..........+n^3= ((n)(n+1)/2))^2 = (S1(n))^2 (4) S4(n)= 1^4+2^4+3^4+..........+n^4= n(n+1)(2n+1)(3n^2+3n-1)/30. Let S3(n)= 1^3+2^3+3^3+............+n^3= (n(n+1)/2)^2= (1+2+3+.......+n)^2 S3(6)=1^3+2^3+3^3+4^3+5^3+6^3 1+8+27+64+125+216= (n(n+1)/2)^2=441.........................(1) S3(1)=1^3=1 S3(2)=1^3+2^3=9 S3(3)=36,S3(4)=100,S3(5)=225,S3(6)=441 1st difference= 8,27,64,125,216. 2nd difference=19,37,61,91. 3rd difference=18,24,30. 4th difference=6,6 In S3(n), n>6 the 4th difference will always be 6 no matter the magnitude of n. Therefore the difference table will always work for n. This suggests S3(n) is a fourth order polynomial, S3(n)=an^4+bn^3+cn^2+dn+e...............................................(2) S3(1)=1=a+b+c+d+e..............................................................(3) S3(2)=9=16a+8b+4C+2d+e...................................................(4) S3(3)=36=81a+27b+9c+3d+e...............................................(5) S3(4)=100=256a+64b+16c+4d+e........................................(6) S3(5)=225=625a+125b+25c+5d+e......................................(7) S3(6)=441=1296a+216b+36c+6d+e....................................(8) Sweep from (3)-(8), 8 =15a+7b+3c+d...................................................................(9) 27=65a+19b+5c+d.................................................................(10) 64=175a+37b+7c+d...............................................................(11) 125=369a+61b+9c+d.............................................................(12) 216=671a+91b+11c+d..........................................................(13) Sweep from (9)-(13), 19=50a+12b+2c....................................................................(14) 37=110a+18b+2c..................................................................(15) 61=194a+24b+2c..................................................................(16) 91=302a+30b+2c..................................................................(17) Sweep from (14)-(17), 18=60a+6b............................................................................(18) 24=84a+6b............................................................................(19) 30=108a+6b..........................................................................(20) Sweep away the bs, 6=24a, >>>>>>>>>>>>>a=1/4,b=1/2,c=1/4,d=0,e=0 Going back to (2), S3(n)=n^4/4+n^3/2+n^2/4=(n^4+2n^3+n^2)/4=(n^2(n+1)^2)/2^2=((n(n+1))/2)^2 =(1^3+2^3+3^3+.................+n^3)=(1+2+3.............+n)^2 case proved. S3(6)=1^3+2^3+3^3+4^3+5^3+6^3 = 441 = (1+2+3+4+5+6)^2)= (21)^2 =441 in the S3(6) particular case. Thanks for the brilliant example of mathematical induction. Well done young man.
Just wondering from some stuff on Perfect Numbers & Mersenne Primes, which has it that a Perfect Number can be expressed as a sum of some number of consecutive odd cubes... e.g. 1^3+3^3+5^3+7^3, ... n^3 etc. Which I guess would basically just be (1+((1-1)*2))^3+(1+((2-1)*2))^3+(1+((3-1)*2))^3+(1+((4-1)*2))^3+...+(1+((n-1)*2))^3, where n is the ordinal of the odd cube (1st odd cube, 2nd odd cube, etc.)?
Would it just be the square of the sum of the [odd] numbers being cubed? Wolfram|Alpha seems to say "no." Not for cubes of first odd numbers. :\ Hmm...
@@MGmirkin you can add and subtract the sum of the first n even cubes: 1³+3³+5³ + ... (2n-1)³= =1³+3³+5³ + ... (2n-1)³ + (2³+4³+...+(2n-2)³)-(0³+2³+4³+...+(2n)³) =(0³+1³+2³+3³+4³+....+(2n)³) -((2(1))³+(2(2))³+....+(2(n))³)= =(sum of cubes till 2n) - (2³2³1³+2³2³+2³3³+...+2³n³) =(sum of cubes till 2n) - 2³ (sum of cubes till n)
Yes. But you will need to know that there exist real numbers A, B, C,....SUCH THAT the sums of any given powers can be expressed in a closed form. As follows: 1^3 + 2^3 + 3^3+...+n^3 = A*n(n+1) + B*n(n+1)(n+2) + C*n(n+1)(n+3) = ( n^2.(n+1)^2 )/4 LHS: In this case, (A,B,C) = (1/2, -1, 1/4), hence the cubes sum to: ( n^2.(n+1)^2 )/4 But we also know that: 1+2+3+....+n = n(n+1)/2 RHS: Placing this value inside the right hand parenthesis gives: [ n(n+1)/2 ]^2 Which simplifies to: n^2.(n+1)^2 / 2^2 And is equal to the LHS.
