Why did my substitution work even though it was incorrect?

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In summary: I'm sorry I didn't mean to imply you were wrong. Just that the algebraic division is something that I think is easier for the OP to use. I'm not even sure the OP knows what an infinite series is.In summary, the conversation discusses a problem involving the expression x^n - y^n and the attempt to solve it using an infinite series. The first factor on the right side of the equation is incorrect, and the terms in the second factor are not an infinite series but instead a finite sum with n terms. The conversation also mentions using division or induction to solve the problem.
  • #1
embphysics
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Hello,

The problem to be solved is [itex]x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y +...+xy^{n-2} + y^{n-1})[/itex]. The first thing I noticed was that the term on the RHS side of the equation could possibly be expressed as an infinite series. And so, I undertook to find the general term:

I noticed that what was being summed was a product, the first term being [itex]x^{n-1}y^0[/itex], and the last [itex]x^0y^{n-1}[/itex], thus my general term needed to at least capture this.

Finally, [itex](x^{n-2} + x^{n-1}y +...+xy^{n-2} + y^{n-1}) = \sum_{i=0}^{n-1} x^{n-1-i}y^i[/itex]

[itex] \sum_{i=0}^{n-1} x^{n-1-i}y^i = x^{n-1} \sum_{i=0}^{n-1} x^{-i}y^i = x_{n-1} \sum_{i=0}^{n-1} (\frac{y}{x})^i[/itex]

This I knew to be the geometric series, the point at which I went wrong:

[itex]x_{n-1} \sum_{i=0}^{n-1} (\frac{y}{x})^i = x^{n-1} (\frac{1-(\frac{y}{x})^n}{1-\frac{y}{x}})[/itex]

I substituted in n rather than n-1. This, however, wasn't the most disconcerting thing, as I am sure you found it disconcerting: even with the incorrect substitution, I was able to properly solve the problem.

Why is that so? How can I fix this?
 
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  • #2
embphysics said:
Hello,

The problem to be solved is [itex]x^n - y^n = (x-y)(x^{n-2} + x^{n-1}y +...+xy^{n-2} + y^{n-1}[/itex]. The first thing I noticed was that the term on the RHS side of the equation could possibly be expressed as an infinite series.
The first and second terms in the second factor have incorrect exponents (and you're missing a right parenthesis). These terms should be xn -1 and xn - 2y.

The terms in the second factor are NOT an infinite series. This is a finite sum with n terms, with the same n that appears on the left side.
embphysics said:
And so, I undertook to find the general term:

I noticed that what was being summed was a product, the first term being [itex]x^{n-1}y^0[/itex], and the last [itex]x^0y^{n-1}[/itex], thus my general term needed to at least capture this.

Finally, [itex](x^{n-2} + x^{n-1}y +...+xy^{n-2} + y^{n-1}) = \sum_{i=0}^{n-1} x^{n-1-i}y^i[/itex]

[itex] \sum_{i=0}^{n-1} x^{n-1-i}y^i = x^{n-1} \sum_{i=0}^{n-1} x^{-i}y^i = x_{n-1} \sum_{i=0}^{n-1} (\frac{y}{x})^i[/itex]

This I knew to be the geometric series, the point at which I went wrong:

[itex]x_{n-1} \sum_{i=0}^{n-1} (\frac{y}{x})^i = x^{n-1} (\frac{1-(\frac{y}{x})^n}{1-\frac{y}{x}})[/itex]

I substituted in n rather than n-1. This, however, wasn't the most disconcerting thing, as I am sure you found it disconcerting: even with the incorrect substitution, I was able to properly solve the problem.

Why is that so? How can I fix this?
 
  • #3
If the purpose of the problem was to show that this is an identity you first have to have a correct identity and I think you have a typo: you wrote ##x^{n-2} + x^{n-1}y ...##. Hopefully you meant ##x^{n-1} + x^{n-2}y ...##

To show it is true you have 2 choices:

1. You can divide ##(x^n -y^n)## by (x-y) and show you get the sum of the terms

2. You can multiply out the right hand side and combine like terms to show that you get the left hand side.

Both these choices involve ordinary algebra, and your fancier computations are not required.

