Need to prove polynomial equality

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sadovnichek

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Hello, World!

Recently I faced with this equality:

загруженное (1).png

Where n > 1, n - natural number

And I don't know how to prove it. I used math induction, but then I broke my brain. Help please!
 
I'd say, at first glance, proof by mathematical induction would be the way to go. Show us where you got to before your brain broke.
 
Harry_the_cat said:
I'd say, at first glance, proof by mathematical induction would be the way to go. Show us where you got to before your brain broke.
Click to expand...
okay,
let n = 2:
[MATH]x^4 - 2x^3 + 2x - 1 = (x-1)^3*(x+1)[/MATH]Correct.
let n = n + 1
[MATH]x^{2n+2}-(n+1)x^{n+2}+(n+1)x^n - 1 = (x-1)^3(x+1)^{2n-1}[/MATH]
What next? I can't understand. Maybe Newton's binominal theorem for [MATH](x+1)^{2n-1}[/MATH]?
 
UPD: This is incorrect equality.
1608206545539.png

Where n > 1, n - natural number

P.S
If you are interested: I want to solve problem:

prove that [MATH]P(x) = x^{2n}-nx^{n+1}+nx^{n -1}- 1 [/MATH] have a root -1 and multiplicity of this root is 3.
 
lookagain said:
What/which is an incorrect equality? The equality in the box is true for n = 2 and 3,
for instance. Of course that does mean it would be true for all values of n > 1
necessarily. What are you referring to?
Click to expand...

So I check if this equality true when n = 4, but it doesn't:

[MATH]x^8 -4x^5 +4x^3 - 1 \neq (x-1)^3(x+1)^5[/MATH]
However:

[MATH]x^{2n} -nx^{n+1} +nx^{n-1} - 1 = (x-1)^3*q(x)[/MATH] for some [MATH]q(x)[/MATH]
 
sadovnichek said:
prove that [MATH]P(x) = x^{2n}-nx^{n+1}+nx^{n -1}- 1 [/MATH] have a root -1 and multiplicity of this root is 3.
Click to expand...
Solved.
How to know a multiplicity of root? We need get a derivative of polynome and check if number a root of this derivative. Continue get a derivative while number - is a root. Count of derivatives will be a multiplicity of root.
[MATH] P(x) = x^{2n} -nx^{n+1} +nx^{n-1}-1 [/MATH][MATH] P(1) = x^{2} -x^{2} +1 -1 = 0[/MATH] (root. 1st multiplicity)

[MATH] P'(x) = 2nx^{2n-1} -n(n+1)x^{n} +n(n-1)x^{n-2} [/MATH][MATH] P'(1) = 2n-n(n+1) +n(n-1) = 0 [/MATH](root. 2nd multiplicity)

[MATH] P''(x) = 2n(2n-1)x^{2n-2} -n^2(n+1)x^{n-1} +n(n-1)(n-2)x^{n-3} [/MATH][MATH] P''(1) = 2n(2n-1) -n^2(n+1) +n(n-1)(n-2) = 0 [/MATH] (root. 3rd multiplicity)

[MATH] P'''(x) = 2n(2n-1)(2n-2)x^{2n-3} -n^2(n+1)*(n-1)x^{n-2} +n(n-1)(n-2)(n-3)x^{n-4} [/MATH][MATH]P'''(1) = 2n(2n-1)(2n-2) -n^2(n+1)(n-1) +n(n-1)(n-2)(n-3) \neq 0 [/MATH] (not root)

Thus 1 is a root of 3rd multiplicity of P(x)
 
Last edited:
You have p(x) then you want to compute p(1). This means that x=1 NOT n=1 So p(1) = 12n-n*1n+1 + n*1n-1 -1 which is nothing like what you wrote.

If you really want help then you need to state the exact problem.
 
Jomo said:
You have p(x) then you want to compute p(1). This means that x=1 NOT n=1 So p(1) = 12n-n*1n+1 + n*1n-1 -1 which is nothing like what you wrote.

If you really want help then you need to state the exact problem.
Click to expand...
Yes! I made a little mistake when computing P(1).

Now it is correct:
[MATH] P(x) = x^{2n} -nx^{n+1} +nx^{n-1}-1 [/MATH][MATH] P(1) = 1^{2n} -n*1^{n+1} +n*1^{n-1} -1 = 1 - n + n -1 = 0 [/MATH] (root. 1st multiplicity)

[MATH] P'(x) = 2nx^{2n-1} -n(n+1)x^{n} +n(n-1)x^{n-2} [/MATH][MATH] P'(1) = 2n-n(n+1) +n(n-1) = 0 [/MATH](root. 2nd multiplicity)

[MATH] P''(x) = 2n(2n-1)x^{2n-2} -n^2(n+1)x^{n-1} +n(n-1)(n-2)x^{n-3} [/MATH][MATH] P''(1) = 2n(2n-1) -n^2(n+1) +n(n-1)(n-2) = 0 [/MATH] (root. 3rd multiplicity)

[MATH] P'''(x) = 2n(2n-1)(2n-2)x^{2n-3} -n^2(n+1)*(n-1)x^{n-2} +n(n-1)(n-2)(n-3)x^{n-4} [/MATH][MATH]P'''(1) = 2n(2n-1)(2n-2) -n^2(n+1)(n-1) +n(n-1)(n-2)(n-3) \neq 0 [/MATH] (not root)

Thus 1 is a root of 3rd multiplicity of P(x)
 
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