Binomials question (1 Viewer)

fancyyyyyy <3 · May 3, 2020

Hi everyone! I came across this binomials question and am a bit stuck.
Express x^8+x^-8 in the form Au^8+Bu^6+Cu^4+Du^2+E, where u=x+x^-1 and A, B, C, D and E are numerical coefficients.

CM_Tutor · May 3, 2020

To derive the result directly, we need to examine $x^n + x^{-n}$ starting with $n = 1$ then $n = 2$ then $n = 4$ :

$\begin{align*} u &= x + \cfrac{1}{x} \\ \text{On squaring} \qquad u^2 &= \bigg(x + \cfrac{1}{x}\bigg)^2 \\ &= x^2 + \cfrac{2x}{x} + \cfrac{1}{x^2} \\ \text{Rearranging, our $n = 2$ case is:} \quad x^2 + \cfrac{1}{x^2} &= u^2 - 2 \qquad \text{(A)} \end{align*}$

Moving on to the $n = 4$ case using the binomial theorem:

$\begin{align*} u^4 &= \bigg(x + \cfrac{1}{x}\bigg)^4 \\ &= x^4 + \cfrac{4x^3}{x} + \cfrac{6x^2}{x^2} + \cfrac{4x}{x^3} + \cfrac{1}{x^4} \\ &= x^4 + 4x^2 + 6 + \cfrac{4}{x^2} + \cfrac{1}{x^4} \\ \text{Rearranging for $n = 4$ case:} \qquad x^4 + \cfrac{1}{x^4} &= u^4 - 4\bigg(x^2 + \cfrac{1}{x^2}\bigg) - 6 \\ &= u^4 - 4(u^2 - 2) - 6 \qquad \text{using (A)} \\ &= u^4 - 4u^2 + 2 \end{align*}$

Aside: the $n = 4$ case could also be found by squaring the $n = 2$ case:

$\begin{align*} x^2 + \cfrac{1}{x^2} &= u^2 - 2 \\ \bigg(x^2 + \cfrac{1}{x^2}\bigg)^2 &= (u^2 - 2)^2 \\ x^4 + \cfrac{2x^2}{x^2} + \cfrac{1}{x^4} &= u^4 - 4u^2 + 4 \\ \text{Rearranging for the $n = 4$ case:} \qquad x^4 + \cfrac{1}{x^4} &= u^4 - 4u^2 + 4 - 2 \\ &= u^4 - 4u^2 + 2 \end{align*}$

As expected, the result for the $n = 4$ case is the same for both methods, $x^4 + x^{-4} = u^4 - 4u^2 + 2$

Now, the (required) $n = 8$ case can be found by squaring the $n = 4$ case (as is done below)... but it could also be obtained by applying the binomial theorem to $(x + x^{-1})^8$ . To use this approach, we would need to first establish that $x^6 + x^{-6} = u^6 - 6u^4 + 9u^2 - 2$ . This could be done by:

using the binomial theorem on $\big(x + x^{-1}\big)^3$ to get that $x^3 + x^{-3} = u^3 - 3u$ and then squaring to get
$x ^6 + x^{-6} = u^6 - 6u^4 + 9u^2 - 2$
using the binomial theorem on $\big(x + x^{-1}\big)^6$ to get that
$\begin{align*} x^6 + x^{-6} &= u^6 - 6(x^4 + x^{-4}) - 15(x^2 + x^{-2}) - 20 \\ &= u^6 - 6(u^4 - 4u^2 + 2) - 15(u^2 - 2) - 20 \\ &= u^6 - 6u^4 +9u^2 - 2 \end{align*}$
expand $(x^4 + x^{-4})(x^2 + x^{-2}) = x^6 + x^2 + x^{-2} + x^{-6}$ to get that
$\begin{align*} x^6 + x^{-6} &= (x^4 + x^{-4})(x^2 + x^{-2}) - (x^2 + x^{-2}) \\ &= (x^4 + x^{-4} - 1)(x^2 + x^{-2}) \\ &= (u^4 - 4u^2 + 1)(u^2 - 2) \\ &= u^6 - 6u^4 + 9u^2 - 2 \end{align*}$

Anyway, taking the squaring approach:

