# How do you find the Taylor polynomial Tn(x) for the function f at the number a f(x)=sqrt(3+x^2), a=1, n=2?

Sep 25, 2017

${T}_{2} \left(x\right) = \frac{3 {x}^{2} + 2 x + 27}{16}$

#### Explanation:

The Taylor polynomial of order $n$ is the $\left(n + 1\right)$-th partial sum of the Taylor series:

f(x) = sum_(n=0)^oo (f^((n))(a))/(n!)(x-a)^n

for $n = 2$ we have:

$\frac{d}{\mathrm{dx}} \sqrt{3 + {x}^{2}} = \frac{x}{\sqrt{3 + {x}^{2}}}$

${d}^{2} / {\mathrm{dx}}^{2} \sqrt{3 + {x}^{2}} = \frac{\sqrt{3 + {x}^{2}} - {x}^{2} / \sqrt{3 + {x}^{2}}}{3 + {x}^{2}} = \frac{3}{3 + {x}^{2}} ^ \left(\frac{3}{2}\right)$

So for $a = 1$:

${f}^{\left(0\right)} \left(1\right) = \sqrt{3 + {1}^{2}} = 2$

${f}^{\left(1\right)} \left(1\right) = \frac{1}{\sqrt{3 + {1}^{2}}} = \frac{1}{2}$

${f}^{\left(2\right)} \left(1\right) = \frac{3}{3 + {1}^{2}} ^ \left(\frac{3}{2}\right) = \frac{3}{8}$

Then:

${T}_{2} \left(x\right) = 2 + \frac{1}{2} \left(x - 1\right) + \frac{3}{16} {\left(x - 1\right)}^{2}$

${T}_{2} \left(x\right) = \frac{32 + 8 \left(x - 1\right) + 3 {\left(x - 1\right)}^{2}}{16}$

${T}_{2} \left(x\right) = \frac{32 + 8 x - 8 + 3 {x}^{2} - 6 x + 3}{16}$

${T}_{2} \left(x\right) = \frac{3 {x}^{2} + 2 x + 27}{16}$ 