So... you used the expression to prove the expression? I dont get how that proves the initial proof if you just claim the main question as the very first step of the solution?
@@PrimeNewtons No I used this while expanding exponential generating function of ChebyshovT polynomial E(x,t) = exp(xt)cos(sqrt(1-x^2)t) and i have seen lately this problem on one of the math forums In my opinion it is good exercise for mathematical induction if we are not allowed to use complex numbers
I used Chebyshov not Chebyshev because your transcription of this name is poor and leads to misreading Name of this guy written in cyrylic has two dots over last e which is read as yo but it is simplified to o (Maybe because it would be difficult to read sh and yo but i dont know why this simplification occured)
well your proposal seems to have a brilliant side . let's do a try . starting from the upper left side you could consider to see the unity, 1 = 1³ in a quarter around this, you could see 3 + 5 , that is twice the main value 4 , 8 = 2³. together with the starting unity you see lying in a square : 1+3+5 = 9 . in a quarter around these 9 you could lay down 7+9+11 squares , that is three times the main value of 9 , 27 = 3³ squares added, so together there's laying a larger square of 6×6 unity squares . in a quarter around these 36 you could add four hooked paths out of 13+15+17+19 unity squares, that is four times the main value of 16 , 32+32 = 64 added, so 4³ added to the already laying 1+8+27 ; the complete sum of squares is now 1+(3+5)+(7+9+11)+(13+15+17+19), laying in a square of 10×10 unities . in a quarter field around the starting unity . well we leave to your imagination how to continue with five hooked paths of unity squares around these 100 . good luck .
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Beauty of Mathmatical Induction
A concise presentation in clear unaffected English and a minimum of theatrics. Bravo!
A very good example ! You do it very clearly in a nice handwriting.
If you simplify the RHS at the very start to [n(n+1)/2]^2 which of course becomes [k(k+1)/2]^2 then it becomes slightly easier to follow maybe since you can state at the WTS part that RHS needs to be [(k+1)(k+2)/2]^2 at that point. You know what ur chasing a bit sooner. Love these videos.
Another superb presentation 😊
Nice & elegant! Congrats, my friend👍
Very nice Induction Proof - 🙏 thanks a lot.
Your teaching style is so amazing
This is my first video I've seen of yours. Subscribed as soon as video was dond
best handwriting ive seen in a while (30+years)!
😮 se sabía que en un momento tenía que utilizar sumatorias conocidas, muy buena demostración!!
Amazing Prof 🎉
Thank you for this, I had seen the relationship and
wondered how it was proved. Your presentations are superb.
Wow. I have studied long ago math at the university of Utrecht (Netherlands) but this was unknown to me. 😊
Nice proof! Also nice handwriting
Bravo professeur. J'aime bien.
That was beautiful!
You shoulda been a teacher...how relaxed and succinct...in other words.. a genius..nice work..
.
Fantastic! Tanks, teacher.
We need to show if 1^3 + 2^3 + 3^3 ... + k^3 = (1 + 2 + 3 ... + k)^2, then 1^3 + 2^3 + 3^3 ... + k^3 + (k + 1)^3 = (1 + 2 + 3 ... + k + (k + 1))^2
First we calculate the difference
(1 + 2 + 3 ... + k + (k + 1))^2 - (1 + 2 + 3 ... + k )^2 = 2(k + 1)(1 + 2 + 3 ... + k ) + (k+1)^2 = 2(k + 1)(k(1+ k)/2 ) + (k+1)^2 = k(1+ k)^2 + (k+1)^2 = (k+1)^2 (k + 1) = (k + 1)^3
Since (1 + 2 + 3 ... + k + (k + 1))^2 = (1 + 2 + 3 ... + k )^2 + (k + 1)^3
and
(1 + 2 + 3 ... + k )^2 = 1^3 + 2^3 + 3^3 ... + k^3
We showed
(1 + 2 + 3 ... + k + (k + 1))^2 = 1^3 + 2^3 + 3^3 ... + k^3 + (k + 1)^3
Very nice mathematics teacher who will not blame or shout at students who got poor grade in Mathematics. Unlike our eastern strict Mathematics teachers.
