# Lagrange Form of the Remainder Term in a Taylor Series

## Key Questions

• By taking the derivatives,

$f \left(x\right) = {e}^{4 x}$
$f ' \left(x\right) = 4 {e}^{4 x}$
$f ' ' \left(x\right) = {4}^{2} {e}^{4 x}$
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${f}^{\left(n + 1\right)} \left(x\right) = {4}^{n + 1} {e}^{4 x}$

So,
R_n(x;3)={f^{(n+1)}(z)}/{(n+1)!}(x-3)^{n+1}={4^{n+1}e^{4z}}/{(n+1)!}(x-3)^{n+1},
where $z$ is between $x$ and $3$.

• Remainder Term of Taylor Series

R_n(x;c)={f^{(n+1)}(z)}/{(n+1)!}(x-c)^{n+1},

where $z$ is a number between $x$ and $c$.

Let us find R_3(x;1) for $f \left(x\right) = \sin 2 x$.

By taking derivatives,

$f ' \left(x\right) = 2 \cos 2 x$
$f ' ' \left(x\right) = - 4 \sin 2 x$
$f ' ' ' \left(x\right) = - 8 \cos 2 x$
${f}^{\left(4\right)} \left(x\right) = 16 \sin 2 x$

So, we have

R_3(x;1)={16sin2z}/{4!}(x-1)^4,

where $z$ is a number between $x$ and $1$.

I hope that this was helpful.

• Taylor remainder term

R_n(x;c)={f^{(n+1)}(z)}/{(n+1)!}(x-c)^{n+1},

where $z$ is between $x$ and $c$.