Math Challenge - November 2020

[songoku]

Spoiler: Question11

Let $f(x)=x(x+1)(x+2)(x+3)=x^4 +6x^3 11x^2 + 6x$

1. Finding the stationary points of $f(x) \rightarrow f'(x)=0$
$4x^3 + 18x^2 + 22x + 6 = 0$
$(2x+3)(4x^2+12x+4)=0$
$x=-\frac{3}{2} , x=\frac{-3 \pm \sqrt{5}}{2}$

Checking the sign diagram of first derivative:
a. For $x<\frac{-3 - \sqrt{5}}{2} \rightarrow f'(x) = (-)$

b. For $\frac{-3 - \sqrt{5}}{2}<x<-\frac{3}{2} \rightarrow f'(x) = (+)$

c. For $-\frac{3}{2} <x<\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (-)$

d. For $x>\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (+)$

$x=\frac{-3 - \sqrt{5}}{2} \rightarrow y=-1$ so $(\frac{-3 - \sqrt{5}}{2} , -1)$ is local minimum

$x=-\frac{3}{2} \rightarrow y=\frac{9}{16}$ so $(-\frac{3}{2} , \frac{9}{16})$ is local maximum

$x=\frac{-3 + \sqrt{5}}{2} \rightarrow y=-1$ so $(\frac{-3 + \sqrt{5}}{2} , -1)$ is local minimum2. Checking the end behaviour of $f(x)$
$\lim_{x \to \pm \infty} {f(x)=\infty}$

So the local minimum found above is also global minimum

3.
$f(x)=a$ will have no solution if $a$ is below global minimum $\rightarrow a < -1$

No $a$ will result in $f(x)$ has one solution

$f(x)=a$ will have two solutions if $a$ is global minimum or $a$ is above global maximum $\rightarrow a=-1$ or $a>\frac{9}{16}$

$f(x)=a$ will have three solutions if $a$ is global maximum $\rightarrow a=\frac{9}{16}$

$f(x)=a$ will have four solutions if $a$ is between local minimum and local maximum $\rightarrow -1<a<\frac{9}{16}$

I am not sure how to write good conclusion for final answer

You should have drawn a graph of the function first. Then, is there a symmetry in the function?

I thought if I drew the graph, no algebra needed anymore because I could just draw horizontal lines and counted the number of intersections?

What happens if you let $z = x + \frac 3 2$?

Sorry I don't understand this part. You mean I equate $f(x)$ with $z=x+ \frac 3 2$?

Can you summarize this to check the answers which are somehow hidden in your text, which is hard to follow, since you didn't say what you are doing there?
No solution: $a = ...$ or $a\in ...$
1 solution: $a = ...$ or $a\in ...$
2 solutions: $a = ...$ or $a\in ...$
etc.

My answer is in part (3) of my working.
a) no solution $\rightarrow a<-1$
b) 1 solution $\rightarrow a=\emptyset$
c) 2 solutions $\rightarrow a=-1 ~\text{or}~a>\frac{9}{16}$
d) 3 solutions $\rightarrow a=\frac{9}{16}$
e) 4 solutions $\rightarrow -1<a<\frac{9}{16}$
f) more than 4 solutions $\rightarrow a=\emptyset$

I know I haven't explicitly asked for, but what are the solutions? At which values of $x$ does the graph intersect the $x-$axis in the five cases?

I need to think about this

... and $14b = 458752$. This would have actually answered the question!

I will edit my post to include this

Edit: sorry can not edit post#49 anymore

 

[Office_Shredder]

Complete #3 solution with more details

Spoiler

solution 1: prove $C^1$ is not complete directly by constructing a Cauchy sequence of functions that converge to a non-differentiable function.

Let $f_n(x) = \sqrt{((x-0.5)^2+1/n^2}$. Since the square root function is concave, in general $|\sqrt{a+b}-\sqrt{a}| \leq \sqrt{b}$.

So this gives us that
$$|f_n(x)-\sqrt{(x-0.5)^2}| \leq \sqrt{1/n^2} = \frac{1}{n}$$.

In particular, since $\sqrt{(x-0.5)^2}=|x-0.5|$ is not differentiable, we have that the sequence $f_n$ converges to a non-differentiable function in $C^0$, the space of continuous functions, with the supremum norm. The sequence itself is Cauchy and therefore demonstrates $C^1$ is not complete.

Method 2: I am going to use this result: https://en.m.wikipedia.org/wiki/Bounded_inverse_theorem

Let $V \subset C^1$ be the set of functions such that $f(0)=0$. This is a subspace of $C^1$ since if $f(0)=0$ and $g(0)=0$, then $\alpha f(0) + \beta g(0)=0$ for any real numbers $\alpha, \beta$. If $C^1$ is complete, $V$ must also be complete, since if $f_n$ is a sequence with $f_n(x)=0$ for all n, and $f_n$ converges to a function $f$, we must have $f(0)=0$.

We will show that $V$ is not complete, so $C^1$ isn't complete either.

Let $T: C^0 \to V$ be the function that takes a continuous function $f(t)$ to $(Tf)(x) = \int_0^{x} f(t)dt$. This is defined for continuous functions, linear, and by the fundamental theorem of calculus returns a continuously differentiable function. Also integrating from 0 to 0 always returns 0, so the range of $T$ is in fact in $V$.

