Math Challenge - March 2021

  • Challenge
  • Thread starter fresh_42
  • Start date
  • Featured
  • #1
fresh_42
Mentor
Insights Author
2021 Award
16,425
15,464
Summary: Lie Algebras, Commutative Algebra, Ordering, Differential Geometry, Algebraic Geometry, Gamma Function, Calculus, Analytic Geometry, Functional Analysis, Units.

1. Prove that all derivations ##D:=\operatorname{Der}(L)## of a semisimple Lie algebra ##L## are inner derivations ##M:=\operatorname{ad}(L).##

2. Give four possible non-isomorphic meanings for the notation ##\mathbb{Z}_p.##

3. (solved by @Office_Shredder ) Let ##T\subseteq (\mathbb{Z}_+^n,\preccurlyeq )## with the partial natural ordering. Then there is a finite subset ##S\subseteq T## such that for every ##t\in T## exists a ##s\in S## with ##s\preccurlyeq t.##
$$
\alpha \preccurlyeq \beta \Longleftrightarrow \alpha_i \leq \beta_i \text{ for all }i=1,\ldots,n
$$

4. (a) Solve the following linear differential equation system:
\begin{align*}
\dot{y}_1(t)=11y_1(t)-80y_2(t)\;&\wedge\;\dot{y}_2(t)=y_1(t)-5y_2(t)\\
y_1(0)=0\;&\wedge\;y_2(0)=0
\end{align*}
(b) Which solutions do ##y_1(0)=\pm \varepsilon\;\wedge\;y_2(0)=\pm \varepsilon## have?
(c) How does the trajectory for ##y_1(0)=0.001\;\wedge\;y_2(0)=0.001## behave for ##t\to \infty##?
(d) What will change if we substitute the coefficient ##-80## by ##-60##?
(e) Calculate (approximately) the radius of the osculating circle at ##t=\pi/12## for both trajectories with initial condition ##\mathbf{y}(0)=(-1,1).##

5. (solved by @bpet ) Consider the ideal ##I =\langle x^2y+xy ,xy^2+1 \rangle \subseteq \mathbb{R}[x,y] ## and compute a reduced Gröbner basis to determine the number of irreducible components of the algebraic variety ##V(I).##

6. (solved by @benorin ) Define the complex function gamma function as
$$
\Gamma(z):=\lim_{n \to \infty}\dfrac{n!\,n^z}{z(z+1)\cdot\ldots\cdot(z+n)}
$$
and prove
(a) ##\Gamma(z)=\displaystyle{\int_0^\infty}e^{-t}\,t^{z-1}\,dt\;;\quad\mathfrak{R}(z)>0##
(b) ##\Gamma(z)^{-1}=e^{\gamma z}z\,\displaystyle{\prod_{n=1}^\infty}\left(1+\dfrac{z}{n}\right)e^{-\frac{z}{n}}##
where ##\gamma :=\displaystyle{\lim_{n \to \infty}\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\ldots+\dfrac{1}{n}-\log(n)\right)}## is the Euler-Mascheroni constant.

7. (solved by @benorin) Let ##u\, : \,[0,1]\times [a,b]\longrightarrow \mathbb{C}## be a continuous function, such that the partial derivative in the first coordinate exists everywhere and is continuous. Define
$$
U(\lambda):=\int_a^b u(\lambda,t)\,dt\; , \;V(\lambda):=\int_a^b \dfrac{\partial u}{\partial \lambda}(\lambda,t)\,dt.
$$
Show that ##U## is continuously differentiable and ##U'(\lambda)=V(\lambda)## for all ##0\leq \lambda\leq 1.##

8. A cardiod is defined as the trace of a point on a circle that rolls around a fixed circle of the same size without slipping.
Kardiode.png


It can be described by ##(x^{2}+y^{2})^{2}+4x(x^{2}+y^{2})-4y^{2}\,=\,0## or in polar coordinates by ##r(\varphi )=2(1-\cos \varphi ).##
Show that:

(a) (solved by @etotheipi ) Given any line, there are exactly three tangents parallel to it. If we connect the points of tangency to the cusp, the three segments meet at equal angles of ##2\pi/3\,.##
Kardiode2.png

(b) (solved by @etotheipi ) The length of a chord through the cusp equals ##4.##
(c) (solved by @etotheipi ) The midpoints of chords through the cusp lie on the perimeter of the fixed generator circle (black one in the first picture).
(d) (solved by @etotheipi ) Calculate length, area and curvature.

9. (solved by @nuuskur ) Let ##A## be a complex Banach algebra with ##1##. Prove that the spectrum
$$
\sigma(a)=\{\lambda\in \mathbb{C}\,|\,\lambda\cdot 1-a \text{ is not invertible }\} \subseteq \{\lambda\in \mathbb{C}\,|\,|\lambda|\leq \|a\| \}
$$
for any ##a\in A## is not empty, bounded and closed.

10. (a) Determine all primes which occur as orders of an element from ##G:=\operatorname{SL}_3(\mathbb{Z}).##
(b) (solved by @Office_Shredder ) Let ##I \trianglelefteq R## be a two-sided ideal in a unitary ring with group of unities ##U##. Show by two different methods that
$$
M:=\{u\in U\,|\,u-1\in I\} \trianglelefteq U
$$
is a normal subgroup.




1606835746499-png-png.png


High Schoolers only (until 26th)


11.
(solved by @Not anonymous ) If ##a,b,c## are real numbers such that ##a+b+c=2## and ##ab+ac+bc=1,## show that ##0\leq a,b,c\leq \dfrac{4}{3}.##

12. Determine all pairs ##(m,n)## of (positive) natural numbers such that ##2022^m-2021^n## is a square.

13. (solved by @archaic , @Not anonymous )
(a) Prove for any ##n\in \mathbb{N},\,n\geq 4##
$$
Q(n):=\dfrac{4^2-9}{4^2-4}\cdot\dfrac{5^2-9}{5^2-4}\cdot\ldots\cdot\dfrac{n^2-9}{n^2-4}>\dfrac{1}{6}.
$$
(b) Does the above statement still hold, if we replace ##1/6## on the right hand side by ##0.1667\,?##

14. (solved by @Not anonymous ) Determine all pairs ##(x,y)\in \mathbb{R}^2## such that
\begin{align*}
5&=\sqrt{1+x+y}+\sqrt{2+x-y}\\
2-x+y&=\sqrt{18+x-y}
\end{align*}

15. (solved by @Not anonymous ) Given a real, continuous function ##f:\mathbb{R}\longrightarrow \mathbb{R}## such that ##f(f(f(x)))=x.##
Prove that ##f(x)=x## for all ##x\in\mathbb{R}.##
 
Last edited:
  • Like
Likes StenEdeback, lekh2003 and Greg Bernhardt

Answers and Replies

  • #2
147
64
2. Give four possible non-isomorphic meanings for the notation Zp.
Well, guess I’ll grab the low-hanging fruit:
  1. The cyclic group of prime order ##C_p##,
  2. The ring ##\mathbb{Z}/p\mathbb{Z}##,
  3. The group of units modulo ##p##, often written as ##(\mathbb{Z}/p\mathbb{Z})^\times##,
  4. The ring of ##p##-adic integers.
Note that ##(1)## and ##(3)## are non-isomorphic (in particular, ##(\mathbb{Z}/p\mathbb{Z})^\times\leqslant C_{p-1}##).