Very nice
amazing video !!!
I'm very glad you covered this identity haha
coincidentally, I was just walking a tutee through this one a couple days ago !!
Beautiful problem ❤
U R a positive energy M. Teacher 🤩Merry Christmas
Thank you very much. You are great.
teacher you are the best
very clear. Bravo!
wow that was beautiful
Wonderful
YOU BEATED MY MATHS TEACHER 😭,THANK YOU FOR GOOD PRESENTATION ❤❤...
Impressive!
It's interesting. It works for other multiples as well (in the video's case, it's a series of multiples of 1) but we have to multiply by an additional value (in the video's case, it's x1, which is why it's not shown):
Here's a series of multiples of 7: 7^3 + 14^3 + 21^3 + 28^3 = (7+14+21+28)^2 x 7
Here's for 14: 14^3 + 28^3 + 42^3 + 56^3 + 70^3 + 84^3 = (14+28+42+56+70+84)^2 x 14
Here's for 166: 166^3 + 332^3 + 498^3 + 664^3 + 830^3 = (166+332+498+664 + 830)^2 x 166
In general it's: r^3 + (2r)^3 + (3r)^3 + (4r)^3 +...+ (nr)^3 = r(r + 2r + 3r + 4r +...+ nr)^2
where r is the common ratio and n is the number of terms. It only works if it follows specifically this general form, if it doesn't start at (r times 1)^3 then you'd have to subtract terms out from the answer.
Chứng minh bằng phương pháp quy nạp. Hay!
Mooi bewijs door twee keer toe te passen: 1^2 +2^2 + … n^2 = n * (n +1) / 2. Fraai!! 👍
Induction is easy. I wanna see someone proving identities like this and 1²+2²+3²+...+n²=n(n+1)(2n+1)/6 by deduction, as those expressions don't come from thin air.
They do:
Let f(x)=e^x + e^2x + e^3x +...+ e^nx.
Geometric sequence, so f(x)=e^x*(e^nx - 1)/(e^x - 1).
f(x)=e^nx - 1 + (e^nx - 1)/(e^x - 1)
What interests us is the limit of k-th derivative of f(x) as x->0.
Examples:
Sum of n 1's (1=1⁰=2⁰=n⁰; k=0) is lim as x->0 of e^x*(e^nx - 1)/(e^x - 1)=lim as x->0 of n*e^nx/e^x=n*e^(0*n)/e^0=n*1/1=n (way to define natural numbers?)
It requires more calculations for bigger k.
@@samueldeandrade8535
Yes.
k-th derivative of f(x)=e^x + e^(2x) +...+e^(nx) is e^x + (2^k)e^(2x) + (3^k)e^(3x) +...+ (n^k)e^(nx); plug in x=0 to get 1+2^k+3^k+...+n^k.
f(x) can be written as (e^x)(1-e^(nx))/(1-e^x), here you cannot plug in x=0 but we can look for the limit as x->0 using d'Hospital's rule.
@@samueldeandrade8535 Yeah, quite cool. If one integrated the right side of the equation, they would get stuff like 1+1/2+1/3+...+1/n or even the Riemann Zeta function in a limit.
If you use Difference Theory these formulae are easy to prove.Note my post on this current proof.
After you do the difference spadework,
You arrive at a=1/3,b=1/2,c= 1/6 and d=0.
S2(n)=1^2+2^2+3^2+..........+n^2= n^3/3+n^2/2+n/6+0*d
=(2n^3+3n^2+n)/6
=n(2n^2+3n+1)/6
=n(n+1)(n+2)/6 which is the formula we need QED.
for the S4(n) you will need to solve a 5X5 simultaneous group.
The hard bit is building up and solving the equations without making mistakes.
