Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Again, I apologize for what looks like a very narrow question. But there's possibly a general principal at work here that I'm not grasping.

I understand the answer provided for exercise 3 in chapter 7 of Spivak's Calculus (4E, p.130), but wonder if another approach might work (and be closer to the spirit of the chapter). For example for (ii), to show that $$\sin x = x-1$$ has a solution, can't I take $$f(x)=\sin x - x+1$$ and argue that for large $|x|$, $x>0 \Rightarrow f(x)<0$ and $x<0 \Rightarrow f(x)>0$, so that I can use Theorem 1 (p. 122: $f(x)$ continuous on $[a,b]$ and $f(a)<0<f(b) \Rightarrow \exists x \in [a,b] : f(x)=0$ )?

share|improve this question
    
You certainly can apply IVT and it is commonly done. But in your case, why is it evident that $f(x) > 0$ for $x < 0$? It's true, but not immediately obvious. Moreover you don't have to worry about all $x < 0$ or $x \leq 0$ for the IVT argument to go through. It suffices to find one $x$ such that $f(x) > 0$ and $x=0$ works! –  Srivatsan Sep 13 '11 at 21:20
    
I was going for generality, partly because of the preceding part of the same problem where $$f(x)=x^{179}+\frac{163}{1+x^{2}+\sin^{2}x}-119,$$which seemed easier to solve using a similar approach. –  raxacoricofallapatorius Sep 13 '11 at 21:34
    
@Sri, $-x\le f(x)\le2-x$ hence $f(x)<0$ for $x>2$ and $f(x)>0$ for $x<0$. Hence "for large $|x|$*" might mean "*for $x>2$ or $x<0$" here. –  Did Sep 13 '11 at 21:42
    
@Didier Oh, I see. :) I took that the large $|x|$ applies only for positive $x$. Yes, if the OP meant what you are saying (and I am now sure s/he did), then s/he is absolutely correct. –  Srivatsan Sep 13 '11 at 22:15

3 Answers 3

up vote 3 down vote accepted

I am currently teaching a course from Spivak's Calculus, and I think your solution to this problem is entirely correct.

You considered the auxiliary function $f(x) = \sin x - x + 1$ and showed that it is continuous and takes negative values for sufficiently large, positive $x$ and also positive values for sufficiently large, negative $x$. So by IVT it must take on the value $0$.

You can be a little more explicit about the sufficiently large business though. For instance, you know that $-1 \leq \sin x \leq 1$ for all $x$, so

$-x \leq f(x) \leq -x + 2$.

From this one sees that $f(x)$ is non-negative for all $x \leq 0$ and non-positive for all $x \geq 2$. In this sort of problem, the more complicated the function gets, the more you want to call on "general principles" in order to give you the estimates you need. For instance, if you had a very complicated polynomial $p(x)$ of degree $19$ in place of $x$, you probably don't want to give explicit values but just use the fact that as $x$ approaches $\pm \infty$, so does $p(x)$, while $\sin x$ stays bounded.

Final comment: to be sure, you don't have to find explicit values for this problem. But in order to be best understood you should probably say something in the way of justification of what happens for sufficiently large $x$. Note that in the above paragraph I gave a less explicit answer for a more general class of problems, and as you know there are other problems in the text which are like the one I made up above. But -- and this is an issue of effective mathematical writing and communication rather than mathematical correctness -- there is a sort of principle of economy at work here. In order to be best understood, it's generally a good idea to use the simplest arguments you can think of that justify a given claim, and a lot of people find more explicit arguments to be simpler than less explicit ones. Anyway, not here but sometimes you do have to be explicit and concrete, so it's a good idea to cultivate the ability to do so...

share|improve this answer
    
Yes, precisely (on all points)! I was perhaps overgeneralizing given the simplicity of the problem at hand, but (again, in the spirit of the chapter) I was trying to come up with an approach that would work more generally (having been intimidated by the first part of the problem). Thanks! –  raxacoricofallapatorius Sep 13 '11 at 22:21
    
