Solution to Rishi's problem

by Seva Lamberdar Mon Apr 29, 2013 7:07 am

Rishi wrote:evaluate

( sqrt(1 + x^2)/x))

When x goes to negative infinity.

Let x = tan y, then function F = sqrt(1+x^2)/x

F = sec y / tan y = 1 / sin y,

Moreover, as x --> - infinity, tan y --> - infinity, i.e. y --> - pi/2

in other words, when x = - infinity, y = - pi/2,

thus F (@ y= - pi/2) = 1/sin(-pi/2) = -1

therefore, sqrt(1+x^2)/x = -1, at x = -infinity

by MaxEntropy_Man Mon Apr 29, 2013 7:13 am

rishi - why did you withdraw your problem?

i am not comfortable with making substitutions like seva has done in limit problems.

my solution:

look at the numerator of your function i.e. sqrt(1+x^2). this is positive for all values of x on the real line. thus as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1. so the limit point you are evaluating as x-> -inf is abs(x)/x which is -1.

by Seva Lamberdar Mon Apr 29, 2013 7:44 am

MaxEntropy_Man wrote:rishi - why did you withdraw your problem?

i am not comfortable with making substitutions like seva has done in limit problems.

my solution:

look at the numerator of your function i.e. sqrt(1+x^2). this is positive for all values of x on the real line. thus as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1. so the limit point you are evaluating as x-> -inf is abs(x)/x which is -1.

"sqrt" of any positive number can be another positive or negative number (not necessarily +abs(x)). To avoid that ambiguity (leading to the answer +1 or -1 in this problem), the trigonometric substitution I made in the above (leading to the precise answer -1) is a better solution.

by Rishi Mon Apr 29, 2013 7:55 am

MaxEntropy_Man wrote:rishi - why did you withdraw your problem?

i am not comfortable with making substitutions like seva has done in limit problems.

my solution:

look at the numerator of your function i.e. sqrt(1+x^2). this is positive for all values of x on the real line. thus as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1. so the limit point you are evaluating as x-> -inf is abs(x)/x which is -1.

Because I saw this video

https://www.youtube.com/watch?v=NzU_-0p79r8

by MaxEntropy_Man Mon Apr 29, 2013 9:51 am

Seva Lamberdar wrote:
MaxEntropy_Man wrote:rishi - why did you withdraw your problem?

i am not comfortable with making substitutions like seva has done in limit problems.

my solution:

look at the numerator of your function i.e. sqrt(1+x^2). this is positive for all values of x on the real line. thus as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1. so the limit point you are evaluating as x-> -inf is abs(x)/x which is -1.

"sqrt" of any positive number can be another positive or negative number (not necessarily +abs(x)). To avoid that ambiguity (leading to the answer +1 or -1 in this problem), the trigonometric substitution I made in the above (leading to the precise answer -1) is a better solution.

i don't agree for the following two reasons.

1) when you write sqrt[f(x)] it usually means the positive root. if the negative root is implied it is explicitly specified as -sqrt[f(x)]. this is the usual convention when writing functions.

2) your own solution doesn't resolve the ambiguity you perceive (albeit a non existent ambiguity) either. i.e. the sqrt[1+(tan(y))^2] can also be taken by the same token as - sec(y). where does that leave you?

anyway as always i am beginning to already sense i might be going down a pointless rabbit hole.

by MaxEntropy_Man Mon Apr 29, 2013 10:05 am

cutting and pasting from a wikilink (http://en.wikipedia.org/wiki/Square_root). basically what i said is tantamount to:

$Solution to Rishi's problem 3314a73542c0786fb2bcc0e3266886c0$

by MaxEntropy_Man Mon Apr 29, 2013 10:10 am

rishi -- your video link makes a very labored argument. IMO it is better to cut through problems like this in the most straightforward and clear minded way and not try to avoid the issue of signs.

by Marathadi-Saamiyaar Mon Apr 29, 2013 11:21 am

MaxEntropy_Man wrote:rishi - why did you withdraw your problem?

i am not comfortable with making substitutions like seva has done in limit problems.

my solution:

look at the numerator of your function i.e. sqrt(1+x^2). this is positive for all values of x on the real line. thus as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1. so the limit point you are evaluating as x-> -inf is abs(x)/x which is -1.

