Michael has a unique style of complicating very easy things. Without any reindexing, all fractions must be brought to the form constant/(..)! We have n^2, so we need a product of two expressions with n to simplify with the ends of the factorial. So we need (n+1)(n+2). So, until now, n^2=(n+1)(n+2)-3n-4. Now we need that n to simplify with the end of the factorial. So, 3n=3(n+2)-6. So n^2-2=(n+1)(n+2)-3(n+2)+2. Therefore, S=S1-3S2+2S3, where S1 is the sum of 1/n! S2 is the sum of 1/(n+1)! S3 is the sum of 1/(n+2)! All three are from 1 to infinity. Therefore, S1=e-1 S2=e-1-1/1! S3=e-1-1/1!-1/2! Therefore, S=(e-1)-3(e-2)+2(e-2-1/2) =0.
5:30 "Utterly obliterates" 1. We need to use this term in mathematical proofs more 2. I love how Michael can say that with a straight face, cause I know I would never be able to.
My Modern Algebra professor, when talking about groups, would refer to the order of that group as the killing power of the generator. Deadpan face n' all. The people that pursue pure mathematics are a little less than sane.
You can do this without using any power series or derivatives. Simply observe that Σ n/n! = Σ 1/(n-1)!, then reindex n -> n+1 and you get that it's the same Σ1/n!.
There's a slight problem that n/n! = 1/(n-1)! somewhat breaks when n=0. You could handwave your way around this by defining n! =infinity for negative n. However, the terms where this would happen are actually from extending the sum to start from zero in the first place. We can avoid this by extending sums _after_ splitting into three sums and making your substitution rather than extending up front.
@@lunstee Even better, we dont have to add the additional terms at all. The end result will be 3 sums which start at 1,2 and 3, respectively so all terms from 3 upwards cancel out and we only have to calculate the remaining 3 terms for the final answer.
You can just reindexing the first 2 sums (noting that the first terms are 0), and cancel n and n-1 with the first factors in the denominator. And you immediate obtain e -3e +2e without using limits and derivations
With a bit of symbol-pushing, you can separate out this into: (n² - 2)/(n+2)! = (n-1)/(n+1)! - n/(n+2)! The whole series then telescopes with most of the terms cancelling out, except for the very first term (which is 0) and the very last term (which is 0 in the limit)
What an elegant solution! Others have pointed out ways this can be done using more purely algebraic techniques, but the Taylor series approach is elegant and very useful for more complex problems!
Anyone else notice a difference in acoustics in the past several weeks? It gives Michael's voice a very tired feeling. I hope its acoustics and not health.
@@alessandrorenna1222 No, to say they are C-infinity just means they are equivalent. Even if that wasn't the case you can use evaluation directly rather than limits even if they're not necessarily the same
If you learn a bit of discrete calculus, there's a really nice shortcut to this! There's an equivalent to integration by parts for sums where you can show that the sum of 1/(n+2)! from 1 to x is equivalent to x/(x+2)! plus the sum of (n^2 - 1)/(n+2)! from 1 to x. Rearranging gives the sum shown in the video up to some value x expressed as the function -x/(x+2)!, and from there it's easy to just take the limit as x goes to infinity to show the sum is equal to 0.
I would have started by rewriting the numerator n^2-2 as (n+2)*(n+1) -3*(n+2) +2. These three terms can then be put over (n+2)! individually, which then reduces 1/n! -3/(n+1)! +2/(n+2)! The sum of these three terms can be split into three sums. The 3/(n+1)! and 2/(n+2)! series can be re-indexed in terms of n! starting with n=1 (or all three sums to n=0) in similar manner as the first step of Michael's solution. The 'new terms' from re-indexing will add to zero, and the sums themselves are also identical aside from factors of 1,-3 and +2 respectively, so the summations also sum to zero.
Much simpler: Seeking to reduce the denominator to n!, begin by re-writing the numerator n²-2 as (n+2)(n+1) - 3(n+2) + 2. Break into three sums. Simplifying factorials, each becomes a constant times an infinite 1/(n+a)! sum. Play the reindex game Dr. Penn started with on the second and third terms separately to get a constant in the numerator and n! in the denominator. As all the sums are equal except for the constant numerators, factor them out. You get (1-3+2) sum_0^∞(1/n!) = 0 * sum(1/n!). You don't even have to recognize the sum as being equal to e to get zero! As an intermediately complicated proof, at 5:58, do the simplification of factorials, re-index the first two terms to again get n! in the denominators, and finish up as I did above. [This was actually my first solution, but I figured I could cut three reindexings to two.)