I think ##(1)## probably comes from the additive group of elements of ##\mathbb{Z}/p\mathbb{Z}##. In fact, I suppose that the notation ##\mathbb{Z}_p## in this case refers to the concrete group ##(\mathbb{Z}/p\mathbb{Z})^+## rather than the abstract group ##C_p## defined by its order. Of course, these are isomorphic as groups — is ##C_p## even a group, and not an isomorphism class thereof? — but perhaps it is more useful in certain contexts to make use of the concrete group.

The only case I can recall where I saw ##(1)## being used was in the definition of a superalgebra: a ##\mathbb{Z}_2##-graded algebra.
It would be a pleasant surprise to me if there were other meanings; I don’t know of any others.
 
  • #3
fresh_42
Mentor
Insights Author
2021 Award
16,425
15,464
Well, guess I’ll grab the low-hanging fruit:
  1. The cyclic group of prime order ##C_p##,
  2. The ring ##\mathbb{Z}/p\mathbb{Z}##,
  3. The group of units modulo ##p##, often written as ##(\mathbb{Z}/p\mathbb{Z})^\times##,
  4. The ring of ##p##-adic integers.
Note that ##(1)## and ##(3)## are non-isomorphic (in particular, ##(\mathbb{Z}/p\mathbb{Z})^\times\leqslant C_{p-1}##).

I think ##(1)## probably comes from the additive group of elements of ##\mathbb{Z}/p\mathbb{Z}##. In fact, I suppose that the notation ##Z_p## in this case refers to the concrete group ##(\mathbb{Z}/p\mathbb{Z})^+## rather than the abstract group ##C_p## defined by its order. Of course, these are isomorphic as groups — is ##C_p## even a group, and not an isomorphism class thereof? — but perhaps it is more useful in certain contexts to make use of the concrete group.

The only case I can recall where I saw ##(1)## being used was in the definition of a superalgebra: a ##\mathbb{Z}_2##-graded algebra.
It would be a pleasant surprise to me if there were other meanings; I don’t know of any others.
(1) and (2) are the same ring here. So (1) and (4) are correct. The units of ##\mathbb{Z}_p## are not written that way. At best it would be ##\mathbb{Z}_{p-1}##.

Thus two solutions are still missing. But I admit that they would possibly be ##\mathbb{Z}_{(p)}## for some authors.
 
  • #4
147
64
(1) and (2) are the same ring here. So (1) and (4) are correct. The units of ##\mathbb{Z}_p## are not written that way. At best it would be ##\mathbb{Z}_{p-1}##.

Thus two solutions are still missing. But I admit that they would possibly be ##\mathbb{Z}_{(p)}## for some authors.
(1) is not a ring. Are you asking for four non-isomorphic rings? If so, I think this should be stated in the question.

I was also iffy on (3), but Wikipedia says the notation is used outside of number theory. So I thought it was worth a shot.
 
  • #5
fresh_42
Mentor
Insights Author
2021 Award
16,425
15,464
(1) is not a ring. Are you asking for four non-isomorphic rings? If so, I think this should be stated in the question.
We can count ##\mathbb{Z}_p## as group, ##\mathbb{Z}-##module, ring, associative algebra, Lie algebra, Jordan algebra, field, vector space, tensor algebra. This would be nine structures, but they are all the same, only with more or less operations.
 
  • #6
martinbn
Science Advisor
2,714
1,075
I suppose for the second question one needs to give references, otherwise I can say that I use ##\mathbb Z_p## to denote real numbers, which of course wouldn't count. If the notation ##\mathbb Z_{(p)}## is allowed, then do localizations count?
 
  • #7
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,508
12,814
Is there something missing from question 15?
 
  • #8
lekh2003
Gold Member
590
339
I think I'm close on Question 11, but I think I may have done something wrong?
$$a(b+c) + bc = 1$$
$$2a-a^2+bc=1$$
The same can be done for the other variables. Then, these equations can be combined.
$$2(a+b+c) - a^2 - b^2 - c^2 + 1= 3$$
Substituting given information:
$$a^2 + b^2 + c^2 = 2$$
This is equivalent to:
$$a^2 + b^2 + c^2 = a + b + c$$
This is a sphere, specifically:
$$\left(a-\frac12\right)^2+\left(b-\frac12\right)^2+\left(c-\frac12\right)^2 = \frac34$$
Here comes my concern. Obviously now I have a set of solutions, that are bounded, and I know information about this because it's a sphere. So, at extreme values of a, b, c can actually reach ##\frac12 \pm \frac{\sqrt3}{4}##. This breaks the given bounds right? Did I mess up a calculation?

Am I misinterpreting the condition? Does it imply that if a and b are bounded by those constraints, c must also be bounded. Or that a, b, c all must always be within the bounds regardless of the conditions? I am confus.
 
  • #9
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
2021 Award
5,008
995
I didn't check your work carefully Lekh, but

Your new equation can't be a full description of the solutions since a+b+c=2 is a plane that can't fully contain that sphere. So you've thrown away information about the constraints.
 
  • #11
fresh_42
Mentor
Insights Author
2021 Award
16,425
15,464
I think I'm close on Question 11, but I think I may have done something wrong?
$$a(b+c) + bc = 1$$
$$2a-a^2+bc=1$$
The same can be done for the other variables. Then, these equations can be combined.
$$2(a+b+c) - a^2 - b^2 - c^2 + 1= 3$$
Substituting given information:
$$a^2 + b^2 + c^2 = 2$$
This is equivalent to:
$$a^2 + b^2 + c^2 = a + b + c$$
This is a sphere, specifically:
$$\left(a-\frac12\right)^2+\left(b-\frac12\right)^2+\left(c-\frac12\right)^2 = \frac34$$
Here comes my concern. Obviously now I have a set of solutions, that are bounded, and I know information about this because it's a sphere. So, at extreme values of a, b, c can actually reach ##\frac12 \pm \frac{\sqrt3}{4}##. This breaks the given bounds right? Did I mess up a calculation?

