If p(x)=x^3, then p'(0)=p''(0)=0 and 0 is a turning point according to your definition.epenguin said:If instead at a point p' = 0 and p'' = 0, that point is not a turning point
I don't understand this. If p(x)=x^4-1, then the point 0 is 'such a point' (p'(0)=p''(0)=0), but the only zero of p' is zero. Do you mean to say that there are at least two roots of p' counted with multiplicity (because the point is already a zero for p of multiplicity at least 2). Anyways, what if the point a where p'(a)=p''(a) and p(a)\neq 0 is not in the interval between two roots (is smaller than the smallest root or larger than the largest root)?epenguin said:Therefore if there is such a point between in the interval between two successive real roots of p there must be at least one other point In this interval where p' = 0, in order for there to be one turning point.
julian said:We also have:
So write the polynomial as:
##
p(x) = a_n (x-a)^n + a_{n-1} (x-a)^{n-1} + \dots + a_4 (x-a)^4 + a_3 (x-a)^3 + a_0
##
where ##a_0 \not= 0##. We have an undulation point at ##x=a## if ##a_3 = 0## and the first non-zero term after it has even power in ##(x-a)##. Say this first non-zero term is ##a_{2m} (x-a)^{2m}##. For ##x## close to ##x=a## we have
##
p (x) \approx a_{2m} (x-a)^{2m} + a_0
##
so the corresponding trough or peak is not intersecting the ##x-##axis and again we must have complex roots?
Ah @timetraveller123 I was on the exact same line of thinking but when I decided to start writing you had already post it.timetraveller123 said:hi i am new to this i am not sure this will work
this is for question 2 of basics
assume the opposite that such a polynomial exists
hence
##
P(x) = \sum_i (x - \alpha_i)
##
where ##\alpha_i## is real
hence
##
P'(x) = P(x)\sum_i \frac{1}{x-\alpha_i}\\
P'(a) = P(a) \sum_i \frac{1}{x - \alpha_i} = 0\\
\sum_i \frac{1}{a - \alpha_i} = 0\\
P''(x) = P'(x)\sum_i \frac{1}{x-\alpha_i} + P(x) \sum_i \frac{-1}{(x-\alpha_i)^2}\\
P''(a) = P(a) \sum_i \frac{-1}{(a-\alpha_i)^2} = 0\\
\sum_i \frac{1}{(a-\alpha_i)^2} = 0
##
if each term were to be real then all term would have to be equally zero else they would have to be complex would this work
fresh_42 said:I didn't use de L'Hôpital but let me just add my solution for the sake of readability:
You could have been a bit more detailed, or at least used ##s=2t## in the second step, but yes, it's correct.eys_physics said:10a) By the change of variables, $$t=\log(x),$$
$$\int_{1}^{\infty}\frac{\log(x)}{x^3}dx=\int_{0}^{\infty}t\exp(-2t)dt=\frac{1}{4}\int_{0}^{\infty}t\exp(-t)dt=\frac{1}{4}\Gamma(2)=\frac{1}{4}$$.
Buzz Bloom said:Here is my partial solution try for #7. I have an upper limit for the number of times between midnight and noon that the clock will give an ambiguous reading...
The solution requires specifying the number of times, N, between 12am and 12 pm that actual time t1 and will be confused with some false time t2.
Buzz Bloom said:The maximum of these two functions of A and B is when A = B = 11, yielding a maximum value of 143. Thus, A=11 and B=11 is not allowed, so there are 143 occurrences when there is a confusion.
Hi StoneTemplePython:StoneTemplePython said:How many times a day is..." so the final answer needs to be stated over an entire day
I do not understand why the length of hands is relevant to anything. I have assumed throughout that the hands are the same length, and in all other ways indistinguishable. I note here that noon and midnight are two occasions when the hour hand and minute hands are in the same place, and this is not a case of confusion since reversing the role of the two hands doesn't change the time.StoneTemplePython said:It's subtle, but shouldn't the length of the hands not matter on occasion?
Buzz Bloom said:Hi StoneTemplePython:
I confess I was fooled by the way the problem was phrased. It would have helped to use the phrase "times a 24 hour day". Now that I understand this, a see that the answer should be 286=2×143 since each of the 143 confused pairs between midnight and noon is exactly the same as another confused pair between noon and midnight.
Buzz Bloom said:I do not understand why the length of hands is relevant to anything. I have assumed throughout that the hands are the same length, and in all other ways indistinguishable. I note here that noon and midnight are two occasions when the hour hand and minute hands are in the same place, and this is not a case of confusion since reversing the role of the two hands doesn't change the time.
Young physicist said:Oh.So do you mean that the clock ticks from one minute to another minute,rather than gradually changing all the time?I know both kinds of non digital clock exist.
The ticking one:
https://tenor.com/view/design-time-clock-tick-tock-gif-3428153
Hi StoneTemplePython:StoneTemplePython said:It has a ticking seconds hand.
I wonder if clocks moving discontinuously second by second were more common when the mathematician created this problem then analogue clocks are today. I confess I vaguely remember that when I was in elementary school in the 1940s, the classrooms had a clock something like that. I don't remember whether it moved incrementally second by second, or minute by minute.StoneTemplePython said:The puzzle was written by a famous mathematician
I apologize for my denseness, but the quote above makes no sense to me in the context of the puzzle statement. Are you describing the puzzle clock as having three hands: hour, minute, second? If time is read by something other than the hour and minute hands, and "The minutes and hours hands don't move at all," then how is time read?StoneTemplePython said:the "ticking one" does not have a ticking minute hand. It has a ticking seconds hand. The minutes and hours hands don't move at all.