$T$ is bounded: if $\|f\|_{\infty} = \epsilon$, then for $0\leq x \leq 1$, we have
$|\int_0^{x} f(t)dt |\leq x \max f(t) \leq\epsilon$.

$T$ is also bijective, since differentiation is its inverse. Hence by the bounded inverse mapping theorem, if $V$ is a Banach space, $T^{-1}$ is bounded as well.

But it is not. Let $f_n=\sin(nx)$. Then $\|f_n\|_{\infty} = 1$ for all $n$, but $\|f'_n\|_{\infty} =n$. Hence $V$ is not a Banach space, and neither is $C^1$.

 

[PeroK]

Sorry I don't understand this part. You mean I equate f(x) with z=x+32?

You can reduce the work if you notice first that $f(x)$ tends to $+\infty$ as $x \rightarrow \pm \infty$ and is symmetric about $x = -\frac 3 2$. This allows you sketch the graph immediately, with a local maximum at $x = -\frac 3 2$ and equal local minina somewhere in $(-3, -2)$ and $(-1, 0)$.

The local maximum is easy to calculate as $\frac 9 4$. Here's a neat way to calculate the local minimum:

First, note that: $$ f(x) = (x^2 + 3x)(x^2 + 3x + 2) \ \ \text{and} \ \ f'(x) = 4(x + \frac 3 2)(x^2 + 3x + 1)$$ Hence, at the local minima we have $$f'(x) = 0 \ \Rightarrow \ x^2 + 3x = -1$$ We can use that directly to get $$f(x_{min}) = (-1)(1) = -1$$
Finally, a quick way to show the symmetry is to note that $$f(x) = (x^2 + 3x)(x^2 + 3x + 2)$$ is the product of two quadratics, both with a symmetry about $x = -\frac 3 2$.

 

[Office_Shredder]

I present to you the most mediocre and non-repeatable solution to #10

Spoiler

since I'm writing this on my phone, I'm going to write $a$ instead of $\alpha$.

First we observe that $f(x)$ has no repeated roots, as $f'(x)=3x^2-3$ which has roots $\pm 1$, which are not roots of $x$.

Furthermore, the rational root test says any rational root of $f$ must be a factor of the constant term divided by a factor of the highest degree term, hence any rational root must be of the form $\pm 1$ which again is not a root.

So $f$ is a separable polynomial, and is irreducible since any factoring must include a polynomial of degree 1 and hence a root.Therefore the automorphisms of the splitting field correspond to permutations of the roots of $f$. If $a$ a root implies $2-a^2$ is a root (to be shown below), $\mathbb{Q}(a)$ has at least two and hence all three of the roots of $f(x)$ since the roots can be factored out leaving only a linear factor left. This shows $\mathbb{Q}(a)$ is the splitting field for $f$. $[\mathbb{Q}(a):\mathbb{Q}]=3$, so the group of automorphisms has magnitude 3. Every permutation on three elements either swaps two and leaves the last unchanged, or cycles all three. Since the order of the permutation must divide the order of the group, the single swaps are not possible, and the Galois group must be $\mathbb{Z}_3$, with the automorphisms mapping roots $a_1$, $a_2$ and $a_3$ as either $a_1\to a_2 \to a_3 \to a_1$ or the inverse.

From here, it suffices to show that if $a$ is a root of $f(x)$, so is $2-a^2$, and that this does not fix a single root while swapping the other two. If it swapped two roots, we would have $2-(2-a^2)^2=a$. Expanding the left hand side and moving $a$ over yields
$$2-4+a^4-4a^2-a=0.$$
$$a^4-4a^2-a-2=0.$$

We know $a^3=3a+1$ and hence $a^4=3a^2+a$ so this gives us
$$-a^2-2=0.$$

But we know that $a$ cannot satisfy a polynomial of degree 2 in $\mathbb{Q}$ since $[\mathbb{Q}(a):\mathbb{Q}]=3$ and by the tower lemma there are no intermediate fields of degree 2 that $\mathbb{Q}(a)$ can contain

So all we need to do now is show if $f(a)=0$, that $2-a^2$ is also a root. This is just algebra:
$$f(2-a^2) = (2-a^2)^3 - 3(2-a^2) -1$$
Expanding gives
$$=8-12a^2 + 6a^4 - a^6 -6+3a^2-1$$
$$=-a^6+6a^4-9a^2+1$$
We use $a^3=3a+1$ to get $a^6=3a^4+a^3$ and hence
$$=3a^4-a^3-9a^2+1$$
Similarly, $a^4=3a^2+a$ so we get
$$=-a^3+3a+1=-f(a)=0$$
Hence the automorphism $\sigma$ exists as desired.

