Discover the Maclaurin Series for f(x) with Derivative Calculations

In summary, the conversation revolves around finding the Maclaurin series for a given function, specifically f(x) = (1+x)^(1/2) – (1-x)^(1/2). The individual has attempted to solve it by combining two Riemann Sums and has arrived at a solution of (7!5!)/16 5! = 315, but upon checking with the fifth derivative, they have gotten a different answer of 105/16. They are seeking advice on where they may have gone wrong in their solution. Suggestions are made to simplify the problem by breaking it into smaller parts and to find the Maclaurin series for each individual function before combining them.
  • #1
defang
5
0
My question is as follows: Let f (x) = (1+x)^(1/2) – (1-x)^(1/2). Find the Maclaurin series for f(x) and use it to find f ^5 (0) and f ^20 (0).

I got: X + Riemann Sum { [ (-1)^(n-1) 1x3x5**x(2n-3) ] / (2^n) x n!} X^n (after combining two Riemann Sums together). And I got (7!5!) / 16 5! = 315. However, I tried to check my answer by taking derivative 5 times, and I got 105/16. Can anyone tell me what I did wrong? thanks
 
Physics news on Phys.org
  • #2
?

Is this question unsolvable? Or is it .....? Please, somebody gives me some advice....
 
  • #3
This really should be put in the homework section, but anyway, I can't tell you what you have done wrong but it'd probably be much easier to try and work out the nth derivative of the series, or at least the nth derivative at x = 0.
 
  • #4
hello there

if I were you I will try to make the problem look more simpler, why don't you split the function, like f(x)=g(x)-h(x) where
h(x)=(1-x)^(1/2) and
g(x)=(1+x)^(1/2)
find the maclarin series for each h(x) and g(x) then find the maclarin series for f(x)
and as for the last parts f^n(x)=g^n(x)-h^n(x) this should most likely work

steven
 

1. What is a MacLaurin series?

A MacLaurin series is a type of Taylor series expansion that is centered around x=0. It is used to approximate a function using a polynomial, and is named after Scottish mathematician Colin Maclaurin.

2. How is a MacLaurin series derived?

A MacLaurin series can be derived using the formula f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ... , where f(0), f'(0), etc. are the derivatives of the function evaluated at x=0. This formula can be generalized for any function, not just polynomials.

3. What is the purpose of a MacLaurin series?

A MacLaurin series is used to approximate a function with a polynomial. This can be useful for simplifying complex functions or for making calculations easier. It can also be used to find the value of a function at a specific point by plugging in the x-value into the polynomial.

4. What is the difference between a MacLaurin series and a Taylor series?

A MacLaurin series is a type of Taylor series where the center of the series is x=0. A Taylor series can be centered around any point, not just x=0. Additionally, a Taylor series may require more terms to accurately approximate a function compared to a MacLaurin series.

5. When is a MacLaurin series the most accurate?

A MacLaurin series is most accurate when the function being approximated is smooth and well-behaved near x=0. This means that the function should have a continuous derivative at x=0 and the higher order derivatives should not drastically change in value near x=0. If the function is not well-behaved near x=0, then a MacLaurin series may not accurately approximate the function.

Similar threads

Replies
1
Views
1K
Replies
1
Views
1K
  • Calculus
Replies
1
Views
1K
  • Calculus
Replies
2
Views
1K
  • Calculus
Replies
1
Views
1K
Replies
1
Views
1K
Replies
6
Views
2K
Replies
2
Views
1K
  • Calculus
Replies
1
Views
3K
Back
Top