Yes...but, most teachers in Indian schools will think you did hanky-panky to get the right answer.

Most of the times sqrt(x), sqrt (1+ x^2), etc calls for trigonometric substitutions and answers readily fall in place.

by Seva Lamberdar Mon Apr 29, 2013 11:25 am

MaxEntropy_Man wrote:
1) when you write sqrt[f(x)] it usually means the positive root. if the negative root is implied it is explicitly specified as -sqrt[f(x)]. this is the usual convention when writing functions.

by stating "as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1", you are basically hand-picking the positive value of answer which is not necessarily true. For example, as x approaches positive or negative infinity, 1+ x^2 = x^2.

In other words, at x = positive or negative infinity, sqrt(1+x^2) = sqrt (x^2) = x, which implies that

sqrt(1+x^2) = - infinity at x = -infinity;

and sqrt(1+x^2) = + infinity at x = +infinity.

MaxEntropy_Man wrote:

2) your own solution doesn't resolve the ambiguity you perceive (albeit a non existent ambiguity) either. i.e. the sqrt[1+(tan(y))^2] can also be taken by the same token as - sec(y). where does that leave you?

I didn't make a statement that sqrt[1+(tan(y))^2] will result in the "absolute" value (abs(sec y)), but simply sec y (without any positive or negative sign before it).

by MaxEntropy_Man Mon Apr 29, 2013 11:38 am

Marathadi-Saamiyaar wrote:

Yes...but, most teachers in Indian schools will think you did hanky-panky to get the right answer.

Most of the times sqrt(x), sqrt (1+ x^2), etc calls for trigonometric substitutions and answers readily fall in place.

there is no hanky panky at all! and the trig substitution does nothing to resolve the perceived problem either. by making the trig substitution you, or whoever else is making this argument, are asking the reader to accept that:

sqrt[1+(tan (y))^2]= sqrt[sec (y)^2] is unambiguously equal to + sec (y). but squaring - sec (y) also produces (sec (y))^2! the answer in both cases is the absolute value of what is inside the square root radical. there is no ambiguity. when a real valued function is positive over the entire real domain, and you take its square root, the positive root is implied.

thus when x is real,

sqrt[x^2] = abs (x)

and when y is real, sqrt[(sec(y))^2] = abs(sec (y)).

if you wanted the negative root, you would explicitly specify it by sticking a negative sign in front of the square root.

by MaxEntropy_Man Mon Apr 29, 2013 11:44 am

Seva Lamberdar wrote:

I didn't make a statement that sqrt[1+(tan(y))^2] will result in the "absolute" value (abs(sec y)), but simply sec y (without any positive or negative sign before it).

but you are, by not putting any sign in front, saying implicitly that sqrt[1+(tan(y))^2] = + sec (y)! to see that i said is true, consider this: when you write any function, say, x, do you mean + x or - x? you can't say neither.

which brings us back to, why is sqrt[1+(tan(y))^2] not equal to - sec (y)?

i am saying both answers, + sec (y) and - sec (y) are incorrect. this is especially dangerous in trig functions which undergo sign changes. in essence you have changed a straightforward specification of a function and added ambiguity to it, when there was none, by substituting a function that changes sign periodically on the real axis!

so in summary, what is correct is that sqrt[1+(tan(y))^2] = abs (sec(y)).

by Marathadi-Saamiyaar Mon Apr 29, 2013 11:45 am

MaxEntropy_Man wrote:
Marathadi-Saamiyaar wrote:

Yes...but, most teachers in Indian schools will think you did hanky-panky to get the right answer.