$\begin{align*} x^4 + \cfrac{1}{x^4} &= u^4 - 4u^2 + 2 \\ \bigg(x^4 + \cfrac{1}{x^4}\bigg)^2 &= (u^4 - 4u^2 + 2)^2 \\ x^8 + \cfrac{2x^4}{x^4} + \cfrac{1}{x^8} &= u^4(u^4 - 4u^2 + 2) - 4u^2 (u^4 - 4u^2 + 2) + 2 (u^4 - 4u^2 + 2) \\ x^8 + 2 + \cfrac{1}{x^8} &= u^8 - 4u^6 + 2u^4 - 4u^6 + 16u^4 - 8u^2 + 2u^4 - 8u^2 + 4 \\ \text{Rearranging for $n = 8$ case:} \qquad x^8 + 2 + \cfrac{1}{x^8} &= u^8 - (4 + 4)u^6 + (2 + 16 + 2)u^4 - (8 + 8)u^2 + 4 - 2 \\ &= u^8 - 8u^6 + 20u^4 - 16u^2 + 2 \end{align*}$

So, I get $A = 1$ , $B = -8$ , $C = 20$ , $D = -16$ , and $E = 2$ .

Edit: Tidying up / expanding working and removing comments now that "equation empty" comments are not appearing.

idkkdi · May 4, 2020

CM_Tutor said:
$u = x + \cfrac{1}{x}$

$\begin{align*} u^2 &= \bigg(x + \cfrac{1}{x}\bigg)^2 \\ &= x^2 + \cfrac{2x}{x} + \cfrac{1}{x^2} \\ x^2 + \cfrac{1}{x^2} &= u^2 - 2 \end{align*}$

$\begin{align*} u^4 &= \bigg(x + \cfrac{1}{x}\bigg)^4 \\ &= x^4 + \cfrac{4x^3}{x} + \cfrac{6x^2}{x^2} + \cfrac{4x}{x^3} + \cfrac{1}{x^4} \\ x^4 + \cfrac{1}{x^4} &= u^4 - 4\bigg(x^2 + \cfrac{1}{x^2}\bigg) - 6 \\ &= u^4 - 4(u^2 - 2) - 6 \\ &= u^4 - 4u^2 + 2 \end{align*}$

$\begin{align*} x^4 + \cfrac{1}{x^4} &= u^4 - 4u^2 + 2 \\ \bigg(x^4 + \cfrac{1}{x^4}\bigg)^2 &= (u^4 - 4u^2 + 2)^2 \\ x^8 + \cfrac{2x^4}{x^4} + \cfrac{1}{x^8} &= (u^4 - 4u^2 + 2)^2 \\ x^8 + \cfrac{1}{x^8} &= (u^4 - 4u^2 + 2)^2 - 2 \\ &= u^8 - 8u^6 + 20u^4 - 16u^2 + 2 \end{align*}$

So, I get $A = 1$ , $B = -8$ , $C = 20$ , $D = -16$ , and $E = 2$ .

Edit: And I have no idea why these are showing as "blank equation", but FYI, A, B, C, D, and E are 1, -8, 20, -16, and 2, respectively.

Nah, you did it wrong. You messed up from C onwards. Not bothered to find it in your massive algebra. Basically, by binomial expansion, x^4 terms are 28x^4, -8*6x^4 thus for C, if we were to make x^4 values total to 0, C has to be +5, rather than 20 since it's 5*4x^2. Probably just a simple error where you didn't see the 4.

CM_Tutor · May 4, 2020

Another approach to this question is to treat the result as an identity, that is, seeking real values A, B, C, D, and E such that:

$x^8 + x^{-8} \equiv Au^8 + Bu^6 + Cu^4 + Du^2 + E \qquad \text{where } u = x + x^{-1}$

So long as the identity is true, we can seek the values of the coefficients. The problem with such an approach is that you need to know the identity in advance - that is, that an expression of the form $x^n + x^{-n}$ for $n \in \mathbb{Z}^+$ must be able to be re-written as a polynomial of degree $n$ in $u$ where $u = x + x^{-1}$ and where the coefficients of all the terms $u^k$ where $k < n$ are zero if $k$ is odd.

Put another way, if I start with $2x + 2 \equiv A(x + 1)^2 + B(x - 1)$ , I can put $x = 0$ to get $2 = A - B$ and put $x = 1$ to get $4 = 4A$ to conclude that $A = 1$ and $B = -1$ , but that doesn't make the result "true" because the identity that I started with is never true.

In any case, the identity we have here is true, and so the approach will work.

