Math Challenge - November 2020

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  • #76
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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.
 
  • #77
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In question 9a) what does the notation $$A/M A$$ mean?
 
  • #78
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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.
 
  • #79
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##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 😬
 
  • #80
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Can you remind me of what the notation ##R_P## means for a prime ideal ##P##?
 
  • #81
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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##.
 
  • #82
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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?
 
  • #83
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9a) is just Nakayama's lemma for local rings, no?
9b) Suppose that [itex]r \in R[/itex] is a non-zero, non-invertible element. Then the ideal [itex](r)[/itex] is contained in some maximal ideal [itex]m[/itex]. But the Krull dimension of the integral domain [itex]R_m[/itex] is zero, so the sequence of inclusions of prime ideals [itex](0) \subseteq m[/itex] cannot be a strict in [itex]R_m[/itex], i.e. [itex]mR_m = 0[/itex]. This means that [itex]r = 0[/itex] i [itex]R_m[/itex], which is a contradiction. Hence [itex]R[/itex] has no non-zero, non-invertible elements, and is therefore a field.
 
  • #84
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What does the [itex]u_t[/itex] notation mean in 8.?
 
  • #85
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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}##
 
  • #87
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9a) is just Nakayama's lemma for local rings, no?
9b) Suppose that [itex]r \in R[/itex] is a non-zero, non-invertible element. Then the ideal [itex](r)[/itex] is contained in some maximal ideal [itex]m[/itex]. But the Krull dimension of the integral domain [itex]R_m[/itex] is zero, so the sequence of inclusions of prime ideals [itex](0) \subseteq m[/itex] cannot be a strict in [itex]R_m[/itex], i.e. [itex]mR_m = 0[/itex]. This means that [itex]r = 0[/itex] i [itex]R_m[/itex], which is a contradiction. Hence [itex]R[/itex] 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.
 
  • #88
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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 [itex]R[/itex] is local, the Jacobson radical (as the intersection of all maximal ideals) is simply the maximal ideal [itex]M[/itex]. An ideal [itex]A[/itex] of [itex]R[/itex] is finitely generated since [itex]R[/itex] is Noetherian. Hence [itex]A = MA[/itex] implies that [itex]A = (0)[/itex] by Nakayama's lemma.
 
  • #89
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My working is a mess, but here it is:

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

(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
--------------------------------------------------------------------------------------------------------------------------------------------------

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##
 
  • #90
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My working is a mess, but here it is:

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

(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##
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)##.
 
  • #91
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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.
 
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  • #92
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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.
 
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  • #93
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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
 
  • #94
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Another mess

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

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 negative


2) 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
 
  • #95
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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.
 
  • #96
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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 negative


2) 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.
 
  • #97
benorin
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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.##
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.
 
  • #98
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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.
 
  • #99
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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
 

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