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HP-50g plots of cube roots (no negative domain displayed)
03-19-2018, 04:35 PM
Post: #1
HP-50g plots of cube roots (no negative domain displayed)
If I plot y1(x)=x^(1/3), I get no negatives. I did a bit of searching and turned up :

http://www.hpmuseum.org/cgi-sys/cgiwrap/...ead=141921

The solution there was kind of awkward as if you want to plot say :

y1(x)=x^(5/3)

you have to plot :

y1(x) = (cube root(x))^5

as both

y1(x)=x^(5/3)

and

y1(x) = cube root(x^5)

Fail to show the negative domain as the calculator is generating the complex roots when x < 0.

I switched from exact to approx, and turned off complex (under modes/cas) but this does not help. Is there a way to force the calculator to return the real root or are you forced to play with the equation entry which is kind of awkward?
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03-21-2018, 07:16 PM (This post was last modified: 03-21-2018 07:19 PM by Dieter.)
Post: #2
RE: HP-50g plots of cube roots (no negative domain displayed)
Cliff,
as you can see you posted this message twice. Please delete the other one. To do so, log in and use the delete button below your post (the one with the red X).

(03-19-2018 04:35 PM)Cliff Stamp Wrote:  If I plot y1(x)=x^(1/3), I get no negatives. I did a bit of searching and turned up :

http://www.hpmuseum.org/cgi-sys/cgiwrap/...ead=141921

The solution there was kind of awkward as if you want to plot say :

y1(x)=x^(5/3)

you have to plot :

y1(x) = (cube root(x))^5

I do not have a 50g, but this behaviour is the way I would expect a decent calculator to react: While x^(1/3) mathematically is the same as the cube root of x, it isn't for a calculator with finite precision: for the cube root of –8 it would calculate –8^0,333333333333 which is not the same as –8^(1/3). So the solution is complex, as the 12-digit-rounded value of 1/3 equals the 3,000000000003rd root of –8.

Then you wrote:

(03-19-2018 04:35 PM)Cliff Stamp Wrote:  ...as both

y1(x)=x^(5/3)

and

y1(x) = cube root(x^5)

Fail to show the negative domain as the calculator is generating the complex roots when x < 0.

It should be clear now why the first method does not work the way you want to.

But I wonder why

y1(x) = cube root(x^5)

should not work while

y1(x) = (cube root(x))^5

gives the desired result.

Are you sure about this?

Dieter
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03-22-2018, 01:58 PM
Post: #3
RE: HP-50g plots of cube roots (no negative domain displayed)
(03-21-2018 07:16 PM)Dieter Wrote:  Cliff,
as you can see you posted this message twice. Please delete the other one. To do so, log in and use the delete button below your post (the one with the red X).

I tried that when I noticed it was duplicated, I don't have permission to do it. I also reported it to a mod for deletion as duplication.


Quote:I do not have a 50g, but this behaviour is the way I would expect a decent calculator to react

Well, my question was how to generate certain behavior, not what is the "right" behavior in a normative sense which is a different question for a different thread. I don't see that as productive as I don't see what kind of categorical imperative you could appeal to so it is going to be nothing more than a statement of preference. However, since you diverted the thread anyway, the reasons why I don't prefer the behavior are many, but here are three obvious ones :

1) In exact mode, if you enter Y1(x) = 3root(x) and Y2(x)=x^(1/3), they are displayed on the screen if you recall them as exactly the same (in root form). Yet they produce different answers. Due to this, you can't know what an equation will produce by looking at it, you have to know the original form it was entered. This breaks mathematical reasoning on a very fundamental level. Plus the whole idea of an exact mode is to prevent the kind of issues where you get things like ((1/7)*7)^1E100 = 0. That doesn't happen in exact mode.

2) Worse still, if you enter one of the functions which won't generate the negative real roots in the plots but you go outside of the graph and manually make a table of values by just evaluating the function, it happily makes the real roots. Thus the calculator won't actually match the graph it makes with the table of values it generates on the same numbers.

Quote:Are you sure about this?

Yes, it can do the cube root of a negative and return a real, then it can raise that to the fifth power. However, if you do the fifth power of the negative, it can't cube root that. Function composition doesn't actually produce the same results as manually running the functions after each other. That likely bothers me the most as I am a functional style programmer and if you break functional composition, well it makes programming that way impossible.
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