# Thread: Possibly Semantic Maclaurin Polynomial Question

1. ## Possibly Semantic Maclaurin Polynomial Question

This comes from the review true/false quiz for chapter 10 of the fifth edition of Calculus with Analytic Geometry by Purcell and Varberg:

$f(x) = x^{5/2}$ has a second-order Maclaurin polynomial.

I answered false because $P_{2}(0) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2$, $f'(x) = \frac{5}{2}x^{3/2}, f''(0) = \frac{15}{4}x^{1/2}$ and $f(0) = f'(0) = f''(0) = 0$. The answer key says true.

2. ## Re: Possibly Semantic Maclaurin Polynomial Question

I would begin by stating:

$\displaystyle f(x)\approx P_2(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2$

Now, we are given:

$\displaystyle f(x)=x^{\Large\frac{5}{2}}$

Hence:

$\displaystyle f'(x)=\frac{5}{2}x^{\Large\frac{3}{2}}$

$\displaystyle f''(x)=\frac{15}{4}x^{\Large\frac{1}{2}}$

And so:

$\displaystyle P_2(x)=a^{\Large\frac{5}{2}}+ \frac{5}{2}a^{\Large\frac{3}{2}}(x-a)+ \frac{15}{8}a^{\Large\frac{1}{2}}(x-a)^2$

Only when $\displaystyle 0<a$ do we have a second order polynomial.

3. ## Re: Possibly Semantic Maclaurin Polynomial Question

Maclaurin polynomial of order 2 is of the form

$\displaystyle P_2(x)=f(0)+f'(0) x + \frac{f \text{''} (0)}{2}x^2$

since $\displaystyle f(0)=f'(0)=f\text{''}(0)=0$

$\displaystyle P_2(x)=0$

4. ## Re: Possibly Semantic Maclaurin Polynomial Question

As Idea said, the second order McLaurin polynomial is "0" just as you got in your first post. What makes you think that "0" is not a polynomial?

5. ## Re: Possibly Semantic Maclaurin Polynomial Question

Originally Posted by HallsofIvy
What makes you think that "0" is not a polynomial?
The fact that when I see "0" the word "polynomial" does not spring to mind.

I suppose that nothing in the definition of polynomial disqualifies it, though.