Am I misinterpreting the condition? Does it imply that if a and b are bounded by those constraints, c must also be bounded. Or that a, b, c all must always be within the bounds regardless of the conditions? I am confus.

You have shown ##a,b,c < 1.37## and ##1.33## was the goal. That's close. The missing margin lies in the last equation: ##a,b,c## cannot simultaneously lie in ##\left[\dfrac{4}{3},\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\right].## What happens in case one equals this limit? Will you need negative values for the other two?

Your estimation is correct. It is simply not the best one.

Edit: You have proven that the given conditions (C) and the sphere (S) are equivalent to (C). From there we cannot conclude that there isn't a property interval (I) which is also equivalent to (C). In formulas:
$$
(\;(C) \wedge (S) \Longleftrightarrow (C)\;) \nRightarrow \nexists (I)\, : \,(\;(C) \wedge (I) \Longleftrightarrow (C)\;)
$$


Hint: Get rid of ##c## and solve your second equation.
 
Last edited:
  • #12
704
513
Liouville, where art thou?!
If [itex]\|a\|<1[/itex], then [itex]1-a\in \mathrm{inv}(A)[/itex].

Pf. Put [itex]x_n := \sum _{k=1}^n a^k,\ n\in\mathbb N[/itex]. It converges due to [itex]\|a\|<1[/itex].Then for every [itex]n[/itex]
[tex]
(1-a)x_n = 1-a^{n+1} = x_n(1-a).
[/tex]
Thus, [itex](1-a) \lim x_n = 1 = \lim x_n(1-a)[/itex].
The subset [itex]\mathrm{inv}(A)[/itex] of invertibles is open.
Pf. [itex]1\in\mathrm{inv}(A)[/itex]. Let [itex]a\in \mathrm{inv}(A)[/itex]. Take [itex]x\in A[/itex] such that [itex]\|x\| < \|a^{-1}\|^{-1}[/itex]. Then [itex]\|a^{-1}x\| < 1[/itex]. By Prop. [itex]a-x = a(1-a^{-1}x) \in \mathrm{inv}(A)[/itex]. Therefore, [itex]B(a, \|a^{-1}\|^{-1}) \subseteq \mathrm{inv}(A)[/itex].
Suppose for a contradiction [itex]\sigma (a) = \emptyset[/itex]. Take [itex]\phi\in A'[/itex] and put
[tex]
f:\mathbb C\to \mathbb C,\quad \gamma\mapsto \phi((a-\gamma\cdot 1)^{-1}).
[/tex]
This is entire. One can work out the derivative of [itex]\gamma\mapsto (a-\gamma\cdot 1)^{-1}[/itex] to be [itex](a-\gamma\cdot 1)^{-2}[/itex]. For [itex]|\gamma| > \|a\|[/itex] we have [itex]\|\gamma ^{-1}a\|<1[/itex] and
[tex]
\begin{align*}
\left\| (a-\gamma \cdot 1)^{-1} \right\| &= \left \| \left (-\gamma(1-\gamma^{-1}\cdot a)\right )^{-1} \right\| \\
&= \left\| (-\gamma)^{-1}\sum (\gamma^{-1}\cdot a)^k \right\| \\
&\leqslant |\gamma|^{-1} \sum \left \|\gamma^{-1}\cdot a \right\|^k \\
&= \frac{1}{|\gamma|} \cdot \frac{1}{1-|\gamma|^{-1}\|a\|} \\
&= \frac{1}{|\gamma|-\|a\|} \xrightarrow[|\gamma|\to\infty]{}0.
\end{align*}
[/tex]
Thus, [itex]f=0[/itex] by Liouville and since [itex]\phi[/itex] was arbitrary, we conclude [itex](a-\gamma\cdot 1)^{-1} = 0[/itex], which is impossible.
Consider [itex]\phi :\mathbb C\to A,\quad \gamma \mapsto a-\gamma \cdot 1[/itex]. It is continuous and [itex]\phi ^{-1}(\mathrm{inv}(A)) = \mathbb C\setminus \sigma (a)[/itex], thus [itex]\sigma (a)[/itex] is closed. Take [itex]\lambda \in \sigma (a)[/itex] and assume [itex]|\lambda|>\|a\|[/itex]. Then [itex]\|\lambda ^{-1}a\|< 1[/itex] and [itex]a-\lambda \cdot 1 = -\lambda (1-\lambda ^{-1}a) \in\mathrm{inv}(A)[/itex]. So the spectrum must be bounded by [itex]\|a\|[/itex].
Funnily enough, the spectral theory works much more nicely in the complex case. In the real case it can be a...real.. nightmare sometimes. One problem is, the spectrum CAN be empty in the real case, e.g take a matrix with characteristic equation [itex]x^2+1 = 0[/itex]..:oldgrumpy:
 
Last edited:
  • #13
benorin
Homework Helper
Insights Author
1,391
147
6. Define the complex function gamma function as
$$
\Gamma(z):=\lim_{n \to \infty}\dfrac{n!\,n^z}{z(z+1)\cdot\ldots\cdot(z+n)}
$$
and prove
(a) ##\Gamma(z)=\displaystyle{\int_0^\infty}e^{-t}\,t^{z-1}\,dt\;;\quad\mathfrak{R}(z)>0##
(b) ##\Gamma(z)^{-1}=e^{\gamma z}z\,\displaystyle{\prod_{n=1}^\infty}\left(1+\dfrac{z}{n}\right)e^{-\frac{z}{n}}##
where ##\gamma :=\displaystyle{\lim_{n \to \infty}\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\ldots+\dfrac{1}{n}-\log(n)\right)}## is the Euler-Mascheroni constant.

This work I have copy+pasted from my insight article, A Path to Fractional Integral Representations of Some Special Functions.

