# Math Questions -- Algebra

Discussion in 'Physics & Math' started by Amadeusboy, Apr 4, 2008.

1. ### AmadeusboyRegistered Member

Messages:
22
This is my first post to SciForums and I'm not sure if this is the proper forum for my question. If this is not the correct forum, please direct me to the proper one and accept my apology for the intrusion.

Recently, I have been reviewing calculus and algebra and came across two problems with which I had difficulty. The questions are from page 7 of Calculus by Boyce and DiPrima (Copyright 1988). They are:

Determine the set $S=${$x\mid x^2+x-s>0 \text{ } \forall x\in \mathbb R^1$ },

and

Determine the set $S=$ {$x\mid x^2-3x+s<0 \text { for some }x\in \mathbb R^1$ }.

Is there a difference between the qualifiers "for all" and "for some"? I thought "for some" was equivalent to "for any" and wouldn't this imply "for all"?

If anyone could assist me with these problems, either mathematical or semantical, it would be appreciated.

2. ### Google AdSenseGuest Advertisement

to hide all adverts.
3. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

Messages:
8,967
You're in the right place...wait for Temur or Quarkhead!

4. ### Google AdSenseGuest Advertisement

to hide all adverts.
5. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
I have the feeling that one of the x in front of the vertical line | and the x in front of $\in$ should be s. Probably the first x should be s.

Suppose P(y) is some statement that may or may not hold depending on y. As an example, let us assume

P(y) = "y is an odd number."

So P(1) holds, but P(2) does not hold, or equivalently, "not P(2)" holds.

The statement

$P(y) \forall y\in\mathbb{N}$

does not hold, because some natural numbers are not odd. But

$P(y) \textrm{ for some } y\in\mathbb{N}$

holds, because some natural numbers are odd. The latter can be read also as "There exists a natural number y such that P(y)".

6. ### Google AdSenseGuest Advertisement

to hide all adverts.
7. ### D HSome other guyValued Senior Member

Messages:
2,257
There is a huge difference between "for all" and "for some". Suppose a given condition is true for one and only one real value. It certainly is not true for all reals, but it is true "for some" reals.

The forms as written don't make sense. Since the sets are written as S, I think these should have been written as

Determine the set $S= \lbrace s\in {\mathbb R}^1 \, \mid \, x^2+x-s>0 \; \forall x\in {\mathbb R}^1 \rbrace$,

and

Determine the set $S=\lbrace s \in {\mathbb R}^1 \, \mid \, x^2-3x+s<0 \text { for some} x\in {\mathbb R}^1 \rbrace$

Last edited: Apr 4, 2008
8. ### QuarkHeadRemedial Math StudentValued Senior Member

Messages:
1,717
Hi Amadeus! Welcome to the bear-pit (just kidding. We're all really nice people who love our Mums).
While I fully agree with temur and D H about your qualifiers, I don't share their reservations about your notation; it looks like perfectly standard set-builder to me.

Let's look at your first inequality. Note straight away that, if x is real, then $x^2 \gt 0$ for all x in R. Then either $x^2 \gt x$ or that $x^2 \lt x$. Do you see when these assertions are true?

Then this implies that in $x^2 + x - s \gt 0$, either $s \lt x$ or $s \gt -x$. Again, do you see why?

Then, remembering that s was not specified as an element in S, and that x is real, I may have for x in R (again remember that this is part of your condition) two real intervals (sets) $[-x, s)$ and $(s, x]$. But now recall that that this must be true for all x in R, so that, since I require a single set I may form $S = (-\infty,s)\cup(s, \infty)$.

See if you can get the other one

Last edited: Apr 4, 2008
9. ### D HSome other guyValued Senior Member

Messages:
2,257
Well, I disagree. First, $x^2+x-s>0$ does not imply $s<x$ or $s>-x$. In terms of $x$, it implies $x<\frac {-1 - \sqrt{1+4s}} 2$ or $x>\frac{1 + \sqrt{1+4s}} 2$. In terms of $s$, it means $s<x^2+x$, period.