To solve for $\sqrt{12-3a^2}$ I did dumb things. Namely, I guessed it was going to involve a $2-a^2$ and just tried to get there. I used wolfram alpha to get $a$, noticed $\sqrt{12-3a^2}$ was a little more than twice $2-a^2$ when $a$ was picked to be the smallest root in magnitude, and then saw adding $a/2$ got me the rest of the way there. Doing a bit of algebra gives the claim:
$$\sqrt{12-3a^2}=4+a-2a^2.$$

First we show that
$$12-3a^2=(4+a-2a^2)^2.$$

Expanding the right hand side yields
$$16+a^2+4a^4+8a-16a^2-4a^3.$$

Moving the $12-3a^2$ over and simplifying terms we need to show
$$0=4+8a-12a^2-4a^3+4a^4.$$

Dividing by 4 leaves us proving
$$0=1+2a-3a^2-a^3+a^4.$$

Plugging in our trusty formula of $a^4= 3a^2+a$, this reduces to
$$0=1+3a-a^3=-f(a)=0.$$

So we have the squared formula. Taking square roots, we now need to only show that $4+a-2a^2>0$. As an aside, we know all the roots are real numbers since complex roots will come in conjugate pairs and then conjugating the roots yields an automorphism that isn't cycling all three roots, so this is actually either a positive or negative number, as are all the roots.

it suffices to show that there exists at least one root with magnitude <1, as then $|4+a-2a^2| > 4 -|a| - 2|a|^2$ and $|a|,|a|^2 < 1$ in this case. But the roots have to be in between the turning points of the polynomial, which we computed to be $\pm 1$ earlier. In particular there is one root smaller than -1, one root larger than 1, and one in between as desired. So $4+a-2a^2 >0$ as required.

You must have had a better idea in mind for how to come up with the polynomial expression for $\sqrt{12-3a^2}$ than my dumb way of doing it.

Edit:I realized you can also just say if there is a polynomial, it must be at most quadratic since you can use the cubic polynomial relation to reduce the degree. Then you just need to find $p(a)^2=12-3a^2$ and use the cubic relation again to reduce the degree of the left hand side. It's still not clear to me how you would have known this is even contained in the field from first principles though.

 

[songoku]

Spoiler: Question 12

Let:
$a$ = number of people speak only English
$b$ = number of people speak only Russian
$c$ = number of people speak only Spanish
$d$ = number of people speak two languages
$e$ = number of people speak three languages

We have:
$(1)~a+b+c+d+e=30$
$(2)~2(a+b+c)<d<3(a+b+c)$
$(3)~a+b+c=e$
$(4)~b<a<c$
$(5)~a<3b$
----------------------------------------------------------------------------------------------------------------
From $(1)$ and $(3)$:
$2e+d=30 ...(6)$

From $(2)$ and $(3)$:
$2e<d<3e ...(7)$

From $(6)$ and $(7)$:
(i) Taking $d=2e$
$4e=30 \to e=7.5$

(ii) Taking $d=3e$:
$5e=30 \to e=6$

So the value of $e$ lies in $6<e<7.5$ and the only possibility is $e=7$

--------------------------------------------------------------------------------------------------------------------
From $(3)$ and $(4)$:
since $a+b+c=7$ and $b<a<c$, the only possible values are $a=2, b=1, c=4$

---------------------------------------------------------------------------------------------------------------------

Final answer:
1) Number of people speak only English = 2 people
2) Number of people speak only Russian = 1 person
3) Number of people speak only Spanish = 4 people
4) Number of people speak all languages = 7 people

 

[fresh_42]

I present to you the most mediocre and non-repeatable solution to #10
since I'm writing this on my phone, I'm going to write $a$ instead of $\alpha$.

First we observe that $f(x)$ has no repeated roots, as $f'(x)=3x^2-3$ which has roots $\pm 1$, which are not roots of $x$.

$f(x)$

Furthermore, the rational root test says any rational root of $f$ must be a factor of the constant term divided by a factor of the highest degree term, hence any rational root must be of the form $\pm 1$ which again is not a root.

Could you elaborate this a bit? What is the rational root test? If $f(x)=(x-\alpha)(x-\beta)(x-\gamma)$, why can't we have $\alpha=2,\beta=\sqrt{2},\gamma=1/2\sqrt{2}$? It is not hard to do it explicitly, but if you use a theorem, can you quote it?

So $f$ is a separable polynomial, and is irreducible since any factoring must include a polynomial of degree 1 and hence a root.Therefore the automorphisms of the splitting field correspond to permutations of the roots of $f$. If $a$ a root implies $2-a^2$ is a root (to be shown below), $\mathbb{Q}(a)$ has at least two and hence all three of the roots of $f(x)$ since the roots can be factored out leaving only a linear factor left. This shows $\mathbb{Q}(a)$ is the splitting field for $f$. $[\mathbb{Q}(a):\mathbb{Q}]=3$, so the group of automorphisms has magnitude 3. Every permutation on three elements either swaps two and leaves the last unchanged, or cycles all three. Since the order of the permutation must divide the order of the group, the single swaps are not possible, and the Galois group must be $\mathbb{Z}_3$, with the automorphisms mapping roots $a_1$, $a_2$ and $a_3$ as either $a_1\to a_2 \to a_3 \to a_1$ or the inverse.

This is all correct, except a tiny gap. You must rule out the case $\alpha=2-\alpha^2$.
I have never heard about the term magnitude in this context. Do you have a reference? Just out of curiosity.