(a)
Theorem 1.2: Gamma Function Integral

(Euler 1730): ## \Gamma ( z ) = \int_{0}^{\infty} e^{ - t}t^{z - 1} dt , \, \Re \left[ z \right] > 0 ##
Proof:
\begin{eqnarray*} \Gamma \left( z \right) &:=& \mathop {\lim }\limits_{\lambda \to \infty } \frac{{\lambda !{\lambda ^{z - 1}}}}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda - 1} \right)}} \\&=& \mathop {\lim }\limits_{\lambda \to \infty } \frac{\lambda ! \lambda ^{z - 1}}{z ( z + 1 ) \cdots ( z + \lambda - 2)}\int_{0}^{1} (1-x)^{z + \lambda - 2} dx\end{eqnarray*}
Now, integrate by parts to get
\begin{eqnarray*}\Gamma ( z) &=& \mathop {\lim }\limits_{\lambda \to \infty } \frac{\lambda ! \lambda ^{z - 1}}{z (z + 1) \cdots ( z + \lambda - 2 )}\left\{ \left[ \left. x ( 1 - x) ^{z + \lambda - 2} \right| \right._{x = 0}^1 + \left( {z + \lambda - 2} \right)\int_{0}^{1} x (1 - x)^{z + \lambda - 3}dx\right\} \\ &=& \mathop {\lim }\limits_{\lambda \to \infty } \frac{\lambda !\lambda ^{z - 1}}{z ( z + 1) \cdots ( z + \lambda - 3 ) } \int_{0}^{1} x (1-x) ^{z + \lambda - 3} dx\end{eqnarray*}
In general, k iterations of integration by parts gives
$$\Gamma (z) = \mathop {\lim }\limits_{\lambda \to \infty } \frac{\lambda !\lambda ^{z - 1}}{z(z + 1) \cdots ( z + \lambda - k - 2 ) k!}\int_{0}^1 ( 1 - x)^{z + \lambda - k - 2} x^k dx $$
in particular, ##\left( {\lambda - 1} \right)## iterations of integration by parts gives
$$\Gamma ( z ) = \mathop {\lim }\limits_{\lambda \to \infty } \, \, \lambda ^{z - 1} \int_{0}^{1} ( 1 - x)^{z - 1} x^{\lambda - 1} dx $$
Substitute ##{x^\lambda } = y \Rightarrow \lambda {x^{\lambda - 1}}dx = dy## , to get
$$\Gamma ( z ) = \mathop {\lim }\limits_{\lambda \to \infty } \lambda ^{z - 1} \int_{0}^{1} \left( 1 - y^{\frac{1}{\lambda }} \right) ^{z - 1} dy $$
Set ##\lambda = \frac{1}{\eta }## , so that ##\eta \to {0^ + }## as ##\lambda \to \infty ## and
$$\Gamma (z) = \mathop {\lim }\limits_{\eta \to {0^{+}}} \int_{0}^{1} \left( \frac{1 - y^{\eta }}{\eta } \right) ^{z - 1} dy \,\mathop = \limits^H \,\;\int_{0}^{1} {\log }^{z - 1}\left( \frac{1}{y} \right) dy $$
where ##\mathop = \limits^H ## denotes the use of L’Hospital’s Rule. Substitute ##y={e^{-t}} \Rightarrow dy = - {e^{-t}}dt## , to get
$$\Gamma \left( z \right) = \;\int_{0}^\infty e^{ - t}t^{z - 1} dt$$
and the theorem is demonstrated.


(b)
Weierstrass took as the definition of the gamma function it’s canonical infinite product representation, the so-called Weierstrass product form of the gamma function. The desired representation of the gamma function is here obtained as a corollary to the Weierstrass Factor Theorem‡, the proof of which shall not be reproduced here[2].
Theorem 1.3: Weierstrass Factor Theorem
(Weierstrass): Let ##f(z)## be an entire (i.e. everywhere analytic) function with simple zeroes at ## z = a_1,a_2,a_3,\ldots## where ##0 < \left| {{a_1}} \right| < \left| a_2\right| < \left| a_3\right| < \ldots## and ##\mathop {\lim }\limits_{M \to \infty}\;{a_M} =\infty##, then
$$f\left( z \right) = f\left( 0 \right){e^{\frac{{f'\left( 0 \right)}}{{f\left( 0 \right)}}z}}\prod\limits_{k = 1}^\infty {\left[ {\left( {1 - \frac{z}{{{a_k}}}} \right){e^{\frac{z}{{{a_k}}}}}}\right]}$$
The Weierstrass product form of the gamma function is then given by
Corollary 1.4: Weierstrass Product Form of the Gamma Function
(Weierstrass): ##\frac{1}{{\Gamma \left( z \right)}} = z{e^{\gamma z}}\prod\limits_{\lambda = 1}^\infty {\left[ {\left( {1 + \frac{z}{\lambda }} \right){e^{ - \frac{z}{\lambda }}}} \right]} ## where ##\gamma ## is Euler’s Constant.[3]
Proof:
Let ##f\left( z \right) = \frac{1}{{\Gamma \left( {z + 1} \right)}}## so that f(z) is an entire function with simple zeroes at ##z = {a_k}: = - k, \forall k \in {\mathbb{Z}^ + }## satisfying the hypotheses necessary to invoke Theorem 1.2, which yields
$$\frac{1}{{\Gamma \left( {z + 1} \right)}} = {e^{f'\left( 0 \right)z}}\prod\limits_{k = 1}^\infty {\left[ {\left( {1 + \frac{z}{k}} \right){e^{ - \frac{z}{k}}}} \right]} $$
Set ##z=1## in the above formula and take natural logarithms of the result to determine
\begin{eqnarray*} f^{\prime} (0) &=& \sum_{k = 1}^\infty \left[ \frac{1}{k} - \log \left( 1 + \frac{1}{k} \right) \right] \\&=& \mathop {\lim }\limits_{M \to \infty } \,\sum_{k = 1}^M \left[ \frac{1}{k} + \log ( k ) - \log ( k + 1) \right] \\&=& \mathop {\lim }\limits_{M \to \infty } \,\left[ H_M - \log ( M + 1) \right] +\log 1\\&=& \mathop {\lim }\limits_{M \to \infty } \,\left[ H_M - \log (M + 1) \right] + \underbrace{\mathop {\lim }\limits_{M \to \infty } \log \left( 1 + \frac{1}{M} \right) }_{ = \log 1}\\&=& \mathop {\lim }\limits_{M \to \infty } \,\left( H_M- \log M \right) = :\gamma \end{eqnarray*}
Where ##H_M## is the ##\text{M}^{th}## harmonic number and theorem is proved upon applying Equation 1.1 and replacing ##f'\left( 0 \right)## with ##\gamma ## , which is Euler’s Constant. Euler’s constant to four decimal places is
$$\gamma : = \mathop {\lim }\limits_{M \to \infty } \left[ {\sum\limits_{k = 1}^M {\left( {\frac{1}{k}} \right) - \ln M} } \right] = 0.5772 \ldots $$
 
