(x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + + x + 1)

  • Thread starter General_Sax
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In summary, the factorization (x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + ... + x + 1) can be obtained by using the distributive law and mathematical induction. The formula for geometric series can be used as well, but it is not necessary and may not be accepted as a rigorous proof by some teachers. The formula holds for all real numbers x, not just those that satisfy |x|<1.
  • #1
General_Sax
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(x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + ... + x + 1)

Where does this factorization come from? I need to know so I can use it in a proof. Thanks.
 
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  • #2
Work out the right side...
 
  • #3
Use mathematical induction:
[tex]
x^{k+1} - 1 = (x^{k+1} - x^k) + (x^k - 1) = (x - 1) x^k +(x^k - 1)
[/tex]
 
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Likes chrisfrag
  • #4
micromass said:
Work out the right side...

So, there is no theorem to use?

@Dickfore

I'm confused as to the next step -- yes I've been trying to work it out.

Should I try to factor (x-1) out of the expression?
 
  • #5
General_Sax said:
@Dickfore

I'm confused as to the next step -- yes I've been trying to work it out.

Should I try to factor (x-1) out of the expression?

According to P.M.I., you should substitute the expression that you're trying to prove for [itex]n = k[/itex], in this case for [itex]x^k - 1[/itex], factor [itex]x - 1[/itex], and see if you get the corresponding expression for [itex]n = k + 1[/itex]. If you do, then the proof is complete.
 
  • #6
General_Sax said:
So, there is no theorem to use?

Could almost say this is the theorem.

Multiply your last bracket by x
on the next line multiply that same last bracket by -1.
Compare. Add.

This is a very useful formula.
That last bracket is the geometrical series for example, hence you calculate it equals (xk - 1)/(x - 1). Heard anything like that before?
 
  • #7
Thanks for the help people. I think I've got it. Just used formula for geometric series and did some algebra -- hope it's good enough.
 
  • #8
General_Sax said:
Thanks for the help people. I think I've got it. Just used formula for geometric series and did some algebra -- hope it's good enough.
Did you really mean that you used the formula for the geometric series? This theorem?
For all real numbers x such that |x|<1, ##\sum_{k=0}^\infty x^k=\frac{1}{1-x}.##​
This theorem is much harder to understand than the formula you're trying to prove, because it involves convergence. Its standard proof uses the formula you're trying to prove, which by the way holds for all real numbers x, not just the ones that satisfy |x|<1.

General_Sax said:
So, there is no theorem to use?
Only the distributive laws a(b+c)=ab+ac, (a+b)c=ac+bc, and if you want to do a rigorous proof that the result of the multiplication is what you (should) think it is, you have to use induction.

General_Sax said:
Should I try to factor (x-1) out of the expression?
Wouldn't that give you two factors of (x-1)? That would only make things worse.
 
Last edited:
  • #9
Thanks for the additional effort/attention Fredrik, but I've already used the geometric "series" -- perhaps it's more accurate to use the term "sum" -- in a proof.

http://en.wikipedia.org/wiki/Geometric_series#Formula

that's the one I used. Just split it up w/ some algebra. It's for a CMPUT course and we haven't much experience with proofs so I don't think they expect much rigour.
 
  • #10
So basically you took the formula ##1+x+\cdots+x^{k-1}=\frac{1-x^{k}}{1-x}## (which doesn't hold for x=1 by the way) and just multiplied it by x-1? If I was your teacher, I wouldn't accept that as an answer. (The formula you found is too similar to what you're supposed to prove). Why don't you just do the multiplication on your right-hand side to see what you get? Do you know how to evaluate a(b+c)? How about (a+b)(b+c)? How about (a+b)(b+c+d)?
 

1. What is the purpose of the equation "(x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + + x + 1)"?

The purpose of this equation is to factorize a polynomial with a degree of k, which can be useful in solving equations or simplifying expressions.

2. How do you solve the equation "(x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + + x + 1)"?

To solve this equation, you can use the distributive property to expand the right side of the equation and then combine like terms. This will result in the original equation, showing that it is an identity and therefore true for all values of x.

3. What is the relationship between the factors on the right side of the equation and the term on the left side?

The factors on the right side of the equation can be seen as a telescoping sum, where the first term is (x-1) and the last term is (x^k - 1). These factors work together to cancel out the (x^k - 1) term on the left side, leaving only the (x-1) term.

4. Can this equation be used to find the roots of a polynomial?

Yes, this equation can be used to find the roots of a polynomial by setting the left side of the equation equal to zero and solving for x. The resulting solutions will be the roots of the polynomial.

5. Are there any restrictions on the values of k in this equation?

Yes, the value of k must be a positive integer, as the equation involves raising x to a power. Negative or non-integer values of k would result in non-real solutions.

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