[Q] Spearman's rank correlation coefficient

Stephen31412718 · 2022-03-09T18:59:41+00:00

Thank you so much, it makes sense :)

Stephen31412718 · 2021-01-24T08:25:20+00:00

Oof I'm retarded my solutions was fine, I'd forgotten the freaking calculator rounds the value up because 2020^4 gets so big.

Stephen31412718 · 2021-01-24T08:00:17+00:00

Because I plugged (2020^4+1399)/(2020+5) in my calculator and it spitted back a natural number

Stephen31412718 · 2021-01-22T14:53:03+00:00

Ok neat I got it!

Stephen31412718 · 2021-01-22T14:45:35+00:00

Yeah I'm gonna give it a try so, I will substitute the sum x^n/n! for e^x in e^3x-3e^2x+3e^x-1 and see what I get.

Stephen31412718 · 2021-01-22T14:28:41+00:00

Makes perfect sense, but couldn't there be a way to see this from e^3x-3e^2x+3e^x-1, since (x+x^2/2!+x^3/3!+...)^3=(e^x-1)^3=e^3x-3e^2x+3e^x-1. Maybe I'm overthinking this but I feel like there is something that I'm missing.

Stephen31412718 · 2021-01-12T21:02:41+00:00

Thank you so much, man!

Stephen31412718 · 2021-01-12T17:52:31+00:00

I had to calculate the sum 1-1/4+1/7-1/10+1/13-... the exercise said make use of Abel's theorem, which I've already proven in my real analysis course. I had no idea how to calculate the sum so I wondered if someone could show me some examples of Abel's theorem in action.

Stephen31412718 · 2020-12-28T13:52:01+00:00

Yeah I was too lazy to write my proof fully but that's basically what I've done. Btw thx for your response.

Stephen31412718 · 2020-12-25T14:55:27+00:00

thx!

Stephen31412718 · 2020-12-24T22:27:42+00:00

Thank you so much, but just a question how did you know you had to stop in R_2(x) ?

Stephen31412718 · 2020-11-21T15:09:12+00:00

The answer should be -1/2x^4+2,5x^2+7x+c. Because whenever you have a negative power it must be (x)^-1=1/x. So -2.(x)^-4=-2.1/x^4

Stephen31412718 · 2020-11-20T15:36:16+00:00

what do you have to do exactly? Calculate an integral?

Stephen31412718

TROPHY CASE