Calc 2 - Taylor Expansion Series of x^(1/2)

In summary, the given conversation discusses the function f(x) = √x and its derivatives, as well as the pattern for the numerator in f^{n}(x). The pattern involves the use of double factorials, with the first term being 1 and every subsequent term alternating between positive and negative numbers. The function f(x) = √x has a derivative of f^{'}(x) = 1/2x^(1/2) and a pattern for its higher order derivatives of (-1)^(n+1) * (1*3*5*...)/(2^n * x^((2n-1)/2)).
  • #1
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Homework Statement



f(x) = [tex]\sqrt{x}[/tex], a = 4

Homework Equations



f(x) = [tex]\sum[/tex]f[tex]^{n}[/tex](a)/n! (x-a)[tex]^{n}[/tex]

The Attempt at a Solution



f(x) = x[tex]^{1/2}[/tex]
f[tex]^{'}[/tex](x) = [tex]\frac{1}{2}[/tex]x[tex]^{1/2}[/tex]
f[tex]^{2}[/tex](x) = -[tex]\frac{1}{2}[/tex]*[tex]\frac{1}{2}[/tex]x[tex]^{-3/2}[/tex]
f[tex]^{3}[/tex](x) = [tex]\frac{1}{2}[/tex]*[tex]\frac{1}{2}[/tex]*[tex]\frac{3}{2}[/tex]x[tex]^{-5/2}[/tex]
f[tex]^{4}[/tex](x) = -[tex]\frac{1}{2}[/tex]*[tex]\frac{1}{2}[/tex]*[tex]\frac{3}{2}[/tex]*[tex]\frac{5}{2}[/tex]x[tex]^{-7/2}[/tex]

f[tex]^{n}[/tex](x) = (-1)[tex]^{n+1}[/tex]*[tex]\frac{1}{2}[/tex][tex]^{n}[/tex]*x[tex]^{-[(2n-1)/2]}[/tex]*?


The problem I am having here is with identifying the pattern. I am able to describe everything except the numbers in the numerator(1, 1*1, 1*1*3, 1*1*3*5...). Any help is greatly appreciated!
 
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  • #2
I have no idea why it looks like that...
 
  • #3
shouldn't the number's in the numerator be (1*-1*-3*-5*...)?
 
  • #4
I left out the signs to simplify and becaus I already identified the pattern w/ (-1)^(n+1)... so if i kept the signs in, the #'s in the numerator would be (1, -1*1, -3*-1*1, -5*-3*-1*1)
 
  • #5
how are you trying to explain it then?
 
  • #6
n! denotes the double factorial of n and is defined recursively for odd numbers,,
eg: 9! = 1 × 3 × 5 × 7 × 9 = 945

does that help?
 

Related to Calc 2 - Taylor Expansion Series of x^(1/2)

1. What is the Taylor Expansion Series of x^(1/2)?

The Taylor Expansion Series of x^(1/2) is a mathematical series that represents the square root of x as an infinite sum of terms. It is derived using the Taylor series expansion method, which is a way to approximate functions using polynomials.

2. What is the purpose of using a Taylor Expansion Series for x^(1/2)?

The purpose of using a Taylor Expansion Series for x^(1/2) is to approximate the value of the square root of x for any given input value. It can be used to calculate the value of the square root function at any point, even if the function is not defined at that point.

3. How accurate is the Taylor Expansion Series for x^(1/2)?

The accuracy of the Taylor Expansion Series for x^(1/2) depends on the number of terms used in the series. The more terms that are included, the more accurate the approximation will be. However, since it is an infinite series, the exact value of the square root cannot be represented.

4. Can the Taylor Expansion Series for x^(1/2) be used for complex numbers?

Yes, the Taylor Expansion Series for x^(1/2) can be used for complex numbers. However, the series will only converge for complex numbers with a magnitude less than 1. For complex numbers with a magnitude greater than 1, the series will diverge.

5. How is the Taylor Expansion Series for x^(1/2) related to other mathematical concepts?

The Taylor Expansion Series for x^(1/2) is related to other mathematical concepts such as derivatives and integrals. It can be used to find the derivative of the square root function, and it can also be used to approximate the value of an integral involving the square root function.

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