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Post by mondo on Dec 14, 2022 3:50:30 GMT
In the second half of page 228 author says "In fact the derivative is itself displayed on the speedometer. However, in the instant that you hit the breaks the second derivative (acceleration) does not exist" This is not very accurate, acceleration is not only a result of breaking (negative acceleration) but also of increasing the speed (positive acceleration). So I think in this example author assumes a constant velocity prior to breaking.
This is followed by an example of a real function that is $0$ for negative values and $x^n$ for positive values. Author says that this is "not $m$ times differentiable at the origin". So how many times is it differentiable on the origin? I would say just once, am I right here?
Thank you.
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Post by Admin on Dec 15, 2022 22:35:21 GMT
Mondo In the second half of page 228 author says "In fact the derivative is itself displayed on the speedometer. However, in the instant that you hit the breaks the second derivative (acceleration) does not exist" This is not very accurate, acceleration is not only a result of breaking (negative acceleration) but also of increasing the speed (positive acceleration). So I think in this example author assumes a constant velocity prior to breaking. No, his point here is that at the instant you put the brakes on there is a discontinuity in the function which defines the velocity. If you think of the left part of $\Lambda$ as a graph representing your velocity against time as it increases then when you put your foot on the break the graph becomes the right hand side of $\Lambda$ where your velocity is decreasing. The discontinuity in the velocity at the top where the point is means that the acceleration is undefined. You have misquoted this from the book. You have written $x^n$ when it should be $x^m$. Check it out again and let me know if you still have a question here. I disagree with your statement above. Vasco |
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Post by mondo on Dec 18, 2022 2:26:46 GMT
In the second half of page 228 author says "In fact the derivative is itself displayed on the speedometer. However, in the instant that you hit the breaks the second derivative (acceleration) does not exist" This is not very accurate, acceleration is not only a result of breaking (negative acceleration) but also of increasing the speed (positive acceleration). So I think in this example author assumes a constant velocity prior to breaking. No, his point here is that at the instant you put the brakes on there is a discontinuity in the function which defines the velocity. If you think of the left part of $\Lambda$ as a graph representing your velocity against time as it increases then when you put your foot on the break the graph becomes the right hand side of $\Lambda$ where your velocity is decreasing. The discontinuity in the velocity at the top where the point is means that the acceleration is undefined. [/quote] I see your description but I can't agree pushing a break pedal will cause a discontinuity in the velocity graph. Here is an example graph which I think is accurate to the situation author is describing:  At the instant A a break pedal was pushed and now (as long as the pedal is kept down) the vehicle is moving with a constant velocity (ignoring all other aspects such as various resistances etc or assuming the time where speed is flat is instantaneous time). So this may be a situation which author calls "second derivative (acceleration) does not exist". But I would definitely not call it a discontinuity. Vasco, do you agree with my interpretation? You have misquoted this from the book. You have written $x^n$ when it should be $x^m$. Check it out again and let me know if you still have a question here. I disagree with your statement above. Yes a typo, sorry. Looks like this aims to generalize the speed-acceleration above. I only don't understand the this part "but not $m$ times at the origin" - how many times is is differentiable at the origin them? Should be $0$ (since it is origin) and $m-1$ everywhere. |
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Post by Admin on Dec 18, 2022 15:35:03 GMT
Mondo
1. Yes I agree with you, the discontinuity is in the acceleration at $A$ and $B$ not the velocity. So Needham is referring to the fact that the derivative of the velocity with respect to time(i.e. acceleration), does not exist at $A$ and $B$.
2. The function we are interested in is defined as $f(x)=x^m$ when $x\geq 0$ and $f(x)=0$(zero) when $x<0$.
It follows that the first $(m-1)$ derivatives of $f$ are $mx^{m-1}, m(m-1)x^{m-2},...,m!x$ when $x\geq 0$ and zero when $x<0$.
So we can see that when $x=0$ or $x<0$ then these $m-1$ derivatives are all equal to zero and so the first $m-1$ derivatives are all continuous at the origin.
However, the $m$th derivative of $x^m$ at the origin is $m!$ when $x>0$ and so the $m$th derivative of $f$ is discontinuous there.
So $f$ is only differentiable $m-1$ times at the origin.
Vasco
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Post by mondo on Dec 18, 2022 20:25:26 GMT
1. Yes I agree with you, the discontinuity is in the acceleration at $A$ and $B$ not the velocity. So Needham is referring to the fact that the derivative of the velocity with respect to time(i.e. acceleration), does not exist at $A$ and $B$. This is because the acceleration graph is equal to $2$ from time $0 - 4$ then we have an abrupt change to $0$ for next $3$ seconds and than we have another abrupt change to $-2$ at second $7$. These "sharp" edges on both sides of $A$ and $B$ make acceleration function not smooth and hence derivatives does not exist (limits from both sides of A and B are different). Is that accurate enough? 2. The function we are interested in is defined as $f(x)=x^m$ when $x\geq 0$ and $f(x)=0$(zero) when $x<0$. It follows that the first $(m-1)$ derivatives of $f$ are $mx^{m-1}, m(m-1)x^{m-2},...,m!x$ when $x\geq 0$ and zero when $x<0$. So we can see that when $x=0$ or $x<0$ then these $m-1$ derivatives are all equal to zero and so the first $m-1$ derivatives are all continuous at the origin. However, the $m$th derivative of $x^m$ at the origin is $m!$ when $x>0$ and so the $m$th derivative of $f$ is discontinuous there. So $f$ is only differentiable $m-1$ times at the origin. Vasco Clear, thank you Vasco! |
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Post by Admin on Dec 18, 2022 21:11:23 GMT
Mondo
That's fine except that the final value of the acceleration is $-(8/3)$
Vasco
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