  • #14
benorin
Homework Helper
Insights Author
1,391
147
7. Let ##u\, : \,[0,1]\times [a,b]\longrightarrow \mathbb{C}## be a continuous function, such that the partial derivative in the first coordinate exists everywhere and is continuous. Define
$$
U(\lambda):=\int_a^b u(\lambda,t)\,dt\; , \;V(\lambda):=\int_a^b \dfrac{\partial u}{\partial \lambda}(\lambda,t)\,dt.
$$
Show that ##U## is continuously differentiable and ##U'(\lambda)=V(\lambda)## for all ##0\leq \lambda\leq 1.##

$$\begin{gathered} \tfrac{dU}{d\lambda}:=\lim_{\Delta \lambda\to 0}\tfrac{\Delta U(\lambda)}{\Delta \lambda} =\int_a^b \lim_{\Delta \lambda\to 0}\tfrac{u(\lambda + \Delta \lambda,t) - u(\lambda ,t)}{\Delta \lambda} \,dt = \int_a^b\tfrac{\partial u}{\partial \lambda }\left( \lambda , t\right)\, dt =: V( \lambda ) \\ \end{gathered}$$

assuming that ##a,b## do not depend on neither ##\lambda## nor ##t##. The function defined by the integral over a fixed interval ## a\leq t \leq b## of a continuous function of ##0\leq \lambda \leq 1## which also exists everywhere is continuous, hence ##U## is continuously differentiable.
 
  • #15
fresh_42
Mentor
Insights Author
2021 Award
16,425
15,464
I knew I would have regretted that one. Well, I wanted to see the ##\varepsilon -\delta ## version without hand waving interchange of integral and limit, or non-transparent argument about continuity.

So here it is for all who are still learning:

Let ##\varepsilon>0.## We have to show that there is a ##\delta>0## such that for all ##\lambda,h\in\mathbb{R}## with ##0<|h|<\delta,\,0\leq \lambda\leq 1,## and ##0\leq \lambda+h\leq 1##
$$
\left|V(\lambda+h)-V(\lambda)\right|<\varepsilon\, , \,\left|\dfrac{U(\lambda+h)-U(\lambda)}{h}-V(\lambda)\right|<\varepsilon.
$$
Every continuous function on a compact interval is uniformly continuous, hence there is a ##\delta>0## such that for all ##(\lambda,t),(\lambda',t')\in[0,1]\times[a,b]##
$$
|\lambda'-\lambda|+|t'-t|<\delta\Longrightarrow\left|\dfrac{\partial u}{\partial \lambda}(\lambda',t')-\dfrac{\partial u}{\partial \lambda}(\lambda,t)\right|<\dfrac{\varepsilon}{b-a}.
$$
By definition of ##V## we have
\begin{align*}
\left|V(\lambda+h)-V(\lambda)\right|&=\left|\int_a^b\left(\dfrac{\partial u}{\partial \lambda}(\lambda+h,t)-\dfrac{\partial u}{\partial \lambda}(\lambda,t)\right)\,dt\right|\\
&\leq \int_a^b\left|\dfrac{\partial u}{\partial \lambda}(\lambda+h,t)-\dfrac{\partial u}{\partial \lambda}(\lambda,t)\right| \,dt\\
&<\varepsilon
\end{align*}
since the integrand is continuous and takes its maximum in ##[a,b].##

Now assume ##0<h<\delta## such that ##0\leq \lambda<\lambda+h\leq 1.## (The case ##h<0## is proven accordingly.) Therefore
\begin{align*}
\left|\dfrac{u(\lambda+h,t)-u(\lambda,t)}{h}-\dfrac{\partial u}{\partial \lambda}(\lambda,t)\right|&=\left|\dfrac{1}{h}\int_{\lambda}^{\lambda+h}\left(\dfrac{\partial u}{\partial \lambda}(\lambda',t)-\dfrac{\partial u}{\partial \lambda}(\lambda,t)\right)d\lambda' \right|\\
&\leq \dfrac{1}{h}\int_{\lambda}^{\lambda+h}\left|\dfrac{\partial u }{\partial\lambda}(\lambda',t)-\dfrac{\partial u}{\partial\lambda}(\lambda,t)\right|d\lambda'\\
&<\dfrac{\varepsilon}{b-a}
\end{align*}
and so
\begin{align*}
\left|\dfrac{U(\lambda+h)-U(\lambda)}{h}-V(\lambda)\right|&=\left|\int_a^b\left(\dfrac{u(\lambda+h,t)-u(\lambda,t)}{h}-\dfrac{\partial u}{\partial \lambda}(\lambda,t)\right)dt\right|\\
&\leq \int_a^b\left|\dfrac{u(\lambda+h,t)-u(\lambda,t)}{h}-\dfrac{\partial u}{\partial \lambda}(\lambda,t)\right|dt\\
&< \int_a^b\dfrac{\varepsilon}{b-a}\,dt = \varepsilon
\end{align*}
 
  • #18
benorin
Homework Helper
Insights Author
1,391
147
@fresh_42 what about #6?
Sorry if this is the second time you see this post: went back hours later to not find my prompting about #6) post (vanished).
 
  • #20
benorin
Homework Helper
Insights Author
1,391
147
Sorry that last post I made twice from my phone because the second time the first post did not appear on the page after a refresh. My bad, I bugged you twice. Sorry. @fresh_42
 
  • #21
585
237
8. A cardiod is defined as the trace of a point on a circle that rolls around a fixed circle of the same size without slipping.


It can be described by ##(x^{2}+y^{2})^{2}+4x(x^{2}+y^{2})-4y^{2}\,=\,0## or in polar coordinates by ##r(\varphi )=2(1-\cos \varphi ).##
Show that:

(a) Given any line, there are exactly three tangents parallel to it. If we connect the points of tangency to the cusp, the three segments meet at equal angles of ##2\pi/3\,.##
(b) The length of a chord through the cusp equals ##4.##
(c) The midpoints of chords through the cusp lie on the perimeter of the fixed generator circle (black one in the first picture).
(d) Calculate length, area and curvature.

