But bprp is just awful. Why would anyone want to learn from him? Not to offend you. I just genuinely do not understand Is he not still that arrogant and irresponsible as he was 2 yrs ago, when I stopped watching him? He used to blunder left and right in his videos, i.e. missing an entile segments from the proof like a cheddar cheese, or not considering negative real numbers when a problem is defined in both +- real numbers-just willy nilly making do with half proof, etc. I mean he used to blunder every two, three videos by the time I finally blocked him and I think that's damn plenty. But fine, blunders, everybody makes them, I make blunders, you make blunders, no one is perfect. What was truly awful about him is he used to delete reasonable comments pointing out faulty logic. Not posting fixes even though the blunder is just so clearly out there for everyone to see. I mean, he could just write some corrections on comment and pin it, the least he could do, or just pin someone else comment that point them out, nope, never. Or just delete the whole faulty video and act as if such didn't happened. That's just disgusting, because he's already injected faulty knowledges to the first few hundred viewers and after he deletes it, the viewers who got faulse knowledge injected won't even realize it. Just,,, list just goes on and on. I just can't call him a teacher after all the irresponsible behaviours he's have shown. I'm surprised he still has viewers even. Does he still does this things?
Simpler solution: n.n.n = n.(n-1).(n-2) +3.n.n -2.n = n.(n-1).(n-2) +3.n.(n-1) +n; each of these is zero for some early terms in the sum, that we can duly drop; in the remaining terms, each of our three terms cancels out the first factors of the n! denominator, leaving a sum from n = k to infinity of 1/(n-k)!, for k in {0, 1, 2} with a factor of 3 on the k = 2 one; we can re-index each of these to m = n-k to make it a sum from m = 0 to infinity of 1/m!, so we have e +3.e +e = 5.e.
This is probably the easiest way of solving it. While in a sense this is what he did in the video, it was in a much more complicated and indirect way. To Tony, I would suggest that after you solve such a problem, try to take a second look at your solution and look for some underlying structure. In this case, you tried to "extract" n-1 and n-2 from the n^3. Once you know that this is what you are looking for, then you should realize that n^3 and n(n-1)(n-2) are pretty "close" in the sense that n^3 - n(n-1)(n-2) has degree 2 instead of 3, and eventually it will lead to the expression in @diddykong3100 comment. Also, a simplifying step to the method in your video you should consider, is that when you try to extract (n-1) from n^2, instead of guessing what to add and remove, simply write it as n^2 = (n-1+1)^2 = (n-1)^2 +2*(n-1) + 1.
The coefficients 1.3,1,0 are called Stirling numbers of order 3 of the first or second kind (I can't remember which). Your method is the way to go. Well done/
Cool solution! Another way you can approach the problem is to see that the summand (n^3/n!) is already very similar to the expansion for e^x, but with n^3 instead of x^n. After some playing around, you might notice that if you plug in "e^x" itself into e^x, you get: e^(e^x) = sum( (e^x)^n / n! ), and if you take the third derivative, you get sum( (n^3 / n!) * (e^nx) ). Evaluating this at x = 0 is precisely the sum. So the solution is the third derivative of e^(e^x) evaluated at 0, which if you compute it, is 5e. And this approach generalizes to higher powers of n in the numerator. Do keep in mind though that this problem was given as the third problem on the calc tiebreaker test, meaning that the students had only 15 minutes to do ALL three problems on the test. The approach above (or similar) might be more efficient or even the only option in those kinds of conditions.
Yeah, as soon as I saw the expression I knew it had something to do with e, and once I saw what answer was about to come out of it (admittedly only at 20 minutes or so), I swore there had to be a more elegant way to do it. Thanks for showing me what it was.
Yeah, I did a similar thing, except I took the derivative of e^x, and multiplied by x, then took the derivative of that, and multiplied it by x, and then took the derivative of that, and evaluated at 1.
@@LogosNigrum's way of solving the problem is much quicker tbh. The only thing I could add, is that this whole approach is quite similar to how work with generating functions is usually being done. (or, to be more specific, moment generating functions)
If you've done some probability you recognise by multiplying by e^-1 and e that you have E(N^3), the third moment of a Poisson distributed random variable. By differentiating the Moment Generating function three times and evaluating at 0 you solve the problem.
If you wanted to generalize this to n^k/n!, I believe OEIS sequence A000110 gives you the coefficients of e. This sequence is the Bell numbers, or the number of ways to partition a set of exactly k elements. Correct me if I’m wrong.
I did similar approach to Tony Wang (the guy in the video) but in the most efficient way in a few minutes. It essentially boils downs to the following decomposition of the numerator: n^3 = n(n^2) = n[(n-1)^2+2n-1] = n[(n-1)((n-2)+1)+2(n-1)+1] = n(n-1)(n-2) + 3n(n-1) + n. Those terms become e+3e+e. I did this while reindexing between each step and replacing the 1/n! terms with e. The reindexing doesn't make any rational constants appear because each time before you cancel with the factorial, you remove the first term which is always zero (so index goes up 1, reindex brings it back down 1). At least for me, this seems to be the most direct and quick calculation. Without the reindexing, my approach can be interpreted as writing n^3 as the sum of falling factorials which are like the power functions (x^n) of discrete calculus. If one has a background in combinatorics, the coffiencients are the third row of the Stirling numbers of the second kind: 1, 3, 1. Each falling factorial when sum with /n! leaves the sum unaffected (for reasons already discussed), so e. The sum of the kth row of Stirling numbers of the 2nd kind is the kth Bell number. So for n^k instead n^3, the result would be e*B_k. This is pretty well known so a combinatorist would know the answer at a glance.
