Question

f(x) = ln(x) - ln(1/x) has cancellation when x approaches x = 1. What can I do about that, and how can I fix it using taylor series/trig identities/rationalization?

Answer 1

I need to remove the cancellation problem, but I don't know how.

Answer 2

Hmm I'm a little confused what you're asking for.
Can't we simply use rules of logarithms:\[\Large f(x)\quad=\quad \ln x + \color{#3366CF}{\ln\frac{1}{x}}\]Becomes:\[\Large f(x)\quad=\quad \ln x + \color{#3366CF}{\ln1-lnx}\]
Which simplifies to,\[\Large f(x)=0, \qquad x\ne0\]
Or is that not what you're trying to do? :o
Maybe a little clarification.

Answer 3

I'm a little confused as well :) So when I graph f(x) = ln(x) + ln(1/x), I get a table that has y=errors for x=infinity to x = 0, and then x =1 gives me y = 0, and x= 2 gives me x = 1.3863. The cancellation I believe is when x approaches x = 1, and y all of the sudden converges to 0. According to my homework problem, there is a way to reduce/eliminate this problem so that I can see more values for smaller values of x, including x = -1, I'm guessing? Except I'm not sure how to do that. the problems says to "propose a reformulation that removes the problem" using Taylor series, trig identities, rationalization, etc.

Answer 4

Getting errors? :O Hmm strange.
My calculator isn't giving me any trouble.
https://www.desmos.com/calculator/qvyht7dgjn

Answer 5

hmm

Answer 6

I think what we did would solve the problem.
It falls under the `etc.` though lol.

It certainly fixes the problem at least :o

Answer 7

x=2 should give you zero also XD Hmm are you putting it in your calculator correctly?

Answer 8

Maybe I'm incorrect in saying where the problem lies? Maybe it's not as x approaches x = 1? I have found the answer online, and it says it cancels near x =1, that I need to use the properties of logs, and that the reformulation would be f = ln x^2. I don't even know how to get there.

Answer 9

Oh oh oh, did you paste the problem incorrectly?
Should it have been:\[\Huge f(x) = \ln(x) \color{red}{-} \ln(1/x)\]
With subtraction in the middle?
Because that would make more sense.

Answer 10

Yes, oh, did I input it wrong?

Answer 11

Yep, I did. So Sorry!

Answer 12

could i get help on my problem please??

Answer 13

The rule of logarithms that we're using is:\[\large \color{orangered}{\log\left(\frac{a}{b}\right) \quad=\quad \log(a)-\log(b)}\]

Answer 14

\[\Large f(x)=\ln x\color{#3366CF}{-\ln\frac{1}{x}}\]Gives us,\[\Large f(x)=\ln x\color{#3366CF}{-\left(\ln1-\ln x\right)}\]Understand how that works? :o

Answer 15

That I do understand.

Answer 16

Oh, and then ln(1) = 1, so you are just left with ln(x^2). But how does this work? What the significance of me having to rewrite it? ln(x^2) just makes and identical graph on the opposite side of the y-axis. How does that solve my cancellation problems (even thought the answer is right?

Answer 17

ln(1)= 0, so you are just left with ln(x^2)*

I think that's what you meant to say lol :)

Answer 18

hmm

Answer 19

yep, sorry lol

Answer 20

`f(x) = ln(x) - ln(1/x) has cancellation when x approaches x = 1.`

This is a weird problem.
Like... they worded it as if there was a discontinuity at x=1 or something.
But it's still giving us f(1)=0 even after we simplify it down.

Not exactly sure what they're wanting here :3 lol weird...

Answer 21

Yeah, I'm getting the math, but there's no point if I don't know what's going on. Thanks for your help though! At least now I can know to refer to my log identities when I get other log problems like this.