Spivak's Calculus problem

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In summary: You can also use the \ operator to put a line break between two terms. For example, you could write(x-y)\sum_{i=0}^{n-1}x^{n-1-i}y^iIf you want to include a space in between two terms, you can use the \space character.
  • #1
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Homework Statement


I am attempting to solve the problems in chapter 1 - Basic Properties of Numbers from Spivak's book.

Problem 1(v) asks the student to prove that
x^n - y^n = (x - y)(x^(n - 1) + (x^(n - 2))y + ... + x(y^(n - 2)) + y^(n - 1))

Homework Equations





The Attempt at a Solution


I solved it by using induction, but induction is not taught until later in the book.
I assume you can solve the exercises solely using the material in the previous chapters, or the hints in the exercise.
After repeatedly applying P9 (distributive law of multiplication) it is obvious that all the terms cancel, except for the first and the last, but proofs with elipses (..) don't strike me as rigorous. So, how can you prove this without using induction?

Thanks
 
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  • #2
If you multiply out the right hand side, all the terms cancel in pairs except x^n and y^n. That doesn't involve induction.

You can't explicitly write all the terms in the general case, so you are forced to use some notation like "...". You could write something like
[tex](x-y)\sum_{i=0}^{n-1}x^{n-1-i}y^i[/tex]
if you feel "..." is a too informal, but remember the main point of notation is to communicate, not to be rigorous for its own sake.
 
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  • #3
AlephZero said:
If you multiply out the right hand side, all the terms cancel in pairs except x^n and y^n. That doesn't involve induction.

You can't explicitly write all the terms in the general case, so you are forced to use some notation like "...". You could write something like
[tex](x-y)\sum_{i=0}^{n-1}x^{n-1-i}y^i[/tex]
if you feel "..." is a too informal, but remember the main point of notation is to communicate, not to be rigorous for its own sake.

Thanks for the reply.
Do you think it is incorrect to use induction in this proof? I agree it is more direct to expand the right hand side, though.
BTW, I am a newbie here, where do I find help for the tex notation for writing formulas in posts like you did with the summation?

Thanks.
 
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  • #4
Can you explain how you used induction to prove this? Then somebody can comment on what you did.

Induction is a perfectly acceptable way to proove things, unless you are doing a homework question which explicitly says "don't use it" of course.

Re Tex, look at the "sticky" threads on the forums, one of them is an introduction to Latex. Also, if you click on a Latex image in a post, you get a pop-up window showing the source code, which is a good learning tool as well a help when doing cut-and-paste editing.
 
  • #5
AlephZero said:
Can you explain how you used induction to prove this? Then somebody can comment on what you did.

Induction is a perfectly acceptable way to proove things, unless you are doing a homework question which explicitly says "don't use it" of course.


Re Tex, look at the "sticky" threads on the forums, one of them is an introduction to Latex. Also, if you click on a Latex image in a post, you get a pop-up window showing the source code, which is a good learning tool as well a help when doing cut-and-paste editing.

Here is my induction proof attempt:
[tex]Let P(n) = x^{n} - y^{n} = (x - y)(x^{n-1} + x^{n-2}y + \dots + x y^{n-2} + y^{n-1})
[/tex]

[tex]For n = 1 [/tex]:
[tex]The equality is trivial.[/tex]
[tex]P(1) = x^{1} - y^{1} = (x - y)
[/tex]

[tex]if P(n) is true then[/tex]
[tex]P(n+1) = (x - y)(x^{n} + x^{n-1}y + \dots + x y^{n-1} + y^{n}) = [/tex]
[tex] (x - y)(x^{n} + y(x^{n-1} + \dots + x y^{n-2} + y^{n-1})) =[/tex]
[tex] (x - y)x^{n} + y(x-y)(x^{n-1} + \dots + x y^{n-2} + y^{n-1}) = [/tex]
[tex] (x - y)x^{n} + y P(n) = [/tex]
[tex] (x - y)x^{n} + y(x^{n} - y^{n}) =[/tex]
[tex]x^{n+1} - y^{n+1}[/tex]
QED

(Sorry but I couldn't format the post properly.
I will read the sticky thread about LaTex when I have more time.)
 
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  • #6
Use ^{n-1} to get the whole thing superscripted.
 
  • #7
ZioX said:
Use ^{n-1} to get the whole thing superscripted.

How do I put line breaks? I get all equalities on the same line...
 
  • #8
That looks OK - though it seems like more work than just multiplying the terms and cancelling them.

