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Solution to Rishi's problem
+4
Marathadi-Saamiyaar
Rishi
MaxEntropy_Man
Seva Lamberdar
8 posters
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Page 2 of 2 • 1, 2
Re: Solution to Rishi's problem
bw wrote:"sqrt(x)" is a function and a function returns only one value for any input "x" which, in this case, is the positive root of "x". a function cannot return two values. that's that.
A+
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: Solution to Rishi's problem
bw wrote:"sqrt(x)" is a function and a function returns only one value for any input "x" which, in this case, is the positive root of "x". a function cannot return two values. that's that.
This is a discussion in mathematics on square roots and not some half-brained imbecilic use of square root function (subroutine) on computers.
Re: Solution to Rishi's problem
Seva Lamberdar wrote:bw wrote:"sqrt(x)" is a function and a function returns only one value for any input "x" which, in this case, is the positive root of "x". a function cannot return two values. that's that.
This is a discussion in mathematics on square roots and not some half-brained imbecilic use of square root function (subroutine) on computers.
In mathematics, a function[1] is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.
are you questioning this, seva?
bw- Posts : 2922
Join date : 2012-11-15
Re: Solution to Rishi's problem
MaxEntropy_Man wrote:Seva Lamberdar wrote:MaxEntropy_Man wrote:Seva Lamberdar wrote:
What does the satement "The real solutions of x^2 = a are written as +,- sqrt(a)" mean to you?
it means that √a is a positive number, and the attaching of a negative sign in front of it changes its sign. the attaching of a positive sign by itself is redundant although not incorrect, but when referring to both roots together, one writes ±√a. the very fact that you have to write ±√a implies that √a cannot be the symbol that refers to both roots as you claimed originally.
No. You are the one who first made a statement "as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1" using abs(x), instead of just x (or +,- x), which led to this entire discussion.
here is what i said, now that i can type in math symbols:
√(1+x²) which is a positive number, tends to √x² also a positive number, and thus equal to |x|, when x is large and negative. i maintain that when you leave the root unsigned as in √x², the implication is that you are referring to the positive root.
Here is what you said in the beginning "as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1" and spent the whole time defending in it the arbitrary use of "abs".
Re: Solution to Rishi's problem
Seva Lamberdar wrote:MaxEntropy_Man wrote:Seva Lamberdar wrote:MaxEntropy_Man wrote:Seva Lamberdar wrote:
What does the satement "The real solutions of x^2 = a are written as +,- sqrt(a)" mean to you?
it means that √a is a positive number, and the attaching of a negative sign in front of it changes its sign. the attaching of a positive sign by itself is redundant although not incorrect, but when referring to both roots together, one writes ±√a. the very fact that you have to write ±√a implies that √a cannot be the symbol that refers to both roots as you claimed originally.
No. You are the one who first made a statement "as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1" using abs(x), instead of just x (or +,- x), which led to this entire discussion.
here is what i said, now that i can type in math symbols:
√(1+x²) which is a positive number, tends to √x² also a positive number, and thus equal to |x|, when x is large and negative. i maintain that when you leave the root unsigned as in √x², the implication is that you are referring to the positive root.
Here is what you said in the beginning "as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1" and spent the whole time defending in it the arbitrary use of "abs".
i still stand by that and it does not contradict anything else i have said. i am done with this.
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: Solution to Rishi's problem
bw wrote:Seva Lamberdar wrote:bw wrote:"sqrt(x)" is a function and a function returns only one value for any input "x" which, in this case, is the positive root of "x". a function cannot return two values. that's that.
This is a discussion in mathematics on square roots and not some half-brained imbecilic use of square root function (subroutine) on computers.
In mathematics, a function[1] is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.
are you questioning this, seva?
If you really want to get on this discussion BW, go through all the comments on this thread right from the beginning. I don't think you have a clue about what exactly is being discussed and why?
Re: Solution to Rishi's problem
Seva Lamberdar wrote:bw wrote:Seva Lamberdar wrote:bw wrote:"sqrt(x)" is a function and a function returns only one value for any input "x" which, in this case, is the positive root of "x". a function cannot return two values. that's that.