Approach 1: Expand the RHS using the binomial theorem and equate coefficients. That is:

$\begin{align*} \text{RHS } &= Au^8 + Bu^6 + Cu^4 + Du^2 + E \qquad \text{where } u = x + x^{-1} \\ &= A\big(x + x^{-1}\big)^8 + B\big(x + x^{-1}\big)^6 + C\big(x + x^{-1}\big)^4 + D\big(x + x^{-1}\big)^2 + E \\ &= A\big(^8 C_0 x^8 + ^8 C_1 x^7 x^{-1} + ^8 C_2 x^6 x^{-2} + ^8 C_3 x^5 x^{-3} + ... + ^8 C_8 x^{-8}\big) \\ &\quad + B\big(^6 C_0 x^6 + ^6 C_1 x^5 x^{-1} + ^6 C_2 x^4 x^{-2} + ... + ^6 C_6 x^{-6}\big) \\ &\quad + C\big(^4 C_0 x^4 + ^4 C_1 x^3 x^{-1} + ^4 C_2 x^2 x^{-2} + ... + ^4 C_4 x^{-4}\big) \\ &\quad + D\big(x^2 + 2x x^{-1} + x^{-2}\big) \\ &\quad + E \\ &= A\big(x^8 + 8x^6 + 28x^4 + 56x^2 + 70 + 56x^{-2} + 28x^{-4} + 8x^{-6} + x^{-8}\big) \\ &\quad + B\big(x^6 + 6x^4 + 15x^2 + 20 + 15x^{-2} + 6x^{-4} + x^{-6}\big) \\ &\quad + C\big(x^4 + 4x^2 + 6 + 4x^{-2} + x^{-4}\big) \\ &\quad + D\big(x^2 + 2 + x^{-2}\big) \\ &\quad + E \\&= A(x^8 + x^{-8}) + (8A + B)(x^6 + x^{-6}) + (28A + 6B + C)(x^4 + x^{-4}) \\ &+(56A + 15B + 4C + D)(x^2 + x^{-2}) + 70A + 20B + 6C + 2D + E \end{align*}$

Equating coefficients, we get:
$\text{Looking at $x^8$:} \quad A = 1$
$\begin{align*} \text{Looking at $x^6$:} \quad &8A + B = 0 \\ \implies B &= -8A \\ &= -8(1) \\ &= -8 \end{align*}$
$\begin{align*} \text{Looking at $x^4$:} \quad &28A + 6B + C = 0 \\ \implies C &= -28A - 6B \\ &= -28(1) - 6(-8) \\ &= -28 + 48 \\ &= 20 \end{align*}$
$\begin{align*} \text{Looking at $x^2$:} \quad &56A + 15B + 4C + D = 0 \\ \implies D &= -56A - 15B - 4C \\ &= -56(1) - 15(-8) - 4(20) \\ &= -56 + 120 - 80 \\ &= -16 \end{align*}$
$\begin{align*} \text{Looking at $x^0$:} \quad &70A + 20B + 6C + 2D + E = 0 \\ \implies E &= -70A - 20B - 6C - 2D \\ &= -70(1) - 20(-8) - 6(20) - 2(-16) \\ &= -70 + 160 - 120 + 32 \\ &= 2 \end{align*}$

which matches the answers found above.

CM_Tutor · May 4, 2020

idkkdi said:
Nah, you did it wrong. You messed up from C onwards. Not bothered to find it in your massive algebra. Basically, by binomial expansion, x^4 terms are 28x^4, -8*6x^4 thus for C, if we were to make x^4 values total to 0, C has to be +5, rather than 20 since it's 5*4x^2. Probably just a simple error where you didn't see the 4.

I don't think I've made a mistake. I agree with you that:

expanding $u^8$ produces $28x^4$ as a term
expanding $u^6$ produces $6x^4$ as a term
expanding $u^4$ produces $x^4$ as a term.

So, the term in $x^4$ from $Au^8 + Bu^6 + Cu^4 + Du^2 + E$ is $(28A + 6B + C)x^4$ .

We then equate coefficients in $x^4$ from the polynomial in $u$ and the expression $x^8 + x^{-8}$ to get:

$\begin{align*} 28A + 6B + C &= 0 \\ C &= -28A - 6B \\ &= -28(1) - 6B \quad \text{as coefficients of $x^8$ give us $A = 1$} \\ &= -28 - 6(-8) \quad \text{as coefficients of $x^6$ give us $B = -8A = -8$} \\ &= -28 + 48 \\ &= 20 \end{align*}$

idkkdi · May 4, 2020

CM_Tutor said:
I don't think I've made a mistake. I agree with you that:

expanding $u^8$ produces $28x^4$ as a term

expanding $u^6$ produces $6x^4$ as a term

expanding $u^4$ produces $x^4$ as a term.

So, the term in $x^4$ from $Au^8 + Bu^6 + Cu^4 + Du^2 + E$ is $(28A + 6B + C)x^4$ .

