Given f(x) = -x - 1, find f(5).
2 answers:
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The Taylor Series expansion of f(x) = sin(x) about a = π i given by
where the c's are contants.
That is
f(x) = c₀ + c₁(x-π) +c (x-π)² + c₃ (x-π)³ + ...,
₂
The first few derivatives of f(x) are
f' = c₁
f'' = 2c₂ = 2! c₂
f''' = 3.2c₃ = 3! c₃
f⁽⁴⁾ = 4.3.2c₄ = 4! c₄
and so on.
The pattern indicates that
The derivatives of f(x) are
f' = cos(x)
f'' = -sin(x)
f''' = -cos(x)
f⁽⁴⁾ = sin(x(
and so on
The pattern indicates that
f⁽ⁿ⁾(x) = cos(x), n=1,5,9, ...,
= -sin(x), n=2,6,10, ...,
= -cos(x), n=3,7,11, ...,
= sin(x), n=4,8,12, ...,
The radius of convergence is |x-π|<1 by the ratio test.
where the c's are contants.
That is
f(x) = c₀ + c₁(x-π) +c (x-π)² + c₃ (x-π)³ + ...,
₂
The first few derivatives of f(x) are
f' = c₁
f'' = 2c₂ = 2! c₂
f''' = 3.2c₃ = 3! c₃
f⁽⁴⁾ = 4.3.2c₄ = 4! c₄
and so on.
The pattern indicates that
The derivatives of f(x) are
f' = cos(x)
f'' = -sin(x)
f''' = -cos(x)
f⁽⁴⁾ = sin(x(
and so on
The pattern indicates that
f⁽ⁿ⁾(x) = cos(x), n=1,5,9, ...,
= -sin(x), n=2,6,10, ...,
= -cos(x), n=3,7,11, ...,
= sin(x), n=4,8,12, ...,
The radius of convergence is |x-π|<1 by the ratio test.