Approximating ln(x): Taylor Series Problem Solution

[jsewell94]

Homework Statement

The first three terms of a Taylor Series centered about 1 for $ln(x)$ is given by:

$\frac{x^{3}}{3}$ - $\frac{3x^{2}}{2}$ + $3x$ - $\frac{11}{6}$

and that

$\int{ln(x)dx}$ = $xlnx - x + c$

Show that an approximation of $ln(x)$ is given by:

$\frac{x^3}{12}$ - $\frac{x^2}{2}$ + $\frac{3x}{2}$ - $\frac{5}{6}$ - $\frac{1}{4x}$


2. The attempt at a solution
I have tried this problem a few times, but it is becoming clear that I am missing some crucial step/idea. Basically, what I have tried is setting lnx equal to the Taylor Series, integrating both sides, and solving for lnx. However, when I do this, I manage to get all of the necessary terms EXCEPT for the 1/4x. Where does that come from, exactly? If someone could help, that'd be awesome! :D

Thanks!

 

[micromass]

I think you likely forgot the +C. C is a constant and should equal some number. What do you think that number is?

 

[jsewell94]

I think you likely forgot the +C. C is a constant and should equal some number. What do you think that number is?

Does c = 1 because the center is at 1?

 

[jsewell94]

Even if that is the case, I'm not sure how I would get 1/4x from that.

 

[micromass]

Does c = 1 because the center is at 1?

No.

What do you get after you subsitute ln(x) with the polynomial and integrate it??

 

[jsewell94]

Please don't judge my slowness :( I'm really not bad at math. I just haven't done any of my homework, lol.

 

[jsewell94]

No.

What do you get after you subsitute ln(x) with the polynomial and integrate it??

After you integrate the polynomial, you get:

$\frac{x^4}{12}-\frac{x^3}{2}+3x^2-\frac{11x}{6} + c$

Which is what I did, and then set it equal to $xln(x)-x+C$

But every time I do that, I get the wrong thing :( (meaning, I don't get the 1/4x)

I feel like I'm misunderstanding the nudges that you are giving me, lol.

 

[micromass]

After you integrate the polynomial, you get:

$\frac{x^4}{12}-\frac{x^3}{2}+3x^2-\frac{11x}{6} + c$

OK, that's good (although the middle term should be $\frac{3x^2}{2}$)

Which is what I did, and then set it equal to $xln(x)-x+C$

But every time I do that, I get the wrong thing :( (meaning, I don't get the 1/4x)

I feel like I'm misunderstanding the nudges that you are giving me, lol.

What you should get is

$$\frac{x^4}{12}-\frac{x^3}{2}+\frac{3x^2}{2}-\frac{11x}{6} + c = x ln(x)-x+C$$

Yes, the constant of integration c and the other constant C are distinct in general, so you can't eliminate them! This is likely your mistake.

The above is equivalent to:

$$\frac{x^4}{12}-\frac{x^3}{2}+\frac{3x^2}{2}-\frac{11x}{6} = x ln(x)-x+(C-c)$$

Can you figure out what number C-c is?? (we're not interested in the specific values of C and c here, we just want a number instead of constants).

 

[jsewell94]

OK, that's good (although the middle term should be $\frac{3x^2}{2}$)

That's what I actually got, I just mistyped it :/

Yes, the constant of integration c and the other constant C are distinct in general, so you can't eliminate them! This is likely your mistake.

I know :D That's why I denoted one as c and the other as C.

Can you figure out what number C-c is?? (we're not interested in the specific values of C and c here, we just want a number instead of constants).

This is where I am confused..Like, I'm not trying to sound annoyingly difficult, but I honestly have no idea what to do with the C-c. :( I understand that it is just some number, but I have absolutely no clue what that number is or how to get it.

My confidence in my math skills is quickly plummeting :(

 

[jsewell94]

I mean, I assume it is 1/4, because that is what I am trying to prove. But what is the reasoning behind that?

 

[micromass]

Maybe you can substitute in some value for x and see what you get??

For example, if I want to determine a constant C such that

$$\sin(x)=x^2+x+C$$

Then I can substitute in x=0 and get

$0=C$.

Can you do something like that to determine C-c?

 

[jsewell94]

Wow..Oh my freaking gosh, I think I am going to go into a corner and cry. I am so stupid :(

Thanks for the help :( I got it now :(

 

[micromass]

Wow..Oh my freaking gosh, I think I am going to go into a corner and cry. I am so stupid :(

Thanks for the help :( I got it now :(

Don't worry about it. Yes, it is a trivial problem, but it's one of these things you had to see before knowing you could do that.

Instead of feeling depressed, you should feel happy because you found a new technique! I bet that next time (on a test perhaps), you won't forget how to do this!

 

[jsewell94]

That's not even a "new technique." That is basic, elementary calculus that everyone should know how to do. Plugging in a given value and solving for C is something that every Calc student learns the moment they learn integration. I am in calculus 2 and didn't even consider it.

AKA, I think my feelings of idiocy are justified :P