Most of the times sqrt(x), sqrt (1+ x^2), etc calls for trigonometric substitutions and answers readily fall in place.

there is no hanky panky at all! and the trig substitution does nothing to resolve the perceived problem either. by making the trig substitution you, or whoever else is making this argument, are asking the reader to accept that:

"I" didn't say there was hanky panky. But, teachers in India will assume there is unless you solve it their "traditional" way.

by MaxEntropy_Man Mon Apr 29, 2013 11:58 am

Marathadi-Saamiyaar wrote: But, teachers in India will assume there is unless you solve it their "traditional" way.

traditional? the trig substitution only muddles it even more! there is nothing traditional about.

by Marathadi-Saamiyaar Mon Apr 29, 2013 12:01 pm

MaxEntropy_Man wrote:
Marathadi-Saamiyaar wrote: But, teachers in India will assume there is unless you solve it their "traditional" way.

traditional? the trig substitution only muddles it even more! there is nothing traditional about.

okie... read closely what I wrote. Also, take a look at the Indian high school and college math books.

by Jeremiah Mburuburu Mon Apr 29, 2013 12:49 pm

Rishi wrote:evaluate
( sqrt(1 + x^2)/x))
When x goes to negative infinity.

let me try this. (i've been so far away from this!)

the limit as x -> -infinity of sqrt(1 + x^2)/x = lim +/-sqrt((1 + x^2)/x^2) = lim +/-sqrt(1/x^2 + 1) = lim +/-sqrt(1) = +/- 1,

i.e. -1 if x < 0, and 1 if x > 0. the required limit is -1.

by Seva Lamberdar Mon Apr 29, 2013 12:56 pm

MaxEntropy_Man wrote:
Seva Lamberdar wrote:

I didn't make a statement that sqrt[1+(tan(y))^2] will result in the "absolute" value (abs(sec y)), but simply sec y (without any positive or negative sign before it).

but you are, by not putting any sign in front, saying implicitly that sqrt[1+(tan(y))^2] = + sec (y)! to see that i said is true, consider this: when you write any function, say, x, do you mean + x or - x? you can't say neither.

which brings us back to, why is sqrt[1+(tan(y))^2] not equal to - sec (y)?

i am saying both answers, + sec (y) and - sec (y) are incorrect. this is especially dangerous in trig functions which undergo sign changes. in essence you have changed a straightforward specification of a function and added ambiguity to it, when there was none, by substituting a function that changes sign periodically on the real axis!

so in summary, what is correct is that sqrt[1+(tan(y))^2] = abs (sec(y)).

This transformation involving x and y (x = tan y) is direct (1 to 1) and has y going uniquely from 0 to -Pi/2 as x goes from 0 to -infinity. There are no overlaps, repetitions or abrupt changes in the values of tan y, sec y or sin y as y varies from 0 to -pi/2 with x going from 0 to -infinity. There is no ambiguity here.

However, in your case to say that sqrt(1+x^2) = abs(x), because x^2 is positive even when x is negative, implies that sqrt(1+x^2) is always positive, including having the value +infinity even at x=-infinity. But, as I indicated earlier, that is not necessarily true, becaues sqrt(1+x^2) can have a value of -infinity at x = -infinity.

by Jeremiah Mburuburu Mon Apr 29, 2013 1:18 pm

Seva Lamberdar wrote:
MaxEntropy_Man wrote:rishi - why did you withdraw your problem?

i am not comfortable with making substitutions like seva has done in limit problems.

my solution:

look at the numerator of your function i.e. sqrt(1+x^2). this is positive for all values of x on the real line. thus as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1. so the limit point you are evaluating as x-> -inf is abs(x)/x which is -1.

"sqrt" of any positive number can be another positive or negative number (not necessarily +abs(x)). To avoid that ambiguity (leading to the answer +1 or -1 in this problem), the trigonometric substitution I made in the above (leading to the precise answer -1) is a better solution.

you should check with rishi whether the expression 1 + x^2 appeared under a radical symbol - the one that looks like a v with a tail on its right - or not, in the original problem statement. if it did, it meant the principal square-root of 1 + x^2, which is defined to be the non-negative root. i have never come across "sqrt()" used to mean anything but the principal square-root.

by Seva Lamberdar Mon Apr 29, 2013 1:25 pm

Jeremiah Mburuburu wrote:
Rishi wrote:evaluate
( sqrt(1 + x^2)/x))
When x goes to negative infinity.
let me try this. (i've been so far away from this!)

