[...]
Op 12/07/2025 om 12:28 schreef Richard Hachel:
[...]
Wolfram Alpha says 5^(3i) = cos(3*ln(5))+i*sin(3*ln(5)), so it seems you still haven't figured out what complex numbers are or how they work.
Not just Wolfram Alpha, any graphic calculator that is able to handle
complex numbers.
https://www.desmos.com/calculator/f3ovmqpg3x
There is not a single online source that will confirm or support your
claim that 5^(3i) = (-3,-123).
Le 12/07/2025 à 23:01, sobriquet a écrit :
Op 12/07/2025 om 12:28 schreef Richard Hachel:
[...]
Wolfram Alpha says 5^(3i) = cos(3*ln(5))+i*sin(3*ln(5)), so it seems you
still haven't figured out what complex numbers are or how they work.
Not just Wolfram Alpha, any graphic calculator that is able to handle
complex numbers.
https://www.desmos.com/calculator/f3ovmqpg3x
There is not a single online source that will confirm or support your
claim that 5^(3i) = (-3,-123).
That's not quite what I said, the value x=5^(3i) is not a point, but a value. x=-123
R.H.
Le 12/07/2025 à 23:03, sobriquet a écrit :
Op 12/07/2025 om 12:28 schreef Richard Hachel:
;[...]
Wolfram Alpha says 5^(3i) = cos(3*ln(5))+i*sin(3*ln(5)), so it seems
you still haven't figured out what complex numbers are or how they work.
Not just Wolfram Alpha, any graphic calculator that is able to handle
complex numbers.
https://www.desmos.com/calculator/f3ovmqpg3x
There is not a single online source that will confirm or support your
claim that 5^(3i) = (-3,-123).
That's not quite what I said, the value x=5^(3i) is not a point, but a
value.
x=-123
R.H.
Op 13/07/2025 om 00:18 schreef Richard Hachel:What mistake? Dropping the 2kπ we get without Wolfram
Le 12/07/2025 à 23:03, sobriquet a écrit :
Op 12/07/2025 om 12:28 schreef Richard Hachel:
;[...]
Wolfram Alpha says 5^(3i) = cos(3*ln(5))+i*sin(3*ln(5)), so it seems
you still haven't figured out what complex numbers are or how they work. >>>
Not just Wolfram Alpha, any graphic calculator that is able to handle
complex numbers.
https://www.desmos.com/calculator/f3ovmqpg3x
There is not a single online source that will confirm or support your
claim that 5^(3i) = (-3,-123).
That's not quite what I said, the value x=5^(3i) is not a point, but a
value.
x=-123
You're right, my mistake.
On 13.07.2025 00:35, sobriquet wrote:
Op 13/07/2025 om 00:18 schreef Richard Hachel:
Le 12/07/2025 à 23:03, sobriquet a écrit :
Op 12/07/2025 om 12:28 schreef Richard Hachel:
;[...]
Wolfram Alpha says 5^(3i) = cos(3*ln(5))+i*sin(3*ln(5)), so it seems
you still haven't figured out what complex numbers are or how they work. >>>>
Not just Wolfram Alpha, any graphic calculator that is able to handle
complex numbers.
https://www.desmos.com/calculator/f3ovmqpg3x
There is not a single online source that will confirm or support your
claim that 5^(3i) = (-3,-123).
That's not quite what I said, the value x=5^(3i) is not a point, but a
value.
x=-123
You're right, my mistake.What mistake? Dropping the 2kπ we get without Wolfram
5^(3i) = e^ln(5^(3i)) = (e^(3i*ln5) = cos(3*ln5) + i*sin(3*ln5)
Regards, WM
On 13.07.2025 00:35, sobriquet wrote:
Op 13/07/2025 om 00:18 schreef Richard Hachel:
Le 12/07/2025 à 23:03, sobriquet a écrit :
Op 12/07/2025 om 12:28 schreef Richard Hachel:
;[...]
Wolfram Alpha says 5^(3i) = cos(3*ln(5))+i*sin(3*ln(5)), so it seems
you still haven't figured out what complex numbers are or how they
work.
Not just Wolfram Alpha, any graphic calculator that is able to
handle complex numbers.
https://www.desmos.com/calculator/f3ovmqpg3x
There is not a single online source that will confirm or support
your claim that 5^(3i) = (-3,-123).
That's not quite what I said, the value x=5^(3i) is not a point, but
a value.
x=-123
You're right, my mistake.What mistake? Dropping the 2kπ we get without Wolfram
5^(3i) = e^ln(5^(3i)) = (e^(3i*ln5) = cos(3*ln5) + i*sin(3*ln5)
Regards, WM
Op 13/07/2025 om 14:08 schreef WM:
On 13.07.2025 00:35, sobriquet wrote:
Op 13/07/2025 om 00:18 schreef Richard Hachel:What mistake? Dropping the 2kπ we get without Wolfram
Le 12/07/2025 à 23:03, sobriquet a écrit :
Op 12/07/2025 om 12:28 schreef Richard Hachel:
;[...]