Buzz Bloom said:Hi StoneTemplePython:
Ah, another wrong assumption on my part. Since there was no specification about the hands moving abruptly second by second, rather than continuously, I assumed continuously.
I wonder if clocks moving discontinuously second by second were more common when the mathematician created this problem then analogue clocks are today. I confess I vaguely remember that when I was in elementary school in the 1940s, the classrooms had a clock something like that. I don't remember whether it moved incrementally second by second, or minute by minute.I apologize for my denseness, but the quote above makes no sense to me in the context of the puzzle statement. Are you describing the puzzle clock as having three hands: hour, minute, second? If time is read by something other than the hour and minute hands, and "The minutes and hours hands don't move at all," then how is time read?
Regards,
Buzz
fresh_42 said:23.
Example: Given a particle of mass ##m## in the potential ##U(\vec{r})=\dfrac{U_0}{\vec{r\,}^{2}}## with a constant ##U_0##. At time ##t=0## the particle is at ##\vec{r}_0## with velocity ##\dot{\vec{r}}_0\,.##
Hint: The Lagrange function with ##\vec{r}=(x,y,z,t)=(x_1,x_2,x_3,t)## of this problem is $$ \mathcal{L}=T-U=\dfrac{m}{2}\,\dot{\vec{r}}\,^2-\dfrac{U_0}{\vec{r\,}^{2}} $$
a) Give a reason why the energy of the particle is conserved, and what is its energy?
b) Consider the following transformations with infinitesimal ##\varepsilon##
$$\vec{r} \longmapsto \vec{r}\,^*=(1+\varepsilon)\,\vec{r}\,\, , \,\,t\longmapsto t^*=(1+\varepsilon)^2\,t$$
and verify the condition (*) to E. Noether's theorem.
c) Compute the corresponding Noether charge ##Q## and evaluate ##Q## for ##t=0##.
Thank you for solving this! I already began to believe it would survive Emmy's centenary!PeroK said:I've just been learning about Noether's theorem, so I thought I'd try this one. The notation is a bit different from what I'm used to.
a) The Lagrangian is independent of time, so energy is conserved (Euler-Lagrange equation for time coordinate):
##\dot{H} = - \frac{\partial L}{\partial t} = 0##
Hence, the total energy ##E## is equal to the Hamiltonian:
##E = H = p_i \dot{x^i} - L = \frac{\partial L}{\partial \dot{x^i}}\dot{x^i} - L = m\dot{x^i}\dot{x^i} - L = \frac12 mv^2 + \frac{U_0}{r^2}##
Hence:
##E = \frac12 mv_0^2 + \frac{U_0}{r_0^2}##
b) We have the transformation:
##x^* = (1 + \epsilon)x; \ y^* = (1 + \epsilon)y; \ z^* = (1 + \epsilon)z; \ t^* = (1 + \epsilon)^2t##
Hence: ##\psi^x = x; \ \psi^y = y; \ \psi^z = z; \ \phi = 2t##
The transformed Lagrangian is:
##L^* = \frac12 m(v^*)^2 - \frac{U_0}{(r^*)^2}##
Where:
##(v^*)^2 = (\frac{dx^*}{dt^*})^2 + \dots = (1+\epsilon)^{-2}(\frac{dx}{dt})^2 + \dots = (1+\epsilon)^{-2}v^2##
And:
##(r^*)^2 = (1+\epsilon)^{2}r^2##
Therefore, we have:
##L^* = L(1+\epsilon)^{-2}##
Hence:
##L^* \frac{dt^*}{dt} = L##
Which is the condition for invariance under the transformation.
c) The corresponding conserved Noether Charge is given by:
##Q = p_i\psi^i - H\phi = m(\dot{x}x + \dot{y}y + \dot{z}z) - 2Et = m\vec{v} \cdot \vec{r} - 2Et##
At ##t=0##, we have ##Q = m\vec{v_0} \cdot \vec{r_0}##, which leads to:
##m\frac{d}{dt}(r^2) = 2m\vec{v} \cdot \vec{r} = 2Q + 4Et##
And
##mr^2 = 2Et^2 + 2Qt + mr_0^2##
Which implies only unstable (circular) orbits.
Your are right, I was a bit lost in the formula jungle. I corrected it now, and ... thanks for reading it!Delta² said:@fresh_42 in post #53 in second line of b) at the end (after the ##L^*##) did you omit a factor of ##\frac{dt^*}{dt}## or is my understanding wrong?
Hi StoneTemplePython:StoneTemplePython said:For this problem: the clock's hands have no jumps!
Buzz Bloom said:Hi StoneTemplePython:
I am still unclear about your meaning regarding the minute and hour hands. If these hands do not jump, then they must move continuously. However, you may mean something else: they each jump by one second intervals synchronized with the second hand jumps.
Continuity is fine. The issue is that there's a bug in your solution (no pun intended!)Buzz Bloom said:My solution assumed that the minute and hour hands move continuously, and this apparently gave me the wrong answer.
Delta² said:Is there some typo in the statement of 14.?
I don't seem to understand the meaning of the expression ##I_A(f')^{-1}##. Is it meant to be ##I_A\circ[(f')^{-1}]##??