I would humbly like to mention problems ##8(a)##, ##8(b)## and ##8(c)## are worth another look, and maybe a re-phrasing.

##8(a)## The claim "given any line...three segments meet at equal angles of ##\frac{2\pi}{3}##" is false. If we take a horizontal line, one of three points of tangency is the cusp, we have only two line segments that meet at an angle of ##\frac{2\pi}{3}##

##8(b)## The length of a chord through the cusp equals ##4## for ##\phi=(2n+1)\pi##. But, its length does not equal ##4## for ##\phi\in R\backslash (2n+1)\pi##, where ##n## is an integer.

##8(c)## The midpoints of chords through the cusp lie on the perimeter of the black circle for ##\phi\in n\pi## where ##n## is an integer. But, not for all ##\phi##. Also, if a line through a cusp intersects the carotid at 3 places, is it still called a chord?

edited for ##8(a)##
 
Last edited:
  • #22
etotheipi
I guess I can try this one...
4. (a) Solve the following linear differential equation system:
\begin{align*}
\dot{y}_1(t)=11y_1(t)-80y_2(t)\;&\wedge\;\dot{y}_2(t)=y_1(t)-5y_2(t)\\
y_1(0)=0\;&\wedge\;y_2(0)=0
\end{align*}
(b) Which solutions do ##y_1(0)=\pm \varepsilon\;\wedge\;y_2(0)=\pm \varepsilon## have?
(c) How does the trajectory for ##y_1(0)=0.001\;\wedge\;y_2(0)=0.001## behave for ##t\to \infty##?
(d) What will change if we substitute the coefficient ##-80## by ##-60##?
(e) Calculate (approximately) the radius of the osculating circle at ##t=\pi/12## for both trajectories with initial condition ##\mathbf{y}(0)=(-1,1).##
For ##y_1(0) = y_2(0) = 0##, ##y_1(t) = y_2(t) = 0## is a solution. Then given ##(\dot{y}_1, \dot{y}_2) = (ay_1 + by_2, cy_1 + dy_2)##, differentiating the second equation and putting it back into the first gives you
$$\ddot{y}_1(t) - (a+d) \dot{y}_1(t) + (ad-bc) y_1(t) = 0$$For ##b = -80## you get a complementary equation ##\lambda^2 - 6\lambda + 25 = 0## which means$$y_1(t) = e^{3t}(A\sin{4t} + B\cos{4t})$$and because ##y_2(t) = \frac{1}{b} \dot{y}_1(t) - \frac{a}{b} y_1(t)##, that also gives$$y_2(t) = \frac{e^{3t}}{20} ( (2A + B)\sin{4t} + (2B - A) \cos{4t})$$For initial conditions ##(y_1(0), y_2(0)) = (p,q)## you get ##B = p## and ##A = 2p - 20q##, i.e.$$\begin{align*}
y_1(t) &= e^{3t} ( (2p - 20q) \sin{4t} + p \cos{4t}) \\

y_2(t) &= \frac{e^{3t}}{20}((5p - 40q)\sin{4t} + 20q \cos{4t})

\end{align*}$$for ##p = q = 0.001## is just oscillatory with ##e^{3t}## and ##\frac{1}{20} e^{3t}## envelope respectively? Anyway now if you change ##b = -60## then instead the complementary equation gives ##\lambda = 1,5## and by the same procedure as before you get$$\begin{align*}
y_1(t) &= Ae^{5t} + Be^t \\
y_2(t) &= \frac{1}{10}Ae^{5t} + \frac{1}{6}B e^t
\end{align*}$$and similarly given ##(y_1(0), y_2(0)) = (p,q)## you get ##A = \frac{5}{2} p - 15q## and ##B = \frac{3}{2}(10q - p)##. For the radius of the osculating circle to the first trajectory, first the ICs ##\mathbf{y}(0) = (-1,1)## imply $$\mathbf{y}(t) = e^{3t} (-22\sin{4t} - \cos{4t}, \cos{4t} - \frac{9}{4} \sin{4t})$$you can work out$$\begin{align*}
\dot{y}_1 \left(\frac{\pi}{12} \right) &= -e^{\frac{\pi}{4}} \left(\frac{91}{2} + 31\sqrt{3} \right) \\
\ddot{y}_1 \left(\frac{\pi}{12} \right) &= e^{\frac{\pi}{4}} \left(89 \sqrt{3} - \frac{521}{2}\right) \\
\dot{y}_2 \left(\frac{\pi}{12} \right) &= -\frac{e^{\frac{\pi}{4}}}{4} \left(12 + \frac{43\sqrt{3}}{2} \right) \\
\ddot{y}_2 \left(\frac{\pi}{12} \right) &= -e^{\frac{\pi}{4}} \left( 244+ 33\sqrt{3} \right)

\end{align*}$$then it's possible to calculate$$R = \left| \frac{(\dot{y}_1^2 + \dot{y}_2^2)^{3/2}}{\dot{y_1}\ddot{y}_2 - \dot{y}_2 \ddot{y}_1} \right|$$same procedure for the other one 😢. Anyway hope I didn't f*ck up too much algebra!
 
Last edited by a moderator:
  • #23
etotheipi
I think for number 8 it's easier to just stay in polar coordinates. Given ##y = r\sin{\varphi}## and ##x = r\cos{\varphi}##, you have$$\begin{align*}
\frac{dy}{d\varphi} &= r\cos{\varphi} + \frac{dr}{d\varphi} \sin{\varphi}\\
\frac{dx}{d\varphi} &= -r\sin{\varphi} + \frac{dr}{d\varphi} \cos{\varphi}
\end{align*}$$You can divide these two to get ##dy/dx##, and writing ##r' := dr/d\varphi##$$\frac{dy}{dx} = \frac{r\cos{\varphi} + r' \sin{\varphi}}{r' \cos{\varphi} - r\sin{\varphi}}$$Given the trajectory ##r(\varphi) = 2(1-\cos{\varphi})##, that becomes$$\frac{dy}{dx} = \frac{\sin^2{\varphi} - \cos^2{\varphi} + \cos{\varphi}}{2\sin{\varphi}\cos{\varphi} - \sin{\varphi}} = \frac{\cos{\varphi} - \cos{2\varphi}}{\sin{2\varphi} - \sin{\varphi}}$$Now, choose an arbitrary line with gradient ##k##. The points on the trajectory with equal gradient will then satisfy, using the "sum-to-product" relations,$$\sin{\left(\frac{3\varphi}{2}\right)}\sin{\left(\frac{\varphi}{2}\right)} = k \cos{\left(\frac{3\varphi}{2}\right)}\sin{\left(\frac{\varphi}{2}\right)}$$Since the gradient's undefined at ##\varphi = 0## we can ignore that solution, and we're left with$$\varphi = \frac{2}{3} \tan^{-1}{k} + \frac{2n\pi}{3}, \quad n \in \mathbb{Z}$$Or in other words, so long as ##k \neq 0## you get 3 non-zero solutions in the interval ##\varphi \in [0, 2\pi]## which are spaced by ##2\pi / 3## in angle.

