Hello every 409645π(72.5).
I hope you are having a great time!
I have presumably found a way to solve
(r1+m)(r2+m)…(rn+m)
if rn are the roots to any equation of the form f(x)=anxn+an−1xn−1+…a1x+a0
and m is any real or complex number.
Claim:
(r1+m)(r2+m)…(rn+m)=(−1)nanf(−m)
if f(x)=anxn+an−1xn−1+…a1x+a0
Simply put an is the coefficient of xn in f(x) or even simpler,
an is the leading coefficient of f(x).
Example
Lets have an explicit example where m=3 and f(x)=2x3+12x2−14x−120
So (r1+3)(r2+3)(r3+3)=r1r2r3+3r1r2+3r1r3+3r2r3+9r1+9r2+9r3+27
Then using Vieta's,
r1r2r3+3r1r2+3r1r3+3r2r3+9r1+9r2+9r3+27=−2−120+3×2−14−9×212+27=12
and indeed (−1)32f(−3)=12
and also the roots of f(x)=2x3+12x2−14x−120 is −5,−4 and 3. Meaning the product is (−5+3)(−4+3)(3+3)=12
Proof 1
(r1+m)(r2+m)…(rn+m)=m0∏y=1nrn+m1(∑cycr1r2…rn−1)+⋯+mn−1(∑cycr1)+mn(∑n=1n1)
Using Vieta's,
m0∏y=1nry+m1(∑cycr1r2…rn−1)+⋯+mn−1(∑cycr1)+mn(∑n=1n1)=(−1)nanm0a0+⋯(−1)1anmn−1an−1+(−1)0anmnan=∑y=0nan(−1)ymn−yan−y=(−1)nanf(−m)
Proof 2
With the basic definition of a polynomial,
anxn+an−1xn−1+…a1x+a0=an(x−r1)(x−r2)⋯(x−rn)
(r1+m)(r2+m)…(rn+m)=(−1)n(−m−r1)(−m−r2)⋯(−m−rn)=(−1)nanf(−m)
Broader definition
(kr1+m)(kr2+m)…(krn+m)
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
(kr1+m)(kr2+m)…(krn+m)=kn(r1+km)(r2+km)⋯(rn+km)=(−1)nknanf(−km)=(−k)nanf(−km)
Phew...
I really do hope this note helped you in one way or another! Well, that's probably it!
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Comments
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Top NewestNice 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 :)
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thanks!
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Perhaps you can add questions into this!
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Try this then
Try this then
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great note @Joel Yip !!!!!
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thank you!
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Should I give more examples?
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nice work Joel. thumbs up!!
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Thank you to all of the positive replies to this note! I just want to say thanks again!
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Good Note. Nice work Joel Yip.
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