We then equate coefficients in $x^4$ from the polynomial in $u$ and the expression $x^8 + x^{-8}$ to get:

$\begin{align*} 28A + 6B + C &= 0 \\ C &= -28A - 6B \\ &= -28(1) - 6B \quad \text{as coefficients of $x^8$ give us $A = 1$} \\ &= -28 - 6(-8) \quad \text{as coefficients of $x^6$ give us $B = -8A = -8$} \\ &= -28 + 48 \\ &= 20 \end{align*}$

Oh, my bad, I was looking at the coefficient of the x^2 value in the polynomial with highest degree 4, which was 5.

CM_Tutor · May 4, 2020

Approach 2: Dealing with the identity without getting bogged down in all the binomials - good approach or not?:

$x^8 + x^{-8} \equiv Au^8 + Bu^6 + Cu^4 + Du^2 + E \qquad \text{where } u = x + x^{-1}$

Firstly, it is obvious that the only term in the polynomial that can produce a term in $x^8$ is $u^8$ :

$\begin{align*} u^8 &= \big(x + x^{-1}\big)^8 \\ &= \binom{8}{0}x^8 + \binom{8}{1}x^7x^{-1} + \binom{8}{2}x^6x^{-2} + ... + \binom{8}{8}x^{-8} \\ &= x^8 + 8x^6 + 28x^4 + ... + x^{-8} \end{align*}$

and the required term will be $x^8$ as $^8C_0 = 1$ , so equating coefficients will give us that $A = 1$ .

Second, we can get simultaneous equations by selective choice of values of $x$ . Taking $x = 1$ gives $u = 2$ :

$\begin{align*} 1^8 + 1^{-8} &= A(2)^8 + B(2)^6 + C(2)^4 + D(2)^2 + E \\ 1 + 1 &= 256 + 64B + 16C + 4D + E \\ 64B + 16C + 4D + E &= 2 - 256 \\ &= -254 \qquad \text{(1)} \end{align*}$

For Extension 2 students, $u = 0 \implies x^2 = -1 \implies x^8 = 1$ gives:

$\begin{align*} 1 + 1^{-1} &= A(0)^8 + B(0)^6 + C(0)^4 + D(0)^2 + E \\ 2 &= E \end{align*}$

This allows (1) to be simplified:

$\begin{align*} 64B + 16C + 4D + E &= -254 \\ 64B + 16C + 4D + 2 &= -254 \qquad \text{using $E = 2$} \\ 64B + 16C + 4D &= -256 \\ 16B + 4C + D &= -64 \qquad \text{(1A)} \end{align*}$

Now, consider $u = 4$ :

$\begin{align*} u^2 = \big(x + x^{-1}\big)^2 &= x^2 + 2 + x^{-2} \\ 4^2 - 2 &= x^2 + x^{-2} \\ x^2 + x^{-2} &= 14 \\ \big(x^2 + x^{-2}\big)^2 &= 14^2 \\ x^4 + 2 + x^{-4} &= 14^2 = 196 \\ x^4 + x^{-4} &= 194 \\ \big(x^4 + x^{-4}\big)^2 &= 194^2 \\ x^8 + 2 + x^{-8} &= 194^2 = 37636 \\ x^8 + x^{-8} &= 37634 \\ \text{So,} \quad 37634 &= A(4)^8 + B(4)^6 + C(4)^4 + D(4)^2 + E \\ 37634 &= 65536 + 4096B + 256C + 16D + 2 \\ -1744 &= 256B +16C + D \\ -1680 &= 240B + 12C \qquad \text{on subtracting (1A)} \\ -140 &= 20B + C \qquad \text{(2)} \end{align*}$

Recognising that further further substitutions will be equally messy, I would turn to the coefficient of $x^6$ as it is obviously zero on the LHS of the identity and arises only fro the terms in $u^6$ and $u^8$ on RHS:

$\begin{align*} u^6 &= \big(x + x^{-1}\big)^6 \\ &= \binom{6}{0}x^6 + \binom{6}{1}x^5x^{-1} + \binom{6}{2}x^4x^{-2} + ... + \binom{6}{6}x^{-6} \\ &= x^6 + 6x^4 + 15x^2 + ... + x^{-6} \end{align*}$

The term in $x^6$ on the RHS is clearly $(8A + B)x^6$ , giving us that $B = -8A = -8$ after equating coefficients. We can then use (2) to get that $C = -140 - 20B = -140 + 160 = 20$ and then use (1A) to get that $D = -64 - 16B - 4C = -64 + 128 - 80 = -16$ .

