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...
A Maclaurin polynomial is a special type of polynomial that is used to approximate a function. It is defined as the Taylor series of a function centered at x=0, and is often used in calculus to approximate functions that are difficult to integrate or differentiate.
To calculate the Maclaurin polynomial of 3rd order for ln(cosx), you will need to use the formula: f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3. In this case, f(x) = ln(cosx), so you will need to find the first three derivatives of ln(cosx) and evaluate them at x=0. Then, plug these values into the formula to get the Maclaurin polynomial.
The Maclaurin polynomial of 3rd order is used for ln(cosx) because it provides a good approximation of the function in the interval around x=0. Since ln(cosx) is a transcendental function, it is difficult to integrate or differentiate, making the Maclaurin polynomial a useful tool for approximating its values.
The 3rd order in the Maclaurin polynomial for ln(cosx) indicates that the polynomial will include terms up to x^3. This means that the polynomial will be a cubic function, and it will provide a better approximation of ln(cosx) compared to a lower order polynomial. However, using a higher order polynomial may not significantly improve the accuracy of the approximation.
No, the Maclaurin polynomial of 3rd order is only an approximation of ln(cosx), not the exact value. It will provide a close estimate of the value of ln(cosx) near x=0, but to find the exact value, you will need to use other methods such as a calculator or numerical methods.