the limit as x -> -infinity of sqrt(1 + x^2)/x = lim +/-sqrt((1 + x^2)/x^2) = lim +/-sqrt(1/x^2 + 1) = lim +/-sqrt(1) = +/- 1,

i.e. -1 if x < 0, and 1 if x > 0. the required limit is -1.

after the simplification you got (sqrt(1+x^2))/x = +/- 1, so how did you end up writing i.e. -1 if x < 0, and 1 if x > 0.

by Jeremiah Mburuburu Mon Apr 29, 2013 1:40 pm

the sign of the original limit is governed by the sign of x in the denominator, because sqrt(1 + x^2), a principal square-root, in the numerator is positive; so if x is negative, the limit is negative, and if x is positive, the limit is positive.

could max post a graph of y = sqrt(1 + x^2)/x from wolfram alpha, plotted for -20 <= x <= 20 or some similar domain?

by MaxEntropy_Man Mon Apr 29, 2013 2:02 pm

Jeremiah Mburuburu wrote:the sign of the original limit is governed by the sign of x in the denominator, because sqrt(1 + x^2), a principal square-root, in the numerator is positive; so if x is negative, the limit is negative, and if x is positive, the limit is positive.

could max post a graph of y = sqrt(1 + x^2)/x from wolfram alpha, plotted for -20 <= x <= 20 or some similar domain?

that is exactly right.
here is that plot:
http://www.wolframalpha.com/input/?i=Plot[Sqrt[1%2Bx^2]%2Fx%2C{x%2C-20%2C20}]

major grant deadline, and this thread is my coffee break.

by Jeremiah Mburuburu Mon Apr 29, 2013 2:14 pm

MaxEntropy_Man wrote:
Jeremiah Mburuburu wrote:the sign of the original limit is governed by the sign of x in the denominator, because sqrt(1 + x^2), a principal square-root, in the numerator is positive; so if x is negative, the limit is negative, and if x is positive, the limit is positive.

could max post a graph of y = sqrt(1 + x^2)/x from wolfram alpha, plotted for -20 <= x <= 20 or some similar domain?

that is exactly right.
here is that plot:
http://www.wolframalpha.com/input/?i=Plot[Sqrt[1%2Bx^2]%2Fx%2C{x%2C-20%2C20}]

major grant deadline, and this thread is my coffee break.

thanks. that does it.

by MaxEntropy_Man Mon Apr 29, 2013 2:15 pm

and here is sqrt[1+x^2]. in written and typed mathematics as JM writes above, we would put that under the square root radical. in a symbolic computation program like mathematica, we would write sqrt. see how mathematca interprets sqrt. this is germane because seva keeps insisting that the sqrt could also include the negative root. not true. if you wanted to specify the negative root you stick a negative sign in front of the square root radical.

http://www.wolframalpha.com/input/?i=plot[sqrt[1%2Bx^2]%2C{x%2C-30%2C30}]

by Seva Lamberdar Tue Apr 30, 2013 5:30 am

MaxEntropy_Man wrote:and here is sqrt[1+x^2]. in written and typed mathematics as JM writes above, we would put that under the square root radical. in a symbolic computation program like mathematica, we would write sqrt. see how mathematca interprets sqrt. this is germane because seva keeps insisting that the sqrt could also include the negative root. not true. if you wanted to specify the negative root you stick a negative sign in front of the square root radical.

%2C{x%2C-30%2C30}]

http://www.wolframalpha.com/input/?i=plot[sqrt[1%2Bx^2]%2C{x%2C-30%2C30}][/quote]

The basic point was to not put "abs" on the right hand side of sqrt expression(e.g. sqrt(1+x^2) = abs(x)) because that can sometimes lead to a confusing answer / deduction, as I indicated. The rest of discussion is meaningless.

by MaxEntropy_Man Tue Apr 30, 2013 5:46 am

Seva Lamberdar wrote:

The basic point was to not put "abs" on the right hand side of sqrt expression(e.g. sqrt(1+x^2) = abs(x)) because that can sometimes lead to a confusing answer / deduction, as I indicated. The rest of discussion is meaningless.

there is no confusion other than in your mind. here is how mathematica interprets sqrt(1+x^2) as in the plot below. it interprets it as always having a +ve value, the same as what i and later JM have said. that is the reason i posted this plot earlier:

