[QUOTE=Xyzzy]Be aware that a large number of avatars in the forum are randomly assigned, and are without meaning.[/QUOTE]

Or just assigned because a Mod thought I would like it. :) It was a certain mods probable final Mac thing he touched. :D

Do peoples avatars affect you too?
 

[QUOTE=Paulie]Or just assigned because a Mod thought I would like it. :) It was a certain mods probable final Mac thing he touched. :D[/QUOTE]
[QUOTE=Paulie]Or just assigned because a Mod thought I would like it. :) It was a certain mods probable final Mac thing he touched. :D[/QUOTE]
:smile:
Some time ago Mike thoughtfully decided that I deserve an avatar and overnight the picture of a young man with a moustache appeared on every post of mine.Because the portrayal was not accurate, as I am a much older man, though with a bushy 'tashe often mistaken for Omar Sharif 'then and now' I objected to him and he promptly deleted it.

In my message I asked him to put instead the formula e^ (i(pi)) - 1 =0 as my avatar. As you know, this represents the five fundamental constants of maths, and even nature. I thought it would truthfully portray me as a man who eats, drinks and sleeps mathematics. With my arsenal of books to back me I dont need to Google much on formulae as I have within my reach maths to the highest level if required. Hence I can be quite a force to reckon with!

In all fairness to him he asked me to send him my formula which he would put down.

Due to my lack of IT skills I could not forward this to him and so it lies.

I hope Mike takes note and does the needful on his own style.

As far as other people's avatars concerned, I get particularly interested in the numbers some use in them and I wonder how female posters portray themselves in their avatars and if they give a clue for recognition.
Mally :coffee:
P.S. BTW mfgoode is my true name with initials M.F. The 'e' at the end means excellent or as one poster put :evil:

[QUOTE=mfgoode]In my message I asked him to put instead the formula e^ (i(pi)) - 1 =0 as my avatar.[/QUOTE]
The correct formula due to Euler is [tex]e^{i\pi} + 1 = 0[/tex]

WRT custom avatars: I have absolutely no art skills at all. I just collect funny or cute avatars. There is a way for users to upload custom avatars I think.

Personally, I like the avatars because at a glance I can tell who is who, probably by the color or something. I never really focus on the picture too much.

I'd prefer the avatars be a bit smaller but a while ago everyone voted for larger ones.

The default layout of the forum software is to place the avatars between the messages vertically. We have to override the software to place them on the left, which I find to be a lot less intrusive.

I haven't assigned any avatars since the incident with mfgoode. I guess I just don't want to chance offending anyone.

Finally, I think BotXXX has one of the coolest avatars. I like the colors but most of all I like how it is original.

Do peoples avatars affect you too?
 

:surprised
Thank you alpertron it was a silly mistake on my part.
Cos (theta) +i sin (theta) = -1 when theta is pi
Mally :coffee:

Another fun 5-constant identity: i[sup]i[/sup]*sqrt(e[sup]pi[/sup])-1 = 0.

[QUOTE=ewmayer]Another fun 5-constant identity: i[sup]i[/sup]*sqrt(e[sup]pi[/sup])-1 = 0.[/QUOTE]

Except for the fact that neither i[sup]i[/sup] nor sqrt(e[sup]π[/sup]) are well-defined. i[sup]i[/sup] really means e[sup]i log(i)[/sup], which gets you in trouble defining the logarithm of a complex number, and with the square root there is a positive and negative root.

Do peoples avatars affect you too?
 

[QUOTE=Jushi]Except for the fact that neither i[sup]i[/sup] nor sqrt(e[sup]π[/sup]) are well-defined. i[sup]i[/sup] really means e[sup]i log(i)[/sup], which gets you in trouble defining the logarithm of a complex number, and with the square root there is a positive and negative root.[/QUOTE]
:smile: HINT for you Jushi.
Try out Euler's formula given by me (and also Alpertron) in the previous posts of this thread.

i^i is a real number and equal to .207 879 576 350 761 908 546 055 ...
If you really can prove it then try deriving this equality using the same formula.

/0\( Phi ) = 2cos (Log (-1/5i) where \o/ is the golden ratio and equal to
1.618033989... or [(sq. rt. (5)) + 1)]/2
Good luck to you
Mally . :coffee:

I think Avatars add another layer of anonymousness to a users pseudonym.

[QUOTE=xilman]He sent me the animated GIF a while later.[/QUOTE]

In case I didn't tell you at teh time, I purloined it off mensanator's website.
I suspect that it was all his own work, as he seems to do all his own graphics.

[QUOTE=ewmayer]I always browse the forum with image rendering disabled, so people's avatars matter not a whit to me.[/QUOTE]Which is a real pity, because Eeyore is one my heros.

Paul