# Proof m^k - 1.

1. Jan 2, 2009

### Gregg

1. The problem statement, all variables and given/known data

$$\Pi_{n=1}^{k} (m + (-1)^n)$$

k
1. (m-1)
2. (m+1)(m-1)
3. (m+1)(m-1)(m+1)
4. (m+1)(m-1)(m+1)(m-1)
5. (m+1)(m-1)(m+1)(m+1)(m-1)
6. (m+1)(m-1)(m+1)(m+1)(m-1)(m+1)

I want it to be a rule to find $$m^k - 1$$ So for each I see the difference between the product and mk - 1

$$\left[ \Pi_{n=1}^{k} (m + (-1)^n) \right] - \left[ m^k - 1 \right]$$

k
1. 0
2. 0
3. m2 - m - 2
4. 2m2 - 2
5. -m4 + 2m3 + 2m2 - 2
6. m4 + m2 - 2

If I can predict the above values couldn't I made a product for $$\left[ \Pi_{n=1}^{k} (m + (-1)^n) \right] - \left[ Difference \right]$$

like for example for k = 3 $$\left[ (m+1)(m-1)(m+1)\right] - \left[ m^2 - m - 2 \right] = m^3 - 1$$

Is the rate that the difference between that product and my value going to be the rate of change of ($$\left[ \Pi_{n=1}^{k} (m + (-1)^n) \right] - \left[ m^k - 1 \right]$$) If so, with respect to what, and how would I differentiate it?

2. Relevant equations

3. The attempt at a solution

This as really as far as I have gotten, I've done little work on proofs and have no clue how to prove this. I'm not even sure I have presented this problem correctly.

Last edited: Jan 3, 2009
2. Jan 2, 2009

### Unco

Hi Gregg,

Plugging k=2 into your formula gives m^2 - 1 = m + 1; have another look at your formula and your results previous (first of all, the right-hand sides are products of two terms, not sums...).

By simply expanding the quadratic you can prove for yourself the more general statement that (a+b)(a-b) = a^2 - b^2.

Then choosing a=m^k and b=1 is the desired formula.

3. Jan 2, 2009

### jgens

Think of it this way: m^k - 1 = (m^k/2 - 1)(m^k/2 + 1). Proof, simply multiply the factors and simplify. Furthermore, when k is even, k/2 is an integer value.

4. Jan 3, 2009

### Gregg

Yes I see what you mean now. Turns out i can't do that though. I need big pi for products don't i.

5. Jan 3, 2009

### Unco

Do you mean you can't expand a quadratic, or you're not allowed to? We're talking about real numbers (or along those lines), right? I'm afraid if you're not allowed to use basic algebra there is not much that can be done. Judging from your original post, this is not a set exercise so I see no reason for a reluctance to use basic algebra. (This is posted in the Calculus and Beyond forum, so I assume this is "basic algebra".)

As Jgens has already been pointed out, $$m^k-1 = (m^{\frac{k}{2}}-1)(m^{\frac{k}{2}}+1)$$ is the product of two factors, always, so there is not need for big Pi notation.

6. Jan 3, 2009

### gabbagabbahey

I think perhaps the OP is looking to completely factorize $m^k-1$ (or possibly $m^{2k}-1$ or maybe even $m^{2^k}-1$ for $k=1,2,3\ldots$); Although the problem statement isn't very clear.

If so, then there may be more than two integer factors (again, I'm assuming he's looking for integer factors). For example; $m^4-1$ factorizes as $(m-1)(m+1)(m^2+1)$ (assuming 'm' is any integer, those factors will also be integers for all 'm')

7. Jan 3, 2009

### HallsofIvy

Staff Emeritus
You want to prove
$$m^k - 1 = \Pi_{n=1}^{k/2} (m^{k/2} + (-1)^{n+1})$$?

You can't prove it, it's not true. For example, if k= 6 that claims that
$$m^6- 1= (m^3+ 1)(m^3-1)(m^3+1)[/itex] and that obviously is not true. More generally, if k is even, you have k/2 factors each of degree k/2 and so their product has degree k2/4, not k. 8. Jan 3, 2009 ### Gregg yeah I know, it doesn't work. I wanted to make one that does work, but can't so I'm not doing it anymore. 9. Jan 3, 2009 ### jgens Rather than work on trying to create a complete set of factors from big pi notation, use what I showed - seperate it into factors, and, if one of the factors is still of even degree you can factor that further using the same general rule. 10. Jan 3, 2009 ### Gregg [tex] \Pi_{n=1}^{k} (m + (-1)^n)$$

k
1. (m-1)
2. (m+1)(m-1)
3. (m+1)(m-1)(m+1)
4. (m+1)(m-1)(m+1)(m-1)
5. (m+1)(m-1)(m+1)(m+1)(m-1)
6. (m+1)(m-1)(m+1)(m+1)(m-1)(m+1)

I want it to be a rule to find $$m^k - 1$$ So for each I see the difference between the product and mk - 1

$$\left[ \Pi_{n=1}^{k} (m + (-1)^n) \right] - \left[ m^k - 1 \right]$$

k
1. 0
2. 0
3. m2 - m - 2
4. 2m2 - 2
5. -m4 + 2m3 + 2m2 - 2
6. m4 + m2 - 2

If I can predict the above values couldn't I made a product for $$\left[ \Pi_{n=1}^{k} (m + (-1)^n) \right] - \left[ Difference \right]$$

like for example for k = 3 $$\left[ (m+1)(m-1)(m+1)\right] - \left[ m^2 - m - 2 \right] = m^3 - 1$$

Is the rate that the difference between that product and my value going to be the rate of change of ($$\left[ \Pi_{n=1}^{k} (m + (-1)^n) \right] - \left[ m^k - 1 \right]$$) If so, with respect to what, and how would I differentiate it?

11. Jan 3, 2009

### gabbagabbahey

If you restrict k to powers of 2 (i.e. $k=2^n \; \forall n=1,2,3\ldots$), there is a very nice solution:

n=1: $m^2-1=(m-1)(m+1)$

n=2: $m^4-1=(m-1)(m+1)(m^2+1)$

n=3: $m^8-1=(m-1)(m+1)(m^2+1)(m^4+1)$

So, $$m^{2^n}-1=(m-1)\Pi_{j=0}^{n-1}(m^{2^{j}}+1)$$

Last edited: Jan 4, 2009
12. Jan 4, 2009

### Gregg

Yeah, I like that.

13. Jan 4, 2009

### Gregg

I think this works

$$m^n - 1 = \left(\prod _{n=1}^a (m+(-1)^{n})-1+m^n-\frac{m^{a+1}}{m+1} \prod_{j=0}^a (1+\frac{1}{m}(-1)^j)\right)$$

Actually i've just went round in a massive circle. I give up.

Last edited: Jan 4, 2009