Find ln using differentials.

In summary, the conversation discusses a method for estimating ln10.3 using differentials, specifically through the use of linear approximations. The process involves finding the tangent line at ln10 and using that to approximate the value of ln10.3. It is also referred to as a linear approximation or a linearization.
  • #1
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So the problem says "Taking ln10=2.30 estimate ln10.3 using differentials."

The only thing I thought of doing was to set an integral from 1 to 10.3 with 1/x being the equation I integrate, but I'm not sure if that is right. Any help?
 
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  • #2
What is a derivative of a function?

It is the limit

[tex]\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex]

So the smaller the [itex]\Delta x[/itex], the nearer the ratio

[tex]\frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex]

actually is to f '(x).

So solving for [itex]f(x+\Delta x)[/itex] in

[tex]f'(x)\approx\frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex]

gives an approximation of [itex]f(x+\Delta x)[/itex] that is as good as [itex]\Delta x[/itex] is small.
 
  • #3
Alright, so I need to find f(x+h) which would be f(10+0.3), with 0.3 being change in x. Using my notes...I see that I could make my equation look like f(x+h)=f(x+(1+(h/x))), where I would break it so that it looks like f(x+h)=f(x)+f(1+(h/x)). I would then substitute 2.30 for f(x), 0.3 for h and 5 for x and get 2.33? It worked on the previous problem (which wasn't assingned) so I gues it should work here as well. Thanks a lot for the help.:smile:
 
  • #4
[tex]f'(x)\Delta x+f(x)\approx f(x+\Delta x)[/tex]

with f=ln(x), f '(x)=1/x so

[tex]\ln(10.3)\approx\frac{1}{10}(0.3)+2.30 =2.33[/tex]

Yep.
 
  • #5
Look up linear approximations, that is what you're doing.

The idea behind it is that the tangent drawn to a curve is very close to the curve at values near the tangent point, so instead of trying to calculate an imposible calculation such as ln 10.3, you calculate the value of the tangent line at ln 10 for x=10.3. So yea, just as quasar said, I just wanted to give you the name of the process so that you can look it up. :smile:
 

1. How do I find ln using differentials?

To find ln using differentials, you can use the formula ln(x+dx) = ln(x) + dx/x. This formula allows you to approximate the value of ln(x) by adding the differential dx to the value of ln(x).

2. What is the purpose of using differentials to find ln?

The purpose of using differentials to find ln is to approximate the value of ln(x) when x is very close to 1. This method is useful for calculating small changes in natural logarithms, which can be difficult to do accurately using traditional methods.

3. Can I use differentials to find ln for any value of x?

No, differentials can only be used to approximate ln for values of x that are very close to 1. If x is too far from 1, the approximation will become less accurate.

4. How do I know if my ln approximation using differentials is accurate?

You can check the accuracy of your ln approximation by calculating the percentage error. This can be done by finding the difference between your approximation and the actual value of ln, and then dividing that difference by the actual value of ln. A smaller percentage error indicates a more accurate approximation.

5. Are there any other methods for finding ln besides using differentials?

Yes, there are other methods for finding ln such as using Taylor series or using the logarithm rules. However, using differentials is a quick and easy method for approximating ln for values close to 1.

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