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Post by mondo on Jun 25, 2023 20:29:52 GMT
Oh sure, I would love to see your derivation!  |
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Post by mondo on Jul 2, 2023 21:57:10 GMT
On page 432 equation (7) for $C_n$ and subsequent one calculate its value at a point $0$ without any comment. Why was a point $0$ chosen? It looks like it was just convenience.
Also later, on page 434, on the very top author says "If $f(z)$ is analytic at $a$ then $F(\xi) \equiv f(a + \xi) = f(z)$" I don't understand it completely. How is $F(\xi)$ identical to $f(a + \xi)$ and this in turn is equal to $f(z)$?
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Post by mondo on Jul 6, 2023 5:02:01 GMT
Does it make sense to you Vasco?
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Post by Admin on Jul 6, 2023 18:10:31 GMT
Mondo
Yes it does make sense.
If you read from the beginning of the paragraph which starts at the bottom of page 433, you will see that Needham i explaining why he chose the origin as the centre of the expansion, and why this involves no loss of generality.
$F(\xi)$ is defined by us to be identically equal to $f(a+\xi)=f(a)$. The origin of the $\xi$-plane is when $z=a$.
So $F(\xi)$ has an origin-centred Taylor expansion which means that $f(z)$ has a Taylor expansion in $(z-a)$ centred at $z=a$
Vasco
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Post by mondo on Jul 7, 2023 7:31:47 GMT
Mondo Yes it does make sense. If you read from the beginning of the paragraph which starts at the bottom of page 433, you will see that Needham i explaining why he chose the origin as the centre of the expansion, and why this involves no loss of generality. $F(\xi)$ is defined by us to be identically equal to $f(a+\xi)=f(a)$. The origin of the $\xi$-plane is when $z=a$. So $F(\xi)$ has an origin-centred Taylor expansion which means that $f(z)$ has a Taylor expansion in $(z-a)$ centred at $z=a$ Vasco $F(\xi)$ is defined by us? I see on page 430, we define $F_a(z)$ to be an amplitwist. How is that identical to $f(a + \xi)$? Even if I substitute $\xi$ for $z$ in $F(z)$ I get $F(\xi) = \frac{f(z-a) - f(a)}{(z-a)} \ne f(a+\xi)$ right? |
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Post by Admin on Jul 7, 2023 14:46:47 GMT
Mondo
For me the function $F$ defined at the top of page 434 is a completely different function from the $F$ defined on page 430.
Vasco
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Post by mondo on Jul 7, 2023 20:48:25 GMT
Mondo For me the function $F$ defined at the top of page 434 is a completely different function from the $F$ defined on page 430. Vasco So you say, on the top of page 434 author defines a new function $F(\xi)$ which is identical to $f(a+\xi)$? If so, how this allows him to do the transformation right below from $F(\xi) => f(z)$. So just because the new function is analytic at $a$ we can say the expansion is also possible at this point? Update: Ok I think I got it - if we substitute $0$ into $F(\xi)$ then we get $f(a)$ and this is our new origin. It sort of makes sense.. So the biggest confusion here came from the fact that I was looking back for a definition of $F(z)$ |
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Post by Admin on Jul 8, 2023 9:44:06 GMT
Mondo
Correct, I agree!
Vasco
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