From here, it suffices to show that if $a$ is a root of $f(x)$, so is $2-a^2$, and that this does not fix a single root while swapping the other two. If it swapped two roots, we would have $2-(2-a^2)^2=a$. Expanding the left hand side and moving $a$ over yields
$$2-4+a^4-4a^2-a=0.$$
$$a^4-4a^2-a-2=0.$$

We know $a^3=3a+1$ and hence $a^4=3a^2+a$ so this gives us
$$-a^2-2=0.$$

But we know that $a$ cannot satisfy a polynomial of degree 2 in $\mathbb{Q}$ since $[\mathbb{Q}(a):\mathbb{Q}]=3$ and by the tower lemma there are no intermediate fields of degree 2 that $\mathbb{Q}(a)$ can contain

So all we need to do now is show if $f(a)=0$, that $2-a^2$ is also a root. This is just algebra:
$$f(2-a^2) = (2-a^2)^3 - 3(2-a^2) -1$$
Expanding gives
$$=8-12a^2 + 6a^4 - a^6 -6+3a^2-1$$
$$=-a^6+6a^4-9a^2+1$$
We use $a^3=3a+1$ to get $a^6=3a^4+a^3$ and hence
$$=3a^4-a^3-9a^2+1$$
Similarly, $a^4=3a^2+a$ so we get
$$=-a^3+3a+1=-f(a)=0$$
Hence the automorphism $\sigma$ exists as desired.

A shortcut after observing $f(\alpha)=f(2-\alpha^2)=0$ and $\alpha\neq 2-\alpha^2$ would have been to note, that we have a Galois extension and thus a transitive automorphism group.

To solve for $\sqrt{12-3a^2}$ I did dumb things. Namely, I guessed it was going to involve a $2-a^2$ and just tried to get there. I used wolfram alpha to get $a$, noticed $\sqrt{12-3a^2}$ was a little more than twice $2-a^2$ when $a$ was picked to be the smallest root in magnitude, and then saw adding $a/2$ got me the rest of the way there. Doing a bit of algebra gives the claim:
$$\sqrt{12-3a^2}=4+a-2a^2.$$

First we show that
$$12-3a^2=(4+a-2a^2)^2.$$

Expanding the right hand side yields
$$16+a^2+4a^4+8a-16a^2-4a^3.$$

Moving the $12-3a^2$ over and simplifying terms we need to show
$$0=4+8a-12a^2-4a^3+4a^4.$$

Dividing by 4 leaves us proving
$$0=1+2a-3a^2-a^3+a^4.$$

Plugging in our trusty formula of $a^4= 3a^2+a$, this reduces to
$$0=1+3a-a^3=-f(a)=0.$$

So we have the squared formula. Taking square roots, we now need to only show that $4+a-2a^2>0$. As an aside, we know all the roots are real numbers since complex roots will come in conjugate pairs and then conjugating the roots yields an automorphism that isn't cycling all three roots, so this is actually either a positive or negative number, as are all the roots.

it suffices to show that there exists at least one root with magnitude <1, as then $|4+a-2a^2| > 4 -|a| - 2|a|^2$ and $|a|,|a|^2 < 1$ in this case. But the roots have to be in between the turning points of the polynomial, which we computed to be $\pm 1$ earlier. In particular there is one root smaller than -1, one root larger than 1, and one in between as desired. So $4+a-2a^2 >0$ as required.

You must have had a better idea in mind for how to come up with the polynomial expression for $\sqrt{12-3a^2}$ than my dumb way of doing it.

Depends on the definition of dumb. I made a long division by $x-\alpha$, observed (via the extrema you already calculated) $\alpha\approx 1/3$ which thus must be the closest to $0$, and got the desired radical for $2-\alpha^2=(x-\varphi \pm \rho)$.

Edit:I realized you can also just say if there is a polynomial, it must be at most quadratic since you can use the cubic polynomial relation to reduce the degree. Then you just need to find $p(a)^2=12-3a^2$ and use the cubic relation again to reduce the degree of the left hand side. It's still not clear to me how you would have known this is even contained in the field from first principles though.

Long division results in something with $\sqrt{\alpha}$, so there is a way to come from one to the other expression. The fact, that they were identical, shortcuts the process a lot.

 

[Incand]

In question 9a) what does the notation $$A/M A$$ mean?

 

[fresh_42]

In question 9a) what does the notation $$A/M A$$ mean?

$A$ and $M$ are ideals in $R$. So $M\cdot A= \{\sum m_\iota a_\iota\}$ is an ideal in $A$ and we can build the quotient $A/AM$ which is an $R-$module. The condition $A/MA \cong_R \{0\}$ only means that $A=MA$ as $R-$modules.

 

[Office_Shredder]

$f(x)$
Could you elaborate this a bit? What is the rational root test? If $f(x)=(x-\alpha)(x-\beta)(x-\gamma)$, why can't we have $\alpha=2,\beta=\sqrt{2},\gamma=1/2\sqrt{2}$? It is not hard to do it explicitly, but if you use a theorem, can you quote it?,

https://en.m.wikipedia.org/wiki/Rational_root_theorem

If an integer polynomial has a rational root, in reduced form the numerator and denominator are constrained to be factors of the constant and the highest degree coefficient.

This is all correct, except a tiny gap. You must rule out the case $\alpha=2-\alpha^2$.
I have never heard about the term magnitude in this context. Do you have a reference? Just out of curiosity.

Whoops! I did forget that case. It can't fix $a$ because then $a$ would be a root to a quadratic polynomial which we ruled out.

As a slightly fancier proof, it would have to fix all three roots of $f(x)$ which we showed were unique, but quadratics only have two roots!

