Understanding Natural Log Problem: Taylor Series Approach

[mjordan]

Homework Statement

y and x are both close to 1.
[tex]ln(y) - ln(x) \approx (y-x)(lnx)'[/tex]

Can someone explain me why this is true?

By the way, (lnx)' is the derivative of lnx, which is just 1/x

The Attempt at a Solution

I guess this is something to do with the taylor series of ln(x). I tried to expand ln(y) and ln(x) by taylor series, but I did not get anything from there.

[tex]((y-1)-\frac{(y-1)^{2}}{2}+\frac{(y-1)^{3}}{3}...)-((x-1)-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{3}...)[/tex]

 

[Mark44]

I would say it's closer to the limit definition of ln(x), using a difference quotient.

[tex]d/dx(ln(x))~=~\lim_{h \rightarrow 0} \frac{ln(x + h) - ln(x)}{h}[/tex]
For suitably small h,
[tex]d/dx(ln(x))~\approx~ \frac{ln(x + h) - ln(x)}{h}[/tex]

Now, let y = x + h, and you're almost where you want to go.

BTW, this should be in the Calculus and Beyond section, not Precalculus.

 

[vorcil]

I would say it's closer to the limit definition of ln(x), using a difference quotient.

[tex]d/dx(ln(x))~=~\lim_{h \rightarrow 0} \frac{ln(x + h) - ln(x)}{h}[/tex]
For suitably small h,
[tex]d/dx(ln(x))~\approx~ \frac{ln(x + h) - ln(x)}{h}[/tex]

Now, let y = x + h, and you're almost where you want to go.

BTW, this should be in the Calculus and Beyond section, not Precalculus.

No it's algebra

ln(x^3) - ln(x) = 10
then x=e^[something]

I don't know how to solve for it
and would also like to know the algebraic approach to solving
ln(x^n) - ln(x) = some integer
then x = e^[some thing]

 

[Mark44]

No it's algebra

If this is an algebra problem, how do you explain the derivative in it?

Don't hijack an existing thread with a new problem. Start a new thread.

ln(x^3) - ln(x) = 10
then x=e^[something]

I don't know how to solve for it
and would also like to know the algebraic approach to solving
ln(x^n) - ln(x) = some integer
then x = e^[some thing]

 

[vorcil]

If this is an algebra problem, how do you explain the derivative in it?

I have no idea you tell me,
All I know is, that it was in my algebra test - university first year mathematics

and I'm pretty sure my university wouldn't accidently put a calculus question in an algebra exam

you're probably right though, there is a calculus and an algebraic approach
and seeing as this was in a first year algebra exam the awnser would be simple and elegant to get too

 

[Mentallic]

seeing as this was in a first year algebra exam the awnser would be simple and elegant to get too

I took another look at the problem after getting some inspiration from this line of yours

ok, simple and elegant? Yes, it struck me.
[tex]ln(y) - ln(x) \approx (y-x)(lnx)'[/tex]

[tex]RHS=\frac{y-x}{x}[/tex] since [tex](lnx)'=\frac{1}{x}[/tex] (yes, derivatives aren't precalculus, but this one was simple)

[tex]=\frac{y}{x}-1[/tex]

therefore, [tex]ln(y)-ln(x) \approx \frac{y}{x}-1[/tex]

[tex]ln\left(\frac{y}{x}\right) \approx \frac{y}{x}-1[/tex]

[tex]\frac{y}{x} \approx e^{\frac{y}{x}-1}[/tex]

For [itex]x,y\approx 1[/itex]

[tex]\frac{y}{x}\approx 1[/tex]

[tex]e^{\frac{y}{x}-1}\approx e^{1-1} \approx 1[/tex]

Is this sufficient?

 

[Mark44]

I took another look at the problem after getting some inspiration from this line of yours

ok, simple and elegant? Yes, it struck me.
[tex]ln(y) - ln(x) \approx (y-x)(lnx)'[/tex]

From your first post, what you were supposed to do was show that ln(y) - ln(x) can be approximated by (y - x) times the derivative of the ln function. Once you have shown that, you're done.

[tex]RHS=\frac{y-x}{x}[/tex] since [tex](lnx)'=\frac{1}{x}[/tex] (yes, derivatives aren't precalculus, but this one was simple)

[tex]=\frac{y}{x}-1[/tex]

therefore, [tex]ln(y)-ln(x) \approx \frac{y}{x}-1[/tex]

[tex]ln\left(\frac{y}{x}\right) \approx \frac{y}{x}-1[/tex]

[tex]\frac{y}{x} \approx e^{\frac{y}{x}-1}[/tex]

For [itex]x,y\approx 1[/itex]

[tex]\frac{y}{x}\approx 1[/tex]

[tex]e^{\frac{y}{x}-1}\approx e^{1-1} \approx 1[/tex]

e^{1 - 1} = e⁰ = 1 exactly. These are not approximations.

Is this sufficient?

It was unnecessary. What was sufficient was showing that ln(y) - ln(x) [itex]\approx[/itex] (y - x)(ln(x))'. After you have done what a problem asks, it's fine to continue exploring different avenues for your own enlightenment or enjoyment, but the problem doesn't require this.

 

[Mark44]

I have no idea you tell me,
All I know is, that it was in my algebra test - university first year mathematics

and I'm pretty sure my university wouldn't accidently put a calculus question in an algebra exam

Then maybe you are mistaken about what is being presented in your university class. I don't know where your university is, but in many U.S. universities, the first-year, college-level (not remedial) math classes start with calculus or precalculus, with the assumption being that a student took algebra and geometry and trig in high school. The straight algebra classes are essentially remedial classes, with class numbers below the 100 level. That's how it is in Washington state, where I live, and I believe that this is true also in many other states.

you're probably right though, there is a calculus and an algebraic approach
and seeing as this was in a first year algebra exam the awnser would be simple and elegant to get too

There is no straight algebraic approach to this problem, because the concept of the derivative is not part of algebra - it is strictly a calculus concept.

 

[mjordan]

I apologize for posting this on a wrong section. But thank you guys, I perfectly understand it now.

 

[Mentallic]

e^{1 - 1} = e⁰ = 1 exactly. These are not approximations.

Yes e⁰=1 exactly, but the fact that the question says for x,y[itex]\approx[/itex]1 then y/x[itex]\approx[/itex]1 therefore e^y/x-1[itex]\approx[/itex]e^1-1

no?

 

[Mark44]

Sure, that part is an approximation.