CY2024 XAO Prediction competition

[debtfree]

@debtfree Oh wise and benevolent one. My musings on FMG, WES, WBC and STX, though somewhat a tad late, have been completed for the month.

Oh @farmerge old mate, I see you have put on your old glasses from earlier in the year and posted in the wrong thread. Get rid of them, the new glasses have worked a treat all year.

I'm updating things this morning, so I'll chase the posts down. Thanks mate.

 

[Dona Ferentes]

Oh @farmerge old mate, I see you have put on your old glasses from earlier in the year and posted in the wrong thread. Get rid of them, the new glasses have worked a treat all year.

I'm updating things this morning, so I'll chase the posts down. Thanks mate.

from another post, (leaving Aust again) also a bit random

farmerge​

...Its been a bit of a struggle to even see the keyboard tonight as I got a piece of grinding stone lodged in the right eyeball this afternoon.
As luck would have the Dr I normally see, next available appointment is on Friday.
Hopefully there may be a cancellation in the next day or two
.
hope you get it fixed. all the best..

 

[farmerge]

Oh @farmerge old mate, I see you have put on your old glasses from earlier in the year and posted in the wrong thread. Get rid of them, the new glasses have worked a treat all year.

I'm updating things this morning, so I'll chase the posts down. Thanks mate.

@debtfree 'tis the piece of grinding stone lodged on the eyeball. Lucky to have got this far with only one eye working!!!!!

 

[debtfree]

@debtfree 'tis the piece of grinding stone lodged on the eyeball. Lucky to have got this far with only one eye working!!!!!

That's for sure @farmerge, I totally understand. I hope things improve very quickly for you and by the way, you're forgiven. All the best mate.

 

[farmerge]

from another post, (leaving Aust again) also a bit random

farmerge​

...Its been a bit of a struggle to even see the keyboard tonight as I got a piece of grinding stone lodged in the right eyeball this afternoon.
As luck would have the Dr I normally see, next available appointment is on Friday.
Hopefully there may be a cancellation in the next day or two
.
hope you get it fixed. all the best..

@Dona Ferentes Vision is somewhat improved from peering through the Left eye and a watery Right eye to almost normal.

 

[Dona Ferentes]

.... We can finally see @wayneL up the very top, waiting, waiting, waiting.

only to be mugged by @Garpal Gumnut , if linearity becomes asymptotic (?)

 

[debtfree]

End of the Week Update and only a bit over 3 weeks to go. The XAO is up 10.98% so far this year.

The XAO is back into @Ferret territory after staying in @houtman territory for most of the week.

Which way will it go, is the Xmas rally done or is it having a day or two rest before shooting up to visit @wayneL and @Garpal Gumnut, time will tell as always.

 

[debtfree]

2024 CY XAO Prediction Comp.
End of the 2nd Week in December Update. With only a bit over 2 weeks to go the XAO is up 9.21% so far this year.

At the end of this week we the XAO sitting in @rcw1 territory. Is it going to go higher or lower, is @brerwallabi going to win another Comp making it two in a row.
I could see @houtman licking his lips at one stage and @wayneL and @Garpal Gumnut looking over the cliff and praying, but all hopes there are lost me thinks.

Good luck all for the rest of the month

 

[Dona Ferentes]

2024 CY XAO Prediction Comp.
End of the 2nd Week in December Update. With only a bit over 2 weeks to go the XAO is up 9.21% so far this year.

At the end of this week we the XAO sitting in @rcw1 territory. Is it going to go higher or lower, is @brerwallabi going to win another Comp making it two in a row.
I could see @houtman licking his lips at one stage and @wayneL and @Garpal Gumnut looking over the cliff and praying, but all hopes there are lost me thinks.

Good luck all for the rest of the month

alas, for the top rungsters, November giveth and December take the away

 

[Garpal Gumnut]

Thanks @debtfree . It is a fascinating comp and if @brerwallabi does it twice in a row it will be an achievement. @rcw1 looks safe enough but this is stock picking writ large.

I'm still predicting the Donald to cark it between now and when he's managed to cause mayhem by changing convention and winning a 3rd term. Anything can happen in human affairs.

Except me winning this comp this year. I give in !! There I said it.

Thanks for your organisation of the comps @debtfree

gg

 

[Garpal Gumnut]

only to be mugged by @Garpal Gumnut , if linearity becomes asymptotic (?)

View attachment 188829

I missed that gem of your's @Dona Ferentes

For the ignorant on ASF, the few, those precious few, I will attempt to explain the expression "if linearity becomes asymptotic"
Asymptotic linearity of M ensures that M∗ is Fréchet differentiable at 0 and so well-known methods involving linearisation at 0, in particular bifurcation theory for parameter dependent situations, can be applied to the equation M∗(v)=0 yielding results about large solutions of M(u)=0. Therefore, since it was first used in this context by Toland [1] and Rabinowitz [2], inversion has become the standard approach used to deal with large solutions of asymptotically linear problems. When treating partial differential equations, the nonlinear terms usually arise as Nemytskii operators, the simplest form being F(u)(x)=f(u(x)) where u:Ω⊂RN→R is a solution and f∈C(R,R). The function f:R→R is asymptotically linear if f(s)/s→l as |s|→∞ for some l∈R. However this does not always ensure that F:X→Y is asymptotically linear in the strict sense as we show in Theorem 5.2 for the case Ω=RN,X=Wk,p(RN) and Y=Lp(RN) with k∈N and 1≤p<∞, which is often required in applications of nonlinear analysis. In fact, in this setting we show that F is asymptotically linear (and F∗ is Fréchet differentiable at 0) only when f is linear. Nonetheless, when f is nonlinear, F clearly has some weaker form of asymptotic linearity and F∗ should inherit some weaker form of differentiability at 0.

gg

Any further discussion on this leave me out, although I am not unfamiliar with Nemytskii operators from when I failed the test to enter ASIO.

gg

 

[Dona Ferentes]

I missed that gem of your's @Dona Ferentes

I remain
Yours.