By magnitude I meant order of the group. I have spent too long thinking about a problem where emphasizing the size is unsigned is important

 

[Office_Shredder]

Can you remind me of what the notation $R_P$ means for a prime ideal $P$?

 

[fresh_42]

Can you remind me of what the notation $R_P$ means for a prime ideal $P$?

If $P \lhd R$ is prime, then $S:=R-P$ is multiplicatively closed.

The short answer is $R_P:=S^{-1}R =\dfrac{R}{S}$.

For the correct answer we define an equivalence relation on $R\times S$ by $(a,s) \sim (b,t) :\Longleftrightarrow (at-bs)u=0$ for some $u\in S$. If $[a/s]$ denotes the equivalence class of $(a,s)$, then $R_P=S^{-1}R=\{[a/s]\, : \,a\in R,s\notin P\}.$

The solution to the problem is simply to calculate the Krull dimension of the local ring $R$.

 

[Office_Shredder]

For number 8, I noticed you wrote the limit of the gradient of u and v, but based on the previous sentence, did you mean the laplacian?

 

[disregardthat]

9a) is just Nakayama's lemma for local rings, no?
9b) Suppose that $r \in R$ is a non-zero, non-invertible element. Then the ideal $(r)$ is contained in some maximal ideal $m$. But the Krull dimension of the integral domain $R_m$ is zero, so the sequence of inclusions of prime ideals $(0) \subseteq m$ cannot be a strict in $R_m$, i.e. $mR_m = 0$. This means that $r = 0$ i $R_m$, which is a contradiction. Hence $R$ has no non-zero, non-invertible elements, and is therefore a field.

 

[disregardthat]

What does the $u_t$ notation mean in 8.?

 

[fresh_42]

For number 8, I noticed you wrote the limit of the gradient of u and v, but based on the previous sentence, did you mean the laplacian?

I mean $\Delta \log u =\dfrac{\Delta u}{u} -\dfrac{|\nabla u|^2}{u^2}$

 

[fresh_42]

What does the ut notation mean in 8.?

$u_t=\dfrac{\partial u(x,t)}{\partial t}$

 

[fresh_42]

9a) is just Nakayama's lemma for local rings, no?
9b) Suppose that $r \in R$ is a non-zero, non-invertible element. Then the ideal $(r)$ is contained in some maximal ideal $m$. But the Krull dimension of the integral domain $R_m$ is zero, so the sequence of inclusions of prime ideals $(0) \subseteq m$ cannot be a strict in $R_m$, i.e. $mR_m = 0$. This means that $r = 0$ i $R_m$, which is a contradiction. Hence $R$ has no non-zero, non-invertible elements, and is therefore a field.

Yes. But maybe you could write a short sentence why Nakayama is applicable.

 

[disregardthat]

Yes. But maybe you could write a short sentence why Nakayama is applicable.

Ok, let's take the most common form of Nakayama's lemma.

Statement 2: If M is a finitely-generated module over R, J(R) is the Jacobson radical of R, and J(R)M = M, then M = 0.

9a) Since $R$ is local, the Jacobson radical (as the intersection of all maximal ideals) is simply the maximal ideal $M$. An ideal $A$ of $R$ is finitely generated since $R$ is Noetherian. Hence $A = MA$ implies that $A = (0)$ by Nakayama's lemma.

 

[songoku]

My working is a mess, but here it is:

Spoiler: Question 15

$$\sin^4 x -\cos^4 x=\lambda(\tan^4 x - \cot^4 x)$$
$$\sin^4 x -\cos^4 x=\lambda \left(\frac{\sin^4 x}{\cos^4 x}-\frac{\cos^4 x}{\sin^4 x} \right)$$
$$\sin^4 x -\cos^4 x=\lambda \left(\frac{\sin^8 x - \cos^8 x}{(\sin x. \cos x)^4}\right)$$
$$(\sin^4 x -\cos^4 x)-\lambda \left( \frac{(\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)}{(\sin x. \cos x)^4}\right) =0$$
$$(\sin^4 x - \cos^4 x )\left(1-\lambda \frac{\sin^4 x + \cos^4 x}{(\sin x. \cos x)^4}\right)=0$$
$$\sin^4 x - \cos^4 x = 0 ~\text{or}~\left(1-\lambda \frac{\sin^4 x + \cos^4 x}{(\sin x. \cos x)^4}\right)=0$$

(i) From $\sin^4 x - \cos^4 x = 0$, we get $x=\frac{\pi}{4}$

(ii) From $1-\lambda \left(\frac{\sin^4 x + \cos^4 x}{(\sin x. \cos x)^4}\right)=0$:

$$(\sin x. \cos x)^4=\lambda (\sin^4 x + \cos^4 x)$$
Because LHS is always positive for $x \in \left(0,\frac{\pi}{2}\right)$, $\lambda \leq 0$ will make the equation has no solution

Since $(\sin x. \cos x)^4 < (\sin^4 x + \cos^4 x)$, I want to compare the maximum value of $(\sin x. \cos x)^4$ and the minimum value of $(\sin^4 x + \cos^4 x)$ by using derivative

