Original post by Student 999
How is it justified that -1/x^2 -3log(x) diverges to -inf?
A simple/intuitive way would be to differentiate to get
2/x^3 - n/x
as x->0+, the deriviative->+inf as the cubic term beats the linear term, so the original curve must be heading south as x->0+.
A bit more maths y=1/x^2 so
-y - n log(y^(-1/2)) = -y + n/2 log(y)
as y->inf, log(y) grows slowly so -y wins.
Just throwing ideas here.
Maybe helpful to look at u^2-nln(u) as u tends to infinity, this thing tends to positive infinity.
(Why? Taking the limit of ln(u)/u^2 as u tends to infinity will tell you that ln(u) is "growing" MUCH slower than u^2)
(Why look at this? Well... I hope you can fill in the gap)
But really the intuition is "the power term" "wins" against "the log term", so let's just "ignore the log".
But you also might notice my rampant usage of quotation marks, so... yeah.
P.S. There is a "list" that goes something like x^x >> x! >> b^x >> x^a >> log(x), with b>1 and a>0. Have you come across this somewhere somehow?
Maybe helpful to look at u^2-nln(u) as u tends to infinity, this thing tends to positive infinity.
(Why? Taking the limit of ln(u)/u^2 as u tends to infinity will tell you that ln(u) is "growing" MUCH slower than u^2)
(Why look at this? Well... I hope you can fill in the gap)
But really the intuition is "the power term" "wins" against "the log term", so let's just "ignore the log".
But you also might notice my rampant usage of quotation marks, so... yeah.
P.S. There is a "list" that goes something like x^x >> x! >> b^x >> x^a >> log(x), with b>1 and a>0. Have you come across this somewhere somehow?
(edited 11 months ago)
Original post by mqb2766
A simple/intuitive way would be to differentiate to get
2/x^3 - n/x
as x->0+, the deriviative->+inf as the cubic term beats the linear term, so the original curve must be heading south as x->0+.
A bit more maths y=1/x^2 so
-y - n log(y^(-1/2)) = -y + n/2 log(y)
as y->inf, log(y) grows slowly so -y wins.
2/x^3 - n/x
as x->0+, the deriviative->+inf as the cubic term beats the linear term, so the original curve must be heading south as x->0+.
A bit more maths y=1/x^2 so
-y - n log(y^(-1/2)) = -y + n/2 log(y)
as y->inf, log(y) grows slowly so -y wins.
I did initially differentiate, and yes it does go south as x->0. When you differentiate you’ll get that in the interval (0,1] it’s increasing.But that doesn’t justify that it goes to -inf as x->0?
(edited 11 months ago)
Original post by Student 999
I did initially differentiate, and yes it does go south as x->0. When you differentiate you’ll get that in the interval (0,1] it’s increasing.But that doesn’t justify that it goes to -inf as x->0?
As the derivative is ~ 1/x^3 it will be the case. But more generally its not as you could have something like sqrt(x) which has derivative ~ 1/sqrt(x). It ->0 but the derivative ->inf.
The second way that transform 0 -> inf is really the way to do it.
(edited 11 months ago)
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