A much shorter (and easier to motivate) solution is to simply write n^3 = n(n-1)(n-2) + 3n(n-1) + n, to make it easily cancel out with the factorials. Now we simply have 5 terms (1+3+1) with the same sum e, giving 5e.
Nice approach, but it is actually a lot easier to use the series expansion of eˣ. eˣ = Σₙ xⁿ/n! Differentiate and multiply by x three times, you will get x(1+3x+x²)eˣ = Σₙ n³xⁿ/n! Plug in x=1, final result: 5e = Σₙ n³/n!
@@zackbarkley7593 Σ(n≥1) nᵏ⁺¹xⁿ⁻¹/n! = ₖFₖ(2,2,…;1,1,…;x) for k∈ℤ≥0 Σ(n≥1) nᵏ⁺¹xⁿ⁻¹/n! = ₋ₖF₋ₖ(1,1,…;2,2,…;x) for k∈ℤ≤0 Another way to write is Σ(n≥0) nᵏxⁿ/n! = (x d/dx)ᵏ {eˣ} for k∈ℤ≥0 Σ(n≥1) nᵏxₖⁿ/n! = ∫(0,xᵢ)ᵏ (e^(x₀)-1) Π(0≤m≤k-1){xₘ⁻¹} (dxᵢ₋₁)ᵏ for k≥i≥1 and i,k ∈ℤ I guess these formulas aren’t really making it simpler though.
Nice problem. 🙂 The only thing I’d do slightly differently is simply reindex some of those sums so the summands are using n! as a factor rather than, say, (n-2)! Or (n-3)! . You end up in the same place but it’s a bit more streamlined.
You can actually use a trick I learned from a physics textbook way back in the day. Notice that: n * x^n = x (d/dx x^n). You can repeatedly apply this operation to get: sum (n^3 x^n / n!) = x (d/dx (x d/dx ( x d/dx (sum x^n /n!)))) . The sum in the inner parens is e^x and then you just take derivatives and find sum(n^3 x^n) = x(e^x x (x + 2) + (x+1) e^x). Evaluate that bad boy at x = 1 and you have your 5e. This trick can be used in all kinds of sum problems!
After some investigation, it seems that the general sum of n^k/n! is e * B(k), where B(n) is the n-th Bell number (sequence A000110 in the OEIS). I suppose you could turn this into a counting problem with that connection in mind.
Interesting. I solved it in a much easier way using telescoping sums. The sum of (n+1)^3/(n+1)! - n^3/n! will be zero. But each term is equal to ((n+1)^2-n^3)/n!, or (n^2+2n+1-n^3)/n! We can then repeat the above, changing the power to 2, then to 1 (the infinite telescoping sum will always be zero). When we make the power 1, we get that one of the terms is the sum of 1/n!, which we know is e, and since it telescopes to zero, the other sum, n/n! is also e (which can be shown in an easier fashion) In the power of 2 case, we have sum[(n+1-n^2)/n!]=0, or in other words, 2e - sum(n^2/n!)=0, so the n^2 sum is 2e In the power of 3 case, we therefore have 2e+2e+e-sum(n^3/n!)=0, so the sum must be 5e This method gives us an easier way to evaluate higher powers in the sum, for example sum[(n+1)^4/(n+1)! - n^4/n! ]=0 sum[(n^3+3n^2+3n+1- n^4)/n! ]=0 5e+6e+3e+e-sum(n^4/n!)=0 sum(n^4/n!)=15e We can also see sum(n^5/n!)=52e I did a few more terms in the sequence, OEIS suggests these coefficients give the "Bell or exponential numbers: number of ways to partition a set of n labeled elements."
Cool video, interesting to see algebraic tricks I will write an alternative solution for those who are interested Everywhere sum will be from n = 0 to infinity Sum(n³/n!) = sum((n+1)²/n!) because of beginning of this video Let's consider f(x) = sum((n+1)²xⁿ/n!) I'll denote integral with const = 0 as int Then int(f(x)dx)/x = sum((n+1)xⁿ/n!), let's denote this sum as g(x) In the same way int(g(x)dx)/x = sum(xⁿ/n!) and it is = e^x Then int(g(x)dx) = x * e^x and g(x) = e^x * (x+1) From this f(x) = e^x * (x²+3x+1) It remains to be noted that our sum = f(1), that is: Sum(n³/n!) = e^1 * (1² + 3*1 + 1) = 5e
i am going to explain my approach: after we get the summationn^2/(n-1)factorial, we can see that if we multiply e^x with x and then differentiate and then after differentiating that if we multiply x again and differentiate again we will get the summation and then if we put the value of x=1 in the resulting equation we will get 5e
Well done presentation. Before I watched it, I knew it would take me a bunch of re-studying power series after 54 year lapse. My hunch was that n! far outpaces n³ so the series will converge. I took the brute force approach, programming a spreadsheet and just let n tick upward. The series converges at 13.59 = 5e Realizing computers have limited precision I could not hsve full confidence, but my answer is confirmed by your diligent work. Thanks
I think this is easier if you use re-indexing. For example, at the step sum_(n = 1)^oo n^2/(n - 1)! you can replace this with sum_(n = 0)^oo (n + 1)^2/n! Then you can expand (n + 1)^2, split up the sum, rinse and repeat.
