Taylor Series

Updated 02 Oct 2026

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  What it is

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    expresses a function's value near a point as an infinite sum of derivative terms: $f(x_0+h) = \sum_{k=0}^{n} \dfrac{f^{(k)}(x_0)}{k!}h^k + R_n(h)$

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  Lagrange remainder

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    $R_n(h) = \dfrac{f^{(n+1)}(\xi)}{(n+1)!}h^{n+1}$ for some $\xi$ between $x_0$ and $x_0+h$

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  Finite approximation

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    $P_n(x)=\sum_{k=0}^{n}f^{(k)}(a)(x-a)^k/k!$ has degree at most $n$ and contains $n+1$ indexed terms.

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    a Maclaurin expansion uses $a=0$.

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  Remainder

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    Under the required differentiability assumptions, $R_n(x)=f^{(n+1)}(\xi)(x-a)^{n+1}/(n+1)!$ for an intermediate $\xi$.

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    An error bound follows by bounding the derivative: $|R_n(x)|\le M|x-a|^{n+1}/(n+1)!$.

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    An infinite Taylor series represents the function only where the remainder tends to zero; smoothness alone is insufficient.

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  Implementation

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    exponential terms can be generated using $t_{k+1}=t_kx/(k+1)$.

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    for sine, $t_{k+1}=-t_kx^2/[(2k+2)(2k+3)]$.

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    argument reduction can avoid adding large alternating terms when evaluating trigonometric functions.

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  Checked examples

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    for $e^2$, degree $16$ is enough for fractional relative error below $10^{-10}$.

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    a threshold of $10^{-10}$ is stricter: degree $18$ is needed. These are different criteria, explaining the report/website distinction.

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    for $\sin(1)$, terms indexed $k=0,\ldots,5$ give degree $11$ and relative error about $1.90\times10^{-10}$.

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  worked example: $e^{0.5}$ via order-4 Maclaurin expansion

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    $P_4(0.5) = 1.6484375$; true value $e^{0.5} = 1.64872127$; actual error $\approx 2.84 \times 10^{-4}$

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    Lagrange bound: $|R_4(0.5)| \leq 7.8125 \times 10^{-4}$ — the actual error falls safely within the bound

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  Useful special cases

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    a polynomial of degree $n$, Taylor-expanded to order $n$, is represented exactly — the remainder $R_n$ is identically zero

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    the Maclaurin series of $f(x) = \dfrac{1}{1-x}$ is exactly the geometric series $\sum_{k=0}^{\infty}x^k$ for $|x|<1$

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  Related

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    Truncation Error

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