Let's start with a little algebra fact:1
(k+1)³-k³=3k²+3k+1
We will add all versions of this equality for k=1,2,...,n
sum[k from 1 to n](k+1)³-k³=sum[k from 1 to n] (3k²+3k+1)
First let's work a little on the Left Hand Side
sum[k from 1 to n](k+1)³-k³=(2³-1³) + (3³-2³) + .... + ((n+1)³-n³)
To make the next thing clearer, I will invert the order of summation:
sum[k from 1 to n](k+1)³-k³= ((n+1)³-n³) + (n³-(n-1)³) + ((n-1)³ - (n-2)³) + .... + (2³-1³)
Note how the -n² can be cancelled with the +n² just after? This goes for almost every term, except (n+1)² and -1²... this is an example of a *telescopic sum*;
sum[k from 1 to n](k+1)³-k³= (n+1)³ + (-n³ + n³) + (-(n-1)³ + (n-1)³) + (-(n-2)³ + (n-2)³) + .... + (-2³ + 2³) - 1³
sum[k from 1 to n](k+1)³-k³=(n+1)³-1=n³+3n²+3n
Take note on that... we will come back to this expression later
Now let's work on the Right Hand Side.
Given that addition and multiplication have some properties(commutativity, associativity and distributive)... we can
sum[k from 1 to n] (3k²+3k+1) = 3(sum[k from 1 to n] k²) + 3(sum[k from 1 to n] k) + (sum[k from 1 to n]1)
sum[k from 1 to n]1 is just 1+1+1+...+1, but n times... so n
sum[k from 1 to n]k is 1 + 2 + 3 + .... + n, which is n(n+1)/2... with this method you must know the sum of the first n 0,1,....(p-1)-powers before getting to the formula for the sum of p-powers
sum[k from 1 to n] k² is what we want.. call it S
After all our work on the sides... make them equal:
n³+3n²+3n = 3S + 3n(n+1)/2 + n
Then it's just solving for S.
Because I did (k+1)³-k³ in the start... this will get us the sum of k²;
If we did (k+1)²-k, we'd have the sum of k's;
If we did for (k+1)⁴-k⁴ in the start... we would have an equality between (n+1)⁴-1⁴, the sum of k³, the sum of k², the sum of k; A bit messy calculations but you can solve for the sum of k³ after you know the sum of k² and the sum of k;
Tanks. Very good !
My favorite surprising result easily proved by induction.
Wonderful!
Awesome!!!!
You inspire me, great!
Wow, thank you!
Love from India 🇮🇳❤
Maravilhoso !!!
Pure Poetry..!
Perfect!
i love it
That's clean as a proof.
thank you air
Excelente
Interesting
You can derive such a formula using telescoping sums (i+1)^4-i^4.
nice
Always note that “it is true for all positive real numbers”!!
Sorry, I got lost in the leap shown at around the 7:00 mark. Gonna have to look at it again more closely.
Awesome lesson. The guy is also very cute. 😅
Thanks
My idea before watching the video:
n = 1: correct
n = 2: correct
Suppose that 1^3 + ... + n^3 = (1 + ... + n)^2. We need to prove that 1^3 + ... + n^3 + (n + 1)^3 = (1 + ... + n + n+1)^2
(1 + ... + n + n+1)^2 = (1 + ... + n)^2 + 2(1 + ... + n)(n + 1) + (n + 1)^2
We need to prove that (n + 1)^3 = 2(1 + ... + n)(n + 1) + (n + 1)^2 which is easy to prove if we know 1 + ... + n = n(n + 1)/2
can you please explain at which moment you verify the assomption that the proposition is true for n=k ? I mean everything is based on that. What if it is wrong for a "random" k ?
You don't. Because it was true for the first few tries, you assume it is true for n=k. If your assumption is false, mathematical induction would fail.
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Wpi;d it be incorrect to treat both sums as integrals and then equate? The you get n^4/4=(n^2/2)^2. This seems to prove it. What am I missing?
😊
掌握技巧是狭隘的，掌握方法是有限的，掌握原理才是终极道理。
where did the cube go
Fir more calrification , when we get tge base case n=1 is true
And assume that for n=k>=1 is true and found that it is true k+1. We done and some might say why.