For (b), just choose an arbitrary angle ##\phi_0## and its corresponding ##\phi' = \phi_0 + \pi## on the opposite side of the chord through the cusp. Since ##r(\phi') = 2(1-\cos{\phi'}) = 2(1+ \cos{\phi_0})##, the length ##L## of the chord is just$$L = r(\phi_0) + r(\phi') = 2(1-\cos{\phi_0}) + 2(1+\cos{\phi_0}) = 4$$I'll try and finish the question later, but I should get on with my actual homework now haha :-p
 
  • #24
fresh_42
Mentor
Insights Author
2021 Award
16,425
15,464
I guess I can try this one...
For ##y_1(0) = y_2(0) = 0##, ##y_1(t) = y_2(t) = 0## is a solution. Then given ##(\dot{y}_1, \dot{y}_2) = (ay_1 + by_2, cy_1 + dy_2)##, differentiating the second equation and putting it back into the first gives you
$$\ddot{y}_1(t) - (a+d) \dot{y}_1(t) + (ad-bc) y_1(t) = 0$$For ##b = -80## you get a complementary equation ##\lambda^2 - 6\lambda + 25 = 0## which means$$y_1(t) = e^{3t}(A\sin{4t} + B\cos{4t})$$and because ##y_2(t) = \frac{1}{b} \dot{y}_1(t) - \frac{a}{b} y_1(t)##, that also gives$$y_2(t) = \frac{e^{3t}}{20} ( (2A + B)\sin{4t} + (2B - A) \cos{4t})$$For initial conditions ##(y_1(0), y_2(0)) = (p,q)## you get ##B = p## and ##A = 2p - 20q##, i.e.$$\begin{align*}
y_1(t) &= e^{3t} ( (2p - 20q) \sin{4t} + p \cos{4t}) \\

y_2(t) &= \frac{e^{3t}}{20}((5p - 40q)\sin{4t} + 20q \cos{4t})

\end{align*}$$for ##p = q = 0.001## is just oscillatory with ##e^{3t}## and ##\frac{1}{20} e^{3t}## envelope respectively? Anyway now if you change ##b = -60## then instead the complementary equation gives ##\lambda = 1,5## and by the same procedure as before you get$$\begin{align*}
y_1(t) &= Ae^{5t} + Be^t \\
y_2(t) &= \frac{1}{10}Ae^{5t} + \frac{1}{6}B e^t
\end{align*}$$and similarly given ##(y_1(0), y_2(0)) = (p,q)## you get ##A = \frac{5}{2} p - 15q## and ##B = \frac{3}{2}(10q - p)##. For the radius of the osculating circle to the first trajectory, first the ICs ##\mathbf{y}(0) = (-1,1)## imply $$\mathbf{y}(t) = e^{3t} (-22\sin{4t} - \cos{4t}, \cos{4t} - \frac{9}{4} \sin{4t})$$you can work out$$\begin{align*}
\dot{y}_1 \left(\frac{\pi}{12} \right) &= -e^{\frac{\pi}{4}} \left(\frac{91}{2} + 31\sqrt{3} \right) \\
\ddot{y}_1 \left(\frac{\pi}{12} \right) &= e^{\frac{\pi}{4}} \left(89 \sqrt{3} - \frac{521}{2}\right) \\
\dot{y}_2 \left(\frac{\pi}{12} \right) &= -\frac{e^{\frac{\pi}{4}}}{4} \left(12 + \frac{43\sqrt{3}}{2} \right) \\
\ddot{y}_2 \left(\frac{\pi}{12} \right) &= -e^{\frac{\pi}{4}} \left( 244+ 33\sqrt{3} \right)

\end{align*}$$then it's possible to calculate$$R = \left| \frac{(\dot{y}_1^2 + \dot{y}_2^2)^{3/2}}{\dot{y_1}\ddot{y}_2 - \dot{y}_2 \ddot{y}_1} \right|$$same procedure for the other one 😢. Anyway hope I didn't f*ck up too much algebra!
I have different coefficients, and a complex solution as the eigenvalues are complex. I have also difficulties to see what answers what and where possible mistakes are, due to lacking calculations. The goal was to examine the trajectories and their stability towards minor changes.
 
  • #25
etotheipi
Okay here's the rest of question 8:

(c) Take a chord at angle ##\phi_0##; by symmetry of the figure around ##\varphi = 0## we can just consider the cases when ##\phi_0 \in [0, \pi)##. Let the midpoint of the chord be ##M##, then using the result of (b) we have$$OM = 2 - r(\phi_0) = 2 - 2(1-\cos{\phi_0}) = 2\cos{\phi_0}$$The directed line segment ##\overrightarrow{OM}## is at an angle ##\phi' = \phi_0 + \pi## to the ray ##\varphi = 0##, so ##OM = -2\cos{\phi'}##. Hence the point ##M## lies on the circle ##r(\varphi) = -2\cos{\varphi}##.

(d) (i) Since ##dx = -r\sin{\varphi} d\varphi + \cos{\varphi} dr## and ##dy = r\cos{\varphi} + \sin{\varphi} dr## the line element satisfies$$dl = \sqrt{dx^2 + dy^2} = \sqrt{r^2 d\varphi^2 + dr^2} = d\varphi \sqrt{r^2 + r_{\varphi}^2}$$That means$$L= \int_0^{2\pi} dl = \int_0^{2\pi} \sqrt{8 - 8\cos{\varphi}} d\varphi = 4 \int_0^{2\pi} \sin \frac{\varphi}{2} d\varphi = 16$$(ii) This is just another integration, given that ##dA = \frac{1}{2}r^2 d\varphi##, $$A = \frac{1}{2} \int_0^{2\pi} (1-\cos{\varphi})^2 d\varphi = 2\left[ \frac{3}{2} \varphi - 2\sin{\varphi} + \frac{1}{4} \sin{4\varphi} \right]_{0}^{2\pi} = 6\pi$$(iii) And the final part is just application of the usual formula$$\begin{align*}

R = \frac{(r^2 + r_{\varphi}^2)^{\frac{3}{2}}}{| r^2 + 2r_{\varphi}^2 -r r_{\varphi \varphi}|} &= \frac{(8 - 8\cos{\varphi})^{\frac{3}{2}}}{12 - 12\cos{\varphi}} \\ \\

&= \frac{8^{\frac{3}{2}} 2^{\frac{1}{2}}}{12} \sin{\frac{\varphi}{2}} = \frac{8}{3} \sin{\frac{\varphi}{2}}\end{align*}$$
 
  • #26
etotheipi
I have different coefficients, and a complex solution as the eigenvalues are complex. I have also difficulties to see what answers what and where possible mistakes are, due to lacking calculations. The goal was to examine the trajectories and their stability towards minor changes.