Also fun fact; there's something called Dobinski's formula that relates this kind of sum to the number ways to partition a set. The answer is 5e because there are 5 ways to partition a set of 3 elements: 1. {a, b, c} 2. {a}, {b, c} 3. {b}, {a, c} 4. {c}, {a, b} 5. {a}, {b}, {c} If we replace n^3 with n^4 then the answer is 15e because there are 15 ways to partition a set of 4 elements. And so on. This is related to srevere7241's comment since the generating function for the number of ways to partition a set is (e^e^x)/e
You can more easily compute this sum and the related sums where n^3 is replaced with n^k by observing these sums are (multiples of) the moments of a Poisson variable of parameter 1 and making use of the law of rare events to compute them. This way we see that the kth moment is exactly the kth Bell number.
Im have not been watching the video yet, but i calculated 5e. My idea was to take the taylor series of e^x, take the derivative multiply by x and repeat. Every time you do this you get an extra n in the numerator up to n^3 then you do the same to e^x in the end you get (x^3+3x^2+x)*e^x the set x=1 and you get 5e
I did it in a different way: Define a function f(x): f(x) = sum{n^3/n!*x^n; 0...inf} Integrate f(x): F(x) = sum{(1/(n-3)!+(n-1)/n!)*x^n; 1...inf} + C = x^3*e^x + sum{n/(n+1)!*x^(n+1); 0...inf} + C Differentiate the result of F(x): f(x) = F'(x) = (x^3+3*x^2)*e^x + sum{n/n!*x^n; 0...inf} = (x^3+3*x^2)*e^x + x*e^x = (x^3+3*x^2+x)*e^x Evaluate f(x) at x=1: f(1) = sum{n^3/n!; 0...inf} = (1^3+3*1^2+1)*e^1 = 5*e If you want to see all the intermediate steps go to this desmos link: www.desmos.com/calculator/ol7h1a2cyx This method is less straight forward but, it is usefull for deriving different vatiations of the sum, for example: sum{n^3/n!*(-(3+√5)/2)^n; 0...inf} = f(-(3+√5)/2) = 0
Alternate approach (and kinda what I was expecting before watching the video) I'm going to denote sum(f(x)) as the sum from 0 to infinite of f(x). So, f(0) + f(1) + f(2) + ... Note that e^x = sum( (x^n)/n! ) Taking the derivative on both sides, we get: e^x = sum( n(x^(n-1))/n! ) Multiplying both sides by x, we get x e^x = sum( n(x^n)/n! ) Differentiating again: xe^x + e^x = sum( n^2 (x^( n-1))/n! ) And multiplying by x again: x^2 e^x + x e^x = sum( n^2 (x^n) / n!) Differentiating one more time, we get x^2 e^x + 3x e^x + e^x = sum( n^3 x^(n-1) / n!) Now, this is true for all x. So, it's true of x = 1 e + 3e + e = sum( n^3 / n! ) 5e = sum( n^3 / n!)
Alternative argument. Insert x^n in the numerator of the original series and observe that it coincides with the sum from n=0 to ∞ of (n+1)² x^n/n!. Expand (n+1)² = n²+2n+1 = n(n-1)+3n+1, then split the series into three portions, simplify the factorials, and relabel indices. Recognise the Maclaurin series of exp(x) times a certain polynomial, then evaluate at x=1. Changing a 3 into a 4 into the original series leads to (n+1)³ = n³+3n²+3n+1 = n(n-1)(n-2)+6n²+n+1 = n(n-1)(n-2)+6n(n-1)+7n+1. Rinse and repeat. In the general case (i.e. changing 3 to any natural r) we see that the polynomial that emerges in front of exp(x) is of order r-1, and its coefficients are the Stirling numbers of the second kind {r,k}. Evaluating at x=1 leads to the r-th Bell number times e. There is probably a simpler combinatorial explanation...
Just notice that if you take e^x = sum(x^n / n!) and then evaluate x(x(x * (e^x)')')' at x=1 this will give you the desired series. (Basically each differentiation drops n from exponent to, but you have to multiply by X so you don't end up with n-1 on the next step. You do it 3 times so you end up with n^3, and you evaluate at 1 to get rid of x)
Yeah bprp covered this solution in an Instagram story, that's why I didn't bother with that and tried something more elementary instead. Very clean solution tho
Define Skm = Sum[(n+k)^m / n!] and by pulling one factor out of the denominator, (easily) show Skm = S(k+1)(m-1) + k*S(k),(m-1). Then note Sk0 = S00 = e. A few uses of the recursion equation and you have S03 = S30 +3*S20 + S10 = 5S00 = 5e.
I tryed doing that manipulating e^x^2. I realised that the sum with n/n! Is just half the derivative of e^x^2, so I saw that n^m/n! Is just (x/2d/dx)^m e^(x^2) evaluated at 1 and computed the dam thing.
I went a step further to solve this for a general exponent p trying to solve a_p = summ(n=0, inf) n^p/n! for which this specific problem is the case p=3. Using simple binomial expansion and some algebra I came to the recurrence relation a_p = summ(r=0, p-1) a_r, with a_0 = e. This recurrence relation is exactly the same as that of the Bell numbers, with a different seed. Because of the linearity of the recurrence, the general solution can simply be written as a_p = e*B_p where B_p is the p-th bell number
Infinite math is so weird, because what this is saying is that the sum of n^3/n! Is equal to the sum of 5/n!, even though n^3 grows significantly faster than the constant term 5… which doesn’t grow at all of course
Son- Mom? Can we watch bprp at the theatres? Mom- No baby, we have bprp at home! -bprp at home- this video :) all jokes aside thank you Tony. I appreciated this video because it brought back nice year one calc memories and makes me look forward to calc 3!
