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Can we find the coefficients of $P(x)=a_4 x^4+a_3 x^3+a_2 x^2 +a_1 x +a_0$ by computing $P(100)$?

Example 1:

Let $$P_1(x)=(x+1)(x+2)(x+3)(x+4).$$ Expanded form of $P_1(x)$ is $$P_1(x)= x^4 + 10 x^3 + 35 x^2 + 50 x + 24.$$

For $x=100$, we get $$P_1(100)=110355024.$$

Divide 110355024 by 2-digit from the right to the left, we get $$P_1(100) \rightarrow 1/10/35/50/24$$ Looking at 1/10/35/50/24, we will find the coefficients of $P_1(x)$: $$P_1(100) \rightarrow 1/10/35/50/24 \rightarrow 1x^4 + 10 x^3 + 35 x^2 + 50 x + 24.$$

Example 2:

Let $$P_2(x)=-(x+1)(x+2)(x+3)(x+4).$$ Expanded form of $P_2(x)$ is $$P_2(x)= -x^4 - 10 x^3 - 35 x^2 - 50 x - 24.$$

For $x=100$, we get $$P_2(100)=-110355024.$$

Divide -110355024 by 2-digit from the right to the left, we get $$P_2(100) \rightarrow -(1/10/35/50/24)$$ Looking at -(1/10/35/50/24), we will find the coefficients of $P_2(x)$: $$P_2(100) \rightarrow -(1/10/35/50/24) \rightarrow -1x^4 - 10 x^3 - 35 x^2 - 50 x - 24.$$

Example 3:

Let $$P_3(x)=(x-1)(x-2)(x-3)(x+4).$$ Expanded form of $P_3(x)$ is $$P_3(x)=x^4 - 2 x^3 - 13 x^2 + 38 x - 24.$$

For $x=100$, we get $$P_3(100)=97873776.$$

Divide 97873776 by 2-digit from the right to the left, we get $$P_3(100) \rightarrow 97/87/37/76$$ Since $76>50$, we transfer $97/87/37/76$ into $$P_3(100) \rightarrow 97/87/(37+1)/(76-100)\rightarrow 97/87/38/-24$$ Since $38<50$, we let $38$ alone. Since $87>50$, we transfer $97/87/38/-24$ into $$P_3(100) \rightarrow (97+1)/(87-100)/38/-24\rightarrow 98/-13/38/-24$$ Since $98>50$, we transfer $98/-13/38/-24 \rightarrow 0/98/-13/38/-24$ into $$P_3(100) \rightarrow (0+1)/(98-100)/-13/38/-24\rightarrow 1/-2/-13/38/-24.$$

Looking at 1/-2/-13/38/-24, we will find the coefficients of $P_3(x)$: $$P_3(100) \rightarrow 1/-2/-13/38/-24 \rightarrow x^4 - 2 x^3 - 13 x^2 + 38 x - 24$$

More generally, we have a proposition.

Proposition:

Let $P(x)=a_4 x^4+a_3 x^3+a_2 x^2 +a_1 x +a_0$.

If $a_i\in \mathbb{Z}\cap [-49;50]$ $\left( i=\overline{1,4} \right)$, then we can find $a_i \left(i=\overline{1,4}\right)$ by computing $P(100)$ and using algorithm which is shown in 3 ealier examples.

Question:

Is the proposition true or false?

If the proposition is false, please give me some counter-examples.

If the proposition is true, please give me some clues to prove the proposition.

Vincenzo Tibullo
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Mai Hoàn Hảo
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    This would mean two polynomials taking the same value at $100$ are equal. – Bernard Jun 18 '17 at 14:37
  • @Winther In the proposition, the coefficients of $P(x)$ are $a_i\in \mathbb{Z}\cap [-49;50]$ $\left( i=\overline{1,4} \right)$, so $P(x)=x-100$ and $P(x)=2(x-100)$ are not counter-examples. – Mai Hoàn Hảo Jun 18 '17 at 14:45
  • @MaiHoànHảo I just noticed that so I removed the comment. – Winther Jun 18 '17 at 14:45
  • @Winther If the proposition is true, please give me some clues to proof the proposition. – Mai Hoàn Hảo Jun 18 '17 at 14:46
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    Write $a_i = -49 + \tilde{a_i}$ where $\tilde{a}_i \in[0,99]$ then $P(100) = [\tilde{a}_4 10^8 + \tilde{a}_3 10^6 + \tilde{a}_2 10^4 + \tilde{a}_1 10^2 + \tilde{a}_0] - 49(10^8+10^6+10^4+100+1)$ and notice that the decimal digits from the different terms in the first bracket will not mix. – Winther Jun 18 '17 at 14:53

1 Answers1

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If you substitute $x=100$ then you get $a_4(100)^4+a_3(100)^3+a_2(100)^2+a_1(100)+a_0$ which is the base 100 expansion of a number, which is unique when the coefficients are less than 100.

Vincenzo Tibullo
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i. m. soloveichik
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  • Thank you for your detailed and clear explanation. It was very helpful in the case of $a_i\in [0;50] \cap \mathbb{Z}$. How about when $a_i <0$ since the condition is that $a_i\in [-49;50]\cap \mathbb{Z}$? – Mai Hoàn Hảo Jun 20 '17 at 14:37