This is a discussion in mathematics on square roots and not some half-brained imbecilic use of square root function (subroutine) on computers.
In mathematics, a function[1] is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.
are you questioning this, seva?
If you really want to get on this discussion BW, go through all the comments on this thread right from the beginning. I don't think you have a clue about what exactly is being discussed and why?
you are claiming that the function sqrt(x) does not represent just the positive root. mathematics, definition of functions, max and some others disagree with you. what else is there to understand?
by the way, abs(x) and sqrt(x^2) are the same.
bw- Posts : 2922
Join date : 2012-11-15
Re: Solution to Rishi's problem
bw wrote:Seva Lamberdar wrote:bw wrote:Seva Lamberdar wrote:bw wrote:"sqrt(x)" is a function and a function returns only one value for any input "x" which, in this case, is the positive root of "x". a function cannot return two values. that's that.
This is a discussion in mathematics on square roots and not some half-brained imbecilic use of square root function (subroutine) on computers.
In mathematics, a function[1] is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.
are you questioning this, seva?
If you really want to get on this discussion BW, go through all the comments on this thread right from the beginning. I don't think you have a clue about what exactly is being discussed and why?
you are claiming that the function sqrt(x) does not represent just the positive root. mathematics, definition of functions, max and some others disagree with you. what else is there to understand?
by the way, abs(x) and sqrt(x^2) are the same.
LOL.
Re: Solution to Rishi's problem
MaxEntropy_Man wrote:Seva Lamberdar wrote:MaxEntropy_Man wrote:Seva Lamberdar wrote:MaxEntropy_Man wrote:
it means that √a is a positive number, and the attaching of a negative sign in front of it changes its sign. the attaching of a positive sign by itself is redundant although not incorrect, but when referring to both roots together, one writes ±√a. the very fact that you have to write ±√a implies that √a cannot be the symbol that refers to both roots as you claimed originally.
No. You are the one who first made a statement "as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1" using abs(x), instead of just x (or +,- x), which led to this entire discussion.
here is what i said, now that i can type in math symbols:
√(1+x²) which is a positive number, tends to √x² also a positive number, and thus equal to |x|, when x is large and negative. i maintain that when you leave the root unsigned as in √x², the implication is that you are referring to the positive root.
Here is what you said in the beginning "as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1" and spent the whole time defending in it the arbitrary use of "abs".
i still stand by that and it does not contradict anything else i have said. i am done with this.
Let's keep in mind what the good book (M.H.) says,
"The real solutions of x^2 = a are written as +,- sqrt(a)"
Re: Solution to Rishi's problem
In general, it is okay to view the sqrt as a (discrete) set-valued mapping, that maps each (strictly) positive real number to a pair of real numbers that differ only in sign. The case of zero is a trivial case.
E.g. the positive number 25 gets mapped to the set {+5, -5}.
If the context requires a point-valued map (i.e. what is commonly understood as a function), then it is convention to choose the positive root, when tacit/unstated.
Also, note that sqrt(x^2) = x is ambiguous, if you want a point-valued map.
Because, it implicitly requires you to solve for a point-value x, whose square is the input argument (which is some positive scalar with value x^2).
Which in turn requires you to make a choice between two real numbers that differ only in sign.
And that choice is unspecified.
Which is why it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x).
Note that the "abs" operator essentially reduces the set of two real numbers that differ in sign, into a single number, and thus it will resolve the ambiguity.
Lastly, the limit question here would get quite a bit complicated if you insist
on interpreting sqrt as a set-valued mapping. In fact, it would not make
any sense unless you also "order" the set. And even then, it needs the notion of point-wise convergence, and some specification of set arithmetic however obvious, and so on.
Definitely not what a Calculus 1 type of question is intending to ask.
E.g. the positive number 25 gets mapped to the set {+5, -5}.
If the context requires a point-valued map (i.e. what is commonly understood as a function), then it is convention to choose the positive root, when tacit/unstated.