Let $p=(\sin x. \cos x)^4$ and $q= (\sin^4 x + \cos^4 x)$

$$p'=4(\sin x. \cos x)^3 . (\cos^2 x - \sin^2 x)$$
$$0=4(\sin x. \cos x)^3 . (\cos^2 x - \sin^2 x)$$
No solution for $(\sin x. \cos x)^3 =0$ and for $(\cos^2 x - \sin^2 x)=0$ the solution is $x=\frac{\pi}{4}$

For $q'=0$:
$$4\sin^3 x . \cos x - 4\cos^3 x . \sin x = 0$$
$$(\sin x . \cos x).(\sin^2 x - \cos^2 x)=0\rightarrow \text{the solution will be}~ x=\frac{\pi}{4}$$

Calculating the minimum value of $(\sin^4 x + \cos^4 x)$, I get $q=0.5$ and the maximum value of $(\sin x. \cos x)^4$ is $p=\frac{1}{16}$.
So for the maximum value of $p$ and minimum value of $q$ coincides, the value of $\lambda$ is $\frac 1 8$

For $0<\lambda<\frac 1 8$, there will be two intersections between graph of $p$ and $q$
--------------------------------------------------------------------------------------------------------------------------------------------------
From (i), it can be seen $x=\frac \pi 4$ will always satisfy the original equation, independent of $\lambda$ so the original question is guaranteed to have at least one solution

Combining (i) and (ii), there can be only cases where original equation has one solution or three solutions
--------------------------------------------------------------------------------------------------------------------------------------------------

Final answer:
a) no solution $\to \lambda = \emptyset$
b) 1 solution $\to \lambda \leq 0 ~\text{or}~ \lambda \geq \frac 1 8$
c) two solutions $\to \lambda = \emptyset$
d) three solutions $\to 0<\lambda<\frac 1 8$

 

[fresh_42]

My working is a mess, but here it is:

Spoiler: Question 15

$$\sin^4 x -\cos^4 x=\lambda(\tan^4 x - \cot^4 x)$$
$$\sin^4 x -\cos^4 x=\lambda \left(\frac{\sin^4 x}{\cos^4 x}-\frac{\cos^4 x}{\sin^4 x} \right)$$
$$\sin^4 x -\cos^4 x=\lambda \left(\frac{\sin^8 x - \cos^8 x}{(\sin x. \cos x)^4}\right)$$
$$(\sin^4 x -\cos^4 x)-\lambda \left( \frac{(\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)}{(\sin x. \cos x)^4}\right) =0$$
$$(\sin^4 x - \cos^4 x )\left(1-\lambda \frac{\sin^4 x + \cos^4 x}{(\sin x. \cos x)^4}\right)=0$$
$$\sin^4 x - \cos^4 x = 0 ~\text{or}~\left(1-\lambda \frac{\sin^4 x + \cos^4 x}{(\sin x. \cos x)^4}\right)=0$$

(i) From $\sin^4 x - \cos^4 x = 0$, we get $x=\frac{\pi}{4}$

(ii) From $1-\lambda \left(\frac{\sin^4 x + \cos^4 x}{(\sin x. \cos x)^4}\right)=0$:

$$(\sin x. \cos x)^4=\lambda (\sin^4 x + \cos^4 x)$$
Because LHS is always positive for $x \in \left(0,\frac{\pi}{2}\right)$, $\lambda \leq 0$ will make the equation has no solution

Since $(\sin x. \cos x)^4 < (\sin^4 x + \cos^4 x)$, I want to compare the maximum value of $(\sin x. \cos x)^4$ and the minimum value of $(\sin^4 x + \cos^4 x)$ by using derivative

Let $p=(\sin x. \cos x)^4$ and $q= (\sin^4 x + \cos^4 x)$

$$p'=4(\sin x. \cos x)^3 . (\cos^2 x - \sin^2 x)$$
$$0=4(\sin x. \cos x)^3 . (\cos^2 x - \sin^2 x)$$
No solution for $(\sin x. \cos x)^3 =0$ and for $(\cos^2 x - \sin^2 x)=0$ the solution is $x=\frac{\pi}{4}$

For $q'=0$:
$$4\sin^3 x . \cos x - 4\cos^3 x . \sin x = 0$$
$$(\sin x . \cos x).(\sin^2 x - \cos^2 x)=0\rightarrow \text{the solution will be}~ x=\frac{\pi}{4}$$

Calculating the minimum value of $(\sin^4 x + \cos^4 x)$, I get $q=0.5$ and the maximum value of $(\sin x. \cos x)^4$ is $p=\frac{1}{16}$.
So for the maximum value of $p$ and minimum value of $q$ coincides, the value of $\lambda$ is $\frac 1 8$

For $0<\lambda<\frac 1 8$, there will be two intersections between graph of $p$ and $q$
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From (i), it can be seen $x=\frac \pi 4$ will always satisfy the original equation, independent of $\lambda$ so the original question is guaranteed to have at least one solution

Combining (i) and (ii), there can be only cases where original equation has one solution or three solutions
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Final answer:
a) no solution $\to \lambda = \emptyset$
b) 1 solution $\to \lambda \leq 0 ~\text{or}~ \lambda \geq \frac 1 8$
c) two solutions $\to \lambda = \emptyset$
d) three solutions $\to 0<\lambda<\frac 1 8$

Correct. You could have shortened it by investigating the function $L(x)=\dfrac{\sin^4x-\cos^4x}{\tan^4x-\cot^4x}$ on $(0,\pi/4)$.