I tell them that if we assume that it is true for k and find out that it is true for k+1
Then we go back to base case k=1
If it is true for n=1 , it is true fir n=2"
And if true for n=2, it true for n=3
And we go on
嬉しそうな良い顔してるねえ。
🌹🙏🌹
😮
Can we prove why induction works as a proof?
no, we can't . we just have to trust the so called 'farmer's logica'.
the matter is :
first: you show that the thesis is true for let's say n = 1 , or like this video,
for n = 3 .
second : you suppose, of course based on some expected outcome, but you just suppose that the thesis would be true for some value n = k .
third : you try to prove mathematically , based on that supposition, that in that case the thesis will also be true for the next value n = (k + 1) .
((if not, then not, of course .))
conclusion: after the proof succeeded, and convinced ,
you can decide that based on the shown validity of the first value of n, i.e.
n = 3, the thesis is also true for n = 4, and therefore for n = 5, because of this also for 6 and for each following natural number .
all this based on the logical steps done from "second" to "third" , which provides a proof *Only* for the follower of the already *proven number* .
in europe we call this kind of proof
a proof by complete, or full, induction,
inductio plena, vollständige Induktion, volledige inductie.
good luck .
@@keescanalfp5143 thank you
There's kinda of a proof using the well ordering principle: any non-empty subset of ℕ has a least number;
Given a statement P(n) over natural numbers n... also given P(0) and P(k) -> P(k+1)...
Consider the set X={n ∊ ℕ | P(n) is false}... this is the set of all natural numbers n for which P(n) is false;
We can prove that X=∅ by contradiction: assume X≠∅and consider p the least number of X(which it must exist under the well ordering principle).
We know 0 ∉ X because it is given that P(0) is true... so 0
This shit fires me up i love math
I doubt anyone at this level still needs to be explained why 1+2+3...+n = n(n+1)/2 but just in case.
First we write the series:
1+2+3...+n
Call this series p
Then beneath it we write the same series but backwards
n+n-1+n-2...1
Call this series q
Now we know both of these series have the same amount of terms, n of them.
So we can add these series term by term, and they will be equal to p+q.
Take the first two terms and add them, that's 1 + n
The second two terms added is 2 + n-1 = 1 + n
The third two terms added is 3 + n-2 = 1+n
And this pattern continues, all the way to the last terms that also add to 1+n
so you have n groups of terms that are all equal to (n+1), leading to a total addition of n(n+1)
You wrote the series twice, once forward and once backwards, and you added them.
Therefore, the sum of one of these series is n(n+1)/2
isnt this circular reasoning?
Not at all, just induction
The teacher first shows that the equation is true for n=1,n=2,n=3
Then he shows that it is always true for n+1/the next integer. So n=4 is true, n=5 is true, etc. equation is true by induction.
If this formula is known, it can indeed be proved. But how did the first person to figure out this formula do it?
think, suppose that when she was young like we were once, she tried on a piece of paper the first cubes
1 _ 8 _ 27 _ 64 _ 125 _ 216 ..
then maybe tried their row of differences ,
7 _ 19 _ 37 _ 61 _ 91 ..
weird somehow , but yet isn't there a kind of regularity between these ,
12 _ 18 _ 24 _ 30 ..
well that is , there could be more than one interesting thing about or between or around them .
then , on another day , trying their sums like in the video
1+8 , which produces 9 . and
1+8+27 , which produces 36 , hey nice.
and
1+8+27+64 , producing 100 . uff, too nice, what's up here . what could be the buzz. all of them seem to be squares . squares of what ? what's between them ? etcetera .. triangle numbers ? how ? why ? and so on .
all hypothetical , out of childish curiosity , nothing more .
I asked chatgpt, and the first person to figure out the formula was Euler, a great mathematician.
@@ccc40476,
so beautiful that it was him !
does c.gpt know *How* the man came upon the formula or even upon the searching for it .
Why does this happen?
3: 50 the assume is sure? by what ?
Giống quy nạp newton nhỉ
{n(n+1)2}^2 is resultant ans
You can use difference theory to prove this.