I first thought about the exp(x) series, but could not be bothered. Simplifying the factorials so I get constants in the numerator then re-indexing is simpler and faster. fn = n^2-2 = (n + 2)(n - 2) + 2 gn = fn / (n + 2)! = (n - 2) / (n + 1)! + 2/(n + 2)! n - 2 = n + 1 - 3, so: gn = 1 / n! - 3 / (n + 1)! + 2 / (n + 2)! Summing and re-indexing while leaving the n=1, 2 for the first term, leaving out the term n=1 for the second term, you get: sn = (1 + 1/2) - 3/2 + 3*sum from n=3 to inf ( 1/n! - 1/n!) = 0
Even after seeing this, my intuition was really bothered by the fact that Sum(n^2/(n+2)! = Sum(2/(n+2)!) =2e-5 (2 times e minus 5) where both sums are over n=1 to infinity. Since n^2 is bigger than 2, my intuition said that an infinity of terms of the n^2 version are larger than the 2 version. (1/3!, 4/4!, 9/5!, 16/6!...) vs (2/3!, 2/4!, 2/5!, 2/6!...). But obviously, the first term of 2/3! > 1/3! makes up for the smaller size on the rest of the infinite number of terms. Interesting one today...
I did it before watching the video but my way was a bit more complicated: evaluate d/dx( 1/x d/dx( 1/x exp(x) ) ) both as a power series and as a product of exp(x) times a rational function of x and then evaluate at x=1 The point was to obtain the sum of (n^2-1)/(n+2)! Taking care of the indices and the first terms I arrived at e-5/2-(e-5/2)=0
Rewrite n^2 - 2 as (n-1) (n+1) - n, and then (n-1) (n+2) / (n+2)! = (n-1) / (n+1)!. Hence, the result is Σₙ₌₁(n-1) / (n+1)! - Σₙ₌₁ n / (n+2)!, where the terms cancel out except from the case of n=1 in the first sum which equals 0.
n^2-2 = (n+2)((n+1)-3)+2, so a_n = 1/n! - 3/(n+1)! + 2/(n+2)!, this thing converges, so we can reorder it as 1/n! - 3/n! + 2/n!, then account for irregularities for n = 1 and 2 and we're done
Needn't have used derivatives, we can leave the series as 3 to infinity and then use the e¹ expansion removing the terms which aren't there (e-1) - 3(e-1-1) + 2(e-1-1-0.5) = 0
wait so a part of this video(when he puts the limit there and it magically turns into all e^x 's) makes me think that the sum from n=0 to inf of (n-k)!/n!= e. is this true? if so then thats very interesting
I guess you mean n(n-1)(n-2)...(n-k)/n! which is n!/((n-k+1)!n!) and obviously simplifies to 1/(n-k+1)! which is just a re-indexed version of 1/n! and so has the same value.
Hi sir i watched all Ur vidéos ....esp lectures on différentiel forms .....i will bé very happy if u Bégin lectures on sobolev spaces and elliptic problems
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Michael has a unique style of complicating very easy things.
Without any reindexing, all fractions must be brought to the form constant/(..)!
We have n^2, so we need a product of two expressions with n to simplify with the ends of the factorial.
So we need (n+1)(n+2).
So, until now,
n^2=(n+1)(n+2)-3n-4.
Now we need that n to simplify with the end of the factorial. So, 3n=3(n+2)-6.
So n^2-2=(n+1)(n+2)-3(n+2)+2.
Therefore, S=S1-3S2+2S3,
where
S1 is the sum of 1/n!
S2 is the sum of 1/(n+1)!
S3 is the sum of 1/(n+2)!
All three are from 1 to infinity.
Therefore,
S1=e-1
S2=e-1-1/1!
S3=e-1-1/1!-1/2!
Therefore,
S=(e-1)-3(e-2)+2(e-2-1/2)
=0.