 

[martinbn]

About 1: It is a very standard maximum principle for the heat equation.

About 8: Using the hint, where $s$ looks more like energy density and it should be integrated over $\Omega$ to get the total energy $E=\int_\Omega s$, doing the usual pde thing, namely differentiate, use the Green's formula, the boundary conditions one gets that

$\frac {dE}{dt}=-aD\int_\Omega \frac1u^2|\nabla u|^2-D\int_\Omega \frac1v^2|\nabla v|^2$

So it is decreasing (well, non-increasing). Easy to see that $E$ is bounded, so the above goes to zero.

One comment: either $\Omega$ should accully be bounded, or if it is just of finite volume, one need sufficient decay of $u$ and $v$ along with the boundary conditions.

 

[martinbn]

For 14: The observations that help is that any real zero has to be negative and that if $a$ is a zero of $f_n(x)$, then $f'_n(a)=-\frac{a^n}{n!}$. In fact for even $n$ there are no zeros, for odd exactly one.

 

[songoku]

Correct. You could have shortened it by investigating the function $L(x)=\dfrac{\sin^4x-\cos^4x}{\tan^4x-\cot^4x}$ on $(0,\pi/4)$.

That can be simplified to $L(x)=\dfrac{(\sin x . \cos x)^4}{\sin^4x+\cos^4x}$

Then how to continue? How to approach it differently from the one I did before? Thanks

 

[songoku]

Another mess

Spoiler: Question 14

$$f_n(x)=1+x+...+\frac{x^n}{n!}$$
$$f'_n(x)=1+x+...+\frac{x^{n-1}}{(n-1)!}=f_n(x)-\frac{x^n}{n!}$$

Let $x=a$ is the zero of $f_n(x)$ so $f_n(a)=0$, then $f'_n(a)=-\dfrac{a^n}{n!}$

Only negative value of $a$ can satisfy $f_n(a)=0$ so the sign of $f'_n(a)$ will depend on whether $n$ is odd or even.1) For odd value of $n$, the sign of $f'_n(a)$ will be positive and for even value of $n$, the sign of $f'_n(a)$ will be negative2) Checking the end behavior of $f_n(x)$
(i) for even value of $n$
$$\lim_{x \to \pm \infty} f_n(x)=+ \infty$$

(ii) for odd value of $n$
$$\lim_{x \to \infty} f_n(x)=+ \infty~\text{and}~\lim_{x \to -\infty} f_n(x) = -\infty$$3) If I "trace" the graph of $f_n(x)$ on interval $(-\infty, \infty)$:

(i) for even value of $n$
There must be at least one turning point to satisfy the end behavior of $f_n(x)$ so there should be at least two zeroes (let say the zeroes are $a$ and $b$ where $a<b<0$).

When the graph crosses x-axis at $x=a$, the slope of $f_n(x)$ will be negative since the graph will be decreasing on interval $(-\infty,a)$ and when the graph crosses the x-axis again at $x=b$, the slope should be positive, which contradicts the sign of $f'_n(x)$ in part (1).

Therefore, for even value of $n$, the graph can not have zero(ii) for odd value of $n$
When the graph crosses x-axis at $x=a$, the slope will be positive since the graph will be increasing on interval $(-\infty,a)$.

Assume $f_n(x)$ has another zero at $x=b$. To have another zero at $x=b$, there must be a turning point and when the graph crosses x-axis at $x=b$, the slope will be negative, which contradicts the sign of $f'_n(x)$ in part (1).

Since a turning point is not a must to satisfy the end behavior of $f_n(x)$, we can rule out the case where $f_n(x)$ has other zero besides $a$

Therefore, for odd value of $n$, the graph only has one zero

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Conclusion:
Since $f_n (x)$ has no zero when $n$ is even and only has one zero when $n$ is odd, it is proven that $f_n (x)$ has at most one real zero

 

[fresh_42]

That can be simplified to $L(x)=\dfrac{(\sin x . \cos x)^4}{\sin^4x+\cos^4x}$

Then how to continue? How to approach it differently from the one I did before? Thanks

You first observe via symmetry considerations why it's sufficient to consider the interval $(0,\pi/4)$. Then you note that $L(x)$ is strictly monotone increasing on that interval, by writing $L(x)=F(\sin(2x))$ as function of the double angle.

 

[fresh_42]

Another mess "Question 14"
$$f_n(x)=1+x+...+\frac{x^n}{n!}$$
$$f'_n(x)=1+x+...+\frac{x^{n-1}}{(n-1)!}=f_n(x)-\frac{x^n}{n!}$$

Let $x=a$ is the zero of $f_n(x)$ so $f_n(a)=0$, then $f'_n(a)=-\dfrac{a^n}{n!}$

Only negative value of $a$ can satisfy $f_n(a)=0$ so the sign of $f'_n(a)$ will depend on whether $n$ is odd or even.1) For odd value of $n$, the sign of $f'_n(a)$ will be positive and for even value of $n$, the sign of $f'_n(a)$ will be negative2) Checking the end behavior of $f_n(x)$
(i) for even value of $n$
$$\lim_{x \to \pm \infty} f_n(x)=+ \infty$$

(ii) for odd value of $n$
$$\lim_{x \to \infty} f_n(x)=+ \infty~\text{and}~\lim_{x \to -\infty} f_n(x) = -\infty$$3) If I "trace" the graph of $f_n(x)$ on interval $(-\infty, \infty)$:

(i) for even value of $n$
There must be at least one turning point to satisfy the end behavior of $f_n(x)$ so there should be at least two zeroes (let say the zeroes are $a$ and $b$ where $a<b<0$).

