Also agreed. Introducing x impressed me. The rest was smooth jazz, but I've never seen introducing an extra variable done this way. Usually it's a generating function trick or some sum where the number of terms in each summand is fixed, here it's brand new because the nth summand has n terms. All that being said, a 3rd way is making the original summand x^n/k! and applying the usual xD twice.
For the first method, one must be aware that changing the order of two summations is allowed in this case since the summands are positive. (thanks to Tonelli) +Method 2 is just CRAZY😂
9:42 That ± sum threw me off. I get what it's supposed to mean in the context, but that's some gnarly abuse of notation. I think it would have made more sense to write + (1-1) * sum[...] if you don't want to spell out the sum twice. It's a nice solution though.
Hmmm. We had this exact d.e. for homework in my junior year Differential Equations course. I solved it and emphatically stated that that is a useless problem...
Truthfully: neither, both and each. All have a relevance that is difficult for me to establish in sense if all are null it is wrong, if one is relevant all are relevant. I put it down to reading Surreal Numbers recently. EDIT: there is beauty in results, methods, questions and surely that beauty is not scientific?
At 9:15 the last term should be x^(n-1) / (n-1)!. The n from the power rule derivative cancels the n of the n!
They both involve splitting the sum to make 1/2 + 1/3 so that's pretty cool
Beautiful example of two different re-visualization techniques to solve problems.
First method is very straightforward and what I would think to do. Second method is nifty.
I thought you said "Second method is shitty" lol. I'm too tired I guess
Method 1 is ez but damn method 2 is creative
Agree!
Also agreed. Introducing x impressed me. The rest was smooth jazz, but I've never seen introducing an extra variable done this way. Usually it's a generating function trick or some sum where the number of terms in each summand is fixed, here it's brand new because the nth summand has n terms.
All that being said, a 3rd way is making the original summand x^n/k! and applying the usual xD twice.
Very nice solution #2 elegant
The 2nd trick is like what you used to derive the Fresnel integrals the other day. Most excellent!
9:15 Rather (n-1)!
You can get the result for n^m instead of n^2 in terms of a sum to m of binomial coefficients, Bell numbers, and Bernoulli numbers.
Nope
You can cancel the k(k-1) to leave (k-2)!.
I sadly understood the title reference
For the first method, one must be aware that changing the order of two summations is allowed in this case since the summands are positive. (thanks to Tonelli)
+Method 2 is just CRAZY😂
9:42 That ± sum threw me off. I get what it's supposed to mean in the context, but that's some gnarly abuse of notation.
I think it would have made more sense to write + (1-1) * sum[...] if you don't want to spell out the sum twice.
It's a nice solution though.
Agree
The second method is cool, if convoluted. 😍
Are there assumptions about the sums being “well-behaved” (in some sense) with some of the manipulations?
X,2×+5=8
Hi,
I prefer the second method.
The thumbnail is wrong, since the differential equation for the second way should be y' = y + x * e^x + x^2 * e^x.
Nice fart!
Your hoodie sleeves gets more dirty.
Hmmm. We had this exact d.e. for homework in my junior year Differential Equations course. I solved it and emphatically stated that that is a useless problem...
The second
Truthfully: neither, both and each. All have a relevance that is difficult for me to establish in sense if all are null it is wrong, if one is relevant all are relevant.
I put it down to reading Surreal Numbers recently. EDIT: there is beauty in results, methods, questions and surely that beauty is not scientific?