Solution to Rishi's problem Gif&s=28&w=262.&h=44

Solution to Rishi's problem Gif&s=28&w=262.&h=44

Solution to Rishi's problem Gif&s=26&w=300.&h=196

so setting sqrt (1+x^2) to abs(x) for large negative values of x is, pardon the pun, absolutely correct.

in fact if you want to get the negative root you have to explicitly insert a negative sign in front, like so:

Solution to Rishi's problem Gif&s=63&w=273.&h=44

Solution to Rishi's problem Gif&s=63&w=273.&h=44

Solution to Rishi's problem Gif&s=44&w=300.&h=187

by MaxEntropy_Man Tue Apr 30, 2013 7:38 am

it appears these images are only stored in a buffer which goes away.

here are more permanent links:

http://www.wolframalpha.com/input/?i=plot[sqrt[1%2Bx^2]%2C{x%2C-30%2C30}]

http://www.wolframalpha.com/input/?i=plot%5B-sqrt%5B1%2Bx%5E2%5D%2C%7Bx%2C-30%2C30%7D%5D

by Seva Lamberdar Tue Apr 30, 2013 8:03 am

MaxEntropy_Man wrote:
Seva Lamberdar wrote:

The basic point was to not put "abs" on the right hand side of sqrt expression(e.g. sqrt(1+x^2) = abs(x)) because that can sometimes lead to a confusing answer / deduction, as I indicated. The rest of discussion is meaningless.

there is no confusion other than in your mind.

I had already indicated earlier the contradition due to adding "abs" unnecessarily, e.g. in your statement "as x -> - inf, sqrt(1+x^2) goes to abs(x), since x^2 >> 1".

For example, when you add "abs" (such as in the above) you end up with the strange looking outcome / result and that has nothing to do with the explicit designation of plus or minus signs for the sqrt of a function.

Let x be any variable, then x^2 = x*x.

Taking square root of both sides in the above has the logical outcome,

sqrt(x^2) = x, (and not as sqrt(x^2) = abs(x)).

Now take another expression 1 + x^2, which for large values of x (such as x approaching +infinity or -infinity) holds the following true

1+ x^2 = x^2 (for x approaching + / - infinity).

Taking the square root of both sides in the above,

sqrt(1+x^2) = sqrt(x^2)

but as indicated above sqrt(x^2) = x,

Therefore, for large x (approaching + / - infinity), sqrt(1 + x^2) = x, and that is not the same thing as writing sqrt(1 + x^2) = abs(x).

Now if a function F(x) is defined as F(x) = sqrt(1 +x^2) and you want to find out the value of F(x) at (i) x = +infinity and (ii) x = -infinity, then according to my solution,

(i) F(x=+infinity) = +infinity, and

(ii) F(x=-infinity) = -infinity;

Whereas, in your case sqrt(1+x^2) = abs(x), the answer in both cases will be

F(x=+infinity, or x= -infinity) = +infinity, which may be acceptable numerically (if the sign + or - does not matter) but not in terms of strict rules of mathematics.

by MaxEntropy_Man Tue Apr 30, 2013 11:14 am

Seva Lamberdar wrote:
Now if a function F(x) is defined as F(x) = sqrt(1 +x^2) and you want to find out the value of F(x) at (i) x = +infinity and (ii) x = -infinity, then according to my solution,

(i) F(x=+infinity) = +infinity, and

(ii) F(x=-infinity) = -infinity;

Whereas, in your case sqrt(1+x^2) = abs(x), the answer in both cases will be

F(x=+infinity, or x= -infinity) = +infinity, which may be acceptable numerically (if the sign + or - does not matter) but not in terms of strict rules of mathematics.

i disagree. furthermore, i have a lurking suspicion that you know i am right, but are unwilling to admit your error because your ego is getting in the way. the other possibility is too depressing and i don't want to go there. seva -- it's the same old monty hall story all over again. this is very unhealthy.

by Seva Lamberdar Tue Apr 30, 2013 11:42 am

MaxEntropy_Man wrote: i disagree. furthermore, i have a lurking suspicion that you know i am right, but are unwilling to admit your error because your ego is getting in the way. the other possibility is too depressing and i don't want to go there. seva -- it's the same old monty hall story all over again. this is very unhealthy.