Thus, we have the results $A = 1$ , $B = -8$ , $C = 20$ , $D = -16$ , and $E = 2$ , as expected.

CM_Tutor · May 4, 2020

Anyone looking for a challenge might consider proving that, for any positive integer $n$ , $x^{2n} + x^{-2n}$ can be re-written as an even degree $2n$ polynomial in $u = x + x^{-1}$ . Can you establish that the coefficients must be integers?

idkkdi · May 4, 2020

CM_Tutor said:
Anyone looking for a challenge might consider proving that, for any positive integer $n$ , $x^{2n} + x^{-2n}$ can be re-written as an even degree $2n$ polynomial in $u = x + x^{-1}$ . Can you establish that the coefficients must be integers?

Similar method as above for placing that equation into a u form. Afterwards, we establish that the coefficient of the term with the highest degree of any one expansion of a u to the exponent term, is always 1. As such we can times the leading term by a certain number to render all terms of that degree as 0.

Now we have proved that A,B,C,D,E,F etc. must be integer.

For the other part of the question, it is given that the equation consists of two terms. x^2n and x^-2n. An even function + another even function will always yield another even degree function. (Any method of proving this?)

Trebla · May 4, 2020

A sneakier approach without as much algebra bashing:

Let
$u_n = x^n + \frac{1}{x^n}$
We can show that
$u_{2n} = (u_n)^2 - 2$
Hence
$u_8 = (u_4)^2 - 2\\= ((u_2)^2 - 2)^2 - 2\\= (((u_1)^2 - 2)^2 - 2)^2 - 2\\= ((u^2 - 2)^2 - 2)^2 - 2\\= (u^4 - 4u^2 + 2)^2 - 2\\= u^8 + 16u^4 + 4 + 2(-4u^6 + 2u^4 - 8u^2) - 2\\= u^8 - 8u^6 + 20u^4 - 16u^2 + 2$

Drongoski · May 4, 2020

Brilliant!

CM_Tutor · May 6, 2020

idkkdi said:
For the other part of the question, it is given that the equation consists of two terms. x^2n and x^-2n. An even function + another even function will always yield another even degree function. (Any method of proving this?)

Easily.

Let $f(x)$ and $g(x)$ be even functions and let $h(x) = f(x) + g(x)$ . We seek to show that $h(x)$ is even.

Since we know that $f(x)$ is even, we know that $f(-x) = f(x)$ . . . Eqn (1)

And since $g(x)$ is also even, we know that $g(-x) = g(x)$ . . . Eqn (2)

Now, as $h(x) = f(x) + g(x)$ it follows that:

$\begin{align*} h(-x) &= f(-x) + g(-x) \\ &= f(x) + g(x) \quad \text{using Eqn (1) and Eqn (2)} \\ &= h(x) \quad \text{using the definition of $h(x)$} \end{align*}$

This shows that $h(-x) = h(x)$ , and so $h(x)$ must be an even function. It is also sufficient to prove that $f(x) - g(x)$ must be even, and a similar approach would allow you to establish that $f(x)g(x)$ and $\cfrac{f(x)}{g(x)}$ are both even.

---

The result I was thinking of, however, is that a polynomial $P(x)$ of degree $2n$ that can only ever have terms in $x^k$ , $k \in \mathbb{N}$ , if $k$ is even. This is relevant to the question that I asked as it proves that there are no terms in $u$ or $u^3$ or $u^5$ , etc, which might intuitively obvious but still needs to be rigorously established. You might also consider how Trebla's excellent solution can be extended to cases where $2n$ is not a power of two, like for finding $u_6$ or $u_{14}$ , for instance.

Pedro123 · May 6, 2020

Trebla said:
A sneakier approach without as much algebra bashing:

Let
$u_n = x^n + \frac{1}{x^n}$
We can show that
$u_{2n} = (u_n)^2 - 2$
Hence
$u_8 = (u_4)^2 - 2\\= ((u_2)^2 - 2)^2 - 2\\= (((u_1)^2 - 2)^2 - 2)^2 - 2\\= ((u^2 - 2)^2 - 2)^2 - 2\\= (u^4 - 4u^2 + 2)^2 - 2\\= u^8 + 16u^4 + 4 + 2(-4u^6 + 2u^4 - 8u^2) - 2\\= u^8 - 8u^6 + 20u^4 - 16u^2 + 2$

This is actually so insanely big-brained it isn't funny. I mean for God's sakes he generalised the damn thing and still had less working than a conventional solution.

Binomials question (1 Viewer)

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