A few of LaTex tips BTW:

(1) Things like superscripts and subscripts will only contain a single character unless you use { } brackets. The { and } don't appear in the generated equation. In other words, x^n-1 and x^{n-1} mean different things, and you wanted x^{n-1}.

(2) If you have several lines of equation, use several different TeX expressions. Otherwise, the result is too wide to fit on the page.

(3) You don't need to use "*" for multiplication, TeX will format the equation in the conventional maths style without it.

[tex]P(n+1) = (x - y)(x^n + x^{n-1}y + \dots + xy^{n-1} + y^n) [/tex]
[tex]= (x - y)(x^n + y(x^{n-1} + \dots + xy^{n-2} + y^{n-1})) [/tex]
[tex]= (x - y)x^n + y(x-y)(x^{n-1} + \dots + xy^{n-2} + y^{n-1}) [/tex]
[tex]= (x - y)x^n + y P(n) [/tex]
[tex]= (x - y)x^n + y(x^n - y^n) [/tex]
[tex]= x^{n+1} - y^{n+1}[/tex]

I also fixed a typo in your original proof :wink:
 
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  • #9
AlephZero said:
That looks OK - though it seems like more work than just multiplying the terms and cancelling them.

A few of LaTex tips BTW:

(1) Things like superscripts and subscripts will only contain a single character unless you use { } brackets. The { and } don't appear in the generated equation. In other words, x^n-1 and x^{n-1} mean different things, and you wanted x^{n-1}.

(2) If you have several lines of equation, use several different TeX expressions. Otherwise, the result is too wide to fit on the page.

(3) You don't need to use "*" for multiplication, TeX will format the equation in the conventional maths style without it.

[tex]P(n+1) = (x - y)(x^n + x^{n-1}y + \dots + xy^{n-1} + y^n) [/tex]
[tex]= (x - y)(x^n + y(x^{n-1} + \dots + xy^{n-2} + y^{n-1})) [/tex]
[tex]= (x - y)x^n + y(x-y)(x^{n-1} + \dots + xy^{n-2} + y^{n-1}) [/tex]
[tex]= (x - y)x^n + y P(n) [/tex]
[tex]= (x - y)x^n + y(x^n - y^n) [/tex]
[tex]= x^{n+1} - y^{n+1}[/tex]

I also fixed a typo in your original proof :wink:

Thanks :-)
 
  • #10
AlephZero said:
If you multiply out the right hand side, all the terms cancel in pairs except x^n and y^n. That doesn't involve induction.

You can't explicitly write all the terms in the general case, so you are forced to use some notation like "...". You could write something like
[tex](x-y)\sum_{i=0}^{n-1}x^{n-1-i}y^i[/tex]
if you feel "..." is a too informal, but remember the main point of notation is to communicate, not to be rigorous for its own sake.

I am having the same problem. Is there an easier way to write it without i's ? You wrote : [tex](x-y)\sum_{i=0}^{n-1}x^{n-1-i}y^i[/tex] When I did the proof, I simply wrote the that the ... signifies the middle terms that are simply inverses of each other and cancel out.
 

1. What is Spivak's Calculus problem?

Spivak's Calculus problem refers to a famously difficult calculus problem proposed by mathematician Michael Spivak in his book "Calculus". It involves finding the limit of a sequence of numbers and has been a point of discussion and debate among mathematicians for many years.

2. What makes Spivak's Calculus problem so challenging?

There are several reasons why Spivak's Calculus problem is considered a difficult problem. Firstly, it requires a deep understanding of calculus concepts and techniques. Secondly, it involves a complex limit that cannot be solved using standard calculus methods. Finally, there is no known definitive solution to the problem, making it an ongoing challenge for mathematicians.

3. Has anyone solved Spivak's Calculus problem?

As of now, there is no known definitive solution to Spivak's Calculus problem. Many mathematicians have attempted to solve it and have proposed different approaches and partial solutions, but the problem remains unsolved. Some have even suggested that the problem may be unsolvable.

4. Why is Spivak's Calculus problem still relevant?

Even though Spivak's Calculus problem has not been solved, it continues to be a point of interest and discussion among mathematicians. It serves as a challenging exercise for students and a test of mathematical abilities for researchers. Additionally, the problem has sparked new developments and insights in the field of calculus.

5. Are there any similar problems to Spivak's Calculus problem?

Yes, there are several similar problems to Spivak's Calculus problem that involve finding the limit of a complex sequence. Some of these problems have been solved, while others remain open challenges. Examples include the Basel problem and the Twin Prime conjecture.

Suggested for: Spivak's Calculus problem

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