You can go anywhere you want to.

by Jeremiah Mburuburu Tue Apr 30, 2013 11:54 am

Seva Lamberdar wrote:
The basic point was to not put "abs" on the right hand side of sqrt expression(e.g. sqrt(1+x^2) = abs(x)) because that can sometimes lead to a confusing answer / deduction, as I indicated...

Let x be any variable, then x^2 = x*x.

Taking square root of both sides in the above has the logical outcome,

sqrt(x^2) = x, (and not as sqrt(x^2) = abs(x)).

sqrt(x^2) = x is false when x < 0. try this: let x = -3. then x^2 = (-3)^2 = 9 and sqrt(x^2) = 3; thus, when x = -3, the equation sqrt(x^2) = x simplifies to 3 = -3, which is patently false.

on the other hand, the equation sqrt(x^2) = abs(x) is true for all real values of x. when x = -3, for example, the equation simplifies to 3 = abs(-3), which is true.

by Seva Lamberdar Tue Apr 30, 2013 12:04 pm

Jeremiah Mburuburu wrote:

sqrt(x^2) = x is false when x < 0.

LOL.

by Jeremiah Mburuburu Tue Apr 30, 2013 12:07 pm

Seva Lamberdar wrote:
Jeremiah Mburuburu wrote:

sqrt(x^2) = x is false when x < 0.

LOL.

please read the example again.

by Seva Lamberdar Tue Apr 30, 2013 12:39 pm

Jeremiah Mburuburu wrote:
Seva Lamberdar wrote:
Jeremiah Mburuburu wrote:

sqrt(x^2) = x is false when x < 0.

LOL.
please read the example again.

That's where the problem lies in your thinking. You think that just because the examples involving numerical calculations by computers etc. are of certain form (e.g. sqrt of a number evaluated as another positive number), that should be considered as the basis for the mathematical operation (sqrt of a function as the absolute value of some variable or another function). That is not right.

by Jeremiah Mburuburu Tue Apr 30, 2013 3:52 pm

Seva Lamberdar wrote:
Jeremiah Mburuburu wrote:
Seva Lamberdar wrote:
Jeremiah Mburuburu wrote:

sqrt(x^2) = x is false when x < 0.

LOL.
please read the example again.

That's where the problem lies in your thinking. You think that just because the examples involving numerical calculations by computers etc. are of certain form (e.g. sqrt of a number evaluated as another positive number), that should be considered as the basis for the mathematical operation (sqrt of a function as the absolute value of some variable or another function). That is not right.

it was not merely a numerical example, but a counterexample; one counterexample is sufficient to prove that a statement is false. here, the case of x = -3, is sufficient to prove that the equation sqrt(x^2) = x is false.

you seem to be misinterpreting "sqrt()". sqrt(x) means the principal square-root of x: the non-negative value r for which r^2 = x. sqrt(x) is the same as x written under the radical symbol; the radical symbol is the one with a v on the left and horizontal tail above and to the right.

there is no key on the keyboard for a radical symbol, so we type "sqrt()" when we mean "the principal square-root of..." just as we type 3^2 when we mean 3-squared. this convention has been in use since the 50s or 60s when the fortran programming language was created.

the principal square-root of x has only one value, a non-negative one; for example, sqrt(9) = 3; by definition, it is never -3; the principal square-root of x is a real-valued function whose domain is the set of all non-negative real numbers, and whose range is the set of all non-negative real numbers. by definition, the range excludes the negative numbers.

a square-root of x - note that i didn't say "the square root of x" - is a number whose square is x; it may be positive or negative. it is a value of r that solves the equation r^2 = x. a square-root of 9 is 3; another is -3. a square-root of 9 is 3 or -3, i.e. the principal square-root of 9 or its opposite (negative). in general, a square-root of x is +/- the principal square-root of x.

the expression in rishi's problem reads sqrt(1 + x^2)/x, and that means: "the principal square-root of 1 + x^2, divided by 2." since the principal square-root is non-negative and 1 + x^2 is strictly positive, the expression is negative when x is negative, and positive when x is positive.