5:30 "Utterly obliterates"
1. We need to use this term in mathematical proofs more
2. I love how Michael can say that with a straight face, cause I know I would never be able to.
My Modern Algebra professor, when talking about groups, would refer to the order of that group as the killing power of the generator. Deadpan face n' all. The people that pursue pure mathematics are a little less than sane.
You can do this without using any power series or derivatives. Simply observe that Σ n/n! = Σ 1/(n-1)!, then reindex n -> n+1 and you get that it's the same Σ1/n!.
There's a slight problem that n/n! = 1/(n-1)! somewhat breaks when n=0. You could handwave your way around this by defining n! =infinity for negative n. However, the terms where this would happen are actually from extending the sum to start from zero in the first place. We can avoid this by extending sums _after_ splitting into three sums and making your substitution rather than extending up front.
@@lunstee Just pull out the n=0 term...which is 0. Similarly, you can pull the first two terms out of the other sum in question.
Show us your work when using this method.
@@lunstee Even better, we dont have to add the additional terms at all. The end result will be 3 sums which start at 1,2 and 3, respectively so all terms from 3 upwards cancel out and we only have to calculate the remaining 3 terms for the final answer.
You can just reindexing the first 2 sums (noting that the first terms are 0), and cancel n and n-1 with the first factors in the denominator. And you immediate obtain e -3e +2e without using limits and derivations
You'll still need to use the limit though.
With a bit of symbol-pushing, you can separate out this into:
(n² - 2)/(n+2)! = (n-1)/(n+1)! - n/(n+2)!
The whole series then telescopes with most of the terms cancelling out, except for the very first term (which is 0) and the very last term (which is 0 in the limit)
What an elegant solution! Others have pointed out ways this can be done using more purely algebraic techniques, but the Taylor series approach is elegant and very useful for more complex problems!
I love that the nearly final answer is "The limit as x approaches 1 of 0"
Anyone else notice a difference in acoustics in the past several weeks? It gives Michael's voice a very tired feeling. I hope its acoustics and not health.
What's the significance of taking the limit? All functions here are C-infinity. You can just say evaluated at x=1 and you're fine
Well. It's the other way round. To state that they are C-infinity you are implicitly evaluating limits
@@alessandrorenna1222 No, to say they are C-infinity just means they are equivalent. Even if that wasn't the case you can use evaluation directly rather than limits even if they're not necessarily the same
If you learn a bit of discrete calculus, there's a really nice shortcut to this! There's an equivalent to integration by parts for sums where you can show that the sum of 1/(n+2)! from 1 to x is equivalent to x/(x+2)! plus the sum of (n^2 - 1)/(n+2)! from 1 to x. Rearranging gives the sum shown in the video up to some value x expressed as the function -x/(x+2)!, and from there it's easy to just take the limit as x goes to infinity to show the sum is equal to 0.
I would have started by rewriting the numerator n^2-2 as (n+2)*(n+1) -3*(n+2) +2. These three terms can then be put over (n+2)! individually, which then reduces 1/n! -3/(n+1)! +2/(n+2)!
The sum of these three terms can be split into three sums. The 3/(n+1)! and 2/(n+2)! series can be re-indexed in terms of n! starting with n=1 (or all three sums to n=0) in similar manner as the first step of Michael's solution. The 'new terms' from re-indexing will add to zero, and the sums themselves are also identical aside from factors of 1,-3 and +2 respectively, so the summations also sum to zero.
Mike has such good blackboard technique with his coloured chalks...I'm envious.
Much simpler: Seeking to reduce the denominator to n!, begin by re-writing the numerator n²-2 as (n+2)(n+1) - 3(n+2) + 2. Break into three sums. Simplifying factorials, each becomes a constant times an infinite 1/(n+a)! sum. Play the reindex game Dr. Penn started with on the second and third terms separately to get a constant in the numerator and n! in the denominator. As all the sums are equal except for the constant numerators, factor them out. You get (1-3+2) sum_0^∞(1/n!) = 0 * sum(1/n!). You don't even have to recognize the sum as being equal to e to get zero!
As an intermediately complicated proof, at 5:58, do the simplification of factorials, re-index the first two terms to again get n! in the denominators, and finish up as I did above. [This was actually my first solution, but I figured I could cut three reindexings to two.)
I first thought about the exp(x) series, but could not be bothered.
