$P(x)$ is a polynomial of least degree such that $P(k) = \frac{1}{k}; k = 1,2,3,..., 9$, then $P(10)$ is equal to?

Question

$P(x)$ is a polynomial of least degree such that $P(k) = \frac{1}{k}; k = 1,2,3,..., 9$, then $P(10)$ is equal to

My try:

Let $P(x) - \frac{1}{x} = Q(x)(x-1)(x-2)...(x-9)$.

But I'm not getting how to transform $P(x)$ into a polynomial .

Yes, $P(x)$ is still a polynomial. The hypotheses tell you something about the roots of $x P(x) - 1$, which is a different polynomial. There's no need to work with rational functions here. — Qiaochu Yuan, Jun 07 '24 at 02:43
The polynomial $xP(x) -1 $ has 9 roots, and by Lagrange, the minimum such $P$ has degree at most 8, hence P has degree 8, now you just match the constant terms — D S, Jun 07 '24 at 02:46

Mike · Accepted Answer · 2024-06-07T16:00:00.437

5

EXTENDED HINT:

So the equation $$xP(x)-1 = Q(x)(x-1)(x-2)\cdots(x-9)$$ holds, for some polynomial $Q(x)$, yielding $$xP(x)\ = \ 1 + Q(x)(x-1)(x-2)\cdots (x-9).$$ Thus, $xP(x)$ must have degree at least $9$, which yields $P(x)$ a polynomial of degree at least $8$.
We claim we can set $Q(x)=K$ for some constant $K$ that makes the RHS [of the 2nd equation in 1.] evalaute to $0$ at $x=0$, and $P(x)$ will still be a polynomial in $x$. [Indeed, if the RHS of the above evaluates to $0$ at $x=0$, then the RHS is a polynomial in $x$ with a root at $x=0$. Thus, the RHS divided by $x$ which is $P(x)$, will still be a polynomial in $x$.] The RHS [of the 2nd equation in 1.] will evaluate to $0$ at $x=0$ however, if $Q(x)=K$ where $K = \frac{(-1)×(-1)^9}{9!} = \frac{1}{9!}$ and thus $$xP(x) \ = \ 1 + \Big(\frac{1}{9!} \times (x-1)\cdots(x-9)\Big).$$ Check that indeed this satisfies both $xP(x)|_{x=0} =0$ and $xP(x)=1$ for all $x=1,2,3,\ldots, 9$. Also, check that $P(x)$ is of degree precisely $8$ which is the lower bound established in 1. above.
Evaluate the RHS of 2. as above at $x=10$ to get $xP(x)|_{x=10} = 2$, and so $P(x)|_{x=10}$ $= \frac{2}{10}$ $= \frac{1}{5}$.

edited Jun 07 '24 at 16:00

answered Jun 07 '24 at 02:47

Mike

Well why can't we simply set $Q(x) = 1$ instead ? – Ash_Blanc Jun 07 '24 at 02:54
The RHS evaluated at $x=0$ must be $0$ for $x$ to divide the RHS. This forces $Q$ to be -1/9!$ – Mike Jun 07 '24 at 02:57
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If $Q(x)=1$ then $1+Q(x)(x-1)\cdots(x-9)$ is not divisive by $x.$ @Ash_Blanc – Thomas Andrews Jun 07 '24 at 02:58
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@mike Should it be $1+$ where you have $(-1)+?$ – Thomas Andrews Jun 07 '24 at 02:59
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Yes @ThomasAndrews just corrected! – Mike Jun 07 '24 at 03:05
Well @Mike how can u claim that $Q(x)$ is of degree 0 not 1 ?? – Ash_Blanc Jun 07 '24 at 03:12
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@Ash_Blanc then can claim $Q(x)$ is degree zero because they demonstrated a degree zero polynomial that satisfies the equation. Since $P$ has minimal degree, so must $Q$. – NaturalLogZ Jun 07 '24 at 03:17
But how exactly , $P(x) - \frac{1}{x}$ was itself divisible by $(x-1)(x-2)...(x-9)$ right ?? Then how $xP(x) - 1$ can be of degree less than 10 ?? – Ash_Blanc Jun 07 '24 at 03:19
@Ash_Blanc what you have assumed is not correct... according to your assumption we can't have $P(0)$ but that's not given in the question – Blue Cat Blues Jun 07 '24 at 03:36
@Ash_Blanc $P(x)$ is a polynomial, but your $P(x) -\frac 1x$ is not a polynomial. – peterwhy Jun 07 '24 at 03:38

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