Also, note that sqrt(x^2) = x is ambiguous, if you want a point-valued map.
Because, it implicitly requires you to solve for a point-value x, whose square is the input argument (which is some positive scalar with value x^2).
Which in turn requires you to make a choice between two real numbers that differ only in sign.
And that choice is unspecified.
Which is why it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x).
Note that the "abs" operator essentially reduces the set of two real numbers that differ in sign, into a single number, and thus it will resolve the ambiguity.
Lastly, the limit question here would get quite a bit complicated if you insist
on interpreting sqrt as a set-valued mapping. In fact, it would not make
any sense unless you also "order" the set. And even then, it needs the notion of point-wise convergence, and some specification of set arithmetic however obvious, and so on.
Definitely not what a Calculus 1 type of question is intending to ask.
aanjaneya- Posts : 15
Join date : 2012-09-17
Re: Solution to Rishi's problem
aanjaneya wrote:In general, it is okay to view the sqrt as a (discrete) set-valued mapping, that maps each (strictly) positive real number to a pair of real numbers that differ only in sign. The case of zero is a trivial case.
E.g. the positive number 25 gets mapped to the set {+5, -5}.
If the context requires a point-valued map (i.e. what is commonly understood as a function), then it is convention to choose the positive root, when tacit/unstated.
Also, note that sqrt(x^2) = x is ambiguous, if you want a point-valued map.
Because, it implicitly requires you to solve for a point-value x, whose square is the input argument (which is some positive scalar with value x^2).
Which in turn requires you to make a choice between two real numbers that differ only in sign.
And that choice is unspecified.
Which is why it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x).
Note that the "abs" operator essentially reduces the set of two real numbers that differ in sign, into a single number, and thus it will resolve the ambiguity.
Lastly, the limit question here would get quite a bit complicated if you insist
on interpreting sqrt as a set-valued mapping. In fact, it would not make
any sense unless you also "order" the set. And even then, it needs the notion of point-wise convergence, and some specification of set arithmetic however obvious, and so on.
Definitely not what a Calculus 1 type of question is intending to ask.
mathematicians! grrr....
nevertheless enjoyed reading that jargon and all.
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: Solution to Rishi's problem
aanjaneya wrote:In general, it is okay to view the sqrt as a (discrete) set-valued mapping, that maps each (strictly) positive real number to a pair of real numbers that differ only in sign. The case of zero is a trivial case.
E.g. the positive number 25 gets mapped to the set {+5, -5}.
If the context requires a point-valued map (i.e. what is commonly understood as a function), then it is convention to choose the positive root, when tacit/unstated.
Also, note that sqrt(x^2) = x is ambiguous, if you want a point-valued map.
Because, it implicitly requires you to solve for a point-value x, whose square is the input argument (which is some positive scalar with value x^2).
Which in turn requires you to make a choice between two real numbers that differ only in sign.
And that choice is unspecified.
Which is why it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x).
Note that the "abs" operator essentially reduces the set of two real numbers that differ in sign, into a single number, and thus it will resolve the ambiguity.
Lastly, the limit question here would get quite a bit complicated if you insist
on interpreting sqrt as a set-valued mapping. In fact, it would not make
any sense unless you also "order" the set. And even then, it needs the notion of point-wise convergence, and some specification of set arithmetic however obvious, and so on.
Definitely not what a Calculus 1 type of question is intending to ask.
You are wrong on most of the ponts in the above, especially "it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x)."
Most of the laws related to the real physical systems, including any evaluations using those laws, need to uphold retracing and reversibility. If a physical system is expressed mathematically in terms of F(x) = sqrt(x^2) which you equate to abs(x), rather than x, you can forget about retracing etc. because by assigning sqrt(x^2) as abs(x) you simply eliminate the entire negative domain.
Re: Solution to Rishi's problem
Seva Lamberdar wrote:
You are wrong on most of the ponts in the above, especially "it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x)."
https://such.forumotion.com/t10137-who-s-up-for-a-nice-meaty-probability-problem#78426
you might also be interested in looking up words such as "convention" and "definition" in the context of mathematics and also the difference between a "definition" and "physical law". you seem very confused.
