Even better is that doing that we done even need to know the classic convergence to e; now it is just a very simple telescoping series where everything cancels out except the first 1, so the sum is 1.
@@bsmith6276 ohh I see, so all terms for n≥2 cancel and we just have 1/1! cool, even cleaner. I'm surprised he didn't see these ways to solve it, but I still think his is interesting as well
An alternative approach: Since the series for x*e^x centered at 0 converges on (-infinity,infinity), and [0,1] is contained in this interval, we can interchange the order of integration and summation for the power series. Applying power rule for antiderivatives at the end, and evaluating from 0 to 1, we have Integrate[x*e^x,{x,0,1}] =Integrate[x*Sum[x^n/n!,{n,0,infinity}],{x,0,1}] =Integrate[Sum[x^(n+1)/n!,{n,0,infinity}],{x,0,1}] =Sum[Integrate[x^(n+1)/n!,{x,0,1}],{n,0,infinity}] =Sum[1/(n!*(n+2)),{n,0,infinity}] Using integration by parts, the value of the original integral is 1, so the series has the same value.
This collapses to a telescoping series after multiplying by n+1 on the top and bottom and then writing n+1 as n+2-1. Differentiating under the sum can also work though it is more involved. There are at least 4 ways of doing this. Either cancel the n+2 in the denominator by differentiating the sum with a t^(n+2) in the numerator and use t=0 as a boundary condition. Similarly we can multiply by n+1 on the top and bottom and then introduce it by differentiating t^(n+1)/(n+2)! as the summand. An exponential can also be employed though it requires more work for no added benefit except perhaps to practise the technique.
Here's another method, using the expansion of e^x. e^x=summation x^n/n! till n tends to infinity. Multiply x on both sides and integrate. Put limits from 0 to 1 and you have the answer.
1/n!(n+2) = (n+1)/(n+2)! = (n+2 - 1)/(n+2)! = (n+2)/(n+2)! - 1/(n+2)! = 1/(n+1)! - 1/(n+2)! So, the partial series from n=0 to n=k can be rewritten as (1/1! - 1/2!) + (1/2! - 1/3!) + (1/3! - 1/4!) + ... + (1/(k+1)! - 1/(k+2)!). And this is a telescoping series, so we can cancel out a lot of common terms, giving us 1/1! - 1/(k+2)!.
Index magic: = n+1 /(n+2)! from 0 = n/(n+2)! from 0+ 1/(n+2)! from 0 = n/(n+2)! from 0+ (e-1) which is equal to from 1 so shift =n-1/(n+1)! from 1 + (e-1) =n/(n+1)! from 1 - (e-1) + (e-1) =n/(n+1)! from 0 which is equal to from 1 so shift = (n-1)/n! from 1 = 1/(n-1)! -1/n! everything cancels except for 1
That was amazing dude. All it was though was me and Peter from Free University of Berlin in a blink of a second without being an absolute psychopath though. There was even a German Lady taking a practice test too and that didn't do anything.
@@rosiefay7283 there are some sums that had gamma function like sum from 1 to infinity of (-1)^n.gamma²(n)/gamma(2n) , and it can be solved by the beta function which is already an integral
It's also solvable through Taylor series for exp(x) and the generating functions method (the value of the series sum ends up being exactly (exp(x)/x - 1/x - 1)' evaluated at x=1)
Getting concerned... why say "good morning" like a regular person... you're better than that. You're 2 seconds of silly often hype me up for the math and I'm getting concerned and sad
Did you mean that one has been a prime number for ages, or one has been a PRIME number for ages? I'm just making sure I write it down in my notes the correct way. 😎
If it's +1 instead of +2, the result is e-1 and if it's +3 instead of +2, the result is e-2. I haven't looked into it much but I'm curious if anybody knows why off the top of their head
Here's one way to understand that, copying from another comment I made: Since the series for x*e^x centered at 0 converges on (-infinity,infinity), and [0,1] is contained in this interval, we can interchange the order of integration and summation for the power series. Applying power rule for antiderivatives at the end, and evaluating from 0 to 1, we have Integrate[x*e^x,{x,0,1}] =Integrate[x*Sum[x^n/n!,{n,0,infinity}],{x,0,1}] =Integrate[Sum[x^(n+1)/n!,{n,0,infinity}],{x,0,1}] =Sum[Integrate[x^(n+1)/n!,{x,0,1}],{n,0,infinity}] =Sum[1/(n!*(n+2)),{n,0,infinity}] Using integration by parts, the value of the original integral is 1, so the series has the same value. If you wanted n+3 instead of n+2, integrate x^2*e^x instead. etc.
