Need Help on Spivak's Calculus Chapter 2

  • #1
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Homework Statement


Hi, I'm having trouble on a question from Spivak's Calculus 4E. It's on page 27, chapter 2 and question number 2.

He asks to find the formula for n Sigma i=1 (2i -1)= 1 + 3 + 5 + .... + (2n-1)


Homework Equations



Unfortunately I don't know how to use the symbols on a thread like this; anyone care to assist? :)

The Attempt at a Solution



Well since this is an induction chapter, I tried coming up with a formula using n+1 but that got me into a mess.
 

Answers and Replies

  • #2
LCKurtz
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Place a dot on a piece of paper. Then add 3 more to make a square with 4 dots. Then add 5 more along the top and right sides of the rectangle. Keep this up. What do you notice?
 
  • #3
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Place a dot on a piece of paper. Then add 3 more to make a square with 4 dots. Then add 5 more along the top and right sides of the rectangle. Keep this up. What do you notice?
Hi, thanks for the help. It ends up being a square so the sum would be n2; however for part 2 of the question, instead of 1+3+5... etc

it is nƩi=1 (2i-1)2 = 12 + 32 + 52 +... + (2n-1)2

So how would I tackle this one? The formula that Spivak comes up with is much more complicated than just n2.
 
  • #4
LCKurtz
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Hi, thanks for the help. It ends up being a square so the sum would be n2; however for part 2 of the question, instead of 1+3+5... etc

it is nƩi=1 (2i-1)2 = 12 + 32 + 52 +... + (2n-1)2

So how would I tackle this one? The formula that Spivak comes up with is much more complicated than just n2.
I don't have a copy of Spivak anymore. I assume he tells you what the sum is. I would just prove it by induction.
 
  • #5
HallsofIvy
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Here's a different way of doing it: let
S=1+ 3+ 5+ ...+ (2n- 5)+ (2n- 3)+ (2n- 1). Reversing the order of addition doesn't change the result so
S= (2n-1)+ (2n-3)+ (2n-5)+ ...+5+ 3+ 1.

Adding vertically, 1+ (2n-1)= 2n. 3+ (2n-3)= 2n, 5+ (2n-5)= 2n, etc. Every one of those sums is 2n. Since there are n such sums (convince yourself of that!) they total to [itex](2n)(n)= 2n^2[/itex]. Because we have, in fact, added twice, the S is equal to [itex]n^2[/itex].
 
  • #6
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I don't have a copy of Spivak anymore. I assume he tells you what the sum is. I would just prove it by induction.
Yes he says the formula would end up being:

n(2n+1)(2n-1) all over 3, but I am not sure how he got to this point.
 
  • #7
LCKurtz
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Yes he says the formula would end up being:

n(2n+1)(2n-1) all over 3, but I am not sure how he got to this point.
But you don't have to know how the formula was discovered to prove it is correct.
 
  • #8
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Here's how he dealt with the problem:

10f2yah.jpg


My question is how did he come up with the difference in the first step?
 
  • #9
Redbelly98
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Here's how he dealt with the problem:

10f2yah.jpg


My question is how did he come up with the difference in the first step?
Well, in between steps 1 & 2 you could have
= 12 + 22 - 22 + 32 + 42 - 42 + ... + (2n)2 - (2n)2
The author probably thought this step was obvious enough and so did not write it explicitly.

EDIT:
Alternatively, you could expand (2i-1)2. If you know the basic sums ∑i and ∑i2, it is possible to write things out in terms of those.
 
  • #10
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Well, in between steps 1 & 2 you could have
= 12 + 22 - 22 + 32 + 42 - 42 + ... + (2n)2 - (2n)2
The author probably thought this step was obvious enough and so did not write it explicitly.

EDIT:
Alternatively, you could expand (2i-1)2. If you know the basic sums ∑i and ∑i2, it is possible to write things out in terms of those.

Hi, thanks for the assistance. That would make sense that he's adding and subtracting the same values, thereby just adding by 0. However, what I'm confused is how he got from the +......+ to just a non-continuous expression with denominator of 6.

I remember watching a video of someone simplifying nƩi=1 j^2 into an expression with denominator of 6, but that was using induction and adding (j+1)^2 to both sides; something I don't think we had to do for a question like this.
 
  • #11
LCKurtz
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He is assuming that you already know this formula:

[tex]\sum_{k=1}^n k^2 = \frac{(n)(n+1)(2n+1)}{6}[/tex]

and using it to derive the result for the odd squares.
 
  • #12
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He is assuming that you already know this formula:

[tex]\sum_{k=1}^n k^2 = \frac{(n)(n+1)(2n+1)}{6}[/tex]

and using it to derive the result for the odd squares.
Ahh, thank you, that formula fits in with simplifying 12 + 22 + 32 + ... + n2

However, a quick question on that. As we see in the first bracket, the addition ends with (2n)2, but the formula only works for upto n2. Could you clarify how 12 + 22 + 32 + ... + (2n)2 derives from [tex]\sum_{k=1}^n k^2 = \frac{(n)(n+1)(2n+1)}{6}[/tex]

And on a similar note, how would such cases that end with (3n)2 or even (4n)2 derive from that same formula?

Thank you.

(Btw, I wish Spivak would introduce or give us these formulas before the problem sets so I wouldn't be so confused; at only question #2; it's really difficult for me being a 1st year student with no proofs knowledge whatsoever.)
 
  • #13
LCKurtz
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Ahh, thank you, that formula fits in with simplifying 12 + 22 + 32 + ... + n2

However, a quick question on that. As we see in the first bracket, the addition ends with (2n)2, but the formula only works for upto n2. Could you clarify how 12 + 22 + 32 + ... + (2n)2 derives from [tex]\sum_{k=1}^n k^2 = \frac{(n)(n+1)(2n+1)}{6}[/tex]

And on a similar note, how would such cases that end with (3n)2 or even (4n)2 derive from that same formula?
That formula works for any positive integer n. 2n is a positive integer so it works for that too. Ditto for any other positive integer.
 

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