 

[Dona Ferentes]

I missed that gem of your's @Dona Ferentes

dang, predictive text... I meant asymptomatic

 

[IvoryWolf]

I missed that gem of your's @Dona Ferentes

For the ignorant on ASF, the few, those precious few, I will attempt to explain the expression "if linearity becomes asymptotic"
Asymptotic linearity of M ensures that M∗ is Fréchet differentiable at 0 and so well-known methods involving linearisation at 0, in particular bifurcation theory for parameter dependent situations, can be applied to the equation M∗(v)=0 yielding results about large solutions of M(u)=0. Therefore, since it was first used in this context by Toland [1] and Rabinowitz [2], inversion has become the standard approach used to deal with large solutions of asymptotically linear problems. When treating partial differential equations, the nonlinear terms usually arise as Nemytskii operators, the simplest form being F(u)(x)=f(u(x)) where u:Ω⊂RN→R is a solution and f∈C(R,R). The function f:R→R is asymptotically linear if f(s)/s→l as |s|→∞ for some l∈R. However this does not always ensure that F:X→Y is asymptotically linear in the strict sense as we show in Theorem 5.2 for the case Ω=RN,X=Wk,p(RN) and Y=Lp(RN) with k∈N and 1≤p<∞, which is often required in applications of nonlinear analysis. In fact, in this setting we show that F is asymptotically linear (and F∗ is Fréchet differentiable at 0) only when f is linear. Nonetheless, when f is nonlinear, F clearly has some weaker form of asymptotic linearity and F∗ should inherit some weaker form of differentiability at 0.

gg

 

[Knobby22]

I missed that gem of your's @Dona Ferentes

For the ignorant on ASF, the few, those precious few, I will attempt to explain the expression "if linearity becomes asymptotic"
Asymptotic linearity of M ensures that M∗ is Fréchet differentiable at 0 and so well-known methods involving linearisation at 0, in particular bifurcation theory for parameter dependent situations, can be applied to the equation M∗(v)=0 yielding results about large solutions of M(u)=0. Therefore, since it was first used in this context by Toland [1] and Rabinowitz [2], inversion has become the standard approach used to deal with large solutions of asymptotically linear problems. When treating partial differential equations, the nonlinear terms usually arise as Nemytskii operators, the simplest form being F(u)(x)=f(u(x)) where u:Ω⊂RN→R is a solution and f∈C(R,R). The function f:R→R is asymptotically linear if f(s)/s→l as |s|→∞ for some l∈R. However this does not always ensure that F:X→Y is asymptotically linear in the strict sense as we show in Theorem 5.2 for the case Ω=RN,X=Wk,p(RN) and Y=Lp(RN) with k∈N and 1≤p<∞, which is often required in applications of nonlinear analysis. In fact, in this setting we show that F is asymptotically linear (and F∗ is Fréchet differentiable at 0) only when f is linear. Nonetheless, when f is nonlinear, F clearly has some weaker form of asymptotic linearity and F∗ should inherit some weaker form of differentiability at 0.

gg

Any further discussion on this leave me out, although I am not unfamiliar with Nemytskii operators from when I failed the test to enter ASIO.

gg

Wow! I studied this and don't remember any if that stuff.

 

[farmerge]

I missed that gem of your's @Dona Ferentes

For the ignorant on ASF, the few, those precious few, I will attempt to explain the expression "if linearity becomes asymptotic"
Asymptotic linearity of M ensures that M∗ is Fréchet differentiable at 0 and so well-known methods involving linearisation at 0, in particular bifurcation theory for parameter dependent situations, can be applied to the equation M∗(v)=0 yielding results about large solutions of M(u)=0. Therefore, since it was first used in this context by Toland [1] and Rabinowitz [2], inversion has become the standard approach used to deal with large solutions of asymptotically linear problems. When treating partial differential equations, the nonlinear terms usually arise as Nemytskii operators, the simplest form being F(u)(x)=f(u(x)) where u:Ω⊂RN→R is a solution and f∈C(R,R). The function f:R→R is asymptotically linear if f(s)/s→l as |s|→∞ for some l∈R. However this does not always ensure that F:X→Y is asymptotically linear in the strict sense as we show in Theorem 5.2 for the case Ω=RN,X=Wk,p(RN) and Y=Lp(RN) with k∈N and 1≤p<∞, which is often required in applications of nonlinear analysis. In fact, in this setting we show that F is asymptotically linear (and F∗ is Fréchet differentiable at 0) only when f is linear. Nonetheless, when f is nonlinear, F clearly has some weaker form of asymptotic linearity and F∗ should inherit some weaker form of differentiability at 0.

gg

Any further discussion on this leave me out, although I am not unfamiliar with Nemytskii operators from when I failed the test to enter ASIO.

gg

@Garpal Gumnut So much wiser for this pile of gobby gook.

 

[Dona Ferentes]

@Garpal Gumnut So much wiser for this pile of gobby gook.

summarising, the key takeout is:

F(u)(x)=f(u(x))​

 

[farmerge]

summarising, the key takeout is:

F(u)(x)=f(u(x))​

@Dona Ferentes Well I never.........

 

[Dona Ferentes]

@Logique2 suddenly back in contention

as low as 8368 early in the day

 

[wayneL]

@Logique2 suddenly back in contention

as low as 8368 early in the day

View attachment 189794

Santa has let me down, the fat c$#@!