Re: Positive/Negative after taking the square root

Joe wrote :

I'm not totally mathematically inept, but I find
I'm
getting confused
by the following:

I'm confused as to what happens to the +/- in the
following examples:

As a simple example, integrate x^2

int{x^2 dx}

Let x^2 = u, then dx = du/(2x) and du = 2x, dx =
+/-
2sqrt{u}

Then the integral becomes
=int{u/(+/-)2sqrt{u} du}
= +/- int{1/2sqrt{u} du}
= +/- 1/3u^1.5
= +/ 1/3x^3 (substitute back in for u)

This is obviously not correct, and the correct
answer
is in taking
only the positive square root when representing dx
interms of u and
du. However, is there a mathematical/logical
reason
for discounting
the negative root? After all, if y^2 = x, then y =
+/-sqrt{x}

A related problem I encountered when deriving the
derivative of
arcsin:

y = arcsinx
x = siny

I have y' = 1/cosy, and wish to write y in terms of
x

Now, sin^2y + cos^2y = 1

siny = x

thus x^2 + cos^2y = 1

cos^2y = 1-x^2

cosy = +/- sqrt{1-x^2}


Thus

y1 = +/- 1/sqrt{1-x^2}

Obviously, again this is incorrect. The correct
answer is obtained by
discounting the negative root - again, what is the
mathematical/
logical reason for this?

There are two common mistakes made here that lead to
the confusion:

One: The expression u^2 = v does NOT imply u = +/-
sqrt(v). This abuse of notation doesn't really even
mean anything mathematically, and it only serves to
create ambiguity. What it DOES imply is |u| = sqrt(v)
or equivalently sgn(u)*u = sqrt(v).

agreed.



Two: You cannot make a substitution u for x which has
u' = 0 for some x; in general if you make such a
substitution the result is not valid. This subtle
condition may or may not be covered in your calculus
class, but the reasoning behind it will be clear when
you take analysis.

can you give examples of that ?

is this about avoiding integral from 0 to 0 ?

i assume here you mean for the given example ...

but this is no longer true for single-valued right ?

also ; can this be used to prove or disprove there are zero's on a line or strip ? ( RH :p )



To resolve problem 1 take the proper expression
sgn(x)*x = sqrt(u) and perform the substitution, then
we have formally dx = 2*sqrt(u)/sgn(x)*du.

this indeed works.


To
resolve problem 2, we need to demand that x =/= 0 on
the interval of integration [a,b], i.e. a and b are
of the same sign. Thus the integral becomes:

int[a,b]{x^2dx}
= int[a^2,b^2]{sgn(x)*u/(2*sqrt(u))du}
= int[a^2,b^2]{sgn(x)*u/(2*sqrt(u))du}
= sgn(x)*int[a^2,b^2]{u/(2*sqrt(u))du}
= sgn(x)*1/3*sqrt(u^3)
= sgn(x)*1/3*sqrt(b^6)-sgn(x)*1/3*sqrt(a^6)
= sgn(x)*1/3*(|b^3|-|a^3|)
= 1/3*(b^3-a^3)

which is the correct evaluation as expected.

regards
tommy1729
.

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