Question

Compute the coefficients for the Taylor series for the following functions about the given point a and then use the first four terms of the series to approximate the given number. $$f(x)=\sqrt{x} \text { with } a=36 ; \text { approximate } \sqrt{39}$$

Short answer

Using the first four terms of the Taylor series expansion for the function $$f(x) = \sqrt{x}$$ about point $$a = 36$$, we approximate the value of $$\sqrt{39}$$ as $$6.1921$$.

Step-by-step solution

Differentiating the function

We will first find the first three derivatives of $$f(x) = \sqrt{x}$$. The first derivative, $$f'(x) = \frac{1}{2\sqrt{x}}$$. The second derivative, $$f''(x) = -\frac{1}{4x\sqrt{x}}$$. The third derivative, $$f^{(3)}(x) = \frac{3}{8x^2\sqrt{x}}$$. Now, we can use Taylor's formula to compute the first four terms in the series.

Using Taylor's formula

Taylor's formula states that a function can be approximated by the following series around point a: $$f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f^{(3)}(a)(x-a)^3}{3!} + ... $$ Computing the terms with our derivatives and $$a = 36:$$ - $$f(36) = \sqrt{36} = 6$$ - $$f'(36) = \frac{1}{2\sqrt{36}} = \frac{1}{12}$$ - $$f''(36) = -\frac{1}{4(36)\sqrt{36}} = -\frac{1}{432}$$ - $$f^{(3)}(36) = \frac{3}{8(36)^2\sqrt{36}} = \frac{1}{3456}$$ Substituting these values into the Taylor series formula, we get: $$f(x) \approx 6 + \frac{1}{12}(x-36) -\frac{1}{432}(x-36)^2 + \frac{1}{3456}(x-36)^3$$

Approximating the function value

The task was to approximate the value of $$f(39) = \sqrt{39}$$ using the first four terms of our Taylor series. Substituting $$x = 39$$ into our series: $$\sqrt{39} \approx 6 + \frac{1}{12}(39-36) -\frac{1}{432}(39-36)^2 + \frac{1}{3456}(39-36)^3$$ Calculating the result: $$\sqrt{39} \approx 6 + \frac{1}{12}(3) - \frac{1}{432}(3)^2 + \frac{1}{3456}(3)^3 = 6+0.25-0.0625+0.0046 \approx 6.1921$$ The approximation of $$\sqrt{39}$$ is $$6.1921$$ using the first four terms of the Taylor series expansion about point $$a = 36$$.