+- HP Forums (https://www.hpmuseum.org/forum)
+-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html)
+--- Forum: General Forum (/forum-4.html)
+--- Thread: Calculator test (/thread-22816.html)
RE: Calculator test - Maximilian Hohmann - Yesterday 02:04 PM
(Yesterday 01:37 PM)AnnoyedOne Wrote: There's really no such thing thing since you can't have -x of anything.
After buying all these calculators, I have plenty of negative Euros on my bank account. And it will require a lot of imaginary money to even out the balance ;-)
But without kidding: I actually do see negative numbers in the real/material world. Wherever there is a natural origin of a coordinate system, e.g. the surface of the ocean. Above is positive, below is negative. At least from a aviator's point of view. For a submariner it will be the opposite.
Regards
Max
RE: Calculator test - AnnoyedOne - Yesterday 02:12 PM
(Yesterday 02:04 PM)Maximilian Hohmann Wrote: ...I actually do see negative numbers in the real/material world.Yes, like I said you see a real relationship between two or more things.
A1
PS: How's your relationship with your bank?
RE: Calculator test - Maximilian Hohmann - Yesterday 02:28 PM
Hello!
(Yesterday 02:12 PM)AnnoyedOne Wrote: PS: How's your relationship with your bank?
This is also one of the big paradoxes of (bank) accounting: The more negative your balance is, the more positive you are seen by your bank manager. Instead of them owing something to you, they will get money from you every month and in the end they will get your house for next to nothing :-)
Regards
Max
RE: Calculator test - AnnoyedOne - Yesterday 03:27 PM
(Yesterday 02:28 PM)Maximilian Hohmann Wrote: This is also one of the big paradoxes of (bank) accounting...
A millionaire/billionaire was once quoted as saying
Quote:If you owe a bank 250k you have a problem.
Owe them 250m and they have a problem.
Banks generally don't like negative numbers. As for imaginary ones...
A1
RE: Calculator test - Commie - Yesterday 06:19 PM
Hi Guys,
Here are complex exponential and natural log:
\[e^{x+iy} = e^{x}cos{y} + {i}e^{x}sin{y}\]
\[ln({x+iy}) = \frac{1}{2}ln({x^{2}+y^{2}})+{i}arctan(\frac{y}{x})\]
Cheers
Darren
RE: Calculator test - klesl - Yesterday 06:25 PM
My favourite example for negative numbers is Dirac's puzzle
https://math.stackexchange.com/questions/495646/math-puzzle-fishing
os similar puzzle. Three persons are living in the house. If 4 persons left this house, how many persons should go in to have empty house?
RE: Calculator test - AnnoyedOne - Yesterday 06:29 PM
(Yesterday 06:25 PM)klesl Wrote: ...how many persons should go in to have empty house?
None. 4 left already. Perhaps more inside. No math needed.
Whatever the case anyone going in (who live there or not) makes the number of occupants > 0.
A1
RE: Calculator test - dm319 - Yesterday 06:35 PM
(Yesterday 02:04 PM)Maximilian Hohmann Wrote: actually do see negative numbers in the real/material world. Wherever there is a natural origin of a coordinate system, e.g. the surface of the ocean. Above is positive, below is negative. At least from a aviator's point of view.
I think you say that easily only because you are familiar and comfortable with the concept of negative numbers. Both locations are simply real, positive distances from a point, in different directions.
RE: Calculator test - cdmackay - Yesterday 07:03 PM
My young daughter couldn't quite grasp the idea of negative numbers, at school, until we sat down and played with money, and IOU/loan notes, and then she said "oh, but that's obvious!".
The bank analogy (for adults) is similarly easy for these odd non-mathematical folk to grasp, it seems.
RE: Calculator test - Idnarn - Yesterday 08:12 PM
(Yesterday 06:19 PM)Commie Wrote: \[ln({x+iy}) = \frac{1}{2}ln({x^{2}+y^{2}})+{i}arctan(\frac{y}{x})\]
This is one logarithm. There are more. The above is missing a term:
\[2k\pi i\]
See: https://en.wikipedia.org/wiki/Complex_logarithm
The above expression is clearer in the polar form where arctan(y/x) is theta and 1/2*ln(x^2+y^2) = ln(sqrt(x^2+y^2)) = ln(r).
\[\ln(re^{i\theta}) = \ln r + i\theta + 2k\pi i\]
RE: Calculator test - Commie - Yesterday 09:22 PM
(Yesterday 08:12 PM)Idnarn Wrote: The above is missing a term:
\[2k\pi i\
\[\ln(re^{i\theta}) = \ln r + i\theta + 2k\pi i\]
I'm sorry, but I don't understand wikipedia's interpretation of k, it seems to be quantized?
The equation, I have shown, is from a pure math prospective, for example assume z=1+2i
Entering this on my ti84+ reveals the answer of 0.8+1.11i
If I calculate this using the equation given, then on my ti30x pro, I get 0.8+1.11i
Doing the same calculation on the hp 35s reveals the same answer. I think you are mixing up pure maths with applied math/physics?I have not specified what the imaginary part actually is physically.
Cheers
Darren
RE: Calculator test - naddy - Yesterday 11:14 PM
(Yesterday 09:22 PM)Commie Wrote:(Yesterday 08:12 PM)Idnarn Wrote: \[\ln(re^{i\theta}) = \ln r + i\theta + 2k\pi i\]
I'm sorry, but I don't understand wikipedia's interpretation of k, it seems to be quantized?
\(k = \dots, -2, -1, 0, 1, 2, \dots\) ; i.e., \(k \in \mathbb Z\)
All of those (infinitely many) values are a correct solution for \(\ln(r e^{i\theta})\). Many complex functions have more than one value for any argument. Typically one is designated the principal value, which in this case is the one you gave a formula for. Section 3 of the HP-15C Advanced Functions Handbook has some helpful notes about this.
Think square roots: \(\sqrt 4 = \pm 2\), but the calculator will only give you the principal value, \(2\).