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Unless I am making a mistake, I am calculating that:

$$\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$$

Here's my reasoning:

  • $\dfrac{x^m + y^m}{x+y} = x^{m-1} - x^{m-2}y - xy^{m-2} + x^{m-3}y^2 + x^2y^{m-3} + \dots + x^{\frac{m-1}{2}}y^{\frac{m-1}{2}} + y^{m-1}$

  • $\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-3} - x^{m-4}y - xy^{m-4} + x^{m-5}y^2 + x^2y^{m-5} + \dots + x^{\frac{m-3}{2}}y^{\frac{m-3}{2}} + y^{m-3}$

  • $(xy)\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-2}y - x^{m-3}y^2 - x^2y^{m-3} + x^{m-4}y^3 + x^3y^{m-4} + \dots + x^{\frac{m-1}{2}}y^{\frac{m-1}{2}} + xy^{m-2}$

  • So that, $\dfrac{x^m + y^m}{x+y} + (xy)\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$

  • I am figuring that $\dfrac{x^{m-2} + y^{m-2}}{x+y}$ has exactly 2 terms less than $\dfrac{x^m + y^m}{x+y}$

Is this reasoning correct? For each value that I test, it seems correct.

Thanks,

-Larry

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    $\begingroup$ It should be noted, this is true for even $m$ as well, but your proof won't work for even $m$. Indeed, it is even true for fractional $m$, and negative $m$, and $m=1$. $\endgroup$ – Thomas Andrews Feb 17 '15 at 19:11
  • $\begingroup$ Very good point. Thanks. $\endgroup$ – Larry Freeman Feb 17 '15 at 19:18
  • $\begingroup$ Also, the typical way to write the expansion is: $$\frac{x^m+y^m}{x+y} = x^{m-1}-x^{m-2}y+x^{m-3}y^2-\cdots - xy^{m-2}+y^{m-1}$$ It makes it easier to put in $\Sigma$ notation: $$\sum_{k=0}^{m-1} (-1)^kx^{m-1-k}y^k$$ Your formula is $$\sum_{k=0}^{\frac{m-1}{2}} (-1)^k (x^{m-1-k}y^k + x^ky^{m-1-k})$$ which is one way to look at the pattern, but a little odd, and has a problem when $k=\frac{m-1}{2}$, so you'll have to adjust that last term. Really not pretty. $\endgroup$ – Thomas Andrews Feb 17 '15 at 19:33
  • $\begingroup$ Thanks for the correction. I appreciate that. I'll keep that in mind. $\endgroup$ – Larry Freeman Feb 17 '15 at 20:11
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Well, I would do it that way : $$ x^m = x\ x^{m-1} $$ So :

$$\begin{align}\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} &= \frac{x^m + y^m+ (xy)\ x^{m-2} + (xy)\ y^{m-2}}{x+y} \\ &=\frac{x^m + y^m+ y\ x^{m-1} + x\ y^{m-1}}{x+y} \\ &=\frac{ x\ x^{m-1} + y\ y^{m-1}+ y\ x^{m-1} + x\ y^{m-1}}{x+y}\\ &=\frac{ (x+y)\ x^{m-1} + (x+y)\ y^{m-1}}{x+y}\\ &= x^{m-1} + y^{m-1}\end{align}$$

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  • $\begingroup$ Because it's easier to see what's going on when you don't have tons of terms. $\endgroup$ – Lery Feb 17 '15 at 1:30
  • $\begingroup$ It's probably even easier to multiply both sides by $(x+y)$ to avoid all the fractions :) $\endgroup$ – Thomas Andrews Feb 17 '15 at 3:10
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Your argument is fine, possibly a bit more complicated than necessary. And the theorem is true for $m$ even, too, so a better proof wouldn't use the closed form for $\frac{x^m+y^m}{x+y}$, which is only true for $m$ odd.

A more advanced way to prove this uses linear recurrences. Show that $a_n=x^n+y^n$ satisfies:

$$a_{n}=(x+y)a_{n-1} - (xy)a_{n-2}$$

It is easy to show this linear recurrence is true for $b_n=x^n$ and $c_n=y^n$ because $z^2-(x+y)z+xy=0$ has roots $z=x,y$. So it is true for $a_n=b_n+c_n$.

The recurrence is also true if $a_n=\alpha\cdot x^n + \beta \cdot y^n$ for any $\alpha,\beta,x,y$. So:

$$\frac{\alpha x^m + \beta y^m}{x+y} + (xy)\frac{\alpha x^{m-2} + \beta y^{m-2}}{x+y} = \alpha x^{m-1} +\beta y^{m-1}$$

That shows that the $x+y$ in the denominator is not part of the original pattern, $a_1$, but related to the $x,y$ independent of the $\alpha,\beta$.

This can then be generalized to more variables. If $a_n=\alpha x^n+\beta y^n+\gamma z^n$ then:

$$a_n = (x+y+z)a_{n-1} -(xy+yz+xz)a_{n-2} + (xyz)a_{n-3}$$

You can go to more variables, with the coefficients of the linear recurrence being the (alternating) basic symmetric polynomials.

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$$\frac{x^m+y^m}{x+y}+(xy)\frac{x^{m-2}+y^{m-2}}{x+y}=\frac{x^m+y^m+(xy)(x^{m-2}+y^{m-2})}{x+y}= \\ =\frac{x^m+y^m+x^{m-1}y+xy^{m-1}}{x+y}=\frac{x^m+x^{m-1}y+xy^{m-1}+y^m}{x+y}\\ =\frac{x^{m-1}(x+y)+y^{m-1}(x+y)}{x+y}=x^{m-1}+y^{m-1}$$

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