A more general approach for sums with terms of the form p(n)/n! where p(n) is a polynomial. Observe that: (x^a * exp(x))' = sum((n+a)/n! * x^(n+a-1)) then you can multiply by some other x^b and then take the derivative to get terms of the form (n+a)*(n+a+b-1)/n! * x^(n+a+b-2) repeat this kind of steps to construct the polynomial p(n)=(n+a)(n+a+b-1)(n+a+b+c-2)... etc. For the particular case of the video, it works with: (x * (x * exp(x))' )' = sum(n^3/n! * x^(n-1)) The function on the left evaluates to (x^2 + 3x + 1) * exp(x) Setting x=1 yields the expected result, sum(n^3/n!) = 5 * e
I tried this sum and found another way of solving this beast of a sum...I choose a generating function (e^(xn)/n!) And summing this from n=0 to infinity we'll get e^(e^(x)) or exp(exp(x))....then if we differentiate that sum thrice and let x->0 we'll get our sum and the answer as 5e....
ain't no way bro thought of all that without thinking about differentiation e^x, then multiplying result by x again and differentiating again rinse and repeat again, then put x=1
Good video but you did some mistakes when dividing an infinite sum into two because you must verify that the two converge. In this case it worked because the resulting sums converge but it can be wrong in other cases.
Everything is eeeeeee Everything is eeeeeee Everything is eeeeeee Chorus translated from a dutch song and 'love' replaced with eeeeeee lol. Also it's fun to set n^2 to (n-1)(n+1)+1 and work from there.
It was carefully explained but too long-winded for me, so I stopped watching after a couple of minutes. You can get the answer much faster using the approach you began with. In the first step, you got it to sum from n=1 of n^2/(n-1)!, then shift n to get sum from n=0 of (n+1)^2/n!; expand the numerator to get n^2+2n+1, then break that up as ( sum from n=0 of n(n+2)/n! ) + e; now apply your original trick again to the first part, so sum from n=1, cancel n, and shift n up, so it becomes ( sum from n=0 of (n+3)/n! ) + e = ( sum from n=0 of n/n! ) +3e +e = e +3e + e=5e.
Dont worry if blackpenredpen retires we have this guy to teach us calculus
LOL
But bprp is just awful.
Why would anyone want to learn from him? Not to offend you. I just genuinely do not understand
Is he not still that arrogant and irresponsible as he was 2 yrs ago, when I stopped watching him? He used to blunder left and right in his videos, i.e. missing an entile segments from the proof like a cheddar cheese, or not considering negative real numbers when a problem is defined in both +- real numbers-just willy nilly making do with half proof, etc.
I mean he used to blunder every two, three videos by the time I finally blocked him and I think that's damn plenty.
But fine, blunders, everybody makes them, I make blunders, you make blunders, no one is perfect. What was truly awful about him is he used to delete reasonable comments pointing out faulty logic. Not posting fixes even though the blunder is just so clearly out there for everyone to see. I mean, he could just write some corrections on comment and pin it, the least he could do, or just pin someone else comment that point them out, nope, never.
Or just delete the whole faulty video and act as if such didn't happened. That's just disgusting, because he's already injected faulty knowledges to the first few hundred viewers and after he deletes it, the viewers who got faulse knowledge injected won't even realize it. Just,,, list just goes on and on. I just can't call him a teacher after all the irresponsible behaviours he's have shown.
I'm surprised he still has viewers even. Does he still does this things?
this is not calculus bro
Here from bprp, amazing video! You went through everything with full clarity.
you are such a chill guy to listen to, also you made sure everyone that wasn’t comfortable with reindexing factorials got comfortable
Simpler solution: n.n.n = n.(n-1).(n-2) +3.n.n -2.n = n.(n-1).(n-2) +3.n.(n-1) +n; each of these is zero for some early terms in the sum, that we can duly drop; in the remaining terms, each of our three terms cancels out the first factors of the n! denominator, leaving a sum from n = k to infinity of 1/(n-k)!, for k in {0, 1, 2} with a factor of 3 on the k = 2 one; we can re-index each of these to m = n-k to make it a sum from m = 0 to infinity of 1/m!, so we have e +3.e +e = 5.e.
I think what he does is ultimately the same thing, but yours is much more straightforward.
This is probably the easiest way of solving it. While in a sense this is what he did in the video, it was in a much more complicated and indirect way.
To Tony, I would suggest that after you solve such a problem, try to take a second look at your solution and look for some underlying structure. In this case, you tried to "extract" n-1 and n-2 from the n^3. Once you know that this is what you are looking for, then you should realize that n^3 and n(n-1)(n-2) are pretty "close" in the sense that n^3 - n(n-1)(n-2) has degree 2 instead of 3, and eventually it will lead to the expression in @diddykong3100 comment.
Also, a simplifying step to the method in your video you should consider, is that when you try to extract (n-1) from n^2, instead of guessing what to add and remove, simply write it as n^2 = (n-1+1)^2 = (n-1)^2 +2*(n-1) + 1.
The coefficients 1.3,1,0 are called Stirling numbers of order 3 of the first or second kind (I can't remember which). Your method is the way to go. Well done/
Cool solution! Another way you can approach the problem is to see that the summand (n^3/n!) is already very similar to the expansion for e^x, but with n^3 instead of x^n. After some playing around, you might notice that if you plug in "e^x" itself into e^x, you get: e^(e^x) = sum( (e^x)^n / n! ), and if you take the third derivative, you get sum( (n^3 / n!) * (e^nx) ). Evaluating this at x = 0 is precisely the sum. So the solution is the third derivative of e^(e^x) evaluated at 0, which if you compute it, is 5e. And this approach generalizes to higher powers of n in the numerator.