Wolfram Alpha says 5^(3i) = cos(3*ln(5))+i*sin(3*ln(5)), so it seems >>>>> you still haven't figured out what complex numbers are or how they
work.
Not just Wolfram Alpha, any graphic calculator that is able to
handle complex numbers.
https://www.desmos.com/calculator/f3ovmqpg3x
There is not a single online source that will confirm or support
your claim that 5^(3i) = (-3,-123).
That's not quite what I said, the value x=5^(3i) is not a point, but
a value.
x=-123
You're right, my mistake.
5^(3i) = e^ln(5^(3i)) = (e^(3i*ln5) = cos(3*ln5) + i*sin(3*ln5)
Regards, WM
My mistake of saying Richard Hachel said that 5^(3i) = (-3,-123) while
he actually said that 5^(3i) = -123
I agree that 5^(3i) = cos(3*ln(5)) + i*sin(3*ln(5))
Le 13/07/2025 à 14:08, WM a écrit :
On 13.07.2025 00:35, sobriquet wrote:Log5=1.609437912
Op 13/07/2025 om 00:18 schreef Richard Hachel:What mistake? Dropping the 2kπ we get without Wolfram
Le 12/07/2025 à 23:03, sobriquet a écrit :
Op 12/07/2025 om 12:28 schreef Richard Hachel:
;[...]
Wolfram Alpha says 5^(3i) = cos(3*ln(5))+i*sin(3*ln(5)), so it
seems you still haven't figured out what complex numbers are or how
they work.
Not just Wolfram Alpha, any graphic calculator that is able to
handle complex numbers.
https://www.desmos.com/calculator/f3ovmqpg3x
There is not a single online source that will confirm or support
your claim that 5^(3i) = (-3,-123).
That's not quite what I said, the value x=5^(3i) is not a point, but
a value.
x=-123
You're right, my mistake.
5^(3i) = e^ln(5^(3i)) = (e^(3i*ln5) = cos(3*ln5) + i*sin(3*ln5)
Log5*3=4.828313737
Wath is cos(4.828313737) ?
Le 13/07/2025 à 14:21, sobriquet a écrit :
Op 13/07/2025 om 14:08 schreef WM:
On 13.07.2025 00:35, sobriquet wrote:
Op 13/07/2025 om 00:18 schreef Richard Hachel:What mistake? Dropping the 2kπ we get without Wolfram
Le 12/07/2025 à 23:03, sobriquet a écrit :
Op 12/07/2025 om 12:28 schreef Richard Hachel:
;[...]
Wolfram Alpha says 5^(3i) = cos(3*ln(5))+i*sin(3*ln(5)), so it
seems you still haven't figured out what complex numbers are or
how they work.
Not just Wolfram Alpha, any graphic calculator that is able to
handle complex numbers.
https://www.desmos.com/calculator/f3ovmqpg3x
There is not a single online source that will confirm or support
your claim that 5^(3i) = (-3,-123).
That's not quite what I said, the value x=5^(3i) is not a point,
but a value.
x=-123
You're right, my mistake.
5^(3i) = e^ln(5^(3i)) = (e^(3i*ln5) = cos(3*ln5) + i*sin(3*ln5)
Regards, WM
My mistake of saying Richard Hachel said that 5^(3i) = (-3,-123) while
he actually said that 5^(3i) = -123
I agree that 5^(3i) = cos(3*ln(5)) + i*sin(3*ln(5))
I agree that, in algebric analysis, 5^(3i) = -123
In complex geometria, I can't understand what term y=5^(3i) is. A round square?
R.H.
3 1/3 usually means 3 + 1/3
Math is full of idiotic nonsense like that, so it's really no wonder
people can't make much sense of math.
Op 13/07/2025 om 15:11 schreef Richard Hachel:
Le 13/07/2025 à 14:21, sobriquet a écrit :
Op 13/07/2025 om 14:08 schreef WM:
On 13.07.2025 00:35, sobriquet wrote:
Op 13/07/2025 om 00:18 schreef Richard Hachel:What mistake? Dropping the 2kπ we get without Wolfram
Le 12/07/2025 à 23:03, sobriquet a écrit :
Op 12/07/2025 om 12:28 schreef Richard Hachel:
;[...]
Wolfram Alpha says 5^(3i) = cos(3*ln(5))+i*sin(3*ln(5)), so it
seems you still haven't figured out what complex numbers are or
how they work.
Not just Wolfram Alpha, any graphic calculator that is able to
handle complex numbers.
https://www.desmos.com/calculator/f3ovmqpg3x
There is not a single online source that will confirm or support >>>>>>> your claim that 5^(3i) = (-3,-123).
