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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

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    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 $c$. – Cade Reinberger Aug 22 '19 at 21:34
  • 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 $Q(x)$. –  Aug 22 '19 at 21:36
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    This may help $\frac{x^{2015} - x^{2014}}{(x-1)^3} = \frac{x^{2014}}{(x-1)^2}$ – Virtuoz Aug 22 '19 at 21:36
  • @Virtuoz tried, the remainder is $a_1 x + b_1$, but how can I get the third coefficient? –  Aug 22 '19 at 21:37
  • It might also help to rewrite $\tilde{x} = x+1$? – Kitter Catter Aug 22 '19 at 21:38
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    @karim-ashli you have only two coefficients then, so you argument with derivatives will work – Virtuoz Aug 22 '19 at 21:39
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    If $x^{2014}=S(x)(x-1)^2+qx+r$ then $x^{2015}-x^{2014}=S(x)(x-1)^3+(qx+r)(x-1)$ – Mark Bennet Aug 22 '19 at 21:42
  • Actually $P''(x) = Q(x) \cdot 6(x-1) + 2 Q'(x) \cdot 3(x-1)^2 + Q''(x) (x-1)^3 + 2a$ so plugging in $x=1$ will still cancel off all the $Q$ contributions. – Daniel Schepler Aug 22 '19 at 21:43
  • Another possible approach: $x^{2015} = [(x-1)+1]^{2015} = \sum_{k=0}^{2015} \binom{2015}{k} (x-1)^{2015-k}$ where all but three terms are divisible by $(x-1)^3$; and similarly for $x^{2014}$. – Daniel Schepler Aug 22 '19 at 21:48

6 Answers6

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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 $$ x^{2015} - x^{2014} = x^{2014} (x-1) = Q(x) (x-1)^3 + (x-1)\cdot(2014x - 2013) $$

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

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

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

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

Peter Szilas
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Using $\ \color{#c00}zf\,\bmod \color{#c00}zg\ = \,\color{#c00}z\, (f \bmod g)\, =\, $ mod Distributive Law to factor out $\, \color{#c00}{z := x\!-\!1}$

$\,\begin{align} \smash[b]{\underbrace{\color{#c00}z(z\!+\!1)^{\large n} \bmod \color{#c00}z^{\large 3}}_{\!\!\!\!\!(x-1)x^n\bmod (x-1)^3}} &=\, \color{#c00}z\,(\color{#0a0}{(1+ z)^{\large n}}\bmod \color{#0a0}{z^{\large 2}})\qquad\ \ \ \ \text{[OP is }\, n = 2014]\\[.4em] &=\,z\,(\color{#0a0}{1+nz})\ \ \text{ by } \color{#0a0}{\text{Binomial or Taylor }}\text{Theorem}\\[.4em] &=\ n\!-\!1 + (1\!-\!2n)\,x + n\, x^{\large 2}\ \ \ \ {\rm by}\,\ \color{#c00}{z=x\!-\!1} \end{align}$

Bill Dubuque
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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 $x-1$

lab bhattacharjee
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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 $$R(x)=(2014x-2013)(x-1)$$

Matteo
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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.

farruhota
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