# Can you prove the power series expansion for (1-2x-x^2) has a specific pattern?

• MHB
• Ackbach
In summary, the power series expansion for (1-2x-x^2) is a series of terms with a specific pattern that can be proven using the Taylor series expansion method. This pattern consists of alternating positive and negative coefficients, and the accuracy of the expansion increases as the number of terms in the series increases. An example of using this expansion to approximate a value is also provided.
Ackbach
Gold Member
MHB
Here is this week's POTW:

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Consider the power series expansion
$\frac{1}{1-2x-x^2} = \sum_{n=0}^\infty a_n x^n.$
Prove that, for each integer $n\geq 0$, there is an integer $m$ such that
$a_n^2 + a_{n+1}^2 = a_m .$

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Re: Problem Of The Week # 250 - Jan 25, 2017

This was Problem A-3 in the 1999 William Lowell Putnam Mathematical Competition.

Congratulations to Kiwi and Theia for their correct solutions! Theia's solution follows:
First, one have to find out the coefficients $$\displaystyle a_n$$. By writing

$$\displaystyle 1 = (1 - 2x - x^2)\sum_{n=0}^{\infty}a_nx^n$$

and separating the sum into three different sums and finally manipulating the summing indices, one can write this as

$$\displaystyle 1 = a_0 + (a_1 - 2a_0)x + \sum_{v=2}^{\infty}(a_v - 2a_{v-1} + a_{v-2})x^v$$.

This means, one can write

\displaystyle \begin{align*} a_0 &= 1 \\ a_1 &= 2 \\ a_v &= 2a_{v-1} + a_{v-2} \ , \end{align*}

where $$\displaystyle v \ge 2$$ (integer).

The last line defines a linear difference equation, whose solution can be found through characteristic equation, namely

$$\displaystyle \lambda ^2 = 2\lambda + 1 \qquad \Rightarrow \qquad \lambda = 1 \pm \sqrt{2} \ ,$$

and thus the general solution is

$$\displaystyle a_n = c_1\cdot (1 - \sqrt{2})^n + c_2\cdot (1 + \sqrt{2})^n$$.

By adding the initial conditions ($$\displaystyle a_0 = 1, a_1 = 2$$) one can write the coefficients of the serie in form ($$\displaystyle n$$ integer, $$\displaystyle \ge 0$$):

$$\displaystyle a_n = \frac{1}{4} \Bigl[\bigl(2 - \sqrt{2}\bigr)\cdot \bigl(1 - \sqrt{2}\bigr)^n + \bigl(2 + \sqrt{2}\bigr)\cdot \bigl(1 + \sqrt{2}\bigr)^n \Bigr]$$.

Now one can write

\displaystyle \begin{align*}a_n^2 + a_{n+1}^2 &= \frac{1}{4} \Bigl[\underbrace{\bigl(10 - 7\sqrt{2}\bigr)}_{=(2 - \sqrt{2})(1 - \sqrt{2})^2}\cdot \bigl(1 - \sqrt{2}\bigr)^{2n} + \underbrace{\bigl(10 + 7\sqrt{2}\bigr)}_{=(2 + \sqrt{2})(1 + \sqrt{2})^2}\cdot \bigl(1 + \sqrt{2}\bigr)^{2n} \Bigr] \\ & = \frac{1}{4} \Bigl[\bigl(2 - \sqrt{2}\bigr)\cdot \bigl(1 - \sqrt{2}\bigr)^{2n + 2} + \bigl(2 + \sqrt{2}\bigr)\cdot \bigl(1 + \sqrt{2}\bigr)^{2n+2} \Bigr] \\ &= a_{2n+2} \\ &= a_m\end{align*}.

Hence for every index $$\displaystyle n \ge 0$$ there is an index $$\displaystyle m$$ such that $$\displaystyle a_n^2 + a_{n+1}^2 = a_m$$.

## 1. What is the power series expansion for (1-2x-x^2)?

The power series expansion for (1-2x-x^2) is: 1 - 2x - x^2 + 0x^3 + 3x^4 + 0x^5 + 5x^6 + 0x^7 + ...

## 2. How do you prove that the power series expansion for (1-2x-x^2) has a specific pattern?

The power series expansion can be proven using the Taylor series expansion method, where the coefficients of the series can be determined by evaluating the derivatives of the function at a specific point.

## 3. Can you explain the pattern of the power series expansion for (1-2x-x^2)?

The pattern of the power series expansion for (1-2x-x^2) is that the coefficients follow the form of 1, -2, 1, 0, 3, 0, 5, 0, ..., where the odd coefficients are positive and the even coefficients are negative. This pattern continues for all terms in the series.

## 4. How do you know that the power series expansion for (1-2x-x^2) is accurate?

The power series expansion for (1-2x-x^2) is considered accurate because it can be used to approximate the function for a certain range of values. As the number of terms in the series increases, the accuracy of the approximation also increases.

## 5. Can you provide an example of using the power series expansion for (1-2x-x^2) to approximate a value?

Sure, for example, if we want to approximate the value of (1-2x-x^2) at x = 0.5, we can use the power series expansion with 5 terms (1 - 2(0.5) - (0.5)^2 + 0(0.5)^3 + 3(0.5)^4), which gives us an approximation of 0.3125. As we increase the number of terms in the series, the approximation will become more accurate.

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