Newton's Sum

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Newton's Sum

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#1 h Apr 25, 2025, 1:51 AM

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If $\sum_{cyc} a = 7, \sum_{cyc} a^2 = 15, \sum_{cyc} a^3 = 19$ and $\sum_{cyc} a^4 = 25$ where each summation runs over $4$ variables $a,b,c$ and $d.$ Then find the value of $\sum_{cyc} a^5.$

This post has been edited 3 times. Last edited by P162008, Apr 25, 2025, 1:52 AM
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#2 h Apr 25, 2025, 2:31 AM

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Since the coefficient of $x^5$ in $\exp(-7x - \frac{15x^2}{2} - \frac{19x^3}{3} - \frac{25x^4}{4})$ is $\frac{279}{10}$ , the answer is $\frac{279}{2}$ .

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#3 h Apr 25, 2025, 2:49 AM

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lbh_qys wrote:

Since the coefficient of $x^5$ in $\exp(-7x - \frac{15x^2}{2} - \frac{19x^3}{3} - \frac{25x^4}{4})$ is $\frac{279}{10}$ , the answer is $\frac{279}{2}$ .

Could you please elaborate your approach?

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#4 h Apr 25, 2025, 11:12 AM

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P162008 wrote:

lbh_qys wrote:

Since the coefficient of $x^5$ in $\exp(-7x - \frac{15x^2}{2} - \frac{19x^3}{3} - \frac{25x^4}{4})$ is $\frac{279}{10}$ , the answer is $\frac{279}{2}$ .

Could you please elaborate your approach?

Assuming $\sum_{\text{cyc}} a^5 = S$ , consider the generating function $\sum_{\text{cyc}} \ln(1-ax)$ . Clearly, this yields $\sum_{k \geq 1} \left(\sum_{\text{cyc}} a^k\right) \cdot \left(-\frac{x^k}{k}\right)$ . Therefore, we obtain

$\sum_{\text{cyc}} \ln(1-ax) \equiv -7x - \frac{15x^2}{2} - \frac{19x^3}{3} - \frac{25x^4}{4} - S \frac{x^5}{5} \pmod{x^6}.$
Since the left-hand side is $\ln\left( \prod_{\text{cyc}} (1-ax) \right)$ , we have

$\ln\left(\prod_{\text{cyc}}(1-ax)\right) + \frac{S x^5}{5} \equiv -7x - \frac{15x^2}{2} - \frac{19x^3}{3} - \frac{25x^4}{4} \pmod{x^6}.$
Taking the exponential of both sides gives

$\prod_{\text{cyc}}(1-ax) \cdot \exp\left(\frac{S x^5}{5}\right) \equiv \exp\left(-7x - \frac{15x^2}{2} - \frac{19x^3}{3} - \frac{25x^4}{4}\right) \pmod{x^6}.$
Since $\exp\left(\frac{S x^5}{5}\right) \equiv 1 + \frac{Sx^5}{5} \pmod{x^6}$ , it is straightforward to observe that the coefficient of $x^5$ in $\prod_{\text{cyc}}(1-ax) \cdot \exp\left(\frac{S x^5}{5}\right)$ is $\frac{S x^5}{5}$ . This leads to the conclusion.

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#5 h Apr 25, 2025, 2:08 PM

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lbh_qys wrote:

P162008 wrote:

lbh_qys wrote:

Since the coefficient of $x^5$ in $\exp(-7x - \frac{15x^2}{2} - \frac{19x^3}{3} - \frac{25x^4}{4})$ is $\frac{279}{10}$ , the answer is $\frac{279}{2}$ .

Could you please elaborate your approach?

Assuming $\sum_{\text{cyc}} a^5 = S$ , consider the generating function $\sum_{\text{cyc}} \ln(1-ax)$ . Clearly, this yields $\sum_{k \geq 1} \left(\sum_{\text{cyc}} a^k\right) \cdot \left(-\frac{x^k}{k}\right)$ . Therefore, we obtain

$\sum_{\text{cyc}} \ln(1-ax) \equiv -7x - \frac{15x^2}{2} - \frac{19x^3}{3} - \frac{25x^4}{4} - S \frac{x^5}{5} \pmod{x^6}.$
Since the left-hand side is $\ln\left( \prod_{\text{cyc}} (1-ax) \right)$ , we have

$\ln\left(\prod_{\text{cyc}}(1-ax)\right) + \frac{S x^5}{5} \equiv -7x - \frac{15x^2}{2} - \frac{19x^3}{3} - \frac{25x^4}{4} \pmod{x^6}.$
Taking the exponential of both sides gives

$\prod_{\text{cyc}}(1-ax) \cdot \exp\left(\frac{S x^5}{5}\right) \equiv \exp\left(-7x - \frac{15x^2}{2} - \frac{19x^3}{3} - \frac{25x^4}{4}\right) \pmod{x^6}.$
Since $\exp\left(\frac{S x^5}{5}\right) \equiv 1 + \frac{Sx^5}{5} \pmod{x^6}$ , it is straightforward to observe that the coefficient of $x^5$ in $\prod_{\text{cyc}}(1-ax) \cdot \exp\left(\frac{S x^5}{5}\right)$ is $\frac{S x^5}{5}$ . This leads to the conclusion.

Isn't there any alternate approach by Newton's Sum?

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#6 h Apr 25, 2025, 2:28 PM

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With Newton's Sums, we find $P(x)=x^{4}-7x^{3}+17x^{2}-11x-\frac{35}{2}$ .

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#7 h Apr 25, 2025, 4:04 PM

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vanstraelen wrote:

With Newton's Sums, we find $P(x)=x^{4}-7x^{3}+17x^{2}-11x-\frac{35}{2}$ .

Nice correct!!

This post has been edited 2 times. Last edited by P162008, Apr 25, 2025, 7:28 PM
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#8 h Apr 25, 2025, 6:38 PM

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Take $P(x)=x^{4}+p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}$ .
Let $P_{k}=a^{k}+b^{k}+c^{k}+d^{k}$ .

1) $P_{1}+p_{3}=0 \Rightarrow p_{3}=-7$ .
2) $P_{2}+p_{3}P_{1}+2p_{2}=0 \Rightarrow p_{2}=17$ .
3) $P_{3}+p_{3}P_{2}+p_{2}P_{1}+3p_{1}=0 \Rightarrow p_{1}=-11$ .
4) $P_{4}+p_{3}P_{3}+p_{2}P_{2}+p_{1}P_{1}+4p_{0}=0 \Rightarrow p_{0}=-\frac{35}{2}$ .
5) $P_{5}+p_{3}P_{4}+p_{2}P_{3}+p_{1}P_{2}+p_{0}P_{1}=0$ .

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