one reason for the misunderstanding may be that we often say "square-root of x" in most cases where we should be saying "the principal square-root of x." in order to avoid such a misunderstanding, many in the u.s. say "radical x" or "rad x" when they mean "the principal square-root of x."

by MaxEntropy_Man Tue Apr 30, 2013 4:18 pm

Jeremiah Mburuburu wrote:
a square-root of x - note that i didn't say "the square root of x" - is a number whose square is x; it may be positive or negative. it is a value of r that solves the equation r^2 = x. a square-root of 9 is 3; another is -3. a square-root of 9 is 3 or -3, i.e. the principal square-root of 9 or its opposite (negative). in general, a square-root of x is +/- the principal square-root of x.

i am quite amazed that you even have to make this simple point. to further make the point that this is a mathematical definition and not simply a computational convenience as seva seems to imagine, he can check "elementary algebra for schools" by hall and knight, a text many of us remember growing up on, written in 1885 (no computers or even a notion of one yet) and freely available online. on page 5 it makes the same point you have made above.

by MaxEntropy_Man Tue Apr 30, 2013 4:32 pm

i wasn't aware of this rad x thing in the US. good to know.

by Jeremiah Mburuburu Tue Apr 30, 2013 4:33 pm

MaxEntropy_Man wrote:
Jeremiah Mburuburu wrote:
a square-root of x - note that i didn't say "the square root of x" - is a number whose square is x; it may be positive or negative. it is a value of r that solves the equation r^2 = x. a square-root of 9 is 3; another is -3. a square-root of 9 is 3 or -3, i.e. the principal square-root of 9 or its opposite (negative). in general, a square-root of x is +/- the principal square-root of x.

i am quite amazed that you even have to make this simple point. to further make the point that this is a mathematical definition and not simply a computational convenience as seva seems to imagine, he can check "elementary algebra for schools" by hall and knight, a text many of us remember growing up on, written in 1885 (no computers or even a notion of one yet) and freely available online. on page 5 it makes the same point you have made above.

i think that the most important part of my post was the first paragraph about counterexamples. one could have destroyed the identity sqrt(x^2) = x as easily by citing the case of x = -1. all one needed was one counterexample.

by MaxEntropy_Man Tue Apr 30, 2013 4:40 pm

Jeremiah Mburuburu wrote:i think that the most important part of my post was the first paragraph about counterexamples. one could have destroyed the identity sqrt(x^2) = x as easily by citing the case of x = -1. all one needed was one counterexample.

indeed, but it is false only if one is ready to accept sqrt as a stand-in for the more wordy "the non-negative root". my prediction is that seva will refuse to accept that. the acceptance of the definition is the problem here.

there are only so many ways one can explain a simple point. if one wants to be deliberately argumentative and opaque, there is nothing to be done.

by Jeremiah Mburuburu Tue Apr 30, 2013 4:44 pm

MaxEntropy_Man wrote:i wasn't aware of this rad x thing in the US. good to know.

yeah dude, rad.

by Propagandhi711 Tue Apr 30, 2013 4:52 pm

Jeremiah Mburuburu wrote:
MaxEntropy_Man wrote:i wasn't aware of this rad x thing in the US. good to know.
yeah dude, rad.

ohhhhohohoho the diapered1 made a vitty. I mean it woulda been vitty in 1995

by Seva Lamberdar Tue Apr 30, 2013 4:54 pm

Max and Jeremiah,

Forget about the online books and the American using "rad x" etc. in trying to justify that sqrt of a number/variable/function is expressed as absolute value.

I have a Mathematical Handbook (by Korn et al., McGraw-Hill, Inc., 1968 edition) which on page 5 (Section 1.2-1) explicitly states the following.

"The real solutions of x^2 = a are written as +,- sqrt(a)".