@Pete: While I'm at it: is it SO-kosher to ask this sort of "did I understand the problem" or "is my answer right too" question about specific textbook problems? It strikes me that many problems (esp. those in excellent texts like Spivak's) are great jumping off points for important discussions and illustrate significant themes and concepts of the sort that have a place here. Rgds. –  raxacoricofallapatorius Sep 13 '11 at 22:26
1  
@rax: you wrote "SO", but I'm assume your talking about math.se. (Anyway, I'm not qualified to answer cultural questions about SO...) Sure, your question is A-OK for this site. Most homework questions are acceptable here, provided the OP shows some level of due diligence at trying to solve them herself. Posting and asking about a correct solution is maximally diligent! And I also agree that Spivak's Calculus provides many opportunities for good discussions on this site. In fact, it turns out that my top-voted question here is closely related to a double-starred problem in that text. –  Pete L. Clark Sep 13 '11 at 22:32

Your approach is correct, but the proof is not water-tight. By which I mean that to complete the proof, one needs the following additional proposition:

Suppose a continuous function $f : \mathbb R \to \mathbb R$ satisfies $f(x) \to \infty$ as $x \to \infty$, and $f(x) \to -\infty$ as $x \to -\infty$. Then there exists some $x_0 \in \mathbb R$ such that $f(x_0) = 0$.

From your question and comments, I feel that you intuitively understand the above statement, and in fact, you are also implicitly using it. However, I am not sure that you know how to prove it. The proof is just a single line, by the way. Nevertheless, I encourage you to do it.

I am emphasizing this point because when one is learning a subject, it is of great pedagogical value to sit down and prove even seemingly obvious statements. Moreover, while one certainly cannot write completely rigorous proofs all the time, one should at least be aware of the little missing details, and also how those details can be taken care of.

share|improve this answer
    
That's a good point. But that proposition is not (I think) needed for the the argument I was making, namely that as $|x| \to \infty$, $f(x) \to -x$; so for such $x<0, f(x)>0$ and for such $x>0, f(x)<0$. I could easily be missing something though. –  raxacoricofallapatorius Sep 14 '11 at 0:51
    
@rax I don't see what your argument is then. Ok, I get it. You do not need that $f(x)$ grows unbounded. You just need to show that there is some $x$ such that $f(x)>0$ and another $x$ such that it is negative. But how are you showing this? –  Srivatsan Sep 14 '11 at 0:54
    
@ Srivatsan: I may not be, so bear with me. But haven't I done so by showing that $f(x) \to -x$ for some $x<0$ and for some $x>0$? –  raxacoricofallapatorius Sep 14 '11 at 1:01
    
@rax Ok, I was waiting for you in the chat room. Anyway.. Your statement that $f(x) \to -x$ doesn't mean anything. How do you define that formally? –  Srivatsan Sep 14 '11 at 1:24
    
@ Srivatsan: That depends on what's given. In the context of the exercise, I'm not sure what more I'd need to provide. Perhaps that $$\mathop{\lim}\limits_{|x| \to \infty}\frac{f(x)}{-x} =1.$$ But I may be missing your point. –  raxacoricofallapatorius Sep 14 '11 at 3:23

Perhaps I'm missing something (given the comments posted thus far), but I don't see what is wrong with your argument, other than being a little more precise in applying Theorem 1, such as:

$$f(3) = \sin(3) - 3 + 1 \leq 1 - 3 + 1 = -1 < 0$$

and

$$f(-1) = -\sin(1) - (-1) + 1 \geq -1 + 1 + 1 = 1 > 0,$$

so there exists $x \in [-1,3]$ such that $f(x) = 0$.

share|improve this answer
    
My argument was the other way round: instead if picking specific values, can't I just say that it's so for large $|x|$? –  raxacoricofallapatorius Sep 13 '11 at 22:13
1  
@raxa, if you just say "for large enough $x$", you're in effect either promising to exhibit a concrete bound such that $x$ larger than that bound will always work (which is a bit more cinvolved than just showing a single $x$ that works), or you're leaving it to the reader to convince himself that there is such a bound. The latter is quite defensible in this case, but in principle what you get out of it here is not a simpler proof, but merely a more compact presentation of a slightly less simple proof. –  Henning Makholm Sep 13 '11 at 22:37
    
@Henning: Yes, and just to be sure I get that point, Pete's demonstration that $f(x)$ is non-negative for all $x \leq 0$ and non-positive for all $x \geq 2$ is an example of how to show such concrete bounds (within the context of an approach that's still "general"). –  raxacoricofallapatorius Sep 13 '11 at 22:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.