Question

Find the third order maclaurin polynomial. Use it to estimate the value of sqrt1.3

Answer 1

$\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$ is the maclaurin polynomial and estimate value of $\sqrt{1.3}$ is 1.14. This can be obtained by using the formula to find the maclaurin polynomial.

Find the third order maclaurin polynomial:

Given the polynomial,

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

The formula to find the maclaurin polynomial,

$f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}$

Next we have to find f'(x), f''(x) and f'''(x),

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

• $f''(x) =-\frac{9}{4}(1+3x)^{-\frac{3}{2} }$

• $f'''(x) = \frac{81}{8}(1+3x)^{-\frac{5}{2} }$

By putting x = 0 , we get,

• $f(0)=(1+3(0))^{\frac{1}{2} }=1$

• $f'(0) = \frac{3}{2}(1+3(0))^{-\frac{1}{2} }=\frac{3}{2}$

• $f''(0) =-\frac{9}{4}(1+3(0))^{-\frac{3}{2} }=-\frac{9}{4}$

• $f'''(0) = \frac{81}{8}(1+3(0))^{-\frac{5}{2} }=\frac{81}{8}$

Therefore the maclaurin polynomial by using the formula will be,

$\sqrt{1+3x}=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}$

$\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$

To find the value of $\sqrt{1.3}$  we can use the maclaurin polynomial,

$\sqrt{1.3}$ is  $\sqrt{1+3x}$ with x = 1/10,

$\sqrt{1+3(1/10)}=1+\frac{3}{2} (1/10)-\frac{9}{8} (1/10)^{2} + \frac{81}{8}(1/10)^{3}$

$\sqrt{1+3(1/10)}=\frac{18247}{16000} = 1.14$

Hence $\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$ is the maclaurin polynomial and estimate value of $\sqrt{1.3}$ is 1.14.

Learn more about maclaurin polynomial here:

brainly.com/question/24188694

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Answer 2

You can use the well-known binomial series,

$\displaystyle (1+x)^\alpha = \sum_{k=0}^\infty \binom \alpha k x^k$

where

$\dbinom \alpha k = \dfrac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-(k-1))}{k!} \text{ and } \dbinom \alpha0 = 1$

Let $\alpha=\frac12$ and replace $x$ with $3x$; then the series expansion is

$\displaystyle (1+3x)^{1/2} = \sum_{k=0}^\infty \binom{\frac12}k (3x)^k$

and the first 4 terms in the expansion are

$\sqrt{1+3x} \approx 1 + \dfrac{\frac12}{1!}(3x) + \dfrac{\frac12\cdot\left(-\frac12\right)}{2!}(3x)^2 + \dfrac{\frac12\cdot\left(-\frac12\right)\cdot\left(-\frac32\right)}{3!}(3x)^3$

which simplify to

$\sqrt{1+3x} \approx \boxed{1 + \dfrac32 x - \dfrac98 x^2 + \dfrac{27}{16} x^3}$

You can also use the standard Maclaurin coefficient derivation by differentiating $f$ a few times.

$f(x) = (1+3x)^{1/2} \implies f(0) = 1$

$f'(x) = \dfrac32 (1+3x)^{-1/2} \implies f'(0) = \dfrac32$

$f''(x) = -\dfrac94 (1+3x)^{-3/2} \implies f''(0) = -\dfrac94$

$f'''(x) = \dfrac{81}8 (1+3x)^{-5/2} \implies f'''(0) = \dfrac{81}8$

Then the 3rd order Maclaurin polynomial is the same as before,

$\sqrt{1+3x} \approx f(0) + \dfrac{f'(0)}{1!} x + \dfrac{f''(0)}{2!} x^2 + \dfrac{f'''(0)}{3!} x^3 = 1 + \dfrac32 x - \dfrac98 x^2 + \dfrac{27}{16} x^3$

Now,

$\sqrt{1.3} = \sqrt{1+3x} \bigg|_{x=\frac1{10}} \\\\ ~~~~~~~~ \approx 1 + \dfrac32 \left(\dfrac1{10}\right) - \dfrac98 \left(\dfrac1{10}\right)^2 + \dfrac{27}{16} \left(\dfrac1{10}\right)^3 \\\\ ~~~~~~~~ = \dfrac{18,247}{16,000} \approx \boxed{1.14044}$

Compare to the actual value which is closer to 1.14018.