Yes, you did some numerical work to evaluate some partial sums, and you spotted a pattern, but you should have shown us the proof. All this is no good without the proof.
I suppose it is left as an exercise to the viewer to prove that each partial sum is indeed equal to [(k+2)!-1]/[(k+2)!]?
It's pretty easy to do with induction.
1/k!(k+2) = (k+1)/(k+2)! = (k+2-1)/(k+2)!
= 1/(k+1)! - 1/(k+2)!
The sum terminates, leaving S_n = 1 - 1/(n+2)!
You can instantly see it can be proved by induction by just verifying that it still holds for k+1.
I was really disappointed when he didn't end up proving that equallity, mainly because, as @iammathwiz7 said, it's pretty easy to do with induction.
ah the good old telescoping sum
what about we multiply n+1 to above and below, we can make it into 1/(n+1)! - 1/(n+2)!, then the equation equal to (e-1)-(e-1-1)=1
This is what I did!
Even better is that doing that we done even need to know the classic convergence to e; now it is just a very simple telescoping series where everything cancels out except the first 1, so the sum is 1.
@@bsmith6276 ohh I see, so all terms for n≥2 cancel and we just have 1/1!
cool, even cleaner.
I'm surprised he didn't see these ways to solve it, but I still think his is interesting as well
I did that too. Quicker, and without any blackboard.
Of course, the answer is the mathematical constant Wau. It's truly everywhere.
indeed
Bro did not induct 😭
8:05 This formula is also very easy to prove via induction.
An alternative approach: Since the series for x*e^x centered at 0 converges on (-infinity,infinity), and [0,1] is contained in this interval, we can interchange the order of integration and summation for the power series. Applying power rule for antiderivatives at the end, and evaluating from 0 to 1, we have
Integrate[x*e^x,{x,0,1}]
=Integrate[x*Sum[x^n/n!,{n,0,infinity}],{x,0,1}]
=Integrate[Sum[x^(n+1)/n!,{n,0,infinity}],{x,0,1}]
=Sum[Integrate[x^(n+1)/n!,{x,0,1}],{n,0,infinity}]
=Sum[1/(n!*(n+2)),{n,0,infinity}]
Using integration by parts, the value of the original integral is 1, so the series has the same value.
This collapses to a telescoping series after multiplying by n+1 on the top and bottom and then writing n+1 as n+2-1. Differentiating under the sum can also work though it is more involved. There are at least 4 ways of doing this. Either cancel the n+2 in the denominator by differentiating the sum with a t^(n+2) in the numerator and use t=0 as a boundary condition. Similarly we can multiply by n+1 on the top and bottom and then introduce it by differentiating t^(n+1)/(n+2)! as the summand. An exponential can also be employed though it requires more work for no added benefit except perhaps to practise the technique.
Here's another method, using the expansion of e^x. e^x=summation x^n/n! till n tends to infinity. Multiply x on both sides and integrate. Put limits from 0 to 1 and you have the answer.
Could you use the maclaurin series of xe^x and integrate?
1/n!(n+2) = (n+1)/(n+2)! = (n+2 - 1)/(n+2)! = (n+2)/(n+2)! - 1/(n+2)! = 1/(n+1)! - 1/(n+2)!