That's not quite what I said, the value x=5^(3i) is not a point,
but a value.
x=-123
You're right, my mistake.
5^(3i) = e^ln(5^(3i)) = (e^(3i*ln5) = cos(3*ln5) + i*sin(3*ln5)
Regards, WM
My mistake of saying Richard Hachel said that 5^(3i) = (-3,-123) while
he actually said that 5^(3i) = -123
I agree that 5^(3i) = cos(3*ln(5)) + i*sin(3*ln(5))
I agree that, in algebric analysis, 5^(3i) = -123
In complex geometria, I can't understand what term y=5^(3i) is. A round
square?
R.H.
I think in a way you're right. Math is a confusing hodgepodge of ideas.
It's basically full of shit. But there is hope that AI will debug all
the math and make it logical and comprehensible.
There are countless examples of math that historically makes sense, but
tends to confuse people. Like mixed numbers.
3 1/3 usually means 3 + 1/3
Le 13/07/2025 à 19:23, sobriquet a écrit :
Op 13/07/2025 om 15:11 schreef Richard Hachel:
Le 13/07/2025 à 14:21, sobriquet a écrit :
Op 13/07/2025 om 14:08 schreef WM:
On 13.07.2025 00:35, sobriquet wrote:
Op 13/07/2025 om 00:18 schreef Richard Hachel:What mistake? Dropping the 2kπ we get without Wolfram
Le 12/07/2025 à 23:03, sobriquet a écrit :
Op 12/07/2025 om 12:28 schreef Richard Hachel:
;[...]
Wolfram Alpha says 5^(3i) = cos(3*ln(5))+i*sin(3*ln(5)), so it >>>>>>>> seems you still haven't figured out what complex numbers are or >>>>>>>> how they work.
Not just Wolfram Alpha, any graphic calculator that is able to >>>>>>>> handle complex numbers.
https://www.desmos.com/calculator/f3ovmqpg3x
There is not a single online source that will confirm or support >>>>>>>> your claim that 5^(3i) = (-3,-123).
That's not quite what I said, the value x=5^(3i) is not a point, >>>>>>> but a value.
x=-123
You're right, my mistake.
5^(3i) = e^ln(5^(3i)) = (e^(3i*ln5) = cos(3*ln5) + i*sin(3*ln5)
Regards, WM
My mistake of saying Richard Hachel said that 5^(3i) = (-3,-123)
while he actually said that 5^(3i) = -123
I agree that 5^(3i) = cos(3*ln(5)) + i*sin(3*ln(5))
I agree that, in algebric analysis, 5^(3i) = -123
In complex geometria, I can't understand what term y=5^(3i) is. A
round square?
R.H.
I think in a way you're right. Math is a confusing hodgepodge of
ideas. It's basically full of shit. But there is hope that AI will
debug all the math and make it logical and comprehensible.
There are countless examples of math that historically makes sense,
but tends to confuse people. Like mixed numbers.
Ok, so you are a kind of "Hachel"-level crank. Didn't know that.
3 1/3 usually means 3 + 1/3
Definitely not. Where have you seen such habit?
Le 13/07/2025 à 19:23, sobriquet a écrit :
3 1/3 usually means 3 + 1/3
No. Never. 3 1/3 is a British/Yankee notation for groceries, nothing
close to mathematics.
Math is full of idiotic nonsense like that, so it's really no wonder
people can't make much sense of math.
Not at all. Mathematical notations are universal, precise and univoque.
Op 13/07/2025 om 19:49 schreef efji:
Le 13/07/2025 à 19:23, sobriquet a écrit :
3 1/3 usually means 3 + 1/3
No. Never. 3 1/3 is a British/Yankee notation for groceries, nothing
close to mathematics.
Math is full of idiotic nonsense like that, so it's really no wonder
people can't make much sense of math.
Not at all. Mathematical notations are universal, precise and univoque.
Wolfram Alpha says arctan(1,2) = 1.10714871779409050301...
Desmos says arctan(1,2) = 0.463647609001...
https://www.wolframalpha.com/input?i=arctan%281%2C2%29
https://www.desmos.com/calculator/dyjkhncmsk
Now of course I know that they simply are using different conventions, arctan(x,y) vs arctan(y,x), but it doesn't take many of such
inconsistencies to cause confusion or errors. It just goes to show
that there are examples of different conventions in math.
Le 13/07/2025 à 19:23, sobriquet a écrit :
3 1/3 usually means 3 + 1/3
No. Never. 3 1/3 is a British/Yankee notation for groceries, nothing
close to mathematics.
Le 13/07/2025 à 19:49, efji a écrit :
Le 13/07/2025 à 19:23, sobriquet a écrit :
3 1/3 usually means 3 + 1/3
No. Never. 3 1/3 is a British/Yankee notation for groceries, nothing
close to mathematics.
