Working through spivak's calculus

  • Thread starter Shylock
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    Calculus
Thank you for the help!In summary, the maximum of two numbers x and y is denoted by max(x,y) and the minimum of x and y is denoted by min(x,y). By using substitution and proof by cases, it can be proven that max(x,y) = (x+y+|y-x|)/2 and min(x,y) = (x+y-|y-x|)/2. The same method can also be applied to derive formulas for max(x,y,z) and min(x,y,z). However, for a more formal and rigorous proof, one may refer to a proof book or "Understanding analysis" by Abbott.
  • #1
Shylock
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i'm looking for some general advice on how to go about solving the problems in spivak's calculus text. I've just finished reading the first chapter, and while I've managed to solve some of the problems, i can't make sense out of most of them without referring to the solutions. how should i be going about this?
 
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  • #2
also, which of these problems do you think I should be able to do? I've taken calcI, intro to logic, symbolic logic, and intermediate logic I, with As in each.
 
  • #3
Well, what question is troubling you? And what's the difficulty??

(I must add that Spivak's exercises can be quite difficult, so don't worry if you don't succeed at first)
 
  • #4
i'm good till problem 13:

The maximum of two numbers x and y is denoted by max(x,y). Thus max(-1,3) = max(3,3) = 3 and max(-1,-4) = max(-4,-1) = -1. The minimum of x and y is denoted by min(x,y,). Prove that

max(x,y) = (x + y + ly - xl) / 2
min(x,y) = (x + y - ly - xl) / 2

Derive a formula for max(x,y,z) and min(x,y,z), using, for example,

max(x,y,z)=max(x,max(y,z)).

i'm pretty lost on where to start.
 
  • #5
Shylock said:
i'm good till problem 13:

The maximum of two numbers x and y is denoted by max(x,y). Thus max(-1,3) = max(3,3) = 3 and max(-1,-4) = max(-4,-1) = -1. The minimum of x and y is denoted by min(x,y,). Prove that

max(x,y) = (x + y + ly - xl) / 2
min(x,y) = (x + y - ly - xl) / 2

Derive a formula for max(x,y,z) and min(x,y,z), using, for example,

max(x,y,z)=max(x,max(y,z)).

i'm pretty lost on where to start.

Well, let's prove that

[tex]\max(x,y)=\frac{x+y+|y-x|}{2}[/tex]

there are three possibilities: x=y, x<y and y<x. What is |x-y| in all these cases??
 
  • #6
0, y-x, -x-y?
 
  • #7
Shylock said:
0, y-x, -x-y?

Yes, and what do you get when you apply this in the formula

[tex]\frac{x+y+|y-x|}{2}[/tex]
 
  • #8
(x+y)/2 , y , 0

how do you know to check for the three possibilities?
 
  • #9
Shylock said:
0, y-x, -x-y?

Oh, sorry, the last one is not correct. That must be -x+y :smile:

So, plugging that in in our formula, we get

(x+y)/2, y, x

So, doesn't that prove our result?
 
  • #10
Shylock said:
how do you know to check for the three possibilities?

Well, the absolute value is annoying, so we must eliminate it. The only way to do that is to do the prove in separate cases.

So, if we had somewhere [itex]|x^2+y+z^2|+z[/itex] for example, then we would have to split it up in the case that [itex]x^2+y+z^2[/itex] is < = or > than 0.
 
  • #11
i still don't understand what we've just done
 
  • #12
Shylock said:
i still don't understand what we've just done

We have just calculated

[tex]\frac{x+y+|y-x|}{2}[/tex]

depending on whether x<y, x=y or y<x...
 
  • #13
i think i understand, but could you perhaps write the proof a bit more formally?
 
  • #15
and does that that mean that the formula for max(x,y,z) =

x + (y+z + ly -zl)/2 ?
 
  • #16
Shylock said:
and does that that mean that the formula for max(x,y,z) =

x + (y+z + ly -zl)/2 ?

How did you find that??
 
  • #17
proof by cases:
(i) if x=y then max(x,y) = (x+y)/2
(ii) if the x<y then max(x,y) = y
(iii) if x>y then max(x,y) = x
thus, for any numbers x,y max(x,y) = (x+y+ly-xl)/2
 
  • #18
micromass said:
How did you find that??

by substitution
although now that it think about it, that doesn't seem right
 
  • #19
Shylock said:
proof by cases:
(i) if x=y then max(x,y) = (x+y)/2
(ii) if the x<y then max(x,y) = y
(iii) if x>y then max(x,y) = x
thus, for any numbers x,y max(x,y) = (x+y+ly-xl)/2

That seems ok. (also note that if x=y, then (x+y)/2=x)

Shylock said:
by substitution

Yes, substitution is the way to solve this. But I can totally not understand, how you arrived at that formula by substitution.

