f(x)=(f(-x))^1, substitute x→-x and +/-a → -/+a yielding:pic.twitter.com/BxWd9YCDD2
PhD student of Theoretical Particle Physics @UCIrvine l @NSF Fellow l Physics & Math Animations l Patreon: https://www.patreon.com/inertialobserver …
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f(x)=(f(-x))^1, substitute x→-x and +/-a → -/+a yielding:pic.twitter.com/BxWd9YCDD2
While taking a shower I realized that the problem also can be solved by solving the functional eqn. for f(x): f(x)f(-x)=1 Take x=0 therefore (f(0))^2=1 so: f(0)=+/-1 but f(0)=/=-1 since 1+f(x)=/=1 (not equal to 1). So f(0)=1, implying f(x)=1 so 1/(1+f(x))=1(1+1)=1/2 yielding "a"
The same approach was used by Einstein in his derivation of the Lorentz transformations with phi(v)phi(-v)=1, yielding phi(v)=1.pic.twitter.com/bFwUoY6jS0
Reminds me to: UU^+=UU^(-1)=1, where U is a unitary matrix, so also: f(x)=exp(ix) and f(-x)=exp(-ix)=(exp(ix))^(-1) (is a solution), have to check ;-)
In general: f(x)=p^x is a solution. (Where p real or complex) and f=/=-1
In general f(x)=p^x is a soln and therefore also f(x)=e^x=exp(x):pic.twitter.com/Mo403ImnyH
if you TeX'd your tweets i probably would have RTd you 50 times by now
Forgotten all about LaTeX that'S more than 25 years ago ;-) Getting too lazy as manager and entrepreneur ;-)
if you can remember all that HEP, you can remember TeX;)
I remember Microsoft Word/Office. (Worked for more than 5 years for Microsoft), LaTex or TeX too confusing and user unfriendly but the nonwysiwig editor TeX has beautiful output ;-) *.tex; *.dvi and , *.ps postscript files compiled ;-)
you studied Theoretical Particle Physics under Gerard `t Hooft, and you're trying to tell me that TeX is too complicated?!?!
At that time I was a LaTex Master/Guru, but nowadays and through the years am getting and have gotten used to Word en PPT with supporting people and team members ;-) A academic friend of mine is still using LaTex.
Oooeepppsss!! .....it is LaTeX and it is also a little "wavy"...
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