Wolfram is accepting groceries notation? This is bad! IMHA it is a not a feature, it is a bug.
Anyway Sobriquet is insincere here, acting as crank.
Op 13/07/2025 om 20:38 schreef Python:
Le 13/07/2025 à 19:49, efji a écrit :
Le 13/07/2025 à 19:23, sobriquet a écrit :
3 1/3 usually means 3 + 1/3
No. Never. 3 1/3 is a British/Yankee notation for groceries, nothing
close to mathematics.
Wolfram is accepting groceries notation? This is bad! IMHA it is a not a
feature, it is a bug.
Anyway Sobriquet is insincere here, acting as crank.
No. I love math and I would like it to be consistent. But I often see
things that just don't make any sense at a conceptual level.
Take for instance:
sum n=0 to infinity x^(2n)/(2n)! = cosh(x)
https://www.wolframalpha.com/input?i=sum+n%3D0+to+infinity+x%5E%282n%29%2F%282n%29%21
You would think oh, that implies (if we take x=0):
sum n=0 to infinity 0^(2n)/(2n)! = cosh(0)
But wolfram says no:
https://www.wolframalpha.com/input?i=sum+n%3D0+to+infinity+0%5E%282n%29%2F%282n%29%21
If you say that 0^0 is 1, you can't say it is 1 sometimes and at other
times it's not 1.. that is simply not a consistent way of dealing with
the meaning of 0^0.
Le 13/07/2025 à 20:06, sobriquet a écrit :
Op 13/07/2025 om 19:49 schreef efji:
Le 13/07/2025 à 19:23, sobriquet a écrit :
3 1/3 usually means 3 + 1/3
No. Never. 3 1/3 is a British/Yankee notation for groceries, nothing
close to mathematics.
Math is full of idiotic nonsense like that, so it's really no wonder
people can't make much sense of math.
Not at all. Mathematical notations are universal, precise and univoque.
Wolfram Alpha says arctan(1,2) = 1.10714871779409050301...
Desmos says arctan(1,2) = 0.463647609001...
https://www.wolframalpha.com/input?i=arctan%281%2C2%29
https://www.desmos.com/calculator/dyjkhncmsk
Now of course I know that they simply are using different conventions,
arctan(x,y) vs arctan(y,x), but it doesn't take many of such
inconsistencies to cause confusion or errors. It just goes to show
that there are examples of different conventions in math.
Nothing above is "maths". You are talking about commercial products and softwares. Mathematical notations in books and article are not
ambiguous. And in maths, there is nothing close to arctan(x,y) used
without a proper definition given before.
Le 13/07/2025 à 20:47, sobriquet a écrit :
Op 13/07/2025 om 20:38 schreef Python:
Le 13/07/2025 à 19:49, efji a écrit :
Le 13/07/2025 à 19:23, sobriquet a écrit :
3 1/3 usually means 3 + 1/3
No. Never. 3 1/3 is a British/Yankee notation for groceries, nothing
close to mathematics.
Wolfram is accepting groceries notation? This is bad! IMHA it is a
not a feature, it is a bug.
Anyway Sobriquet is insincere here, acting as crank.
No. I love math and I would like it to be consistent. But I often see
things that just don't make any sense at a conceptual level.
Take for instance:
sum n=0 to infinity x^(2n)/(2n)! = cosh(x)
https://www.wolframalpha.com/input?
i=sum+n%3D0+to+infinity+x%5E%282n%29%2F%282n%29%21
You would think oh, that implies (if we take x=0):
sum n=0 to infinity 0^(2n)/(2n)! = cosh(0)
But wolfram says no:
https://www.wolframalpha.com/input?
i=sum+n%3D0+to+infinity+0%5E%282n%29%2F%282n%29%21
If you say that 0^0 is 1, you can't say it is 1 sometimes and at other
times it's not 1.. that is simply not a consistent way of dealing with
the meaning of 0^0.
Oh, come on... Not this old 0^0 stuff.
Grow up.
Op 13/07/2025 om 20:49 schreef Python:
Le 13/07/2025 à 20:47, sobriquet a écrit :
Op 13/07/2025 om 20:38 schreef Python:
Le 13/07/2025 à 19:49, efji a écrit :
Le 13/07/2025 à 19:23, sobriquet a écrit :
3 1/3 usually means 3 + 1/3
No. Never. 3 1/3 is a British/Yankee notation for groceries, nothing >>>>> close to mathematics.
Wolfram is accepting groceries notation? This is bad! IMHA it is a
not a feature, it is a bug.
Anyway Sobriquet is insincere here, acting as crank.
No. I love math and I would like it to be consistent. But I often see
things that just don't make any sense at a conceptual level.