You know that

[tex]max{x,u}=\frac{x+u+|x-u|}{2}[/tex]

So, what if I substitute u=max(x,y)?
 
  • #20
i think i tried to do it in my head and i got lucky, haha
but now you've lost me
 
  • #21
well, not lucky
 
  • #22
do you mean u = max(y,z)?
 
  • #23
so hopefully i did this right

{[x+(y+z+ly-zl)/2]+[l(y+z+ly-zl)/2 - x]l}/2 ?

would this solution be acceptable or would i have to simplify it?
 
Last edited:
  • #24
Shylock said:
so hopefully i did this right

{[x+(y+z+ly-zl)/2]+[l(y+z+ly-zl)/2 - x]l}/2 ?

would this solution be acceptable or would i have to simplify it?

Yes, that's good! I don't think simplifying will give you much more nicer results, so I guess that this is acceptable :smile:
 
  • #25
Could you show me what the formal proof would look like? It seems that the proofs I was taught to do in logic were more rigorous with the language and explanation. For example, do I need to cite the rules?
 
  • #26
Does Spivak go into more detail about doing this, or is this something I need to work with on my own?
 
  • #27
Shylock said:
Could you show me what the formal proof would look like? It seems that the proofs I was taught to do in logic were more rigorous with the language and explanation. For example, do I need to cite the rules?

Formal proofs, like the ones you did in logic, are not being done here. It can be done, but it would be far too tedious. There are simply too many rules here.
A proof for this would look like this:


There are two possibilities: [itex]x\leq y[/itex] and [itex]y\leq x[/itex].

If [itex]x\leq y[/itex], then max(x,y)=y. Also, we have

[tex]\frac{x+y+|y-x|}{2}=\frac{x+y+y-x}{2}=y[/tex]

If [itex]y\leq x[/itex], then max(x,y)=x. Also, we have that

[tex]\frac{x+y+|y-x|}{2}=\frac{x+y-y+x}{2}=x[/tex]

So in both cases, we have

[tex]\frac{x+y+|y-x|}{2}=max(x,y)[/tex]

this proves the result.

I think that's as formal as you want it. If you want it more formal, you can, but it would be tougher to read.

Shylock said:
Does Spivak go into more detail about doing this, or is this something I need to work with on my own?

No, he doesn't. You may want to pick up a proof book (like Velleman's) or "Understanding analysis" by Abbott to see how to do proofs.
 
  • #28
so then i shouldn't be worried about having to write formal proofs for the problems in the book?
 
  • #29
Shylock said:
so then i shouldn't be worried about having to write formal proofs for the problems in the book?

No, not at all. Formal proofs are not being done here :smile:
 
  • #30
cool, thanks for the help
 
  • #31
Shylock said:
and does that that mean that the formula for max(x,y,z) =

x + (y+z + ly -zl)/2 ?

See for yourself: try (x,y,z) = (1,2,2).

Anyway, max(x,y,z) = max(x,max(y,z)) = (1/2)*[x + max(y,z) + |max(y,z) - x|]. Now substitute the expression for max(y,z).


RGV
 

1. What is "Working through Spivak's Calculus"?

"Working through Spivak's Calculus" refers to the process of studying and completing the exercises in the textbook "Calculus" by Michael Spivak. This book is often used as a rigorous introduction to calculus for students in mathematics and science fields.

2. Is "Working through Spivak's Calculus" necessary for learning calculus?

No, there are many other textbooks and resources available for learning calculus. However, "Working through Spivak's Calculus" is highly recommended for those seeking a deeper understanding of the subject and for those preparing for advanced courses in mathematics or physics.

3. How difficult is "Working through Spivak's Calculus"?

The difficulty level of "Working through Spivak's Calculus" may vary for each individual. However, it is generally considered to be a challenging textbook, especially for those with little background in mathematics. It requires a strong understanding of algebra and basic mathematical concepts.

4. What makes "Working through Spivak's Calculus" different from other calculus textbooks?

"Working through Spivak's Calculus" is known for its rigorous and theoretical approach to calculus. It emphasizes mathematical proofs and logical reasoning, rather than just memorizing formulas and procedures. It also covers a wide range of topics, including limits, derivatives, integrals, and more advanced concepts such as differential equations and multivariable calculus.

5. How can I make the most out of "Working through Spivak's Calculus"?

To make the most out of "Working through Spivak's Calculus," it is important to actively engage with the material and complete all the exercises and problems. It is also helpful to seek out additional resources, such as online tutorials or study groups, to supplement your learning. It may also be beneficial to work through the book with a tutor or mentor who can provide guidance and clarification on difficult concepts.

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