10. ### AmadeusboyRegistered Member

Messages:
22
I'm sorry for the lateness of my response; normally I do not have much free time during the work day.

Temur and D H are correct--I inadvertently substituted a "$x$" for "$s$" in the beginning of both statements. My thanks to both for pointing out the error. The questions should read:

Determine the set $S=\left\{s\mid x^2+x-s>0\text{ }\forall x\in\mathbb{R}^1\right\}$,

and

Determine the set $S=\left\{s\mid x^2-3x+s<0\text{ for some } x\in\mathbb{R}^1\right\}$.

I had never used Tex and was most likely more concerned with getting the Tex command correct at the expense of the content. I regret the error and any time wasted due to this oversight.

After reading D H's second post and remembering that one is looking for only real values of x, it is clear that the radicand must be at least zero for real values of x. Thus,
$1+4s\geq 0 \text { } \Rightarrow \text { } s\geq-\frac{1}{4}$.
However, $s>-\frac{1}{4}$ are the only values that meet the above stipulations implying $S=\left$$-\frac14,\infty\right$$$.
Please correct me if I'm wrong, but I believe this answers the first question, after one takes the correction into account. I'll use this technique to examine the second question again.

Also, my thanks to temur and D H for correcting my misconceptions about "for all" and "for some".

Last edited: Apr 5, 2008
11. ### D HSome other guyValued Senior Member

Messages:
2,257
Almost, but exactly wrong. You want the quadratic $x^2+x-s$ to be positive definite. A quadratic form is positive definite if and only if the coefficient of the $x^2[/itex] term is positive (which is true here) and the radicand is [i]negative[/i]. A negative radicand ensures that the quadratic has no zeros in the reals. In other words, it won't change signs.$

12. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
Basically the first set is the set of s for which the minimum of the quadratic polynomial is positive, and the second set is that for which the minimum is negative.

13. ### AmadeusboyRegistered Member

Messages:
22
D H, thanks for helping me with this. I see my error now. A zero or positive radicand implies a real solution(s) for $x$, which was not what was desired.

14. ### AmadeusboyRegistered Member

Messages:
22
I have been reviewing some more and came across another question which has been giving me difficulties. The question comes from page forty of Calculus by Boyce and DiPrima (1988)

"The cost of a taxi is fifty cents for the first fifth of a mile and ten cents for each additional fifth of a mile. Give a formula for the cost $C$ in dollars as a funtion of the miles traveled $x$ for $0\leq x\leq 2$."

The first question I have is interpreting the problem. Are fractional increments of the costs allowed? For example, does riding in the taxi for one tenth of a mile cost a quarter and riding for three tenths of a mile cost fifty-five cents?

If fractional increments of the costs are allowed, the solution would be

$C(x)=\left\{\frac{5x}{2} \tex {, } 0 \leq x \leq \frac{1}{5} \\ \frac{1}{2} + \frac{x-\frac{1}{5}}{2}\tex{, } \frac{1}{5}<x\tex{.}$

In this case $C(\frac{1}{10})=0.25$ and $C(\frac{3}{10})=0.55$.

Or does riding in the taxi for one tenth of a mile cost cost fifty and riding for three tenths of a mile cost sixty cents? (Alternatively does riding in the taxi for three tenths of a mile cost fifty cents and four tenths of a mile cost sixty cents?) In other words, are only whole multiples of the costs are allowed?

