3
$\begingroup$

Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$.

Let $P(x)=x^{2015}-x^{2014}=Q(x)(x-1)^3+ax^2+bx+c.$ If we put $x=1$ in $P(x)$ and $P'(x)$, we get $a+b+c=0$ and $2a+b=1$. Then: $c=a-1$. The second derivative won't help in finding $b$, so, what should I do? Thank you

$\endgroup$
  • 1
    $\begingroup$ Admittedly I haven't done out the full calculation, but doesn't the second derivative give you $a$? Then you could substitute to find $b$ and $c$. $\endgroup$ – Cade Reinberger Aug 22 at 21:34
  • $\begingroup$ No, because $P''(x)=Q''(x)(x-1)^3+Q'(x)3(x-1)^2+Q'(x)3(x-1)+Q(x)3+2a$, and we don't know any value of $Q(x)$. $\endgroup$ – user672596 Aug 22 at 21:36
  • 1
    $\begingroup$ This may help $\frac{x^{2015} - x^{2014}}{(x-1)^3} = \frac{x^{2014}}{(x-1)^2}$ $\endgroup$ – Virtuoz Aug 22 at 21:36
  • 2
    $\begingroup$ @karim-ashli you have only two coefficients then, so you argument with derivatives will work $\endgroup$ – Virtuoz Aug 22 at 21:39
  • 2
    $\begingroup$ If $x^{2014}=S(x)(x-1)^2+qx+r$ then $x^{2015}-x^{2014}=S(x)(x-1)^3+(qx+r)(x-1)$ $\endgroup$ – Mark Bennet Aug 22 at 21:42
6
$\begingroup$

Rewrite expression as $$ \frac{x^{2015} - x^{2014}}{(x-1)^3} = \frac{x^{2014}}{(x-1)^2} $$ so we have $$ x^{2014} = Q(x) (x-1)^2 + ax + b. $$

Now you need to find coefficients $a$ and $b$. Your argument with derivatives should work: $$ 1^{2014} = 1 = a + b $$ $$ 2014\cdot 1^{2013} = 2014 = a. $$

Hence, $b = -2013$ and the final answer is $$ x^{2015} - x^{2014} = x^{2014} (x-1) = Q(x) (x-1)^3 + (x-1)\cdot(2014x - 2013) $$

$\endgroup$
3
$\begingroup$

$\dfrac{x^{2014}(x-1)}{(x-1)^3}= \dfrac{x^{2014}}{(x-1)^2}$;

$x^{2014}= Q(x)(x-1)^2+ ax+b$;

Binomial expansion:

$x^{2014}=(1+(x-1))^{2014}=$

$\sum_{k=0}^{n}\binom{2014}{k}1^{n-k}(x-1)^k$

Remainder $ax+b$:

$\binom{2014}{0}1+\binom{2014}{1}(x-1)=$

$1+2014x-2014=2014x-2013$.

Originally:

$x^{2014}(x-1)$ is divided by $(x-1)^3$:

Hence

$x^{2014}(x-1)=Q(x)(x-1)^3+(x-1)(2014x-2013),$

with quadratic remainder

$(x-1)(2014x-2013)$.

$\endgroup$
2
$\begingroup$

Using $\ zf\,\bmod zg\ = z\, (f \bmod g)\, =\, $ mod Distributive Law to factor out $\, z = x\!-\!1$

$\begin{align} z(z\!+\!1)^{\large n} \bmod z^{\large 3} &=\, z\,(\color{#c00}{(1+ z)^{\large n}}\bmod \color{#c00}{z^{\large 2}})\\[.4em] &=\,z\,(\color{#c00}{1+nz})\ \ \text{ by } \color{#c00}{\text{Binomial or Taylor }}\text{Theorem}\\[.4em] &=\ n\!-\!1 + (1\!-\!2n)x + n x^{\large 2}\ \ \ \ \ \text{[OP is }\, n = 2104] \end{align}$

$\endgroup$
1
$\begingroup$

Another way:

Set $x-1=y$

$x^r=(1+y)^r\equiv1+\binom r1y+\binom r2y^2\pmod{y^3}$

$x^m-x^n\equiv y(m-n)+y^2\left(\binom m2-\binom n2\right)\pmod{y^3}$

Replace $y$ with $x-1$

$\endgroup$
  • $\begingroup$ (+1) This is the way I did this. You can do it in your head this way. $\endgroup$ – robjohn Aug 23 at 9:09
0
$\begingroup$

I can use the division algorithm of a polynomial by another polynomial: in particular set $A(x)=x^{2014}$ and $B(x)=x^2-2x+1$. Proceeding with successive division, I obtain as quotient $$Q(x)=x^{2012}+2x^{2011}+3x^{2010}+\cdots + 2012x+2013$$ The general term of this polynomial is $nx^{2013-n}$. When $n=2013$, I obtain as polynomial rest: $R(x)=-2012x+4026x-2013=2014x-2013$. Before starting this process I have cancel out $(x-1)$ from the numerator and denominator of the fraction, so: $$Q(x)=(x^{2012}+2x^{2011}+3x^{2010}+\cdots + 2012x+2013)(x-1)$$ and $$R(x)=(2014x-2013)(x-1)$$

$\endgroup$
0
$\begingroup$

It is: $$\frac{x^{2015}-x^{2014}}{(x-1)^3}=\frac{(x-1+1)^{2015}-(x-1+1)^{2014}}{(x-1)^3}=\\ =\small{\frac{\left[S(x)+{2015\choose 2}(x-1)^2+{2015\choose 1}(x-1)+1\right]-\left[T(x)-{2014\choose 2}(x-1)^2-{2014\choose 1}(x-1)-1\right]}{(x-1)^3}}=\\ =\frac{S(x)-T(x)+2014(x-1)^2+(x-1)}{(x-1)^3}=Q(x)+\frac{(x-1)(2014x-2013)}{(x-1)^3}.$$ Note: It is more expanded and direct version of lab bhattacharjee's answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy