Absolute value question

  • Thread starter asdf1
  • Start date
  • #1
asdf1
734
0
if h=ln(absolute x)
then how do you calculate e^h?
 

Answers and Replies

  • #2
TD
Homework Helper
1,022
0
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?
 
  • #3
asdf1
734
0
that's what i originally thought,
but the correct answer doesn't need absolute value...
i think that's strange...
 
  • #4
VietDao29
Homework Helper
1,426
3
It must be |x|. Have you ever seen:
[tex]e ^ {\mbox{something}} = \mbox{a negative number}[/tex]??
Viet Dao,
 
  • #5
lurflurf
Homework Helper
2,453
149
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]
 
  • #6
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.
 
  • #7
VietDao29
Homework Helper
1,426
3
lurflurf said:
yes

[tex]e^{\pi i}=-1[/tex]
:rolleyes:
Okay, my something is [tex]\in \mathbb{R}[/tex], not C :biggrin:.
Viet Dao,
 
  • #8
lurflurf
Homework Helper
2,453
149
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 a lot 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|
 
  • #9
TD
Homework Helper
1,022
0
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]
 
  • #10
master_coda
591
0
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.
 
  • #11
TD
Homework Helper
1,022
0
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.
 
  • #12
lurflurf
Homework Helper
2,453
149
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).
 
  • #13
asdf1
734
0
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...
 
  • #14
lurflurf
Homework Helper
2,453
149
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.
 
  • #15
HallsofIvy
Science Advisor
Homework Helper
43,021
970
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.
 
  • #16
TD
Homework Helper
1,022
0
Sorry, my bad then although I nowhere saw asdf1 posting before me that it concerned the reals, only Vietdao.
 
  • #17
VietDao29
Homework Helper
1,426
3
So isn't it obvious that:
[tex]\ln|x| \in \mathbb{R} \mbox{, \forall x \in \mathbb{R} - \{0\}}[/tex]?
Viet Dao,
 
  • #18
asdf1
734
0
so llurflurf you mean that there is supposed to be an absolute value sign, except that it's already included in "c"?
 
  • #19
asdf1
734
0
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~
 
  • #20
lurflurf
Homework Helper
2,453
149
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
 
  • #21
asdf1
734
0
crud! the "c" does absob the positive/negative sign!
sorry~
i made a mistake in the calculations...
 
  • #22
asdf1
734
0
thanks for correcting my mistake~
 

Suggested for: Absolute value question

  • Last Post
Replies
3
Views
590
  • Last Post
Replies
2
Views
781
Replies
22
Views
878
  • Last Post
Replies
1
Views
405
  • Last Post
Replies
19
Views
485
  • Last Post
Replies
2
Views
526
  • Last Post
Replies
5
Views
614
Replies
3
Views
453
  • Last Post
Replies
15
Views
1K
  • Last Post
Replies
4
Views
592
Top