For this case, I have come up with a solution that somewhat works:

$C(x)=\left\{\frac{1}{2} \tex {, } 0 \leq x \leq \frac{1}{5} \\ \frac{1}{2} + \frac{[5x]}{10}\tex{, } \frac{1}{5}<x<2\tex{.}$

(Here the brackets [] denote the greatest integer function.) In this case $C(\frac{1}{10})=0.50$ and $C(\frac{3}{10})=0.60$. But $C(\frac{2}{5})=0.70$

The two preceding problems in the text involve the definition and use of the Heaviside unit step function, which they define as follows:

$H(t)=\left\{1\tex{, } t \geq 0\\0\tex{, }t<0\tex{.}$

Since my question immediately follows the questions regarding the Heaviside unit step function, I was given the impression that this function would be part of the answer. The solution I found using this function had many cases. Starting from the beginning, for $0<x\leq\frac{1}{5}$,

$C(x)=\frac{1}{2}H(x)$.

Here$C(\frac{1}{10})=0.50$.

For $\frac{1}{5}<x\leq \frac{2}{5}$,

$C(x)=\frac{1}{2}+\frac{1}{10}H(x-\frac{1}{5})H(\frac{2}{5}-x)$

Here $C(\frac{3}{10})=0.60$.

For $\frac{2}{5}<x\leq \frac{3}{5}$,

$C(x)=\frac{1}{2}+\frac{2}{10}H(x-\frac{2}{5})H(\frac{3}{5}-x)$

Here $C(\frac{1}{2})=0.70$.

Continuing on is this fashion, one could write for $\frac{1}{5}<x$,

$C(x)=\frac{1}{2}+\frac{n}{10}H(x-\frac{n}{5})H(\frac{n+1}{5}-x)$, where $n=[5x]$.

Or alternatively, for $0\leq x\leq2$

$C(x)=\frac{1}{2}H(x)H(2-x)+\frac{n}{10}H(x)H(2-x)$, where $n=[5x]$.

$C(x)=0 \tex{ for }x<0 \tex{ and } x>2$. (It also means that just sitting in the taxi and going nowhere costs you fifty cents.)

Does this adequately and accurately answer the question? Any comments and suggestions would be appreciated.

Thanks,

Amadeusboy

15. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
I think your second interpretation is correct, since it is how taxi meters work. For the second part, you should be able to write your function without any conditional statements if you cook up an appropriate combination of Heaviside functions.

16. ### AmadeusboyRegistered Member

Messages:
22
Thanks, Temur
Since the function cuts out at x<0 and x>2, would the last expression given work?

Amadeusboy

17. ### camilusthe villain with x-ray glassesRegistered Senior Member

Messages:
895
sorry for the off-topic post, but damn, I need to learn set theory! and my college offers no more math classes after differential equations.. I need to change schools i know..

and I could probably teach myself but I've been too busy reading science fiction, H.P. Lovecraft is a god!

18. ### ReikuBannedBanned

Messages:
11,238
Same here! Set theory is going to be a topic i am going to have to learn.

19. ### AmadeusboyRegistered Member

Messages:
22
I came across two other problems about which I had questions.

Firstly, how does one prove that $sin(\theta+2n\pi)=sin\theta$? In other words, how does one prove that $sin 2n\pi=0$?

Since induction is introduced much later in the text, how does one prove this without induction.

Lastly, I had a question about the fundamental period of a function.

First, I will give their definition of the fundamental period and secondly, I will give the problem and my attempt to solve it.

A function $f$ is said to be periodic with period if$p$ if for each $x$ in the domain of $f$ the point $x+p$ is also in the domain of $f$, and $f(x+p)=f(x)$. The smallest positive number $p$ with this property is called the fundamental period.​

Show that the functions $f(x)=ax+b$ and $f(x)=ax^2+bx+c \tex{, }a\neq0$ are not periodic with any period. More generally, it can be shown that the only rational function that is periodic is the constant function $f(x)=c$, which can have any period. This problem is from page forty-nine of the aforementioned text.
I proceeded to solve the first part of the question as follows. I need to find $p$such that $f(x+p)=f(x)$. Thus,

$f(x+p)=a(x+p)+b=ax+b=f(x) \tex{ or } ax+ap+b=ax$.

This implies $ap=0$ or, since $a\neq0\tex{, } p=0$. Hence, there is no p which satisfies the given definition.

