Expert: Paul Klarreich - 2/3/2011

Question
In trying to refresh for a course in advanced math going back to my calc 2 Taylor Series math.  However, I appear a bit rusty.  I can do simple lineaizations of functions, but not more complicated ones.  For example, take the function:

(3 +5ln(1 + 4sin2x))^(1/2)   

Just so its understood, (3 +5ln(1 + 4sin2x)) is UNDER a square root.

I want to linearize this for SMALL x.

I want to do this step-by-step linearization.

Here’s what I do so far:  

Sin2x ~ x

But I get confused now what to do with the ln(1+x), and ultimately what I do afterwards to get at least the linearized function.  The Taylor series part of it is easy.

Hope you can help.

Answer
Questioner: Mike
Country: United States
Category: Calculus
Private: No
Subject: Calculus Linearization of a Function at Small x - Taylor Series
Question: In trying to refresh for a course in advanced math going back to my calc 2 Taylor Series math.  However, I appear a bit rusty.  I can do simple lineaizations of functions, but not more complicated ones.  For example, take the function:

(3 +5ln(1 + 4sin2x))^(1/2)   

Just so its understood, (3 +5ln(1 + 4sin2x)) is UNDER a square root.

I want to linearize this for SMALL x.

I want to do this step-by-step linearization.

Here’s what I do so far:  

Sin2x ~ x

But I get confused now what to do with the ln(1+x), and ultimately what I do afterwards to get at least the linearized function.  The Taylor series part of it is easy.

Hope you can help.
..........................................
I thought that you are really supposed to do:

f(x) = f(0) + f'(0) dx, with the dx being your small x.

But I see that you want to use standard approximations for certain things instead.

OK, then ---

For starters,  sin x <-> x, so,  sin 2x <-> 2x, not x.

Then:

(3 +5ln(1 + 4sin2x))^(1/2) -->

(3 +5ln(1 + 4(2x)))^(1/2) -->

(3 +5ln(1 + 8x))^(1/2) -->


Now for ln(1 + 8x) do:


f(x) = f(0) + f'(0) dx

ln(1 + 8x) = ln(1 + 0) + 8/(1 + 0) x

ln(1 + 8x) = 0 + 8x = 8x

So you are up to:

(3 +5ln(1 + 8x))^(1/2) -->

(3 +5(8x))^(1/2) -->

(3 +40x)^(1/2) -->

At this point, do it again: (I would use the binomial expansion, myself.)

f(x) = (3 +40x)^(1/2)

f'(x) = 1/2 (3 +40x)^(-1/2)(40)

f'(x) = 20 / (3 +40x)^(1/2)

f'(0) = 20 / (3)^(1/2)

I think that does it -- just do your

f(x) = f(0) + f'(0) dx

stuff again.