Take for instance:
sum n=0 to infinity x^(2n)/(2n)! = cosh(x)
https://www.wolframalpha.com/input?
i=sum+n%3D0+to+infinity+x%5E%282n%29%2F%282n%29%21
You would think oh, that implies (if we take x=0):
sum n=0 to infinity 0^(2n)/(2n)! = cosh(0)
But wolfram says no:
https://www.wolframalpha.com/input?
i=sum+n%3D0+to+infinity+0%5E%282n%29%2F%282n%29%21
If you say that 0^0 is 1, you can't say it is 1 sometimes and at other
times it's not 1.. that is simply not a consistent way of dealing with
the meaning of 0^0.
Oh, come on... Not this old 0^0 stuff.
Grow up.
So you claim math is perfectly consistent and if people point out clear inconsistencies in math, they are childish?
So you claim math is perfectly consistent and if people point out clear inconsistencies in math, they are childish?
Op 13/07/2025 om 20:49 schreef Python:
Le 13/07/2025 à 20:47, sobriquet a écrit :
Op 13/07/2025 om 20:38 schreef Python:
Le 13/07/2025 à 19:49, efji a écrit :
Le 13/07/2025 à 19:23, sobriquet a écrit :
3 1/3 usually means 3 + 1/3
No. Never. 3 1/3 is a British/Yankee notation for groceries, nothing >>>>> close to mathematics.
Wolfram is accepting groceries notation? This is bad! IMHA it is a
not a feature, it is a bug.
Anyway Sobriquet is insincere here, acting as crank.
No. I love math and I would like it to be consistent. But I often see
things that just don't make any sense at a conceptual level.
Take for instance:
sum n=0 to infinity x^(2n)/(2n)! = cosh(x)
https://www.wolframalpha.com/input?
i=sum+n%3D0+to+infinity+x%5E%282n%29%2F%282n%29%21
You would think oh, that implies (if we take x=0):
sum n=0 to infinity 0^(2n)/(2n)! = cosh(0)
But wolfram says no:
https://www.wolframalpha.com/input?
i=sum+n%3D0+to+infinity+0%5E%282n%29%2F%282n%29%21
If you say that 0^0 is 1, you can't say it is 1 sometimes and at other
times it's not 1.. that is simply not a consistent way of dealing with
the meaning of 0^0.
Oh, come on... Not this old 0^0 stuff.
Grow up.
So you claim math is perfectly consistent and if people point out clear inconsistencies in math, they are childish?
Le 13/07/2025 à 20:58, sobriquet a écrit :
Op 13/07/2025 om 20:49 schreef Python:
Le 13/07/2025 à 20:47, sobriquet a écrit :
Op 13/07/2025 om 20:38 schreef Python:
Le 13/07/2025 à 19:49, efji a écrit :
Le 13/07/2025 à 19:23, sobriquet a écrit :
3 1/3 usually means 3 + 1/3
No. Never. 3 1/3 is a British/Yankee notation for groceries,
nothing close to mathematics.
Wolfram is accepting groceries notation? This is bad! IMHA it is a
not a feature, it is a bug.
Anyway Sobriquet is insincere here, acting as crank.
No. I love math and I would like it to be consistent. But I often see
things that just don't make any sense at a conceptual level.
Take for instance:
sum n=0 to infinity x^(2n)/(2n)! = cosh(x)
https://www.wolframalpha.com/input?
i=sum+n%3D0+to+infinity+x%5E%282n%29%2F%282n%29%21
You would think oh, that implies (if we take x=0):
sum n=0 to infinity 0^(2n)/(2n)! = cosh(0)
But wolfram says no:
https://www.wolframalpha.com/input?
i=sum+n%3D0+to+infinity+0%5E%282n%29%2F%282n%29%21
If you say that 0^0 is 1, you can't say it is 1 sometimes and at
other times it's not 1.. that is simply not a consistent way of
dealing with the meaning of 0^0.
Oh, come on... Not this old 0^0 stuff.
Grow up.
So you claim math is perfectly consistent and if people point out
clear inconsistencies in math, they are childish?
NOTHING you pointed you never leaded to a mistake. People are clever
enough to consider the context, are you?
Op 13/07/2025 om 22:08 schreef efji:
Le 13/07/2025 à 20:58, sobriquet a écrit :
So you claim math is perfectly consistent and if people point out
clear inconsistencies in math, they are childish?
Yes. At least in maths that you are able to understand.
Childish or trolls.
But you are welcome to point what you call "inconsistencies" :)
Ok, I clearly overstated my case by claiming that math is full of shit.
But what I'm trying to get at is the feeling that math probably isn't
always defined in the way that provides optimal conceptual clarity.
Historically we can point to examples like Roman numerals, which were
a kind of primitive number system that was clearly suboptimal in many respects compared to the later Hindu-Arabic number system.
