Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Feb 28, 2017 16:07:29 GMT
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Post by Admin on Feb 28, 2017 19:25:00 GMT
Gary
This is just a problem with notation I think. The $(n-1)$th derivative is $(z-a)^1=(z-a)$ and so the $n$th derivative is $[(z-a)]'=1$. To use the notation $(z-a)^n$ and allow $n$ to take on the value $0$ and $z$ to take on the value $a$ at the same time we need to define $(z-a)^0$ as being equal to $1$ for all values of $z$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 28, 2017 22:13:02 GMT
Gary This is just a problem with notation I think. The $(n-1)$th derivative is $(z-a)^1=(z-a)$ and so the $n$th derivative is $[(z-a)]'=1$. To use the notation $(z-a)^n$ and allow $n$ to take on the value $0$ and $z$ to take on the value $a$ at the same time we need to define $(z-a)^0$ as being equal to $1$ for all values of $z$. Vasco Vasco, Thank you. I didn't know how to deal with $P^{(n)}$. I will fix it. Gary |
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Post by mondo on Apr 3, 2022 22:21:32 GMT
Gary, where did you get the file "nh.ch7.notes.p346.pdf" from?
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Post by Admin on Apr 3, 2022 23:01:36 GMT
Gary, where did you get the file "nh.ch7.notes.p346.pdf" from? Mondo That's one of Gary's own files. You have a link to it and can now download it. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 3, 2022 23:17:27 GMT
Gary, where did you get the file "nh.ch7.notes.p346.pdf" from? Mondo,
I incorporated that note into my chapter notes. I remember that as a challenging chapter. Vasco has the last word on this material. I have no credentials in math.
Gary
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Post by mondo on Apr 3, 2022 23:19:25 GMT
Vasco, yes I know he uploaded it here but it probably comes from some website that host exercises solutions/hints. Do you know where this fine originates from? I think I found your second website that also has a nice set of exercises solution -> sites.google.com/site/vascoprat/needham?authuser=0 UPDATE: Ok thank you Gary, it may be useful as well! |
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Post by Admin on Apr 4, 2022 16:00:25 GMT
Mondo
Yes, there is link to my site on the first page of the forum.
The philosophy of this forum is to not post solutions explicitly on the forum but to provide links to solutions that you want to share with the world. This is so that anyone using the forum can choose whether and when to look at the answers. So if they are working on their own solution they would probably choose not to look at someone else's solution until they had tried to find their own solution.
Publishing discussions of exercises etc on the forum are encouraged as you of course already know.
I do not understand why you cannot see the maths properly on your mobile phone. I will try to find a solution and if I do I will let you know of course.
Vasco
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