Calculator test

[Maximilian Hohmann]

(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

[AnnoyedOne]

(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?

[Maximilian Hohmann]

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

[AnnoyedOne]

(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

[Commie]

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

[klesl]

My favourite example for negative numbers is Dirac's puzzle
https://math.stackexchange.com/questions...le-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?

[AnnoyedOne]

(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

[dm319]

(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.

[cdmackay]

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.

[Idnarn]

(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).

\[\ln(re^{i\theta}) = \ln r + i\theta + 2k\pi i\]

[Commie]

(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

[naddy]

(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, \(2\).

[Albert Chan]

(Yesterday 01:56 PM)Commie Wrote:  Here is one of the equations for tan(x+iy):
\[\tan(x+iy) = \frac{sin{2x} + {i}sinh{2y}}{cosh{2y}+cos{2x}}\]

Proof is trivial, using sum to product formula. Let Y = i*y

RHS
= (sin(2x) + sin(2Y)) / (cos(2x) + cos(2Y))
= (2*sin(x+Y)*cos(x-Y)) / (2*cos(x+Y)*cos(x-Y))
= sin(x+Y) / cos(x+Y)
= tan(x+i*y)

Note that cos(2x) has range [-1,1], cosh(2y) has range [1,inf)
RHS denominator may cause catastrophic cancellation.

This was the reason Free42 tan(z) later switched to this:

tan(x+i*y) = (sin(x)*cos(x) + i * cosh(y)*sinh(y)) / (cos(x)^2 + sinh(y)^2)

[AnnoyedOne]

(Yesterday 11:14 PM)naddy Wrote:  

(Yesterday 09:22 PM)Commie Wrote:  I'm sorry, but I don't understand...

...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.

naddy beat me to it. In particular read p69 of the AFH.

The HP-15C AFH was written by people that knew their stuff. IMO worth reading even if you don't use a 15C.

A1

[Albert Chan]

(Yesterday 08:12 PM)Idnarn Wrote:  \[\ln(re^{i\theta}) = \ln r + i\theta + 2k\pi i\]

(Yesterday 09:22 PM)Commie Wrote:  I'm sorry, but I don't understand wikipedia's interpretation of k, it seems to be quantized?

Perhaps it is better to rewrite LHS with k included, instead of from thin air.

e^(pi * i) = -1

Multiply LHS ln argument by 1 = e^(2*k*pi * i), integer k, for all solutions.
Note: ln(z) may mean only principal value, not all solutions.

\[\ln(r\;e^{\theta \,i}) = \ln(r\; e^{\theta\,i}\;e^{2k\pi\,i}) = \ln r + \theta\,i + 2k\pi\,i \]

[Commie]

Hi,

Can someone please tell me how calculators calculate:

sin(2000.x)? and cos(2000.x)?

Try them in your calculator and then take the inverse to get back to 2000x? what happened to all the 2.pi circles?

Whether using Cordics or poly's makes no difference, and using rads only.

Cheers
Darren

[Idnarn]

(12-08-2024 08:38 PM)carey Wrote:  A similar workaround applies to the Casio 9860gii and CG50 calculators as they also have complex argument limitations for trig functions but not for e^.

KhiCAS has its quirks compared to regular giac (xcas) on a computer (e.g., there are some quirks when running it with higher precision), but for the most part, it delivers things that the fx-CG50 is missing.

[Albert Chan]

(Today 03:21 PM)Commie Wrote:  Can someone please tell me how calculators calculate:

sin(2000.x)? and cos(2000.x)?

Try them in your calculator and then take the inverse to get back to 2000x? what happened to all the 2.pi circles?

2000/pi ≈ 636.619772 = 637 - 0.380228

We divide radians by pi, not (2*pi), so that reduced angle is within ±pi/2

sin(pi - x) = sin(x)

asin(sin(2000)) ≈ asin(sin(0.380228*pi)) ≈ asin(0.930040) ≈ 1.19452

637*pi - 1.19452 ≈ 2000.00000

cos(pi - x) = -cos(x)

acos(cos(2000)) ≈ acos(-cos(0.380228*pi)) ≈ acos(-0.367458) = pi - acos(0.367458) ≈ 1.94707

637*pi - (pi - 1.94707) = 637*pi - 1.19452 ≈ 2000.00000

[Commie]

(Today 04:48 PM)Albert Chan Wrote:  2000/pi ≈ 636.619772 = 637 - 0.380228

We divide radians by pi, not (2*pi), so that reduced angle is within ±pi/2

sin(pi - x) = sin(x)

asin(sin(2000)) ≈ asin(sin(0.380228*pi)) ≈ asin(0.930040) ≈ 1.19452

637*pi - 1.19452 ≈ 2000.00000

cos(pi - x) = -cos(x)

acos(cos(2000)) ≈ acos(-cos(0.380228*pi)) ≈ acos(-0.367458) = pi - acos(0.367458) ≈ 1.94707

637*pi - (pi - 1.94707) = 637*pi - 1.19452 ≈ 2000.00000

Hi Albert,

First off, I was hoping Naddy or A1 or both could give the answer since they are both very smart guys, sigh...

I do it slightly differently to you, but first let me just mention that the sine algorithm which is either using cordics or poly's in both cases the ranges are limited, usually between -pi/2...pi/2.
So the calculator is reducing the 2000.x to get it in range for the Cordic or poly algorithm.

Assuming x is one then the sine(2000) gives 0.93 and the arcsine(0.93....) gives 1.19452.... but
sine(1.19452...) gives 0.93 now where has the 637*pi gone?
What does the calculator do with it?

I do it slightly different, 2000/(2.pi)=318.3098862, now remove 318 to leave 0.3098862
Then 0.3098862*2.pi=1.947072317 which then goes to the sine poly to yield
sine(2000)=0.9300395044.This commonly known in the trade as removing the sine circles.
Now, I started with 2000 right?, what does the calculator do with the excess sine circles? On any calculator for that matter, because evidently there is a big difference between sine(2000) and sine(1.947072317)? My sine algorithm is limited to just -pi...pi.?

Cheers
Darren

[AnnoyedOne]

(Today 06:10 PM)Commie Wrote:  ...what does the calculator do with the excess sine circles? On any calculator for that matter...

First, thanks for the compliment.

Second trigonometric functions (and some others) are both cyclical and symmetrical. Ask any EE/mathematician/physicist.

2.pi is the cycle length for trigonometric functions but -pi/2 to pi/2 works better for some purposes. Its that simple.

In theory calculating from 0 to pi/2 is sufficient because of symmetry.

A1