Absolute value question

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if h=ln(absolute x)
then how do you calculate e^h?
 

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TD
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If [itex]h = \ln \left| x \right|[/itex] then [itex]e^h = e^{\ln \left| x \right|} = \left| x \right|[/itex].

Seems simple, do you mean this?
 
734
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that's what i originally thought,
but the correct answer doesn't need absolute value...
i think that's strange...
 
VietDao29
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It must be |x|. Have you ever seen:
[tex]e ^ {\mbox{something}} = \mbox{a negative number}[/tex]??
Viet Dao,
 
lurflurf
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VietDao29 said:
It must be |x|. Have you ever seen:
[tex]e ^ {\mbox{something}} = \mbox{a negative number}[/tex]??
Viet Dao,
yes

[tex]e^{\pi i}=-1[/tex]
 
rachmaninoff
Which brings up an important point, [itex]e^{ \ln x }[/itex] depends on whether we're working in the Reals or the Complexes. The real logarithm is undefined for x < 0.
 
VietDao29
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lurflurf said:
yes

[tex]e^{\pi i}=-1[/tex]
:rolleyes:
Okay, my something is [tex]\in \mathbb{R}[/tex], not C :biggrin:.
Viet Dao,
 
lurflurf
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asdf1 said:
that's what i originally thought,
but the correct answer doesn't need absolute value...
i think that's strange...
What were the preceding steps? This type of thing happens alot when what you were finding was the absolute value of something and droping it gives the something.
for instance if
xy'=y y(0)=0 (the solution being y=x or y=-x)
one might get
|y|=|x| as an answer but though error have
y=|x|
 
TD
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asdf1 said:
that's what i originally thought,
but the correct answer doesn't need absolute value...
i think that's strange...
Well, you can always take a numerical example if you're in doubt...:

[tex]\begin{array}{l}
e^{\ln \left( { - 1} \right)} = - 1 \\
e^{\ln \left| { - 1} \right|} = 1 \\
\end{array}[/tex]
 
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TD said:
Well, you can always take a numerical example if you're in doubt...:

[tex]\begin{array}{l}
e^{\ln \left( { - 1} \right)} = - 1 \\
e^{\ln \left| { - 1} \right|} = 1 \\
\end{array}[/tex]
This is incorrect. Assuming we're working in the reals [itex]e^{\ln(-1)}[/itex] isn't defined, because [itex]\ln[/itex] is undefined for negative numbers and so any expression including it is also undefined.
 
TD
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My example wasn't necessarily over the reals, I don't see why we'd have to exclude to complex numbers - at least not for this example which is only intended to show that abs values are needed.
 
lurflurf
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TD said:
My example wasn't necessarily over the reals, I don't see why we'd have to exclude to complex numbers - at least not for this example which is only intended to show that abs values are needed.
Again where how did this arise. One way absolute values often arise in arguments of logarithums is when one wants an antiderivative of a logarothmic derivative that is restricted to reals and valid almost every where.
ie
xy'=1 y(e)=1
y=log(|x|)
clearly if we are working with complex numbers
y=log(x) would do
and sometimes errors arise as when
xy'=y y(1)=1
clearly
y=x
when restricted to real one often finds
|y|=exp(log(|x|))=|x|
but a common error is to get
y=|x| which satisfys the differential equation every where except the origin
if you log(|x|) did not arise though integrating a logarithmic derivative perhaps some other error was made to cause you to get |h| when h was wanted (or mayber h>0 so it does not matter).
 
734
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Originally, I was trying to solve this problem:
dy/dx + ytanx = sinx, y(0)=1

The first step I did was to integrate tanx and then there came the absolute value dilema...
 
lurflurf
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asdf1 said:
Originally, I was trying to solve this problem:
dy/dx + ytanx = sinx, y(0)=1

The first step I did was to integrate tanx and then there came the absolute value dilema...
So it was due to integrating a logarimic derivative.
Say you were to solve th associated homogeneous problem in preparation for variation of parameters.
y'+y tan(x)=0
y'/y=cos'(x)/cos(x)
log(|y|)=a+log(|cos(x)|)
|y|=|C||cos(x)| then let C absorb the sign
y=C*cos(x)
which is clearly a solution
of course if we were working with complex numbers we would never have introduced the absolute value. This problem arises though finding the function. It has nothing to do with the function itself.
 
HallsofIvy
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TD said:
My example wasn't necessarily over the reals, I don't see why we'd have to exclude to complex numbers - at least not for this example which is only intended to show that abs values are needed.
Because the original poster said his question only concerned real numbers and to go outside that without specifically saying so can be confusing.
 
TD
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Sorry, my bad then although I nowhere saw asdf1 posting before me that it concerned the reals, only Vietdao.
 
VietDao29
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So isn't it obvious that:
[tex]\ln|x| \in \mathbb{R} \mbox{, \forall x \in \mathbb{R} - \{0\}}[/tex]?????
Viet Dao,
 
734
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so llurflurf you mean that there is supposed to be an absolute value sign, except that it's already included in "c"?
 
734
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the constant in this question doesn't seem to absorb...

lurflurf said:
So it was due to integrating a logarimic derivative.
Say you were to solve th associated homogeneous problem in preparation for variation of parameters.
y'+y tan(x)=0
y'/y=cos'(x)/cos(x)
log(|y|)=a+log(|cos(x)|)
|y|=|C||cos(x)| then let C absorb the sign
y=C*cos(x)
which is clearly a solution
of course if we were working with complex numbers we would never have introduced the absolute value. This problem arises though finding the function. It has nothing to do with the function itself.
that question seems to work, but in this question:
y`= -y/x, y(1)=1
the same problem arises, but the constant doesn't absorb the negative sign...
it's weird~
 
lurflurf
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asdf1 said:
that question seems to work, but in this question:
y`= -y/x, y(1)=1
the same problem arises, but the constant doesn't absorb the negative sign...
it's weird~
What do you mean
y'=-y/x y(1)=1
y'/y=-1/x
log(|y|)=-log(|x|)+a=log(1/|x|)+a
|y|=exp(a)/|x|
|y|=|c|/|x|
y=c/x
y(1)=c/1=c=1
y=1/x
 
734
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crud! the "c" does absob the positive/negative sign!!!
sorry~
i made a mistake in the calculations...
 
734
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thanks for correcting my mistake~
 

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