noformalities
New Member
- Joined
- Oct 16, 2018
- Messages
- 3
- Gender
- Male
- HSC
- N/A
In the HSC, would we ever be marked down if we forget to include the absolute values?
As integral95 said, you extend the log to the complex domain to solve the problem. The complex argument will always cancel out, leading to a real solution. So it essentially just a hack to fix the issue of extending the limits to the negative domain without using complex numbers. But the problem can be solved without absolute values, so there is technically no need for them.Hack??
1/x is defined for all real x except x=0, ln(x) is defined only for x>0. You need the abs value. Try the integral I of 1/x from x=-3 to x=-4 and see how it goes without the abs value. As an interesting question is I negative or positive?
If we used the above definition, we'd just get I = ln(-3) - ln(-4) = ln(3) + i*pi - (ln(4) + i*pi) = ln(3) - ln(4), which is the correct answer.Yes, we can extend the log function to C but what about the concept of the integral? Restricting contour integrals back down to the real line is fraught with technical issues. Let's see the evaluation of the simple integral I above using your method? It is certainly trivial using ln|x|.
What would the "proper" way of integrating be, without leaving ?schools justify it by saying "we want to incorporate x<0 so add absolute values", which is essentially a hack since their explanation is more or less along the lines of "it works and it fixes the issue".
I always justify things to my students. By understanding where it comes from, I think it helps them appreciate and understand the material better.It's not a hack. It is just a fact!
d/dx ln|x|=1/x for all non-zero x in exactly the same way as d/dx 1/x=-1/x^2 for all non zero x.
You need the abs value. There is no funny business here, just reality.
It doesn't need to be "justified"....it is just true.
Yes I'm not saying it's not true! I'm just saying that the way they teach it at school is by adding absolute values and showing it works, therefore it's true. I don't particularly agree with that type of teaching lol.how about
It doesn't work because it "fixes the issue" .......it works because it's true.
Yeh I think that's an interesting one haha. It's kind of like the harmonic series: 1 + 1/2 + 1/3 + ... , which looks like it converges since 1/n -> 0.A nice thing to think about while we are here:
The derivative of ln(x) approaches zero as x approaches infinity.
Therefore the ln(x) function has a tangent with arbitrarily small gradient as x goes to infinity.
Therefore ln(x) is horizontal in the limit.
Therefore ln(x) has a horizontal asymptote?
My understanding of the issue is that the integral of 1/x = log(x) = log|x| upon simplification after passing through the complex field (see discussion where the imaginary terms cancel out). So the result log|x| is correct, and the final result is in R (even though the proof to show this went outside of R).What would the "proper" way of integrating be, without leaving ?
This is something I noticed recently while working on a problem.A nice thing to think about here:
The derivative of ln(x) approaches zero as x approaches infinity.
Therefore the ln(x) function has a tangent with arbitrarily small gradient as x goes to infinity.
Therefore ln(x) is horizontal in the limit.
Therefore ln(x) has a horizontal asymptote?
I found that it was a more intuitive way to try to understand the paradox.Yes but it is still true that the tangent is arbitrarily close to the horizontal for sufficiently large x. There is no need to consider sums.
Yes, but if I differentiate ln(x), I also get 1/x, so one can argue that ln(x) is also the integral of 1/x. It will also have the same properties you described: ln(x) is also a real function, it can also solve for negative x values as I demonstrated earlier. It fits all the properties you describe, the only difference is that the absolute value version can be derived using complex numbers from the more general case of ln(x).No! not after simplification passing through the complex field. It has NOTHING to do with the complex field.
The derivative of ln|x| is 1/x therefore the integral of 1/x is ln|x|+C.
There is nothing here...it is a simple fact.
ln|x| is a real function...it is a real function whose derivative is 1/x...therefore the integral of 1/x is ln|x|+C
No complex field, no contour integration no nothing. The derivative of x^3 is 3x^2. Therefore the integral of 3x^2 is x^3. That is all that is going on. You really are doing your students a disservice by ploughing through complex numbers when it has nothing to do with complex numbers.
How about this!
Let y=ln|x|. Then y=ln(x) for x>0 and ln(-x) for x<0. Thus y'=1/x for x>0. For x<0 and via the chain rule
y'=1/(-x)(-1)=1/x. That is y'=1/x ALWAYS for non zero x. So the derivative of ln|x| is 1/x and thus the integral of 1/x is ln|x|
It... .has ..nothing.... to ....do... with .....complex..... numbers!.
Could you elaborate on why the domain restriction is necessary?The derivative of ln(x) is not 1/x much as you would like this to be true. If you want to say something along these lines you will need to define a function h with domain x>0 and definition h(x)=1/x for x>0. The the derivative of ln(x) is most certainly h(x).