Do keep in mind though that this problem was given as the third problem on the calc tiebreaker test, meaning that the students had only 15 minutes to do ALL three problems on the test. The approach above (or similar) might be more efficient or even the only option in those kinds of conditions.
Yeah, as soon as I saw the expression I knew it had something to do with e, and once I saw what answer was about to come out of it (admittedly only at 20 minutes or so), I swore there had to be a more elegant way to do it. Thanks for showing me what it was.
cool solution!
Yeah, I did a similar thing, except I took the derivative of e^x, and multiplied by x, then took the derivative of that, and multiplied it by x, and then took the derivative of that, and evaluated at 1.
@@LogosNigrum same.
@@LogosNigrum's way of solving the problem is much quicker tbh. The only thing I could add, is that this whole approach is quite similar to how work with generating functions is usually being done. (or, to be more specific, moment generating functions)
The series equals (x (d/dx))^3 e^x evaluated at x=1. The left hand side expands to x e^x + 3x e^x + x^2 e^x. Setting x=1 gives 5e.
This was also my approach 😅
This is so underrated the explanation was so clear, you just earned another sub
This was excellent! I appreciate you explicitly explaining what you did at 11:30 and again at end. Looking forward to your next video!
If you've done some probability you recognise by multiplying by e^-1 and e that you have E(N^3), the third moment of a Poisson distributed random variable. By differentiating the Moment Generating function three times and evaluating at 0 you solve the problem.
If you wanted to generalize this to n^k/n!, I believe OEIS sequence A000110 gives you the coefficients of e. This sequence is the Bell numbers, or the number of ways to partition a set of exactly k elements. Correct me if I’m wrong.
I love generalizations!
thats awesome
I did similar approach to Tony Wang (the guy in the video) but in the most efficient way in a few minutes. It essentially boils downs to the following decomposition of the numerator:
n^3
= n(n^2)
= n[(n-1)^2+2n-1]
= n[(n-1)((n-2)+1)+2(n-1)+1]
= n(n-1)(n-2) + 3n(n-1) + n.
Those terms become e+3e+e. I did this while reindexing between each step and replacing the 1/n! terms with e. The reindexing doesn't make any rational constants appear because each time before you cancel with the factorial, you remove the first term which is always zero (so index goes up 1, reindex brings it back down 1). At least for me, this seems to be the most direct and quick calculation.
Without the reindexing, my approach can be interpreted as writing n^3 as the sum of falling factorials which are like the power functions (x^n) of discrete calculus. If one has a background in combinatorics, the coffiencients are the third row of the Stirling numbers of the second kind: 1, 3, 1. Each falling factorial when sum with /n! leaves the sum unaffected (for reasons already discussed), so e. The sum of the kth row of Stirling numbers of the 2nd kind is the kth Bell number. So for n^k instead n^3, the result would be e*B_k. This is pretty well known so a combinatorist would know the answer at a glance.
A much shorter (and easier to motivate) solution is to simply write n^3 = n(n-1)(n-2) + 3n(n-1) + n, to make it easily cancel out with the factorials. Now we simply have 5 terms (1+3+1) with the same sum e, giving 5e.
Nice approach, but it is actually a lot easier to use the series expansion of eˣ.
eˣ = Σₙ xⁿ/n!
Differentiate and multiply by x three times, you will get
x(1+3x+x²)eˣ = Σₙ n³xⁿ/n!
Plug in x=1, final result:
5e = Σₙ n³/n!
Smart. I saw this alternate approach elsewhere
Is there then the general formula for sum for n to the k power for k an integer or real?
Thats clever
@@zackbarkley7593
Σ(n≥1) nᵏ⁺¹xⁿ⁻¹/n! = ₖFₖ(2,2,…;1,1,…;x) for k∈ℤ≥0
Σ(n≥1) nᵏ⁺¹xⁿ⁻¹/n! = ₋ₖF₋ₖ(1,1,…;2,2,…;x) for k∈ℤ≤0
Another way to write is
Σ(n≥0) nᵏxⁿ/n! = (x d/dx)ᵏ {eˣ} for k∈ℤ≥0
Σ(n≥1) nᵏxₖⁿ/n! = ∫(0,xᵢ)ᵏ (e^(x₀)-1) Π(0≤m≤k-1){xₘ⁻¹} (dxᵢ₋₁)ᵏ for k≥i≥1 and i,k ∈ℤ
I guess these formulas aren’t really making it simpler though.
Nice problem. 🙂 The only thing I’d do slightly differently is simply reindex some of those sums so the summands are using n! as a factor rather than, say, (n-2)! Or (n-3)! . You end up in the same place but it’s a bit more streamlined.
Good point
You can actually use a trick I learned from a physics textbook way back in the day.
Notice that: n * x^n = x (d/dx x^n). You can repeatedly apply this operation to get:
sum (n^3 x^n / n!) = x (d/dx (x d/dx ( x d/dx (sum x^n /n!)))) .
The sum in the inner parens is e^x and then you just take derivatives and find sum(n^3 x^n) = x(e^x x (x + 2) + (x+1) e^x).
Evaluate that bad boy at x = 1 and you have your 5e. This trick can be used in all kinds of sum problems!
what textbook was this?