The above implies that the expression for x in terms of square root of a is not just the absolute or positive value of sqrt(a) but includes values in both positive and negative domains (as + and -, respectively). Btw this notation (using both + and -, and not just + or "abs") for sqrt expressions is used all over the book.

by MaxEntropy_Man Tue Apr 30, 2013 7:20 pm

Seva Lamberdar wrote:
I have a Mathematical Handbook (by Korn et al., McGraw-Hill, Inc., 1968 edition) which on page 5 (Section 1.2-1) explicitly states the following.
"The real solutions of x^2 = a are written as +,- sqrt(a)".

in that case, let's try this again. earlier you were claiming:

"sqrt" of any positive number can be another positive or negative number (not necessarily +abs(x)).

in other words you were implying that writing "sqrt" is sufficient to refer to both roots. however now you are saying you are agreeing with a book that says:

"The real solutions of x^2 = a are written as +,- sqrt(a)".

in other words, you are agreeing with the view that it is necessary to insert a negative sign in front of sqrt(a) to refer to its negative square root. have you changed your mind?

your book does not contradict what i have always been saying, i.e. it is necessary to put a negative sign in front of sqrt(a) to refer to its negative root.

by the same token, the absence of a sign in front of the sqrt, as is the convention in writing any number, refers to the positive root, i.e. the principal root. for example, to refer to positive five, i don't have to write "+5"; it is sufficient to write "5".

by MaxEntropy_Man Tue Apr 30, 2013 7:33 pm

the book you referred to by korn & korn on page 5 says (http://www.amazon.com/Mathematical-Handbook-Scientists-Engineers-Definitions/dp/0486411478/ref=sr_1_1?ie=UTF8&qid=1367364279&sr=8-1&keywords=mathematical+handbook+korn):

if a is real and positive, many authors specifically denote the real positive solution values of x^2=a ......as √a....

by Seva Lamberdar Tue Apr 30, 2013 8:01 pm

MaxEntropy_Man wrote:the book you referred to by korn & korn on page 5 says (http://www.amazon.com/Mathematical-Handbook-Scientists-Engineers-Definitions/dp/0486411478/ref=sr_1_1?ie=UTF8&qid=1367364279&sr=8-1&keywords=mathematical+handbook+korn):

if a is real and positive, many authors specifically denote the real positive solution values of x^2=a ......as √a....

"if a is real and positive, many authors specifically denote the real positive solution values of x^2=a ......as √a... "

The above statement on page (5) of the Handbook for positive solution (value) of sqrt(a) is immediately followed (on page 5 of the Handbook) by a more general statement, as indicated earlier and in the following (covering both positive and negative domains),

"The real solutions of x^2 = a are written as +,- sqrt(a)".

Incidentally, note also on page (10) of the Handbook, the square root of 1 is indicated both as +1 and -1, i.e., sqrt(1) = +,- 1 (and not just +1 or abs(1))

by MaxEntropy_Man Tue Apr 30, 2013 8:06 pm

Seva Lamberdar wrote:
"The real solutions of x^2 = a are written as +,- sqrt(a)".

right, but earlier you were claiming that:

"sqrt" of any positive number can be another positive or negative number (not necessarily +abs(x)).

so have you changed your mind? in other words if "sqrt" itself stood for a positive or a negative number, why the need to write +,- sqrt(a)?

by Seva Lamberdar Tue Apr 30, 2013 8:31 pm

MaxEntropy_Man wrote:
Seva Lamberdar wrote:
"The real solutions of x^2 = a are written as +,- sqrt(a)".

right, but earlier you were claiming that:
"sqrt" of any positive number can be another positive or negative number (not necessarily +abs(x)).

so have you changed your mind? in other words if "sqrt" itself stood for a positive or a negative number, why the need to write +,- sqrt(a)?

What does the satement "The real solutions of x^2 = a are written as +,- sqrt(a)" mean to you? Does it cover both postive and negative values of x or only the positive (or absolute) values?

by MaxEntropy_Man Tue Apr 30, 2013 8:38 pm

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.

by Seva Lamberdar Tue Apr 30, 2013 8:52 pm

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.

by Marathadi-Saamiyaar Tue Apr 30, 2013 8:55 pm

What took you guys so long ?

by MaxEntropy_Man Tue Apr 30, 2013 9:02 pm

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.

by bw Tue Apr 30, 2013 10:04 pm

"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.

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