Simplifying the factorials so I get constants in the numerator then re-indexing is simpler and faster.
fn = n^2-2 = (n + 2)(n - 2) + 2
gn = fn / (n + 2)! = (n - 2) / (n + 1)! + 2/(n + 2)!
n - 2 = n + 1 - 3, so:
gn = 1 / n! - 3 / (n + 1)! + 2 / (n + 2)!
Summing and re-indexing while leaving the n=1, 2 for the first term, leaving out the term n=1 for the second term, you get:
sn = (1 + 1/2) - 3/2 + 3*sum from n=3 to inf ( 1/n! - 1/n!) = 0
Even after seeing this, my intuition was really bothered by the fact that Sum(n^2/(n+2)! = Sum(2/(n+2)!) =2e-5 (2 times e minus 5) where both sums are over n=1 to infinity. Since n^2 is bigger than 2, my intuition said that an infinity of terms of the n^2 version are larger than the 2 version. (1/3!, 4/4!, 9/5!, 16/6!...) vs (2/3!, 2/4!, 2/5!, 2/6!...). But obviously, the first term of 2/3! > 1/3! makes up for the smaller size on the rest of the infinite number of terms. Interesting one today...
That was fun.
Thank you, professor.
0 sure was a surprise. Penn really is a joy and entertainment par none! Hope he gives me a heart someday, I feel like a the rusty tin man.
Lovely result!
Nothing more annoying than when you discover you have to go through major elaboration math techniques only to find out the answer is 0
Lmao 😂
we can use a expential of linear transformation sigma P to polynome P(X+1) actine at X2-4X+2 evaluate at 0
I did it before watching the video but my way was a bit more complicated:
evaluate d/dx( 1/x d/dx( 1/x exp(x) ) ) both as a power series and as a product of exp(x) times a rational function of x and then evaluate at x=1
The point was to obtain the sum of (n^2-1)/(n+2)!
Taking care of the indices and the first terms I arrived at e-5/2-(e-5/2)=0
Rewrite n^2 - 2 as (n-1) (n+1) - n, and then (n-1) (n+2) / (n+2)! = (n-1) / (n+1)!. Hence, the result is Σₙ₌₁(n-1) / (n+1)! - Σₙ₌₁ n / (n+2)!, where the terms cancel out except from the case of n=1 in the first sum which equals 0.
Go, Hokies! I never attended VA Tech, but several friends did, and we watched _Fatal Attraction_ in a theater in Blacksburg. Does that count? 😉
Smoothly done!
8:56
Why you stopped suggesting problems for Michael
In short, not enough time for that 😂 But maybe I should take few hours for that 👀
n^2-2 = (n+2)((n+1)-3)+2, so a_n = 1/n! - 3/(n+1)! + 2/(n+2)!, this thing converges, so we can reorder it as 1/n! - 3/n! + 2/n!, then account for irregularities for n = 1 and 2 and we're done
The description is impeccable as always 👌
Really clever solution
Needn't have used derivatives, we can leave the series as 3 to infinity and then use the e¹ expansion removing the terms which aren't there
(e-1) - 3(e-1-1) + 2(e-1-1-0.5) = 0
My nephew just graduated from his and wanted to go to VT but got a bigger scholarship to go to SLU instead.
I admit, I was stumped😮
Why so complicated
Just see that n^2-2=(n+2)(n-1)-n
Then you have a telescopic sum
(n-1)/(n+1)! - n/(n+2)!
That is egal to 0
wait so a part of this video(when he puts the limit there and it magically turns into all e^x 's) makes me think that the sum from n=0 to inf of (n-k)!/n!= e. is this true? if so then thats very interesting
I guess you mean n(n-1)(n-2)...(n-k)/n! which is n!/((n-k+1)!n!) and obviously simplifies to 1/(n-k+1)! which is just a re-indexed version of 1/n! and so has the same value.
There are problems for the first few terms where n-k
@@GandalfTheWise0002 The terms where n-k
I give questions like this in my second semester calculus class.
How can it be zero if the first three terms add up to cero and the rest of the terms are all positive? I missed something I guess...
The rest of the terms aren't all positive. Check out n=3.
@@TedHopp You're right. As I said, I was missing something... Thanks.
Hi sir i watched all Ur vidéos ....esp lectures on différentiel forms .....i will bé very happy if u Bégin lectures on sobolev spaces and elliptic problems
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Σ inf/n=1 - 0