@@BellamyJohn-g3j I believe it was the Stat Mech textbook from Harvey Mudd College. I don't recall the author
I did it much more simply. You just have to rearrange:
n^3 = n(n-1)(n-2) + 3n(n-1) + n.
Then you see immediately that the answer is (1+3+1)e = 5e.
Awesome explanation, I could understand every step! 👍🏻
I did it in 3 minutes using change of index variable. Replace n with n+1 and break off sums equal to e a few times and you get 5e in 6 steps.
Loved your video ,hoping to see more .
I love these videos you learn a lot and you understand the math behind it.
A great video topic, with a really satisfying proof, really nicely explained... Top tier video!
After some investigation, it seems that the general sum of n^k/n! is e * B(k), where B(n) is the n-th Bell number (sequence A000110 in the OEIS). I suppose you could turn this into a counting problem with that connection in mind.
Oh.... thx
Great video!!! More please!!!
excellently explained and excellent video, you need more subs
Interesting. I solved it in a much easier way using telescoping sums.
The sum of
(n+1)^3/(n+1)! - n^3/n! will be zero.
But each term is equal to ((n+1)^2-n^3)/n!, or (n^2+2n+1-n^3)/n!
We can then repeat the above, changing the power to 2, then to 1 (the infinite telescoping sum will always be zero).
When we make the power 1, we get that one of the terms is the sum of 1/n!, which we know is e, and since it telescopes to zero, the other sum, n/n! is also e (which can be shown in an easier fashion)
In the power of 2 case, we have sum[(n+1-n^2)/n!]=0, or in other words, 2e - sum(n^2/n!)=0, so the n^2 sum is 2e
In the power of 3 case, we therefore have 2e+2e+e-sum(n^3/n!)=0, so the sum must be 5e
This method gives us an easier way to evaluate higher powers in the sum, for example
sum[(n+1)^4/(n+1)! - n^4/n! ]=0
sum[(n^3+3n^2+3n+1- n^4)/n! ]=0
5e+6e+3e+e-sum(n^4/n!)=0
sum(n^4/n!)=15e
We can also see sum(n^5/n!)=52e
I did a few more terms in the sequence, OEIS suggests these coefficients give the "Bell or exponential numbers: number of ways to partition a set of n labeled elements."
Extremely well explained,thank you very much
Cool video, interesting to see algebraic tricks
I will write an alternative solution for those who are interested
Everywhere sum will be from n = 0 to infinity
Sum(n³/n!) = sum((n+1)²/n!) because of beginning of this video
Let's consider f(x) = sum((n+1)²xⁿ/n!)
I'll denote integral with const = 0 as int
Then int(f(x)dx)/x = sum((n+1)xⁿ/n!), let's denote this sum as g(x)
In the same way int(g(x)dx)/x = sum(xⁿ/n!) and it is = e^x
Then int(g(x)dx) = x * e^x and g(x) = e^x * (x+1)
From this f(x) = e^x * (x²+3x+1)
It remains to be noted that our sum = f(1), that is:
Sum(n³/n!) = e^1 * (1² + 3*1 + 1) = 5e
i am going to explain my approach: after we get the summationn^2/(n-1)factorial, we can see that if we multiply e^x with x and then differentiate and then after differentiating that if we multiply x again and differentiate again we will get the summation and then if we put the value of x=1 in the resulting equation we will get 5e
That's smart
Really thorough explanation, well done!
Well done presentation. Before I watched it, I knew it would take me a bunch of re-studying power series after 54 year lapse. My hunch was that n! far outpaces n³ so the series will converge. I took the brute force approach, programming a spreadsheet and just let n tick upward. The series converges at 13.59 = 5e
Realizing computers have limited precision I could not hsve full confidence, but my answer is confirmed by your diligent work. Thanks
You could have a career in math education, if you are not there already. Your presentation is outstanding.
I think this is easier if you use re-indexing. For example, at the step
sum_(n = 1)^oo n^2/(n - 1)!
you can replace this with
sum_(n = 0)^oo (n + 1)^2/n!
Then you can expand (n + 1)^2, split up the sum, rinse and repeat.
Also fun fact; there's something called Dobinski's formula that relates this kind of sum to the number ways to partition a set. The answer is 5e because there are 5 ways to partition a set of 3 elements:
1. {a, b, c}
2. {a}, {b, c}
3. {b}, {a, c}
4. {c}, {a, b}
5. {a}, {b}, {c}
If we replace n^3 with n^4 then the answer is 15e because there are 15 ways to partition a set of 4 elements. And so on.
This is related to srevere7241's comment since the generating function for the number of ways to partition a set is (e^e^x)/e
Definitely worth my sub
Write n^3=n(n-1)(n-2)+an(n-1)+bn+c and successively get a, b, c. Thresulting split is various multiples of e.
Here from bprp, keep going your are amazing
Amazing how I was able to solve it even after 12 years after IITJEE
You can more easily compute this sum and the related sums where n^3 is replaced with n^k by observing these sums are (multiples of) the moments of a Poisson variable of parameter 1 and making use of the law of rare events to compute them. This way we see that the kth moment is exactly the kth Bell number.
Interesting solution
Very detailed descriptions thank you
Beautiful problems ❤❤❤
Please bring some more Olympiad math problems 😊😊
Love from India ❤
Im have not been watching the video yet, but i calculated 5e.
My idea was to take the taylor series of e^x, take the derivative multiply by x and repeat.