If the problem is something else, you were not clear about that.

A note: When you are using summation signs you must be very careful to keep your indices straight. Even your first sum is not written correctly.
 
  • #4
embphysics said:
Hello,

The problem to be solved is [itex]x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y +...+xy^{n-2} + y^{n-1})[/itex]. The first thing I noticed was that the term on the RHS side of the equation could possibly be expressed as an infinite series. And so, I undertook to find the general term:

I noticed that what was being summed was a product, the first term being [itex]x^{n-1}y^0[/itex], and the last [itex]x^0y^{n-1}[/itex], thus my general term needed to at least capture this.

Finally, [itex](x^{n-2} + x^{n-1}y +...+xy^{n-2} + y^{n-1}) = \sum_{i=0}^{n-1} x^{n-1-i}y^i[/itex]

[itex] \sum_{i=0}^{n-1} x^{n-1-i}y^i = x^{n-1} \sum_{i=0}^{n-1} x^{-i}y^i = x_{n-1} \sum_{i=0}^{n-1} (\frac{y}{x})^i[/itex]

This I knew to be the geometric series, the point at which I went wrong:

[itex]x_{n-1} \sum_{i=0}^{n-1} (\frac{y}{x})^i = x^{n-1} (\frac{1-(\frac{y}{x})^n}{1-\frac{y}{x}})[/itex]

I substituted in n rather than n-1. This, however, wasn't the most disconcerting thing, as I am sure you found it disconcerting: even with the incorrect substitution, I was able to properly solve the problem.

Why is that so? How can I fix this?

Let
[tex]t_n = x^{n-1} + x^{n-2} y + \cdots + x y^{n-2} + y^{n-1},[/tex]
so you want to prove that ##x^n - y^n = (x-y) t_n.## This almost cries out for the use of induction on n.
 
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  • #5
Ray Vickson said:
Let
[tex]t_n = x^{n-1} + x^{n-2} y + \cdots + x y^{n-2} + y^{n-1},[/tex]
so you want to prove that ##x^n - y^n = (x-y) t_n.## This almost cries out for the use of induction on n.

One could certainly use induction, but division is easier.
 
  • #6
brmath said:
One could certainly use induction, but division is easier.

Yes, you already mentioned division. I am mentioning another method that the OP may, or may not prefer. I do not know what the OP knows or does not know. If he does not know about induction my suggestion is not helpful; if he does not know how to do (algebraic) division, your suggestion will not be helpful to him.
 
  • #7
Yes, Ray, all true.
 

FAQ: Why did my substitution work even though it was incorrect?

1. What is the proof of x^n - y^n?

The proof of x^n - y^n is a mathematical process that shows the validity of the statement x^n - y^n = (x-y)(x^(n-1) + x^(n-2)y + x^(n-3)y^2 + ... + xy^(n-2) + y^(n-1)). It involves using algebraic manipulation and mathematical principles to demonstrate that the equality holds true for all possible values of x and y.

2. Why is the proof of x^n - y^n important?

The proof of x^n - y^n is important because it provides a fundamental understanding of the relationship between two variables raised to a power. It also serves as a basis for more complex mathematical concepts and equations involving exponents, such as the binomial theorem and polynomial division.

3. What are the applications of x^n - y^n?

The equation x^n - y^n has many applications in fields such as physics, engineering, and economics. It can be used to model exponential growth and decay, calculate compound interest, and solve problems involving rates of change. In physics, it can be applied to study the behavior of particles and waves, while in engineering, it can be used to design efficient structures and systems.

4. How is the proof of x^n - y^n related to other mathematical proofs?

The proof of x^n - y^n is related to other mathematical proofs in that it utilizes similar techniques and principles, such as the laws of exponents, the distributive property, and factoring. It is also closely connected to the proofs of other fundamental equations, such as the Pythagorean theorem and the quadratic formula.

5. Are there any variations of the proof of x^n - y^n?

Yes, there are variations of the proof of x^n - y^n depending on the specific conditions and constraints given. For example, there are proofs that focus on specific values of x and y, such as positive integers or real numbers. There are also proofs that use different methods, such as mathematical induction or geometric proofs. However, the fundamental principle remains the same in all variations of the proof.

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