Working through spivak's calculus

[Shylock]

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?

 

[Shylock]

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.

 

[micromass]

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)

 

[Shylock]

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.

 

[micromass]

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

$$\max(x,y)=\frac{x+y+|y-x|}{2}$$

there are three possibilities: x=y, x<y and y<x. What is |x-y| in all these cases??

 

[Shylock]

0, y-x, -x-y?

 

[micromass]

0, y-x, -x-y?

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

$$\frac{x+y+|y-x|}{2}$$

 

[Shylock]

(x+y)/2 , y , 0

how do you know to check for the three possibilities?

 

[micromass]

0, y-x, -x-y?

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

So, plugging that in in our formula, we get

(x+y)/2, y, x

So, doesn't that prove our result?

 

[micromass]

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 $|x^2+y+z^2|+z$ for example, then we would have to split it up in the case that $x^2+y+z^2$ is < = or > than 0.

 

[Shylock]

i still don't understand what we've just done

 

[micromass]

i still don't understand what we've just done

We have just calculated

$$\frac{x+y+|y-x|}{2}$$

depending on whether x<y, x=y or y<x...

 

[Shylock]

i think i understand, but could you perhaps write the proof a bit more formally?

 

[micromass]

Well, if you understand, how would you write the proof then?

 

[Shylock]

and does that that mean that the formula for max(x,y,z) =

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

 

[micromass]

and does that that mean that the formula for max(x,y,z) =

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

How did you find that??

 

[Shylock]

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

 

[Shylock]

How did you find that??

by substitution
although now that it think about it, that doesn't seem right

 

[micromass]

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)

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

$$max{x,u}=\frac{x+u+|x-u|}{2}$$

So, what if I substitute u=max(x,y)?

 

[Shylock]

i think i tried to do it in my head and i got lucky, haha
but now you've lost me

 

[Shylock]

well, not lucky

 

[Shylock]

do you mean u = max(y,z)?

 

[Shylock]

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?

 

[micromass]

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

 

[Shylock]

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?

 

[Shylock]

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

 

[micromass]

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: $x\leq y$ and $y\leq x$.

If $x\leq y$, then max(x,y)=y. Also, we have

$$\frac{x+y+|y-x|}{2}=\frac{x+y+y-x}{2}=y$$

If $y\leq x$, then max(x,y)=x. Also, we have that

$$\frac{x+y+|y-x|}{2}=\frac{x+y-y+x}{2}=x$$

So in both cases, we have

$$\frac{x+y+|y-x|}{2}=max(x,y)$$

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.

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.

 

[Shylock]

so then i shouldn't be worried about having to write formal proofs for the problems in the book?

 

[micromass]

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

 

[Shylock]

cool, thanks for the help

 

[Ray Vickson]

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