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?

Question

Cameron Alexander · Accepted Answer

#T_2(x) = (3x^2+2x+27)/16#

The Taylor polynomial of order #n# is the #(n+1)#-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:

#d/dx sqrt(3+x^2) = x/sqrt(3+x^2)#

#d^2/dx^2 sqrt(3+x^2) = (sqrt(3+x^2) - x^2/sqrt(3+x^2))/(3+x^2) = 3/(3+x^2)^(3/2)#

So for #a=1#:

#f^((0))(1) = sqrt(3+1^2) = 2#

#f^((1))(1) = 1/sqrt(3+1^2) = 1/2#

#f^((2))(1) = 3/(3+1^2)^(3/2) = 3/8#

Then:

#T_2(x) = 2+1/2(x-1)+3/16(x-1)^2#

#T_2(x) = (32+8(x-1)+3(x-1)^2)/16#

#T_2(x) = (32+8x-8+3x^2-6x+3)/16#

#T_2(x) = (3x^2+2x+27)/16#

Ezra Adams · Answer

undefined To find the Taylor polynomial $ T_n(x) $ for the function $ f(x) = \sqrt{3 + x^2} $ at the number $ a = 1 $ with $ n = 2 $, follow these steps:

1. Find the $ n $-th derivative of $ f(x) $, denoted as $ f^{(n)}(x) $.
2. Evaluate each derivative at $ x = a $.
3. Write down the general formula for the Taylor polynomial $ T_n(x) $ using the derivatives and the terms of the Maclaurin series.
4. Substitute the evaluated derivatives into the Taylor polynomial formula.

First, find the first and second derivatives of $ f(x) $:

$$ f(x) = \sqrt{3 + x^2} $$

$$ f'(x) = \frac{1}{2\sqrt{3 + x^2}}(2x) = \frac{x}{\sqrt{3 + x^2}} $$

$$ f''(x) = \frac{d}{dx} \left(\frac{x}{\sqrt{3 + x^2}}\right) $$

Use the quotient rule:

$$ = \frac{(1)(\sqrt{3 + x^2}) - (x)(\frac{1}{2}(3 + x^2)^{-\frac{1}{2}}(2x))}{(3 + x^2)} $$

$$ = \frac{\sqrt{3 + x^2} - \frac{x^2}{\sqrt{3 + x^2}}}{3 + x^2} $$

Next, evaluate the derivatives at $ x = 1 $:

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

$$ f''(1) = \frac{\sqrt{3 + 1^2} - \frac{1^2}{\sqrt{3 + 1^2}}}{3 + 1^2} = \frac{\sqrt{4} - \frac{1}{\sqrt{4}}}{4} = \frac{2 - \frac{1}{2}}{4} = \frac{3}{8} $$

Now, write down the Taylor polynomial $ T_2(x) $ using the derivatives:

$$ T_2(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 $$

$$ = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 $$

$$ = \sqrt{3 + 1^2} + \frac{1}{2\sqrt{2}}(x - 1) + \frac{3}{8 \times 2!}(x - 1)^2 $$

$$ = \sqrt{4} + \frac{1}{2\sqrt{2}}(x - 1) + \frac{3}{16}(x - 1)^2 $$

$$ = 2 + \frac{1}{2\sqrt{2}}(x - 1) + \frac{3}{16}(x - 1)^2 $$

So, the Taylor polynomial $ T_2(x) $ for the function $ f(x) = \sqrt{3 + x^2} $ at $ a = 1 $ with $ n = 2 $ is $ 2 + \frac{1}{2\sqrt{2}}(x - 1) + \frac{3}{16}(x - 1)^2 $.