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I'm having some trouble with the following question:

Determine the Taylor polynomial of degree $n$ at $0$ of the function $f(x) = \frac{x^5}{1 + x^2}$.

So, let $P_{n,a,f}$ denote the Taylor polynomial of degree $n$ at $a$ of a function $f$.

My first idea was to use two properties that my teacher taught us about Taylor series:

If, for some functions $f,g$, $P_{n,a,f}(x)P_{n,a,g}(x) = \sum_{k=0}^{2n} c_k (x - a)^k$, then:

$$P_{n,a,fg}(x) = \sum_{k=0}^n c_k (x - a)^k \ \ (1)$$

If $g$ is a function defined as: $g(x) = f(x^k)$, then: $$P_{kn,0,g}(x)=P_{n,0,f}(x^k) \ \ (2)$$

So, $f(x) = h(x)g(x)$, where $h(x) = x^5$ and $g(x) = \frac{1}{1 + x^2}$.

Furthermore, $g(x) = k(x^2)$ where $k(x) = \frac{1}{1 + x}$.

So, my plan was to use $(2)$ and find $P_{n,0,g}$, and then using $(1)$ find $P_{n,0,hg} =P_{n,0,f}$.

The thing is that I'm having some trouble doing this because if we apply proposition $(2)$, we can only calculate $P_{2n,0,g}$ and not $P_{n,0,g}$.

Is it possible to avoid this problem but still use these properties to solve the problem? If not, how can this problem be solved?

Bernard
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Eduardo Magalhães
  • 4,878

3 Answers3

7

As @Rodrigo de Azevedo said, first simply write

$$F(x) = x^5\left(\dfrac{1}{1 - (-x^2)}\right)$$

Notice that the latter (in the parenthesis) is sum of an infinite geometric series. Now just use the geometric series to evaluate the polynomial. Hope this helps.

BooleanCoder
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The idea with power series works fine.

Let set $S(x)=\sum\limits_{n=0}^{\infty} a_nx^n$ and let it verify $(1+x^2)S(x)=x^5$

Note that the RHS $x^5$ is the series $\sum b_nx^n$ with $b_5=1$ and $b_i=0,\forall i\neq 5$, we will use this when identifying the coefficients.

$\begin{align}(1+x^2)S(x) &=(1+x^2)\sum\limits_{n=0}^{\infty} a_nx^n\\ &=\sum\limits_{n=0}^{\infty} a_nx^n+\sum\limits_{n=0}^{\infty} a_nx^{n+2}&\text{shifting indexes of second series}\\ &=\sum\limits_{n=0}^{\infty} a_nx^n+\sum\limits_{n=2}^{\infty} a_{n-2}x^n&\text{isolating first terms and regrouping the rest}\\ &=a_0+a_1x+\sum\limits_{n=2}^{\infty} (a_n+a_{n-2})x^n\\ \end{align}$

Identifying coefficients gives us

$\begin{cases} a_0=0\\a_1=0\\a_2+a_0=0\\a_3+a_1=0\\a_4+a_2=0\\a_5+a_3=1\\a_n+a_{n-2}=0&\forall n\ge 6\\\end{cases}\iff\begin{cases} a_0=a_1=a_2=a_3=a_4=0\\a_5=1\\a_{2n}=0\\a_{2n+1}=(-1)^{n-2}\end{cases}$

And there you have it $$\frac{x^5}{1+x^2}=x^5-x^7+x^9-x^{11}+\cdots$$

zwim
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0

Write $$f(x) = x^5 \left(\frac{1}{1 - (- x^2)}\right)$$ Does $\frac{1}{1 - (- x^2)}$ remind you of something? Using SymPy,

>>> from sympy import *
>>> x = symbols('x', real=True)
>>> f = x**5 / (1 + x**2)
>>> f.series(x, 0, 20)
x**5 - x**7 + x**9 - x**11 + x**13 - x**15 + x**17 - x**19 + O(x**20)

Thus,

$$\frac{x^5}{1 + x^2} = x^{5} - x^{7} + x^{9} - x^{11} + x^{13} - x^{15} + x^{17} - x^{19} + O\left(x^{20}\right)$$

Rodrigo de Azevedo
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