So, the partial series from n=0 to n=k can be rewritten as (1/1! - 1/2!) + (1/2! - 1/3!) + (1/3! - 1/4!) + ... + (1/(k+1)! - 1/(k+2)!). And this is a telescoping series, so we can cancel out a lot of common terms, giving us 1/1! - 1/(k+2)!.
10:10 ... I know that 1 is not prime...
Index magic:
= n+1 /(n+2)! from 0
= n/(n+2)! from 0+ 1/(n+2)! from 0
= n/(n+2)! from 0+ (e-1)
which is equal to from 1 so shift
=n-1/(n+1)! from 1 + (e-1)
=n/(n+1)! from 1 - (e-1) + (e-1)
=n/(n+1)! from 0
which is equal to from 1 so shift
= (n-1)/n! from 1
= 1/(n-1)! -1/n!
everything cancels except for 1
I was literally solving this question a few minutes ago and now this popped up on my feed lmao
nice :'D
That was amazing dude. All it was though was me and Peter from Free University of Berlin in a blink of a second without being an absolute psychopath though. There was even a German Lady taking a practice test too and that didn't do anything.
we can transform "1/(n+2) = int (0,1) of x^(n+1)" it gives the integral from (0,1) of xe^x which gives "1"
But that's irrelevant. We have a sum here, not an integral.
@@rosiefay7283 there are some sums that had gamma function like sum from 1 to infinity of (-1)^n.gamma²(n)/gamma(2n) , and it can be solved by the beta function which is already an integral
It's also solvable through Taylor series for exp(x) and the generating functions method (the value of the series sum ends up being exactly (exp(x)/x - 1/x - 1)' evaluated at x=1)
I read pokemon instead of problem. Still excited
cool way to approximate 1😮
This series telescopes.
Oh fuck yeah
this is seriesly cool!
When will there be a new flammy wood vid?
In would be cool to prove that the pattern found is indeed correct for each k € N
Getting concerned... why say "good morning" like a regular person... you're better than that. You're 2 seconds of silly often hype me up for the math and I'm getting concerned and sad
🔥
Did you mean that one has been a prime number for ages, or one has been a PRIME number for ages? I'm just making sure I write it down in my notes the correct way. 😎
If it's in physics, it's convergent!
(n+1)/(n+1), -> (n+1)/(n+2)!, ->(n+1+1-1)/(n+2)! -> 1/(n+1)! - 1/(n+2)! ->reindexing -> 1+sum-sum ->=1
If it's +1 instead of +2, the result is e-1 and if it's +3 instead of +2, the result is e-2. I haven't looked into it much but I'm curious if anybody knows why off the top of their head
Here's one way to understand that, copying from another comment I made: Since the series for x*e^x centered at 0 converges on (-infinity,infinity), and [0,1] is contained in this interval, we can interchange the order of integration and summation for the power series. Applying power rule for antiderivatives at the end, and evaluating from 0 to 1, we have
Integrate[x*e^x,{x,0,1}]
=Integrate[x*Sum[x^n/n!,{n,0,infinity}],{x,0,1}]
=Integrate[Sum[x^(n+1)/n!,{n,0,infinity}],{x,0,1}]
=Sum[Integrate[x^(n+1)/n!,{x,0,1}],{n,0,infinity}]
=Sum[1/(n!*(n+2)),{n,0,infinity}]
Using integration by parts, the value of the original integral is 1, so the series has the same value.
If you wanted n+3 instead of n+2, integrate x^2*e^x instead. etc.
Yes, you did some numerical work to evaluate some partial sums, and you spotted a pattern, but you should have shown us the proof. All this is no good without the proof.
I dont understand ,why dont we just integrate x*e^x from 0 to 1 , and get result 1 ?
That's pretty
The series is just x*e^x that we are integrating from 0 to 1
Yes daddy more
e^x=Sum(x^n/n!) ,two side times x and integrate from 0 to 1 ,easy to get the result ,is 1 . Why dont just do it like this way?
first