For the solution to the quadratic, I took a similar approach. Again, I need to find $p$such that $f(x+p)=f(x)$. Thus,

$f(x+p)=a(x+p)^2+b(x+p)+c=ax^2+bx+c=f(x)$

or

$ax^2+2xpa+ap^2+bx+bp+c=ax^2+bx+c$.

This implies $ap^2+p(2xa+b)=0$ or, since $a\neq0\tex{, } p(p+2x+\frac{b}{a})=0$.

Thus $p=\left\{0,-2x-\frac{b}{a}\right\}$.

The first solution for $p$ is identical to the solution in the previous problem. However, the second solution for $p$ is a function of $x$.

Is it because of the fact that $p$ depends on $x$, it is a function and not a number, that the second solution fails to give a period? Or is my interpretation of the definition incorrect? It makes sense that the period cannot be dependent on $x$, but I'm trying to make my solution fit the definition. Thus, I'm looking for the reason why $f(x)=a^2+bx+c$ has no period.

Any assistance is appreciated.

Amadeusboy

Last edited: Aug 10, 2008
20. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
Remember that the relation should hold for each x. This means that if you get a polynomial expression in x then all the coefficients for the powers of x must be 0. In general you will get an expression of form f(p,x)=0 and so the "internal" parameters of f must be such that f(p,x) = F(p) for some F.

ps: I did not check your earlier result very carefully, but from the skills you have demonstrated here I am confident that your results are correct.

Messages:
227
22. ### AmadeusboyRegistered Member

Messages:
22
Temur and Paulfr, thank you for your replies.

Paulfr, indeed one can use the identity

$sin(\alpha + \beta)=sin\alpha cos\beta +sin\beta cos\alpha$

and find

$sin(\theta +2n\pi)=sin\theta cos2n\pi+sin2n\pi cos\theta$.

I know that

$sin2n\pi=0$ and $cos2n\pi=1 \tex{; (1)}$

thus,

$sin\theta cos2n\pi+sin2n\pi cos\theta=sin\theta$.

However, I am not sure how to prove statement (1) without using induction. I can construct a geometric proof similar to the one at the site you mentioned for the identity, but I am uncertain how I would construct such a proof for $cos2n\pi \tex{ or }sin2n\pi$.

Temur, I had a question about your response. Firstly,

Does this mean that $p$ is independent of $x$ or is $p\neq p(x)$?

Secondly,
Does this imply that, given the expression

$ap^2+2apx+bp=0\tex{, (2)$

previously posted, the coefficient $2ap=0$; hence, $p=0$, since $a\neq 0$?

Finally,
If I am understanding your statement (please correct me if I am wrong), in this case, the $f(x,p)=0$ you mention is statement (2). Furthermore, combining $p\neq p(x)$ and the implications from statement (2) one finds $F(p)=0$.

Thanks again for your responses.

Amadeusboy

23. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
I don't think you can get away without induction altogether. I suspect the idea of the problem was to reduce it to a simpler statement whose rigorous proof requires induction, or prepare you to the next lecture by forcing you to feel the need for something like induction.

$p$ is a constant, so it is independent of x by definition.

Yes.

I have to correct some of my statements here. In some cases we should interpret the requirement that f(x,p) does not depend on x as an equation for p (in general as an equation relating p and the "internal" parameters as I said before, but in some problems the internal parameters are fixed with the problem and we cannot change them). Specifically in this case we have $f(x,p)=ap^2+2apx+bp$, and requiring that $f(x,p)=F(p)$ (i.e., $f$ cannot depend on $x$) we have $p=0$. We get this not even solving the equation for the period (which is F(p)=0), but just from the requirement that the period should be a constant. So in general, the equation f(x,p)=0 will decompose into two equations on p, one requiring that f(x,p) does not depend on x, call it G(p)=0, and another one which is just f(x,p)=0 (here f(x,p)=F(p)).