Graph Forums — Negative Roots

Re: Negative Roots

null@example.com (Ivan Johansen) — Wed, 01 Jun 2016 20:52:11 +0000

jsnc wrote:

I created the relation for the equation of a circle in terms of y (also works for x^2 + y^2 = 1).
It would be good to correct the FAQ that asks about graphing a circle, where the Reply directs one to use Parametrics.

The FAQ mentions both parametric functions and relations as a way to plot a circle.

jsnc wrote:

As f(x) = (1-x^2)^.5 is mathematically equivalent to the above formula, why isn't the same graph produced?

Because Graph works by calculating one result for every input, so one value of x gives one value of f(x). I actually plan to redesign this, but I lack the time to actually implement it.

jsnc wrote:

Also, .25^.5 = +.5 and -.5, Why doesn't the graph of f(x)=(1-x^2)^.5 show both roots (specific example of the same question)?

Same answer. Every value of x gives at most one value of f(x). And to be honest every calculator I have seen does the same.

jsnc wrote:

Perhaps add support for the character ± (ASCII ALT-241), this will produce multiple roots (and multiples of multiples with multiple ± occurrences in the same equation). But, this is math after all.

Yes, but it requires a complete redesign as Graph currently can only handle one value at a time. I will do that when I find the time.

jsnc wrote:

As an aside, in the Forum, it would be good to be able to select text, and then copy and paste (helpful for discussion where one needs to repeat information/formulas).

I have no trouble copy-pasting text in the forum. I use that a lot.

jsnc wrote:

The only problem with Relations, is that the Evaluate Tool does not work for Relations (would have to deal with multiple y values for single values of x when an equation is not formally a function). For the negative root issue, one can create f(x) = +(function), and f(x) = -(function), and then each Function can be Evaluated (a work around).

Yes, that is very unfortunately. It is also one of the things I would like to implement when I find the time.

Re: Negative Roots

null@example.com (jsnc) — Tue, 31 May 2016 21:11:48 +0000

The only problem with Relations, is that the Evaluate Tool does not work for Relations (would have to deal with multiple y values for single values of x when an equation is not formally a function). For the negative root issue, one can create f(x) = +(function), and f(x) = -(function), and then each Function can be Evaluated (a work around).

Also, being able to show π (multiples) is great (helping my son with his TRIG), but it would be great to be able to show both Integers, and π (multiples).

Re: Negative Roots

null@example.com (jsnc) — Tue, 31 May 2016 20:22:47 +0000

Thank you for the quick reply.
I created the relation for the equation of a circle in terms of y (also works for x^2 + y^2 = 1).
It would be good to correct the FAQ that asks about graphing a circle, where the Reply directs one to use Parametrics.
As f(x) = (1-x^2)^.5 is mathematically equivalent to the above formula, why isn't the same graph produced?

Also, .25^.5 = +.5 and -.5, Why doesn't the graph of f(x)=(1-x^2)^.5 show both roots (specific example of the same question)?

Perhaps add support for the character ± (ASCII ALT-241), this will produce multiple roots (and multiples of multiples with multiple ± occurrences in the same equation). But, this is math after all.

As an aside, in the Forum, it would be good to be able to select text, and then copy and paste (helpful for discussion where one needs to repeat information/formulas).

Re: Negative Roots

null@example.com (Ivan Johansen) — Tue, 31 May 2016 17:44:00 +0000

This is the normal way functions work. f(x) gives one result for a given value of x. However you can create a relation y^2=1-x^2 to plot your circle.

Negative Roots

null@example.com (jsnc) — Tue, 31 May 2016 15:45:55 +0000

Why not show the negative roots for square roots, cubic roots, etc.
One could then use standard and not parametric equations to graph a circle.
f(x) = (1-x^2)^.5 would then show the entire circle.