I need help finding the taylor series expansion for f(x)=5/(2x+3) using c=0
I tried to do this problem but now I am wondering if it is even possible.
when you take the multiple derivatives of the function and sub in the c=0 you end up with a pattern that is changing sign (which is just (-1)^n) and over 3,9,27,81, and so on(which is 3^(n+1)) but the other pattern in the numerator goes 5,10,40,240,... which when put into wolfram alpha as a sequence says there is nothing that can give that pattern of numbers.
anyone have any ideas on what is happening here?
I tried to do this problem but now I am wondering if it is even possible.
when you take the multiple derivatives of the function and sub in the c=0 you end up with a pattern that is changing sign (which is just (-1)^n) and over 3,9,27,81, and so on(which is 3^(n+1)) but the other pattern in the numerator goes 5,10,40,240,... which when put into wolfram alpha as a sequence says there is nothing that can give that pattern of numbers.
anyone have any ideas on what is happening here?
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Don't multiply the numbers together.
f '(x) = 5 * (-1)(2x+3)^(-2) * 2
f ''(x) = 5 * (-1)(-2)(2x+3)^(-3) * 2^2
f '''(x) = 5 * (-1)(-2)(-3)(2x+3)^(-4) * 2^3
...
f^(n)(x) = 5 * (-1)(-2)(-3)...(-n)(2x+3)^(-(n+1)) * 2^n = 5 * (-1)^n * 2^n * n! / (2x+3)^(n+1)
==> f^(n)(0) = 5 * (-1)^n * 2^n * n! / 3^(n+1) [this includes n = 0].
So, f(x) = Σ(n = 0 to ∞) [5 * (-1)^n * 2^n * n! / 3^(n+1)] x^n/n!
............= Σ(n = 0 to ∞) 5 * (-1)^n * 2^n x^n / 3^(n+1)
............= (5/3) Σ(n = 0 to ∞) (-2/3)^n x^n.
I hope this helps!
f '(x) = 5 * (-1)(2x+3)^(-2) * 2
f ''(x) = 5 * (-1)(-2)(2x+3)^(-3) * 2^2
f '''(x) = 5 * (-1)(-2)(-3)(2x+3)^(-4) * 2^3
...
f^(n)(x) = 5 * (-1)(-2)(-3)...(-n)(2x+3)^(-(n+1)) * 2^n = 5 * (-1)^n * 2^n * n! / (2x+3)^(n+1)
==> f^(n)(0) = 5 * (-1)^n * 2^n * n! / 3^(n+1) [this includes n = 0].
So, f(x) = Σ(n = 0 to ∞) [5 * (-1)^n * 2^n * n! / 3^(n+1)] x^n/n!
............= Σ(n = 0 to ∞) 5 * (-1)^n * 2^n x^n / 3^(n+1)
............= (5/3) Σ(n = 0 to ∞) (-2/3)^n x^n.
I hope this helps!
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when finding a series expansion, it's best to not simplify.
So the question as for the Maclaurin series of f(x) = 5/(2x+3).
y = C0 + C1x + C2 x^2 + C3 x^3 + C4 x^4 + ...
the coefficients of a taylor series are represented by
Cn = f^(n) (a) / n! , where f^(0)(a) simply means f(a)
C0 = f(a) = 5/3
C1 = f'(0)/1!
= -5*2 / (2x+3)^2 {at 0} = -5*2 / 3^2
C2 = f''(0)/2!
= 5*2*2^2 / (2x + 3)^3 {at 0} = 5*2*2^2 / (2! 3^3)
C3 = f'''(0) / 3!
= -5*2^3*(2*3) / (2x + 3)^4 {at 0} = -5*2^3*(2*3) / (3! 3^4)
C4 = f^4(0)/4!
= 5*2^4*(2*3*4) / (2x + 3)^5 {at 0} = 5*2^4*(2*3*4) / (4! 3^5)
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it seems to me that,
Cn = (-1)^n (5) (2^n) (n!) / (n! 3^(n+1)) = (-1)^n (5) (2^n) / 3^(n+1) where n = 0,1,2,3,...
therefore the series is:
y = Σ{0,inf} [(-1)^n (5) (2^n) / 3^(n+1)] x^n
plug some n values to check:
n = 0 gives 5/3
n = 1 gives (-10/9)x
n = 2 gives (20/27)x^2
n = 3 gives (-40/81)x^3
n = 4 gives (80/234)x^4
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So the question as for the Maclaurin series of f(x) = 5/(2x+3).
y = C0 + C1x + C2 x^2 + C3 x^3 + C4 x^4 + ...
the coefficients of a taylor series are represented by
Cn = f^(n) (a) / n! , where f^(0)(a) simply means f(a)
C0 = f(a) = 5/3
C1 = f'(0)/1!
= -5*2 / (2x+3)^2 {at 0} = -5*2 / 3^2
C2 = f''(0)/2!
= 5*2*2^2 / (2x + 3)^3 {at 0} = 5*2*2^2 / (2! 3^3)
C3 = f'''(0) / 3!
= -5*2^3*(2*3) / (2x + 3)^4 {at 0} = -5*2^3*(2*3) / (3! 3^4)
C4 = f^4(0)/4!
= 5*2^4*(2*3*4) / (2x + 3)^5 {at 0} = 5*2^4*(2*3*4) / (4! 3^5)
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it seems to me that,
Cn = (-1)^n (5) (2^n) (n!) / (n! 3^(n+1)) = (-1)^n (5) (2^n) / 3^(n+1) where n = 0,1,2,3,...
therefore the series is:
y = Σ{0,inf} [(-1)^n (5) (2^n) / 3^(n+1)] x^n
plug some n values to check:
n = 0 gives 5/3
n = 1 gives (-10/9)x
n = 2 gives (20/27)x^2
n = 3 gives (-40/81)x^3
n = 4 gives (80/234)x^4
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