The discussion revolves around finding the Maclaurin polynomial of the third order for the function ln(cos(x)). Participants explore the nature of Maclaurin and Taylor series, the implications of polynomial order, and the infinite series representations of cos(x) and ln(x).
The discussion is active, with participants providing hints and guidance on how to approach the problem. There are multiple interpretations of how to derive the series, and some participants question the necessity of certain steps while others suggest alternative methods.
Participants note that the problem is posed for educational purposes, and there is a mention of the derivatives of ln(cos(x)) at x=0 as a potential alternative approach. The discussion includes considerations of what constitutes a third-degree polynomial representation.
MacLaurin is a Taylor series centered about x=0.carlodelmundo said:[*]What is the difference between MacLaurin vs. Taylor Series? (Hint: Is MacLaurin centered at any point?)
MacLaurin is an infinite sum, when we talk about order 3 we take only first 3 sums.carlodelmundo said:[*]What does it mean for a polynomial to have order 3? (Hint: Look at the exponent)
I know the series for cos(t).carlodelmundo said:[*]What is the infinite series equivalent of cos (x) and ln (x)? (Hint: Look it up in a textbook)
carlodelmundo said:[*]What does it mean when a function is a composite of two functions? (Hint: f(g(x)) is a composition of two functions. The input of f(x) is g(x).)
[*]Can we multiply basic infinite series (such as sin(x), cos(x)) to create a new function? (Hint: yes)
[*]Can we take the composition of two series and create a new series? (Hint: yes)
From this, the problem is trivial. You can arrive to the solution in no time.
As you said, if you let t = cos x - 1. Then naturally you'll get r3(p3(x)). Note that you forgot to cancel out the -1 (ie: 1 + -1 = 0 when substituting t for cos x - 1).This is where I stuck, suppose p_3(x) is 3rd degree series of cos(x) and r_3(t) is 3rd degree series for ln(1+t).
It seems logical for a 3rd order series for ln(cosx) to be something like this:
r_3(p_3(x)-1) but from this equation I get very complicated expression that seems to me wrong.
carlodelmundo said:Note that you forgot to cancel out the -1 (ie: 1 + -1 = 0 when substituting t for cos x - 1).
carlodelmundo said:True. The expression is indeed complicated, but remember: the question asks you to write the series to the third order.
Hint: Write the first 4 or 5 terms of cos (x). Then use composition to transform the polynomial values of cos(x) for ln(x).
carlodelmundo said:Keep this in mind also: The question doesn't specify to find the composite of two MacLaurin Series (ie: ln(cos(x))). It just says to find the third degree polynomial that represents this function. Technically, you can just write the first 4 terms of cos(x) and natural log each term.
Char. Limit said:Wouldn't it be easier just to compute the values of the derivatives of ln(cos(x)) at x=0 than do all of that rigomarole? It's just four equations...