So in historical developments it is often the case that math evolves
and ideas in math tend to shift where certain things that were generally assumed to be meaningless or nonsensical tend to get accepted later on,
like negative numbers or complex numbers. Or things that seemed obvious
on a conceptual level, like the shortest distance between two points
being a straight line, that later became doubtful (in curved geometry).
So ultimately we strive towards approximating a kind of ideal conceptual
math framework where we'd like a minimal conceptual basis (no redundant concepts) that provides the most versatile and powerful conceptual toolbox. In that respect I'm interested in the psychology of crackpots that might claim that 1*1=2 or that 5^(3i)=-123.
Is there a mathematical fact about such concepts, or is it just the
majority opinion (whatever the math community claims must be right) or
is it that we can potentially come up with a conceptual framework that underpins concepts like numbers or functions such that we can easily
provide a rigorous argument of why one claim like 1*1=1 makes sense and another claim like 1*1=2 is nonsense.
Or is it like relativity where one person can claim event A preceded
event B and another person can claim event B preceded event A and they
could both be right despite holding contradictory views, because they
witness the events from different perspectives?
With certain nonsense claims, like the sun revolving around the earth,
it's easy to debunk it with evidence because we can empirically observe
the solar system from a point in space where it's obvious that planets
like the earth revolve around the sun. But with other nonsense claims
about abstractions like numbers, sets or gods, it's not always so easy
to categorically refute it. Like the claim that there exists a set that contains all sets or the claim that god can do anything.
Can god create a stone that is so heavy that not even god can lift it?
Le 13/07/2025 à 20:58, sobriquet a écrit :
So you claim math is perfectly consistent and if people point out
clear inconsistencies in math, they are childish?
Yes. At least in maths that you are able to understand.
Childish or trolls.
But you are welcome to point what you call "inconsistencies" :)
Op 13/07/2025 om 22:08 schreef efji:
In that respect I'm interested in the psychology of crackpots that might claim that 1*1=2 or that 5^(3i)=-123.
Le 14/07/2025 à 14:20, sobriquet a écrit :
In that respect I'm interested in the psychology of crackpots that might
claim that 1*1=2 or that 5^(3i)=-123.
Don't spend to much time on them :)
Actually it's a mystery to me...
Mathematical notations in books and article are not
ambiguous.
Le 14/07/2025 à 14:20, sobriquet a écrit :
Op 13/07/2025 om 22:08 schreef efji:
In that respect I'm interested in the psychology of crackpots that might
claim that 1*1=2 or that 5^(3i)=-123.
In imaginary algebria (not in complex trigonomztria) :
5^(3i)=-123
y=5^(3i)
f(y)=5^(x) with y₀=f(0)=1
g(x)=-f(-x)+2y₀=-5^(-3i)+2
y=-125+2=-123
R.H.
Le 14/07/2025 à 14:20, sobriquet a écrit :
Op 13/07/2025 om 22:08 schreef efji:
Le 13/07/2025 à 20:58, sobriquet a écrit :
So you claim math is perfectly consistent and if people point out
clear inconsistencies in math, they are childish?
Yes. At least in maths that you are able to understand.
Childish or trolls.
But you are welcome to point what you call "inconsistencies" :)
Ok, I clearly overstated my case by claiming that math is full of shit.
But what I'm trying to get at is the feeling that math probably isn't
always defined in the way that provides optimal conceptual clarity.
Precise definition is one thing, independent of the "observer".
Clarity is another thing. What is perfectly clear for me could be
obscure for you :)
Historically we can point to examples like Roman numerals, which were
a kind of primitive number system that was clearly suboptimal in many
respects compared to the later Hindu-Arabic number system.
So in historical developments it is often the case that math evolves
and ideas in math tend to shift where certain things that were
generally assumed to be meaningless or nonsensical tend to get
accepted later on, like negative numbers or complex numbers. Or things
that seemed obvious on a conceptual level, like the shortest distance
between two points being a straight line, that later became doubtful
(in curved geometry).
Yes, history of maths is chaotic. Concepts and notations have been fixed
in a rigorous way only lately, in the late XIXth and in the XXth
century. Have you heard about Bourbaki ?
[..]
With certain nonsense claims, like the sun revolving around the earth,
it's easy to debunk it with evidence because we can empirically
observe the solar system from a point in space where it's obvious that
planets like the earth revolve around the sun. But with other nonsense
claims
It has been established way before the space conquest!
about abstractions like numbers, sets or gods, it's not always so easy
to categorically refute it. Like the claim that there exists a set
that contains all sets or the claim that god can do anything.
Can god create a stone that is so heavy that not even god can lift it?
Sets or gods?
What a strange comparison. Sets theory is perfectly clear !
Op 14/07/2025 om 20:11 schreef Richard Hachel:
Le 14/07/2025 à 14:20, sobriquet a écrit :
Op 13/07/2025 om 22:08 schreef efji:
In that respect I'm interested in the psychology of crackpots that might >>> claim that 1*1=2 or that 5^(3i)=-123.
