Formula for Root Product

In solving this problem, I found a formula for a product $(1+x_{1})(1+x_{2})(1+x_{3})(1+x_{4})...(1+x_{d})$ up until the last root of a polynomial of degree d. I challenge Brilliant members to find this formula, with proof.

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

it would be kind enough if u tell us......

Max B - 7 years, 2 months ago

There is a pattern.....for example...............

( a + 1 ) ( b + 1 ) ( C + 1 ) = 1 + a + b + c + a b + b c + c a + a b c .

Hope you want this!

Satvik Golechha - 7 years, 2 months ago

Yes, this is the pattern, but it can be simplified even further. Think about the certain terms in your example.

Tristan Shin - 7 years, 2 months ago

Well, Tristan, I can't get it......please post the answer.....

Satvik Golechha - 7 years, 2 months ago

Let the polynomial be $f(x)$ with degree $n$ and roots $r_1, r_2, \dots, r_{n-1}, r_{n}$

Suppose $f(x) = a_0x^n + a_1x^{n-1} + \dots + a_{n-1}x + a_n = a_0(x-r_1)(x-r_2)\dots(x-r_{n-1})(x-r_{n})$

Then $f(-1) = a_0(-1)^n + a_1(-1)^{n-1} + \dots + a_{n-1}(-1) + a_n = a_0(-1-r_1)(-1-r_2)\dots(-1-r_{n-1})(-1-r_n)$

$f(-1) = a_0(-1)(1+r_1)(-1)(1+r_2)\dots(-1)(1+r_{n-1})(-1)(1+r_n)$

$\dfrac{f(-1)}{a_0} = (-1)^n(1+ r_1)(1 + r_2)\dots(1+ r_{n-1})(1+ r_n)$

$(1+ r_1)(1 + r_2)\dots(1+ r_{n-1})(1+ r_n) = \left\{ \begin{array}{lr} \frac{-f(-1)}{a_0} & : n \equiv 1 \pmod2 \\ \frac{f(-1)}{a_0} & : n \equiv 0 \pmod2 \end{array} \right.$

Siddhartha Srivastava - 7 years, 2 months ago

The solution is almost complete. The third time you state $f\left(-1\right)$ , this can be simplified after in the next step. In addition, what if there is a coefficient of the $x^{n}$ term? Consider these and make an edit.

Tristan Shin - 7 years, 2 months ago

I don't understand what you meant in the first line. I've edited for the case when there is a coefficient of the leading term.

Siddhartha Srivastava - 7 years, 2 months ago

@Siddhartha Srivastava – Your solution is just one thing away. Instead of using modular arithmetic in your final formula, what can you do with the sign instead?

Tristan Shin - 7 years, 1 month ago

@Tristan Shin – Are you implying $(-1)^n$ ?

Daniel Liu - 7 years, 1 month ago

I'll let some more people try, then I'll post this answer sometime next week(April 14 to 18).

Tristan Shin - 7 years, 2 months ago

It seems that Siddhartha Srivastava has come very close to solving this problem. Congratulations!

Tristan Shin - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$