leehuan's All-Levels-Of-Maths SOS thread (5 Viewers)

leehuan · Mar 4, 2016

InteGrand said:
$\noindent Another way to write the domain is $D = \left \{x \in \mathbb{R}: -\frac{7\pi}{6}+2k\pi \leq x \leq \frac{\pi}{6}+2k\pi, \, k\in \mathbb{Z}\right \}$ (just to avoid writing something like $+2k\pi$ next to an interval, which I'm not sure whether or not is a standard notation).$

To be honest I completely forgot about this. Thanks

seanieg89 · Mar 4, 2016

InteGrand said:
$(just to avoid writing something like $+2k\pi$ next to an interval, which I'm not sure whether or not is a standard notation).$

$I+2\pi\mathbb{Z}$ is pretty standard. More generally $A+B$ to denote the set of all sums of elements in $A$ and elements in $B$. This is pretty unambiguous and common notation, although the corresponding product notation sometimes has different meanings (eg the product of two ideals in a ring.) $$

InteGrand · Mar 4, 2016

seanieg89 said:
$I+2\pi\mathbb{Z}$ is pretty standard. More generally $A+B$ to denote the set of all sums of elements in $A$ and elements in $B$. This is pretty unambiguous and common notation, although the corresponding product notation sometimes has different meanings (eg the product of two ideals in a ring.) $$

Ah OK, cheers.

leehuan · Mar 5, 2016

MX2 problem. Required - c and d but all parts are given

$$Consider the polynomial $f\left(x\right)={x}^{3}-3x+k$, where $k$ is an integer greater than 2. \\ a) Show that $f\left(x\right)$ has exactly one real zero $r$, and explain why $r<-1$. \\ b) Give a reason as to why the two complex roots form a conjugate pair. \\ c) If the complex zeroes are $a \pm bi$, use $\sum \alpha \beta$ to prove that ${b}^{2}=3\left({a}^{2}-1\right) \\ $d) Find all three zeroes of $f(x)$ given that $k=0$ and $a, b \in \mathbb Z.$

InteGrand · Mar 5, 2016

leehuan said:
MX2 problem. Required - c and d but all parts are given

$$Consider the polynomial $f\left(x\right)={x}^{3}-3x+k$, where $k$ is an integer greater than 2. \\ a) Show that $f\left(x\right)$ has exactly one real zero $r$, and explain why $r<-1$. \\ b) Give a reason as to why the two complex roots form a conjugate pair. \\ c) If the complex zeroes are $a \pm bi$, use $\sum \alpha \beta$ to prove that ${b}^{2}=3\left({a}^{2}-1\right) \\ $d) Find all three zeroes of $f(x)$ given that $k=0$ and $a, b \in \mathbb Z.$

$\noindent (c) Let the roots be $r$ (real) and $a\pm bi$ ($a,b$ real). By sum of roots being 0, we have $2a + r = 0 \Rightarrow r = -2a$. Now the pairwise product of roots is easily found: $r(a+bi) + r(a-bi) + (a+bi)(a-bi) =- 3$. The LHS simplifies to: $LHS= ra + ra + a^2 + b^2 = 2ra + a^2 + b^2 = -4a^2 + a^2 + b^2 = -3a^2 + b^2$ (we used $r=-2a$). Equating this to $-3$ and rearranging gives us the desired result.$

$\noindent (d) Are your sure it said $k=0$ (They said originally that $k>2$)? This is easy to solve since it's just asking for the roots of $x^3 -3x =0$, the left-hand side of which we can factorise easily.$

leehuan · Mar 5, 2016

Terrible. I meant k=2702. Sorry

InteGrand · Mar 5, 2016

leehuan said:
Terrible. I meant k=2702. Sorry

$\noindent We are given the roots are $r$ (real) and $a\pm bi$, where $a$ and $b$ are integers. By the product of roots, we have $r\left(a^2 + b^2\right) = -2702$. Since $b^2 = 3a^2 - 3$ and $r=-2a$, we have$

$\begin{align*}-2a\left(a^2 + 3a^2 - 3\right) &= -2702 \\ \Longleftrightarrow -2a\left(4a^2 - 3\right) &= -2702 \\ \Longleftrightarrow a\left(4a^2 - 3\right) &= 1351\quad (\text{dividing through by }-2). \\ \end{align*}$

$\noindent Now observe that $1351 = 7 \times 193$. Note that $193 = 4 \times 7^2 - 3$. So the solution is $a=7$. So $b^2 = 3\left(a^2 -1\right) = 3\times 48 = 144$, so take $b=12$. Also, $r=-2a = -14$. So the roots are: $-14, 7+12i,7-12i$.$

Carrotsticks · Mar 5, 2016

Test

http://i0.kym-cdn.com/photos/images/original/000/002/116/Longcat_War.jpg

leehuan · Mar 5, 2016

I will admit this is just a bit of confusion in my head and laziness to sort it out.

$\\ $What is the condition for $a>b \Rightarrow \frac{1}{a}<\frac{1}{b}$?$$

InteGrand · Mar 5, 2016

leehuan said:
I will admit this is just a bit of confusion in my head and laziness to sort it out.