In imaginary algebria (not in complex trigonomztria) :
5^(3i)=-123
y=5^(3i)
f(y)=5^(x) with y₀=f(0)=1
g(x)=-f(-x)+2y₀=-5^(-3i)+2
y=-125+2=-123
R.H.
Yes, and what is 1/i in your crackpot approach?
Le 15/07/2025 à 16:53, sobriquet a écrit :
Op 14/07/2025 om 20:11 schreef Richard Hachel:
Le 14/07/2025 à 14:20, sobriquet a écrit :
Op 13/07/2025 om 22:08 schreef efji:
In that respect I'm interested in the psychology of crackpots that might >>>> claim that 1*1=2 or that 5^(3i)=-123.
In imaginary algebria (not in complex trigonomztria) :
5^(3i)=-123
y=5^(3i)
f(y)=5^(x) with y₀=f(0)=1
g(x)=-f(-x)+2y₀=-5^(-3i)+2
y=-125+2=-123
R.H.
Yes, and what is 1/i in your crackpot approach?
Merci de bien vouloir respecter votre interlocuteur, ce mépris à la con devient intolérable.
Le 15/07/2025 à 17:38, Richard Hachel a écrit :
Le 15/07/2025 à 16:53, sobriquet a écrit :
Posting in French in an English-speaking group IS a lack of respect.
Sets theory is perfectly clear !
Le 15/07/2025 à 16:53, sobriquet a écrit :[...]
Yes, and what is 1/i in your crackpot approach?
Merci de bien vouloir respecter votre interlocuteur, ce mépris à la con devient intolérable.
Je réponds quand même non pour vous, mais pour ceux que ça pourrait intéresser.
Qu'avons nous dit? Que i n'était pas un nombre, mais un opérateur.
Et que lorsqu'il était écrit sous forme simple i, il voulait dire i=1*i=-1.
Puisque i^x=-1 quelque soit x, et que (-i)^x dépend de la parité de x.
Ici, on pose la question : y=1/i.
La réponse est d'une dramatique simplicité.
1/i=i^(-1)
Nous venons de dire que i^x=-1 quelque soit x.
1/i=-1
Attention au piège, certains rigolos se sentant malins pourrait dire que l'on va multiplier par i les deux termes du quotient, mais c'est une
erreur de concept. On aurait alors i/i²=-1/-1=1 SAUF QUE i est un opérateur, et qu'il ne sert à rien de le répéter deux fois. C'est pas comme ça que ça marche.
R.H.
Op 15/07/2025 om 17:38 schreef Richard Hachel:
Le 15/07/2025 à 16:53, sobriquet a écrit :[...]
Yes, and what is 1/i in your crackpot approach?
Merci de bien vouloir respecter votre interlocuteur, ce mépris à la con
devient intolérable.
Je réponds quand même non pour vous, mais pour ceux que ça pourrait
intéresser.
Qu'avons nous dit? Que i n'était pas un nombre, mais un opérateur.
Et que lorsqu'il était écrit sous forme simple i, il voulait dire i=1*i=-1.
Puisque i^x=-1 quelque soit x, et que (-i)^x dépend de la parité de x.
Ici, on pose la question : y=1/i.
La réponse est d'une dramatique simplicité.
1/i=i^(-1)
Nous venons de dire que i^x=-1 quelque soit x.
1/i=-1
Attention au piège, certains rigolos se sentant malins pourrait dire que
l'on va multiplier par i les deux termes du quotient, mais c'est une
erreur de concept. On aurait alors i/i²=-1/-1=1 SAUF QUE i est un
opérateur, et qu'il ne sert à rien de le répéter deux fois. C'est pas
comme ça que ça marche.
R.H.
Ok, and what is tan(i + pi/2)?
Le 15/07/2025 à 22:17, sobriquet a écrit :
Op 15/07/2025 om 17:38 schreef Richard Hachel:
Le 15/07/2025 à 16:53, sobriquet a écrit :[...]
Yes, and what is 1/i in your crackpot approach?
Merci de bien vouloir respecter votre interlocuteur, ce mépris à la
con devient intolérable.
Je réponds quand même non pour vous, mais pour ceux que ça pourrait
intéresser.
Qu'avons nous dit? Que i n'était pas un nombre, mais un opérateur.
Et que lorsqu'il était écrit sous forme simple i, il voulait dire
i=1*i=-1.
Puisque i^x=-1 quelque soit x, et que (-i)^x dépend de la parité de x. >>> Ici, on pose la question : y=1/i.
La réponse est d'une dramatique simplicité.
1/i=i^(-1)
Nous venons de dire que i^x=-1 quelque soit x.