Every time you do this you get an extra n in the numerator up to n^3
then you do the same to e^x
in the end you get
(x^3+3x^2+x)*e^x
the set x=1 and you get 5e
I saw the video. My approach seems a lot easier tbh, i took like 3 minutes lmao.
Right, bprp already posted an approach like this so I wanted something more elementary
amazing man..keep doing this..
I did it in a different way:
Define a function f(x):
f(x) = sum{n^3/n!*x^n; 0...inf}
Integrate f(x):
F(x) = sum{(1/(n-3)!+(n-1)/n!)*x^n; 1...inf} + C
= x^3*e^x + sum{n/(n+1)!*x^(n+1); 0...inf} + C
Differentiate the result of F(x):
f(x) = F'(x) = (x^3+3*x^2)*e^x + sum{n/n!*x^n; 0...inf}
= (x^3+3*x^2)*e^x + x*e^x
= (x^3+3*x^2+x)*e^x
Evaluate f(x) at x=1:
f(1) = sum{n^3/n!; 0...inf}
= (1^3+3*1^2+1)*e^1
= 5*e
If you want to see all the intermediate steps go to this desmos link:
www.desmos.com/calculator/ol7h1a2cyx
This method is less straight forward but, it is usefull for deriving different vatiations of the sum, for example:
sum{n^3/n!*(-(3+√5)/2)^n; 0...inf}
= f(-(3+√5)/2)
= 0
Alternate approach (and kinda what I was expecting before watching the video)
I'm going to denote sum(f(x)) as the sum from 0 to infinite of f(x). So, f(0) + f(1) + f(2) + ...
Note that e^x = sum( (x^n)/n! )
Taking the derivative on both sides, we get:
e^x = sum( n(x^(n-1))/n! )
Multiplying both sides by x, we get
x e^x = sum( n(x^n)/n! )
Differentiating again:
xe^x + e^x = sum( n^2 (x^( n-1))/n! )
And multiplying by x again:
x^2 e^x + x e^x = sum( n^2 (x^n) / n!)
Differentiating one more time, we get
x^2 e^x + 3x e^x + e^x = sum( n^3 x^(n-1) / n!)
Now, this is true for all x. So, it's true of x = 1
e + 3e + e = sum( n^3 / n! )
5e = sum( n^3 / n!)
stroking to this tn big bro
Alternative argument. Insert x^n in the numerator of the original series and observe that it coincides with the sum from n=0 to ∞ of (n+1)² x^n/n!. Expand (n+1)² = n²+2n+1 = n(n-1)+3n+1, then split the series into three portions, simplify the factorials, and relabel indices. Recognise the Maclaurin series of exp(x) times a certain polynomial, then evaluate at x=1.
Changing a 3 into a 4 into the original series leads to (n+1)³ = n³+3n²+3n+1 = n(n-1)(n-2)+6n²+n+1 = n(n-1)(n-2)+6n(n-1)+7n+1. Rinse and repeat.
In the general case (i.e. changing 3 to any natural r) we see that the polynomial that emerges in front of exp(x) is of order r-1, and its coefficients are the Stirling numbers of the second kind {r,k}. Evaluating at x=1 leads to the r-th Bell number times e. There is probably a simpler combinatorial explanation...
Yep, definitely the easiest way to do this.
i really liked your way of explaining, great video!
Just notice that if you take e^x = sum(x^n / n!) and then evaluate x(x(x * (e^x)')')' at x=1 this will give you the desired series. (Basically each differentiation drops n from exponent to, but you have to multiply by X so you don't end up with n-1 on the next step. You do it 3 times so you end up with n^3, and you evaluate at 1 to get rid of x)
Yeah bprp covered this solution in an Instagram story, that's why I didn't bother with that and tried something more elementary instead. Very clean solution tho
Define Skm = Sum[(n+k)^m / n!] and by pulling one factor out of the denominator, (easily) show Skm = S(k+1)(m-1) + k*S(k),(m-1). Then note Sk0 = S00 = e. A few uses of the recursion equation and you have S03 = S30 +3*S20 + S10 = 5S00 = 5e.
Now this is going to be a questions in jee advance 😂
Nah, not possible
These kind of questions are not asked in jee, even though it was lastly asked a long time ago in 2007.
@@Dharun-ge2foThis is asked in mains 2021 ,2023,2024
I tryed doing that manipulating e^x^2.
I realised that the sum with n/n! Is just half the derivative of e^x^2, so I saw that n^m/n! Is just (x/2d/dx)^m e^(x^2) evaluated at 1 and computed the dam thing.
I went a step further to solve this for a general exponent p trying to solve a_p = summ(n=0, inf) n^p/n! for which this specific problem is the case p=3. Using simple binomial expansion and some algebra I came to the recurrence relation a_p = summ(r=0, p-1) a_r, with a_0 = e. This recurrence relation is exactly the same as that of the Bell numbers, with a different seed. Because of the linearity of the recurrence, the general solution can simply be written as a_p = e*B_p where B_p is the p-th bell number
Interesting
Scared me for a sec lol. This problem is from BMT 2023 Calculus Tiebreaker.
the calculus goat himself
Ohhh 😅😅😅
Infinite math is so weird, because what this is saying is that the sum of n^3/n! Is equal to the sum of 5/n!, even though n^3 grows significantly faster than the constant term 5… which doesn’t grow at all of course
dude
Your markers technique are from BPRP
I should consider follow this channel seriously
(Really, not sarcastic, I am talking extremly seriously)
this guy is actually cool as hell, bro🗿
his pen-game is immaculate sheeeesh
Son- Mom? Can we watch bprp at the theatres? Mom- No baby, we have bprp at home! -bprp at home- this video :) all jokes aside thank you Tony. I appreciated this video because it brought back nice year one calc memories and makes me look forward to calc 3!