$\\ $What is the condition for $a>b \Rightarrow \frac{1}{a}<\frac{1}{b}$?$$

$\noindent That's true iff $a$ and $b$ have the same sign (obviously neither can be $0$).$

Paradoxica · Mar 5, 2016

InteGrand said:
$\noindent That's true iff $a$ and $b$ have the same sign (obviously neither can be $0$).$

More concisely... ab>0

leehuan · Mar 5, 2016

Mhm makes sense. Thanks

leehuan · Mar 5, 2016

Only the case for (-1,0) but the whole question is again given (with a previous part)

$\\ $a) Prove that $f\left(x\right)=1+x+{x}^{2}$ is positive for all $x. \\ $b) By considering the cases $x\ge 0, \, \bf{-1<x<0}, \, $and $x\le -1\\ $prove that $1+x+{x}^{2}+{x}^{3}+{x}^{4} $ is always positive.$$

Paradoxica · Mar 5, 2016

leehuan said:
Only the case for (-1,0) but the whole question is again given (with a previous part)

$\\ $a) Prove that $f\left(x\right)=1+x+{x}^{2}$ is positive for all $x. \\ $b) By considering the cases $x\ge 0, \, \bf{-1<x<0}, \, $and $x\le -1\\ $prove that $1+x+{x}^{2}+{x}^{3}+{x}^{4} $ is always positive.$$

$1+x+x^2+x^3+x^4 \equiv (1+x+x^2)^2 - x(1+x)^2$

This is probably the identity in question required to prove that statement given the nature of the question.

Paradoxica · Mar 5, 2016

The most straightforward way, however, would be to re-write the polynomial as a sum of perfect squares with zeroes that occur at different x-values.

$x^4+x^3+x^2+x+1\equiv x^2(x+\frac{1}{2})^2 + \frac{x^2}{2} + (\frac{x}{2}+1)^2$

as none of the terms all simultaneously share the same zero point over the reals, it is impossible for this expression to ever be zero, and hence it is strictly positive over the reals.

InteGrand · Mar 5, 2016

leehuan said:
Only the case for (-1,0) but the whole question is again given (with a previous part)

$\\ $a) Prove that $f\left(x\right)=1+x+{x}^{2}$ is positive for all $x. \\ $b) By considering the cases $x\ge 0, \, \bf{-1<x<0}, \, $and $x\le -1\\ $prove that $1+x+{x}^{2}+{x}^{3}+{x}^{4} $ is always positive.$$

$\noindent Another way for part (b) (though I considered intervals differently to them, I find this way easier). Let $g(x) \equiv 1 + x + x^2 + x^3 + x^4$. Clearly $g(x) >0$ for all $x\geq 0$. Now consider $x<0$. Note that we can use the G.P. sum formula to write $g(x) \equiv \frac{x^5 -1}{x-1}$. For negative $x$, clearly $x^5$ and $x$ are both negative, so $x^5-1$ and $x-1$ are also both negative, which implies $g(x)$ is positive (as it is the quotient of two negative quantities). Thus $g(x) >0$ for all real $x$. Q.E.D.$

seanieg89 · Mar 5, 2016

Alternatively alternatively:

P(x)=(x^5-1)/(x-1) for x =/= 1.

But x^5 is increasing, so the numerator has the same sign as the denominator and P(x) > 0 apart from possibly at x=1.

(And P(1)=5 so we are positive here too.)

Edit: Haha, Integrand beat me to it.

Paradoxica · Mar 5, 2016

InteGrand said:
$\noindent Another way for part (b) (though I considered intervals differently to them, I find this way easier). Let $g(x) \equiv 1 + x + x^2 + x^3 + x^4$. Clearly $g(x) >0$ for all $x\geq 0$. Now consider $x<0$. Note that we can use the G.P. sum formula to write $g(x) \equiv \frac{x^5 -1}{x-1}$. For negative $x$, clearly $x^5$ and $x$ are both negative, so $x^5-1$ and $x-1$ are also both negative, which implies $g(x)$ is positive (as it is the quotient of two negative quantities). Thus $g(x) >0$ for all real $x$. Q.E.D.$

seanieg89 said:
Alternatively alternatively:

$\noindent$ P(x) \equiv \frac{x^5-1}{x-1} \quad \forall x \neq 1 \\ $But $x^5$ is monotonically increasing, so the numerator has the same sign as the denominator and $P(x) > 0$ apart from possibly at $x=1$. But $P(1)=5$ so it is true $\forall x \in \mathbb{R}$.$

Variety is the spice of life.

leehuan · Mar 5, 2016

Lol thanks everyone. I ended up doing this:

Let x=-b, 0<b<1
RTP:1-b+b^2-b^3+b^4

1>b implies 1-b>0
b>0 allows multiplying by b^3: b^4-b^3>0
b>0 also implies b^2>0
Sum them up.
________________

Getting confused with what I learnt. Someone please break down surjective functions for me, and injective just means one-to-one both x and y (therefore inverse function exists for the maximum domain) right?

seanieg89 · Mar 5, 2016

leehuan said:
Lol thanks everyone. I ended up doing this:

Let x=-b, 0<b<1
RTP:1-b+b^2-b^3+b^4

1>b implies 1-b>0
b>0 allows multiplying by b^3: b^4-b^3>0
b>0 also implies b^2>0
Sum them up.
________________

Getting confused with what I learnt. Someone please break down surjective functions for me, and injective just means one-to-one both x and y (therefore inverse function exists for the maximum domain) right?

When you specify a function f: X->Y you are also specifying the domain X and the co-domain Y.

With this in mind:

f is said to be injective if any two distinct elements of X get sent to distinct elements of Y by f.

f is said to be surjective if for every y in the codomain Y there exists an x in X with f(x)=y. Ie f "hits everything" in the codomain.

f is said to be bijective if it is both injective and surjective.

leehuan's All-Levels-Of-Maths SOS thread (5 Viewers)

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