(1) S1(n)= 1^1+2^1+3^1+..........+n^1= n(n+1)/2 (n is a member of natural numbers)
(2) S2(n)= 1^2+2^2+3^2+..........+n^2=n(2n+1)(2n+2)/6
(3) S3(n)= 1^3+2^3+3^3+..........+n^3= ((n)(n+1)/2))^2 = (S1(n))^2
(4) S4(n)= 1^4+2^4+3^4+..........+n^4= n(n+1)(2n+1)(3n^2+3n-1)/30.
Let S3(n)= 1^3+2^3+3^3+............+n^3= (n(n+1)/2)^2= (1+2+3+.......+n)^2
S3(6)=1^3+2^3+3^3+4^3+5^3+6^3 1+8+27+64+125+216= (n(n+1)/2)^2=441.........................(1)
S3(1)=1^3=1
S3(2)=1^3+2^3=9
S3(3)=36,S3(4)=100,S3(5)=225,S3(6)=441
1st difference= 8,27,64,125,216.
2nd difference=19,37,61,91.
3rd difference=18,24,30.
4th difference=6,6
In S3(n), n>6 the 4th difference will always be 6 no matter the magnitude of n.
Therefore the difference table will always work for n.
This suggests S3(n) is a fourth order polynomial,
S3(n)=an^4+bn^3+cn^2+dn+e...............................................(2)
S3(1)=1=a+b+c+d+e..............................................................(3)
S3(2)=9=16a+8b+4C+2d+e...................................................(4)
S3(3)=36=81a+27b+9c+3d+e...............................................(5)
S3(4)=100=256a+64b+16c+4d+e........................................(6)
S3(5)=225=625a+125b+25c+5d+e......................................(7)
S3(6)=441=1296a+216b+36c+6d+e....................................(8)
Sweep from (3)-(8),
8 =15a+7b+3c+d...................................................................(9)
27=65a+19b+5c+d.................................................................(10)
64=175a+37b+7c+d...............................................................(11)
125=369a+61b+9c+d.............................................................(12)
216=671a+91b+11c+d..........................................................(13)
Sweep from (9)-(13),
19=50a+12b+2c....................................................................(14)
37=110a+18b+2c..................................................................(15)
61=194a+24b+2c..................................................................(16)
91=302a+30b+2c..................................................................(17)
Sweep from (14)-(17),
18=60a+6b............................................................................(18)
24=84a+6b............................................................................(19)
30=108a+6b..........................................................................(20)
Sweep away the bs,
6=24a, >>>>>>>>>>>>>a=1/4,b=1/2,c=1/4,d=0,e=0
Going back to (2),
S3(n)=n^4/4+n^3/2+n^2/4=(n^4+2n^3+n^2)/4=(n^2(n+1)^2)/2^2=((n(n+1))/2)^2
=(1^3+2^3+3^3+.................+n^3)=(1+2+3.............+n)^2 case proved.
S3(6)=1^3+2^3+3^3+4^3+5^3+6^3 = 441 = (1+2+3+4+5+6)^2)= (21)^2 =441 in the S3(6) particular case.
Thanks for the brilliant example of mathematical induction.
Well done young man.
What about "the first n odd cubes"?
Just wondering from some stuff on Perfect Numbers & Mersenne Primes, which has it that a Perfect Number can be expressed as a sum of some number of consecutive odd cubes...
e.g. 1^3+3^3+5^3+7^3, ... n^3 etc.
Which I guess would basically just be (1+((1-1)*2))^3+(1+((2-1)*2))^3+(1+((3-1)*2))^3+(1+((4-1)*2))^3+...+(1+((n-1)*2))^3, where n is the ordinal of the odd cube (1st odd cube, 2nd odd cube, etc.)?
Would it just be the square of the sum of the [odd] numbers being cubed?
Wolfram|Alpha seems to say "no." Not for cubes of first odd numbers. :\ Hmm...
@@MGmirkin you can add and subtract the sum of the first n even cubes:
1³+3³+5³ + ... (2n-1)³=
=1³+3³+5³ + ... (2n-1)³ + (2³+4³+...+(2n-2)³)-(0³+2³+4³+...+(2n)³)
=(0³+1³+2³+3³+4³+....+(2n)³) -((2(1))³+(2(2))³+....+(2(n))³)=
=(sum of cubes till 2n) - (2³2³1³+2³2³+2³3³+...+2³n³)
=(sum of cubes till 2n) - 2³ (sum of cubes till n)
It is common knowledge that 1+2+...+n=n(n+1)/2 and 1^3+3^3+...+n^3=n^2(n+1)^2/4.