😁
Why are we not trying to see whether the series is convergent or divergent?
Because factorial grows at a rate faster than cubed, so we already know it is convergent
n^3 = n(n-1)(n-2) +3n(n-1) + 5n and the answer follows
I did it in less than 6 minutes.
Why ain't you using taylors expansions
It is a 2 liner question max
Amazing video
Good work
Great explanation! Wishing you will get big enough room in near future, for landscape view, I have viewed the whole video by zooming & scrolling 😊
HAHA yeahhh 😅😅😅
best video ever. i love maths
A more general approach for sums with terms of the form p(n)/n! where p(n) is a polynomial. Observe that:
(x^a * exp(x))' = sum((n+a)/n! * x^(n+a-1))
then you can multiply by some other x^b and then take the derivative to get terms of the form
(n+a)*(n+a+b-1)/n! * x^(n+a+b-2)
repeat this kind of steps to construct the polynomial p(n)=(n+a)(n+a+b-1)(n+a+b+c-2)... etc.
For the particular case of the video, it works with:
(x * (x * exp(x))' )' = sum(n^3/n! * x^(n-1))
The function on the left evaluates to (x^2 + 3x + 1) * exp(x)
Setting x=1 yields the expected result, sum(n^3/n!) = 5 * e
I tried this sum and found another way of solving this beast of a sum...I choose a generating function (e^(xn)/n!) And summing this from n=0 to infinity we'll get e^(e^(x)) or exp(exp(x))....then if we differentiate that sum thrice and let x->0 we'll get our sum and the answer as 5e....
ain't no way bro thought of all that without thinking about differentiation e^x, then multiplying result by x again and differentiating again rinse and repeat again, then put x=1
Valid solution, but bprp already covered it
From bprp. Love the pen switch
Sum does not change whether n starts from 0 or 1, since first (n=0th) term = 0^3/0! = 0/1=0. So, n can start from 1.
n^3 / n! = n^2/(n-1)! = [(n-1)+1]^2 / (n-1)! = (n-1) / (n-2)! + 2 / (n-2)! + 1/(n-1)! = 1/(n-3)! + 1/(n-2)! + 2/(n-2)! + 1/(n-1)! = 1/(n-3)! + 3/(n-2)! + 1/(n-1)!
Therefore, sum = (first 3 terms) + (remaining n>=3 terms) = (0^3/0! + 1^3/1! + 2^3/2!) + (1/0! + 1/1! + ...) + 3 (1/1! + 1/2! + ...) + (1/2! + 1/3! + ...)
= (0 + 1 + 4) + (e) + 3 (e-1) + (e - 1 - 1/1!) = 5 + e + 3e - 3 + e - 2 = (1 + 3 + 1) e + (5 - 3 - 2) = 5 e + 0 = 5 e
Hence, sum = 5 e
interesting solution
Bro we need integrals bring video ❤
Soon
Theres a slight error at 15:30 where you missed the ! on the left side. Could be useful to point it out.
Good one!
Esto demuestra la hipótesis de riman
We can start with n=0, that's an easy one down
Nice solution!!!
Nice problem! Nice job!
I'd put n-2 +3 + 1/(n-1) imediately, this way you have the sum of 1/(n-3)! + 3/(n-2)! + 1/(n-1)! Which is faster delt with
so nice!
excellent
This is a jee question they have copied from jee mains
Who?
Hoping i could be like this one day (i understand nothing)
Buenisimo análisis
Aura + 1000
keep going broo
You should use landscape whiteboard, not portrait
Unfortunately we don’t have room for that ☹️
from Morocco thank you son...but around mn10-30 you split the sum....had we the right to do so????
I think so because the sum converges absolutely
5e ?
Limit as ab integral ?
2024???
Apparently it’s from 2023 BMT Calc tiebreaker round
Too small i cant read
don't worry, you'll grow and learn how to read in no time!
Good job 👍
Good video but you did some mistakes when dividing an infinite sum into two because you must verify that the two converge. In this case it worked because the resulting sums converge but it can be wrong in other cases.
Good point, I'll keep try to keep things rigorous enough in the future
nice
tony
How you can get questions of the Berkeley Math tournament?
Nice, elegant and very clear explanation. Good job!
Everything is eeeeeee
Everything is eeeeeee
Everything is eeeeeee
Chorus translated from a dutch song and 'love' replaced with eeeeeee lol.
Also it's fun to set n^2 to (n-1)(n+1)+1 and work from there.
Average jee mains question
*@ Tony Wang* -- The camera needs a better close-up of the board. What you are writing is too far away, and it is an effort to make it out.
You're right, but with the space limitations I have it's hard to make that happen
It was carefully explained but too long-winded for me, so I stopped watching after a couple of minutes.
You can get the answer much faster using the approach you began with. In the first step, you got it to sum from n=1 of n^2/(n-1)!, then shift n to get sum from n=0 of (n+1)^2/n!; expand the numerator to get n^2+2n+1, then break that up as ( sum from n=0 of n(n+2)/n! ) + e; now apply your original trick again to the first part, so sum from n=1, cancel n, and shift n up, so it becomes ( sum from n=0 of (n+3)/n! ) + e = ( sum from n=0 of n/n! ) +3e +e = e +3e + e=5e.
Ooh this is elegant