Estimating ln using Linear Approximations

[ludi_srbin]

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?

 

[quasar987]

What is a derivative of a function?

It is the limit

$$\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

So the smaller the $\Delta x$, the nearer the ratio

$$\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

actually is to f '(x).

So solving for $f(x+\Delta x)$ in

$$f'(x)\approx\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

gives an approximation of $f(x+\Delta x)$ that is as good as $\Delta x$ is small.

 

[ludi_srbin]

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.

 

[quasar987]

$$f'(x)\Delta x+f(x)\approx f(x+\Delta x)$$

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

$$\ln(10.3)\approx\frac{1}{10}(0.3)+2.30 =2.33$$

Yep.

 

[dontdisturbmycircles]

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.