Why should there be a zero at all? You have to make two cases here: no zero and we are done, so second case, we assume a zero. If there is one, there have to be at least two with a turning point in between.

When the graph crosses x-axis at $x=a$, the slope of $f_n(x)$ will be negative since the graph will be decreasing on interval $(-\infty,a)$ and when the graph crosses the x-axis again at $x=b$, the slope should be positive, which contradicts the sign of $f'_n(x)$ in part (1).

Therefore, for even value of $n$, the graph can not have zero

It would have been far easier, if you only concentrated on $b$. $a$ is irrelevant and confusing, the more as there could be many more zeroes in between. You only have at least two, and did not specify which of the possibly many you are referring to. Just take the rightmost zero $b$ and forget about the rest.

(ii) for odd value of $n$
When the graph crosses x-axis at $x=a$, the slope will be positive since the graph will be increasing on interval $(-\infty,a)$.

Assume $f_n(x)$ has another zero at $x=b$. To have another zero at $x=b$, there must be a turning point and when the graph crosses x-axis at $x=b$, the slope will be negative, which contradicts the sign of $f'_n(x)$ in part (1).

Same as before. Just consider the second to the leftmost zero $b$.

Since a turning point is not a must to satisfy the end behavior of $f_n(x)$, we can rule out the case where $f_n(x)$ has other zero besides $a$

I neither understand this nor do I see where you need it. The turning points are irrelevant.

Therefore, for odd value of $n$, the graph only has one zero

Your idea is the same in all cases:

Whenever the function crosses the line from negative to positive, the slope has to be positive. This rules out even $n$ immediately. Now for odd $n$ it means, we can only cross the line once, since positive to negative would imply a negative slope, again contradicting (1).

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Conclusion:
Since $f_n (x)$ has no zero when $n$ is even and only has one zero when $n$ is odd, it is proven that $f_n (x)$ has at most one real zero

Conclusion:
You should concentrate on your basic idea and omit all the confusing irrelevant rest. Once you had (1) you were basically done. Of course purists might perform an induction instead, but this is not necessary.

 

[benorin]

1. Let $u(x,t)$ satisfy the one dimensional diffusion equation $u_t=Du_{xx}$ in a space-time rectangle $R=\{0\leq x\leq l,0\leq t\leq T\}$, then the maximum value of $u(x,t)$ is assumed either on the initial line $(t=0)$ or on the boundary lines $(x=0 \,or\,x=l )$. $D > 0.$

Spoiler: #1) hand-wavy proof

Well it seems obvious that the maximum of $u(x,t)$ would either occur on the boundary lines $t=0$ or $x=0$ or $x=l$ since suppose it were the one dimensional heat equation of a rod of length $l$, either the temperature $u(x,t)$ is maximized at the radiating point source to begin with or one end of the rod inevitably becomes the hottest point. But I suppose you want a math proof? Well, I'm still thinking...
I found this nifty 1-D heat equation slides for a worked out example problem.

 

[fresh_42]

Spoiler: #1) hand-wavy proof

Well it seems obvious that the maximum of $u(x,t)$ would either occur on the boundary lines $t=0$ or $x=0$ or $x=l$ since suppose it were the one dimensional heat equation of a rod of length $l$, either the temperature $u(x,t)$ is maximized at the radiating point source to begin with or one end of the rod inevitably becomes the hottest point. But I suppose you want a math proof? Well, I'm still thinking...
I found this nifty 1-D heat equation slides for a worked out example problem.

You have to rule out the possibility that the maximum is somewhere in the inner of the spacetime diagram. It could heat up and cool down at some point. If we knew $u_{xx}<0$ at the maximum, we would have a contradiction for inner points. But $u_{xx}=0$ cannot be ruled out.

My solution uses a standard technique: Assume a maximum at the three boundaries, vary that value, and show that it is impossible to get a lower value. Good old calculus of variations.

 

[songoku]

Why should there be a zero at all? You have to make two cases here: no zero and we are done, so second case, we assume a zero. If there is one, there have to be at least two with a turning point in between.

It would have been far easier, if you only concentrated on $b$. $a$ is irrelevant and confusing, the more as there could be many more zeroes in between. You only have at least two, and did not specify which of the possibly many you are referring to. Just take the rightmost zero $b$ and forget about the rest.

Same as before. Just consider the second to the leftmost zero $b$.

I neither understand this nor do I see where you need it. The turning points are irrelevant.

Your idea is the same in all cases:

Whenever the function crosses the line from negative to positive, the slope has to be positive. This rules out even $n$ immediately. Now for odd $n$ it means, we can only cross the line once, since positive to negative would imply a negative slope, again contradicting (1).

Conclusion:
You should concentrate on your basic idea and omit all the confusing irrelevant rest. Once you had (1) you were basically done. Of course purists might perform an induction instead, but this is not necessary.

Ah I see. Thank you very much