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In Reply to: Your papa member of tse-ka? posted by Victor Khomenko on July 27, 2000 at 06:23:52:
Hi, Victor
'Tsa-gi' is Central AeroHydrodynamic Institute, dad is just
a normal :) soviet engineer.
>>>
You must have one BIG dacha...
<<<Yep, but it was NOT as big for my great-granfather family.
>>>
***As I understand, your fast considerations are true only
if that center point is equal or bigger than e.I didn't know that. I knew quickly that getting close to
1 was dangerous - 1^64 vs. 64^1 - hmmm... But I didn't see
the e as the break point, of course. However both numbers
being close to some center point gave me some assurance.
Enough to bet my dollar.
<<<It's all in derivatives to me.
f(x)=a^x, g(x)=x^a,
f'(x)=ln(a)*a^x, g'(x)=a*x^(a-1)
When x = a, we get f'(a)=ln(a)*a^a, g'(a)=a*a^(a-1)=a^a
f'(a) is less than g'(a) when ln(a) is less than 1, i.e
when a is less than eThis determines functions behaviour near the intersection
point, including which one increases faster.
>>>
***Otherwise
it seems to be wrong, sorry. :) I didn't calculate, but
I SWEAR that say, 2.69^2.71 is LESS than 2.71^2.69No questions, such quick evaluation is not always going to
work. But as I said - often the decisions in life call for
one being "secure" rather than right. Especially when the
information is not present.
<<<Yes. Another reason for it is that even these sloooooooow
decisions aren't error-free. At least IME. :) And when time
is restricted, I prefer ANY decision over the brain-lock.
>>>
Since this subject interests me, I would be very curious if
someone could provide another way of doing quick estimation
on a fly.
<<<That's not to me. I'm also familiar with a few guys like your
friend Leon. And I always admired their ability to provide not
just proper, but ELEGANT solutions - thing impossible to me.
>>>
***If honestly, I'd not be able to make a quick decision for
this question. Well, I tried to evaluate it "by fingers"
first, but after a few attempts I gave up and decided just
to do all this math.I suspect this is all in the mindset. You are obviously trained
in exact science and are usually able to arrive at precise answer
(and very effectively, thank you). That is great, but has a small
byproduct that smart people had discovered some time ago - absence
of such firm ground sometimes stops the process. But in many cases
the process can not tolerate being stopped - something MUST be done
and such cases call for a different approach.
We all can come up with many situations like that.I have to admit that the first time I got exposed to this concept
of decision making as science I was very fascinated.
Many professions - astronauts, nuclear plant operators, commandos,
even government officials are trained in this.
Most people never get to see it.This resonated in me for some reason.
<<<Gee I NEED such a training!
>>>
Which reminds me... I need to send him $300...
<<<Hey Victor, can you pass to Leon this related problem:
when a > 1 and a not equal e, a^x and x^a have ANOTHER
intersection besides x=a. How to find it analytically???? :)
I've spent last 3 or 4 days masturbating my brain over it,
but I still "can't get no satisfaction", darn.
regards, gnat
a^x=x^a
x*ln(a)=a*ln(x)
x/a=ln(x-a)
(e^x)/(e^a)=e^(ln(x-a))
(e^x)/(e^a)=x-a
(e^x)*(e^(-a))=x-a
e^(-a*x)=x-a
now take da/dx for both sides
(-a)*e^(-a*x)=1
e^(-a*x)=-a^-1
ln(e^(-a*x))=ln(-a^-1)
-a*x=-ln(-a)
a*x=ln(-a)
x=ln(-a)/a
since a> =0, ln(-a)=infinity
therefore x=infinity.
checking our work
a^x=x^a
a^infinity=infinity^a for all a> 1
yes, x=infinity is other solution.
make that take d/dx NOT da/dx
'Tsa-gi' is Central AeroHydrodynamic Institute, dad is just
a normal :) soviet engineer.I presume - retired now? Is TsAGI still strong? I believe they have lost a lot of their personnel.
< < <
It's all in derivatives to me.
f(x)=a^x, g(x)=x^a,
f'(x)=ln(a)*a^x, g'(x)=a*x^(a-1)
When x = a, we get f'(a)=ln(a)*a^a, g'(a)=a*a^(a-1)=a^a
f'(a) is less than g'(a) when ln(a) is less than 1, i.e
when a is less than e***This determines functions behaviour near the intersection
point, including which one increases faster.
Yes, absolutely true. The simple graphical analysis doesn't tell you where they intersect, just that they do.> > >
***Otherwise
it seems to be wrong, sorry. :) I didn't calculate, but
I SWEAR that say, 2.69^2.71 is LESS than 2.71^2.69No need, your formula tells it all.
***That's not to me. I'm also familiar with a few guys like your
friend Leon. And I always admired their ability to provide not
just proper, but ELEGANT solutions - thing impossible to me.
***I have to admit that the first time I got exposed to this concept
of decision making as science I was very fascinated.
Many professions - astronauts, nuclear plant operators, commandos,
even government officials are trained in this.
Most people never get to see it.***This resonated in me for some reason.
< < <***Gee I NEED such a training!
All you would need (and I am not for a second suggesting this - BTW!) would be few weeks with my wife. She would put relentless pressure on you - C'mon now, decide already! She firmly believes that all major decisions (and the resulting responsibility for the wrong ones) must fall on my shoulders. She will go with most of them but reserves the right to crucify me for bad ones. Stalin - Zhukov relationship, if you will. Freedom to make decisions for as long as you make good ones...
I'have managed to stay from the family Gulag for this long...
***Hey Victor, can you pass to Leon this related problem:***when a > 1 and a not equal e, a^x and x^a have ANOTHER
intersection besides x=a. How to find it analytically???? :)
I've spent last 3 or 4 days masturbating my brain over it,
but I still "can't get no satisfaction", darn.I gave it to him. I don't see much of him anymore, unfortunately, but just as random chance would dictate, he stopped by my office yesterday, right after you and I talked here.
BTW - explaining things like that to him is NOT trivial - he always starts asking questions that seem orthogonal to your problem.
I suspect we may get kicked out of the Film forum soon.
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