Hmm when I check with WolframAlpha it seems to be the same as what I wrote down. But I wouldn't be surprised if I messed up. I'll check it tomorrow 😜
 
  • #27
fresh_42
Mentor
Insights Author
2021 Award
16,425
15,464
Okay here's the rest of question 8:

(c) Take a chord at angle ##\phi_0##; by symmetry of the figure around ##\varphi = 0## we can just consider the cases when ##\phi_0 \in [0, \pi)##. Let the midpoint of the chord be ##M##, then using the result of (b) we have$$OM = 2 - r(\phi_0) = 2 - 2(1-\cos{\phi_0}) = 2\cos{\phi_0}$$The directed line segment ##\overrightarrow{OM}## is at an angle ##\phi' = \phi_0 + \pi## to the ray ##\varphi = 0##, so ##OM = -2\cos{\phi'}##. Hence the point ##M## lies on the circle ##r(\varphi) = -2\cos{\varphi}##.
And why is this the "black" circle? How about a Cartesian representation, since it shows the location of the center, which ##r(\varphi )## does not?
(d) (i) Since ##dx = -r\sin{\varphi} d\varphi + \cos{\varphi} dr## and ##dy = r\cos{\varphi} + \sin{\varphi} dr## the line element satisfies$$dl = \sqrt{dx^2 + dy^2} = \sqrt{r^2 d\varphi^2 + dr^2} = d\varphi \sqrt{r^2 + r_{\varphi}^2}$$That means$$L= \int_0^{2\pi} dl = \int_0^{2\pi} \sqrt{8 - 8\cos{\varphi}} d\varphi = 4 \int_0^{2\pi} \sin \frac{\varphi}{2} d\varphi = 16$$(ii) This is just another integration, given that ##dA = \frac{1}{2}r^2 d\varphi##, $$A = \frac{1}{2} \int_0^{2\pi} (1-\cos{\varphi})^2 d\varphi = 2\left[ \frac{3}{2} \varphi - 2\sin{\varphi} + \frac{1}{4} \sin{4\varphi} \right]_{0}^{2\pi} = 6\pi$$(iii) And the final part is just application of the usual formula$$\begin{align*}

R = \frac{(r^2 + r_{\varphi}^2)^{\frac{3}{2}}}{| r^2 + 2r_{\varphi}^2 -r r_{\varphi \varphi}|} &= \frac{(8 - 8\cos{\varphi})^{\frac{3}{2}}}{12 - 12\cos{\varphi}} \\ \\

&= \frac{8^{\frac{3}{2}} 2^{\frac{1}{2}}}{12} \sin{\frac{\varphi}{2}} = \frac{8}{3} \sin{\frac{\varphi}{2}}\end{align*}$$
... and the curvature is the reciprocal of the radius!
 
  • #28
fresh_42
Mentor
Insights Author
2021 Award
16,425
15,464
Hmm when I check with WolframAlpha it seems to be the same as what I wrote down. But I wouldn't be surprised if I messed up. I'll check it tomorrow 😜
I haven't checked whether some scaling via the free parameter gives similar solutions. What are your eigenvectors?
 
  • #29
etotheipi
And why is this the "black" circle? How about a Cartesian representation, since it shows the location of the center, which r(φ) does not?

I'm not sure I completely understand what you're after, but let me try :smile:. First consider ##S^1 \subseteq \mathbb{R}^2## to be the black circle drawn in the diagram, around which the cardioid will be drawn. Let the centre of the "rolling" circle have a position ##\mathbf{x}_c = (2\cos{\varphi}, 2\sin{\varphi})##, and let the angle turned by the "rolling" circle be ##\xi = 2\varphi##. Hence the position of the point on the "rolling" circle which is tracing out the cardioid is$$\mathbf{x}_p = (2\cos{\varphi} - \cos{2\varphi}, 2\sin{\varphi} - \sin{2\varphi})$$Now do a coordinate transformation ##\mathbf{x}_p' = \mathbf{x}_p - (1,0)##, so that the new origin coincides with the cusp. The norm of this vector satisfies$$\begin{align*}

\lVert \mathbf{x}_p' \rVert^2 &= 6 + 2\cos{2\varphi} - 4\cos{\varphi} - 4\cos{\varphi}(2\cos^2{\varphi} - 1) - 4\sin{\varphi}(2\sin{\varphi} \cos{\varphi}) \\

&= 4 + 4\cos^2{\varphi} - 8 \cos{\varphi} = [2(1-\cos{\varphi})]^2
\end{align*}
$$so that ##\lVert \mathbf{x}_p' \rVert = 2(1-\cos{\varphi})## is the equation of the cardioid. And also in these coordinates, the equation of the original black circle of unit radius is just ##r(\varphi) = -2\cos{\varphi}##. It's just because you can e.g. write the Cartesian form$$(x+1)^2 + y^2 = 1 \implies (r\cos{\varphi} + 1)^2 + r^2\sin^2{\varphi} = 1 \implies r = -2\cos{\varphi}$$
 
  • #30
etotheipi
I haven't checked whether some scaling via the free parameter gives similar solutions. What are your eigenvectors?

By this do you mean the functions ##e^{(3 \pm 4i)t}##?
 

Related Threads on Math Challenge - March 2021

Replies
67
Views
6K
Replies
86
Views
7K
  • Last Post
4
Replies
114
Views
3K
Replies
93
Views
4K
Replies
102
Views
4K
Replies
100
Views
4K
Replies
61
Views
4K
Replies
42
Views
3K
Replies
88
Views
5K
Replies
77
Views
10K
Top