1/i=-1
Attention au piège, certains rigolos se sentant malins pourrait dire
que l'on va multiplier par i les deux termes du quotient, mais c'est
une erreur de concept. On aurait alors i/i²=-1/-1=1 SAUF QUE i est un
opérateur, et qu'il ne sert à rien de le répéter deux fois. C'est pas >>> comme ça que ça marche.
R.H.
Ok, and what is tan(i + pi/2)?
This is another complex trigonometry problem.
That's not the point here.
Can we please actually address the specifics?
R.H.
Op 15/07/2025 om 23:06 schreef Richard Hachel:
Le 15/07/2025 à 22:17, sobriquet a écrit :
Op 15/07/2025 om 17:38 schreef Richard Hachel:
Le 15/07/2025 à 16:53, sobriquet a écrit :[...]
Yes, and what is 1/i in your crackpot approach?
Merci de bien vouloir respecter votre interlocuteur, ce mépris à la
con devient intolérable.
Je réponds quand même non pour vous, mais pour ceux que ça pourrait >>>> intéresser.
Qu'avons nous dit? Que i n'était pas un nombre, mais un opérateur.
Et que lorsqu'il était écrit sous forme simple i, il voulait dire
i=1*i=-1.
Puisque i^x=-1 quelque soit x, et que (-i)^x dépend de la parité de x. >>>> Ici, on pose la question : y=1/i.
La réponse est d'une dramatique simplicité.
1/i=i^(-1)
Nous venons de dire que i^x=-1 quelque soit x.
1/i=-1
Attention au piège, certains rigolos se sentant malins pourrait dire
que l'on va multiplier par i les deux termes du quotient, mais c'est
une erreur de concept. On aurait alors i/i²=-1/-1=1 SAUF QUE i est un >>>> opérateur, et qu'il ne sert à rien de le répéter deux fois. C'est pas >>>> comme ça que ça marche.
R.H.
Ok, and what is tan(i + pi/2)?
This is another complex trigonometry problem.
That's not the point here.
Can we please actually address the specifics?
R.H.
How about log(i)? Can we do that in your crackpot system?
Le 15/07/2025 à 23:27, sobriquet a écrit :
Op 15/07/2025 om 23:06 schreef Richard Hachel:
Le 15/07/2025 à 22:17, sobriquet a écrit :
Op 15/07/2025 om 17:38 schreef Richard Hachel:
Le 15/07/2025 à 16:53, sobriquet a écrit :[...]
Yes, and what is 1/i in your crackpot approach?
Merci de bien vouloir respecter votre interlocuteur, ce mépris à la >>>>> con devient intolérable.
Je réponds quand même non pour vous, mais pour ceux que ça pourrait >>>>> intéresser.
Qu'avons nous dit? Que i n'était pas un nombre, mais un opérateur. >>>>> Et que lorsqu'il était écrit sous forme simple i, il voulait dire
i=1*i=-1.
Puisque i^x=-1 quelque soit x, et que (-i)^x dépend de la parité de x. >>>>> Ici, on pose la question : y=1/i.
La réponse est d'une dramatique simplicité.
1/i=i^(-1)
Nous venons de dire que i^x=-1 quelque soit x.
1/i=-1
Attention au piège, certains rigolos se sentant malins pourrait
dire que l'on va multiplier par i les deux termes du quotient, mais
c'est une erreur de concept. On aurait alors i/i²=-1/-1=1 SAUF QUE
i est un opérateur, et qu'il ne sert à rien de le répéter deux
fois. C'est pas comme ça que ça marche.
R.H.
Ok, and what is tan(i + pi/2)?
This is another complex trigonometry problem.
That's not the point here.
Can we please actually address the specifics?
R.H.
How about log(i)? Can we do that in your crackpot system?
Log i = 0
If y=Log(1) then y=0
Log(1)=Log(-i)
g(x)=-f(-x)+2yo
Log i = 0
R.H.
Op 16/07/2025 om 02:02 schreef Richard Hachel:
But for F(x) = 5^x, you mirrored the curve in the point (x,y) = (0,1), because
F(0) = 5^0 = 1
so if we try to follow that same crackpot logic,
we have an analogous
curve f(x) = log(x) and if you try to plug in 0, we get
f(0) = log(0) = - infinity
https://www.wolframalpha.com/input?i=log%280%29
So we are somehow supposed to mirror the curve for f(x) in the point
(0, - infinity).
No crakpot logic here.
we have an analogous curve f(x) = log(x) and if you try to plug in 0,
we get
f(0) = log(0) = - infinity
f(x)=a^x is différent that f(x)=Log x
Here g(x) become -Log(-x) and the curve is in miroir of point $(0,0).
Le 16/07/2025 à 05:25, Richard Hachel a écrit :
f(x)=Log x
Here g(x) become -Log(-x) and the curve is in miroir of point $(0,0).
What about f(x) = 1+Log(x) ?
It becomes g(x) = -(1+Log(-x)) with your "logic", right ?
and then ?
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