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: Solution to Rishi's problem
MaxEntropy_Man wrote:Seva Lamberdar wrote:
You are wrong on most of the ponts in the above, especially "it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x)."
you might also be interested in looking up words such as "convention" and "definition" in the context of mathematics and also the difference between a "definition" and "physical law". you seem very confused.
Convention should be in agreement and must not violate the basic premise contained in the definition. By assigning abs(x) as the equivalent to F(x) or sqrt(x^2) is not the right / sensible convention, because as indicated earlier it cuts off the negative domain thus hampering the exact retracing between F(x) and x.
Re: Solution to Rishi's problem
Seva Lamberdar wrote:aanjaneya wrote:In general, it is okay to view the sqrt as a (discrete) set-valued mapping, that maps each (strictly) positive real number to a pair of real numbers that differ only in sign. The case of zero is a trivial case.
E.g. the positive number 25 gets mapped to the set {+5, -5}.
If the context requires a point-valued map (i.e. what is commonly understood as a function), then it is convention to choose the positive root, when tacit/unstated.
Also, note that sqrt(x^2) = x is ambiguous, if you want a point-valued map.
Because, it implicitly requires you to solve for a point-value x, whose square is the input argument (which is some positive scalar with value x^2).
Which in turn requires you to make a choice between two real numbers that differ only in sign.
And that choice is unspecified.
Which is why it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x).
Note that the "abs" operator essentially reduces the set of two real numbers that differ in sign, into a single number, and thus it will resolve the ambiguity.
Lastly, the limit question here would get quite a bit complicated if you insist
on interpreting sqrt as a set-valued mapping. In fact, it would not make
any sense unless you also "order" the set. And even then, it needs the notion of point-wise convergence, and some specification of set arithmetic however obvious, and so on.
Definitely not what a Calculus 1 type of question is intending to ask.
You are wrong on most of the ponts in the above, especially "it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x)."
ouch!
bw- Posts : 2922
Join date : 2012-11-15
Re: Solution to Rishi's problem
bw wrote:Seva Lamberdar wrote:aanjaneya wrote:In general, it is okay to view the sqrt as a (discrete) set-valued mapping, that maps each (strictly) positive real number to a pair of real numbers that differ only in sign. The case of zero is a trivial case.
E.g. the positive number 25 gets mapped to the set {+5, -5}.
If the context requires a point-valued map (i.e. what is commonly understood as a function), then it is convention to choose the positive root, when tacit/unstated.
Also, note that sqrt(x^2) = x is ambiguous, if you want a point-valued map.
Because, it implicitly requires you to solve for a point-value x, whose square is the input argument (which is some positive scalar with value x^2).
Which in turn requires you to make a choice between two real numbers that differ only in sign.
And that choice is unspecified.
Which is why it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x).
Note that the "abs" operator essentially reduces the set of two real numbers that differ in sign, into a single number, and thus it will resolve the ambiguity.
Lastly, the limit question here would get quite a bit complicated if you insist
on interpreting sqrt as a set-valued mapping. In fact, it would not make
any sense unless you also "order" the set. And even then, it needs the notion of point-wise convergence, and some specification of set arithmetic however obvious, and so on.
Definitely not what a Calculus 1 type of question is intending to ask.
You are wrong on most of the ponts in the above, especially "it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x)."
ouch!
All this for sqrt(X^2)???
At least if it is for sqrt (-X^2) I can understand.
Marathadi-Saamiyaar- Posts : 17675
Join date : 2011-04-30
Age : 110
Re: Solution to Rishi's problem
Marathadi-Saamiyaar wrote:bw wrote:Seva Lamberdar wrote:
You are wrong on most of the ponts in the above, especially "it is a sensible convention is to take the positive root, and use sqrt(x^2) = abs(x)."
ouch!
All this for sqrt(X^2)???
At least if it is for sqrt (-X^2) I can understand.
LOL.
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