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Generalisation of $$\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right)$$

Hello every $$\frac { 45\pi }{ 4096 } \left( \begin{matrix} 7 \\ 2.5 \end{matrix} \right)$$.

I hope you are having a great time!

I have presumably found a way to solve

$$\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right)$$

if $${ r }_{ n }$$ are the roots to any equation of the form $$f\left( x \right)={ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }+\dots { a }_{ 1 }{ x }+{ a }_{ 0 }$$

and $$m$$ is any real or complex number.

Claim:

$$\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right) ={ \left( -1 \right) }^{ n }\frac { f\left( -m \right) }{ { a }_{ n } }$$

if $$f\left( x \right)={ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }+\dots { a }_{ 1 }{ x }+{ a }_{ 0 }$$

Simply put $${ a }_{ n }$$ is the coefficient of $${ x }^{ n }$$ in $$f\left( x \right)$$ or even simpler,

$${ a }_{ n }$$ is the leading coefficient of $$f\left( x \right)$$.

Example

Lets have an explicit example where $$m=3$$ and $$f\left( x \right) =2{ x }^{ 3 }+12{ x }^{ 2 }-14x-120$$

So $$\left( { r }_{ 1 }+3 \right) \left( { r }_{ 2 }+3 \right) \left( { r }_{ 3 }+3 \right) ={ r }_{ 1 }{ r }_{ 2 }{ r }_{ 3 }+3{ r }_{ 1 }{ r }_{ 2 }+3{ r }_{ 1 }{ r }_{ 3 }+3{ r }_{ 2 }{ r }_{ 3 }+9{ r }_{ 1 }+9{ r }_{ 2 }+9{ r }_{ 3 }+27$$

Then using Vieta's, $${ r }_{ 1 }{ r }_{ 2 }{ r }_{ 3 }+3{ r }_{ 1 }{ r }_{ 2 }+3{ r }_{ 1 }{ r }_{ 3 }+3{ r }_{ 2 }{ r }_{ 3 }+9{ r }_{ 1 }+9{ r }_{ 2 }+9{ r }_{ 3 }+27=-\frac { -120 }{ 2 } +3\times \frac { -14 }{ 2 } -9\times \frac { 12 }{ 2 } +27\\ =12$$

and indeed $${ \left( -1 \right) }^{ 3 }\frac { f\left( -3 \right) }{ 2 } =12$$

and also the roots of $$f\left( x \right) =2{ x }^{ 3 }+12{ x }^{ 2 }-14x-120$$ is $$-5$$,$$-4$$ and $$3$$. Meaning the product is $$\left( -5+3 \right) \left( -4+3 \right) \left( 3+3 \right) =12$$

Proof 1

$$\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right) ={ m }^{ 0 }\prod _{ y=1 }^{ n }{ { r }_{ n } } +{ m }^{ 1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 }{ r }_{ 2 }\dots { r }_{ n-1 } } \right) +\cdots +{ m }^{ n-1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 } } \right) +{ m }^{ n }\left( \sum _{ n=1 }^{ n }{ 1 } \right)$$

Using Vieta's, $${ m }^{ 0 }\prod _{ y=1 }^{ n }{ { r }_{ y } } +{ m }^{ 1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 }{ r }_{ 2 }\dots { r }_{ n-1 } } \right) +\cdots +{ m }^{ n-1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 } } \right) +{ m }^{ n }\left( \sum _{ n=1 }^{ n }{ 1 } \right) ={ \left( -1 \right) }^{ n }\frac { { m }^{ 0 }{ a }_{ 0 } }{ { a }_{ n } } +\cdots { \left( -1 \right) }^{ 1 }\frac { { m }^{ n-1 }{ a }_{ n-1 } }{ { a }_{ n } } +{ \left( -1 \right) }^{ 0 }\frac { { m }^{ n }{ a }_{ n } }{ { a }_{ n } } \\ =\sum _{ y=0 }^{ n }{ \frac { { \left( -1 \right) }^{ y }{ m }^{ n-y }{ a }_{ n-y } }{ { a }_{ n } } } \\ ={ \left( -1 \right) }^{ n }\frac { f\left( -m \right) }{ { a }_{ n } }$$

Proof 2

With the basic definition of a polynomial,

$${ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }+\dots { a }_{ 1 }{ x }+{ a }_{ 0 }={ a }_{ n }\left( x-{ r }_{ 1 } \right) \left( x-{ r }_{ 2 } \right) \cdots \left( x-{ r }_{ n } \right)$$

$$\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right) ={ \left( -1 \right) }^{ n }\left( -m-{ r }_{ 1 } \right) \left( -m-{ r }_{ 2 } \right) \cdots \left( -m-{ r }_{ n } \right) \\ ={ \left( -1 \right) }^{ n }\frac { f\left( -m \right) }{ { a }_{ n } }$$

$$\left( k{ r }_{ 1 }+m \right) \left( k{ r }_{ 2 }+m \right) \dots \left( { kr }_{ n }+m \right)$$

for $$k$$ being any real or complex number.

We too also can solve this by take $$k$$ out in each term and then use the generalisation to solve

$$\left( k{ r }_{ 1 }+m \right) \left( k{ r }_{ 2 }+m \right) \dots \left( { kr }_{ n }+m \right) ={ k }^{ n }\left( { r }_{ 1 }+\frac { m }{ k } \right) \left( { r }_{ 2 }+\frac { m }{ k } \right) \cdots \left( { r }_{ n }+\frac { m }{ k } \right) \\ ={ \left( -1 \right) }^{ n }{ k }^{ n }\frac { f\left( -\frac { m }{ k } \right) }{ { a }_{ n } } \\ ={ \left( -k \right) }^{ n }\frac { f\left( -\frac { m }{ k } \right) }{ { a }_{ n } }$$

Phew...

I really do hope this note helped you in one way or another! Well, that's probably it!

Note by Joel Yip
1 year, 10 months ago

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Nice note! @Joel Yip

Yes, I think this is a good generalization for such sums. No wonder we'll get to see more good problems from many others now :)

- 1 year, 10 months ago

thanks!

- 1 year, 10 months ago

Good Note. Nice work Joel Yip.

- 1 year, 10 months ago

Thank you to all of the positive replies to this note! I just want to say thanks again!

- 1 year, 10 months ago

nice work Joel. thumbs up!!

- 1 year, 10 months ago

Should I give more examples?

- 1 year, 10 months ago

great note @Joel Yip !!!!!

- 1 year, 10 months ago

thank you!

- 1 year, 10 months ago

Perhaps you can add questions into this!

- 1 year, 10 months ago