Thus, [n(n+1)/2]^2=n^2(n+1)^2/4, and equality is verified.
I make de formula in another side of equation and get same result
It can be done without induction
12:26 The face :))))
Hummm
数学归纳法嘛
Begendım kibar
It's called Nicomachus Identity
I don't know why but i thing you don't need to put thos intro in you're videos
Мат. Индукцией доказывается элементмрно
数学归纳法秒解
Is there an alternative proof without using induction?
Yes. But you will need to know that there exist real numbers A, B, C,....SUCH THAT the sums of any given powers can be expressed in a closed form. As follows:
1^3 + 2^3 + 3^3+...+n^3
= A*n(n+1) + B*n(n+1)(n+2) +
C*n(n+1)(n+3) = ( n^2.(n+1)^2 )/4
LHS:
In this case, (A,B,C) = (1/2, -1, 1/4), hence the cubes sum to:
( n^2.(n+1)^2 )/4
But we also know that:
1+2+3+....+n = n(n+1)/2
RHS:
Placing this value inside the right hand parenthesis gives:
[ n(n+1)/2 ]^2
Which simplifies to:
n^2.(n+1)^2 / 2^2
And is equal to the LHS.
@@tonybantu9427Why the strange choice of polynomials like n(n+1), n(n+1)(n+2)…? are they orthogonal or something?
So... you used the expression to prove the expression? I dont get how that proves the initial proof if you just claim the main question as the very first step of the solution?
That is called mathematical induction. You make a claim and then show your claim is true.
Prove that (cos(x))^{(n)} = cos(x+n*pi/2)
where n means nth derivative
Is this from differential equations?
@@PrimeNewtons No I used this while expanding exponential generating function of ChebyshovT polynomial E(x,t) = exp(xt)cos(sqrt(1-x^2)t) and i have seen lately this problem on one of the math forums
In my opinion it is good exercise for mathematical induction if we are not allowed to use complex numbers
I used Chebyshov not Chebyshev because your transcription of this name is poor and leads to misreading
Name of this guy written in cyrylic has two dots over last e which is read as yo but it is simplified to o
(Maybe because it would be difficult to read sh and yo but i dont know why this simplification occured)
I'm looking at this video soon
the way you write a term and then go back to wrap in in parentheses (instead of making parentheses right away) is anxiety-inducing 😮💨
@ ... better use [ abc ( def ) ghi ] ... @
🇵🇸🇵🇸
Lets join you and me in a social media. I wanna share mathematical content with you. I am a permanent mathematics teacher of government of Nepal.
We can correspond by email. I also have Instagram @primenewtons.
준석아
笨！变成积分两步就证明了🤣
There is a graphical solution.
Unfortunately, cannot be shown on comments.
Оно доказывается иначе и гораздо быстрее через вывод общей суммы рядов
🌆🏙️🌃🌌🌉
🏙️🏙️🌃🌌🌉
🌃🌃🌃🌌🌉
🌌🌌🌌🌌🌉
🌉🌉🌉🌉🌉
Maybe able to prove visibly by using ↑
But I’m not sure how to explain cubed terms…
well your proposal seems to have a brilliant side .
let's do a try . starting from the upper left side you could consider to see the unity,
1 = 1³
in a quarter around this, you could see
3 + 5 , that is twice the main value 4 ,
8 = 2³.
together with the starting unity you see lying in a square :
1+3+5 = 9 .
in a quarter around these 9 you could lay down 7+9+11 squares , that is three times the main value of 9 ,
27 = 3³ squares added, so together there's laying a larger square of 6×6 unity squares .
in a quarter around these 36 you could add four hooked paths out of
13+15+17+19 unity squares, that is four times the main value of 16 ,
32+32 = 64 added, so 4³ added to the already laying
1+8+27 ;
the complete sum of squares is now
1+(3+5)+(7+9+11)+(13+15+17+19),
laying in a square of 10×10 unities .
in a quarter field around the starting unity . well we leave to your imagination how to continue with five hooked paths of unity squares around these 100 .
good luck .