Post by Gary on Jun 30, 2017 23:24:38 GMT
Vasco,
My question is why a function requires analyticity and amplitwist in order to obtain equal Riemann integrals on two contours between two points, so long as the contours do not include the center. I begin with the idea illustrated first in [9] and then in [11] and [13], that $R_M$ can be thought of as the vector from F(A) to F(B) (START to FINISH) that is equivalent to the sum of the $\tilde{\Delta}$'s (which are not the same as the $f'\Delta$'s on an image $f(K)$). Now suppose that an $f$, analytic or merely continuous, traverses two contours between points $A$ and $B$, resulting in two new contours between $f(A)$ and $f(B)$. If $f$ is analytic, we know that a $\Delta$ that points in the direction of one of the original contours, say $K_1$ or $K_2$ will be amplitwisted to $\tilde{\Delta} = f(z) \Delta$ that points in the direction of $\tilde{K_1}$ or $\tilde{K_2}$ (Again, not the same as $f(K_1)$ and $f(K_2)$). If we stipulate that the $\Delta_i$ have equal length, then $\tilde{\Delta}_i$ also have equal length. Then it is clear by the definition of vector sums that the $\mathbf{vector}$ sums are equal in the limit, so their Riemann sums are equal.
$\hspace{5em}\sum_\tilde{K_1} \tilde{\Delta}_i = \sum_\tilde{K_2} \tilde{\Delta}_j$
But how do we know that a merely continuous $f$ would not produce the same equality? Is it that the $\tilde{\Delta}_i$ might be of varying length on $\tilde{K_1}$? Is it that they might point away from $\tilde{K_1}$ or $\tilde{K_2}$ as $f$ traverses $K_1$ or $K_2$, so that in the end $\tilde{K_1}$ does not meet $\tilde{K_2}$ at $F(B)$? (Imagine Figure [24] with a second $\tilde{K}.$) Are there any merely continuous $f$ that do vanish on traversing any two joined contours not including the origin?
Gary
My question is why a function requires analyticity and amplitwist in order to obtain equal Riemann integrals on two contours between two points, so long as the contours do not include the center. I begin with the idea illustrated first in [9] and then in [11] and [13], that $R_M$ can be thought of as the vector from F(A) to F(B) (START to FINISH) that is equivalent to the sum of the $\tilde{\Delta}$'s (which are not the same as the $f'\Delta$'s on an image $f(K)$). Now suppose that an $f$, analytic or merely continuous, traverses two contours between points $A$ and $B$, resulting in two new contours between $f(A)$ and $f(B)$. If $f$ is analytic, we know that a $\Delta$ that points in the direction of one of the original contours, say $K_1$ or $K_2$ will be amplitwisted to $\tilde{\Delta} = f(z) \Delta$ that points in the direction of $\tilde{K_1}$ or $\tilde{K_2}$ (Again, not the same as $f(K_1)$ and $f(K_2)$). If we stipulate that the $\Delta_i$ have equal length, then $\tilde{\Delta}_i$ also have equal length. Then it is clear by the definition of vector sums that the $\mathbf{vector}$ sums are equal in the limit, so their Riemann sums are equal.
$\hspace{5em}\sum_\tilde{K_1} \tilde{\Delta}_i = \sum_\tilde{K_2} \tilde{\Delta}_j$
But how do we know that a merely continuous $f$ would not produce the same equality? Is it that the $\tilde{\Delta}_i$ might be of varying length on $\tilde{K_1}$? Is it that they might point away from $\tilde{K_1}$ or $\tilde{K_2}$ as $f$ traverses $K_1$ or $K_2$, so that in the end $\tilde{K_1}$ does not meet $\tilde{K_2}$ at $F(B)$? (Imagine Figure [24] with a second $\tilde{K}.$) Are there any merely continuous $f$ that do vanish on traversing any two joined contours not including the origin?
Gary
