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数学代写|黎曼曲面代写Riemann surface代考|MAT565 Now when is moved around

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数学代写|黎曼曲面代写Riemann surface代考|Now when is moved around s, the closed path a can be dragged along

Now when $\zeta$ is moved around $s$, the closed path $\alpha$ can be dragged along, so the only effect is to continue the integrands around their singularities. Thus
$$T^k(T-I) h(\zeta)=\int_\alpha\left(T^k f\right)(x)\left(T^{-k} g\right)(\zeta-x) d x .$$
Formally, the tensor product of two quasi-unipotent transformations is again quasi-unipotent. So if $\left(T_1^{N_1}-I\right)^{K_1} f=0$ and $\left(T_2^{N_2}-I\right)^{K_2} g=0$ then there are $N$ and $K$ such that $\left(\left(T_1 \otimes T_2\right)^N-I\right)^K f \otimes g=0$. The numbers $N$ and $K$ depend universally on $N_1, N_2, K_1$, and $K_2$, independently of the spaces on which $T_1$ and $T_2$ act. The latter identity is a formal consequence of the first two. We can apply the proof of this identity to the case where $T_1$ represents the action of $T$ on $f$ and $T_2$ represents the action of $T^{-1}$ on $g$. Both of these operations are quasi-unipotent by hypothesis, so there are $N$ and $K$ with
$$\left(T^N-I\right)^K(T-I) h(\zeta)=0$$

数学代写|黎曼曲面代写Riemann surface代考|Ezercise:

Exercise: Show that if $f$ and $g$ have quasi-regular singularities, then $f * g$ has quasi-regular singularities.

Let $Q(\Gamma)$ denote the field generated over $Q$ by the roots of unity, $\pi$, and the values and derivatives of the Gamma function at rational points. The new coefficients arising from convolutions of powers of $\zeta$ and $\log \zeta$ lie in $Q(\Gamma)$. This may be seen as in the proof 2.3 .

Proposition 11.2 Suppose that $f=\sum f_n$ and $g=\sum g_n$ satisfy conditions (2.5.0) through (2.5.4) from 2 . Then the convolution $f * g=\sum f_m * g_n$ satisfies those conditions. If $f$ and $f$ satisfy condition (2.5.5) with respect to vector spaces $H_n^{\prime}$ and $H_n^{\prime \prime}$, and if there are vector spaces $H_k$, independent over $\mathrm{Q}(\Gamma)$, such that $H_m^{\prime} H_n^{\prime \prime} \subset H_{n+m}$, then $f * g$ satisfies condition (2.5.5).

数学代写|黎曼曲面代写Riemann surface代考|Now when is moved around s, the closed path a can be dragged along

$$T^k(T-I) h(\zeta)=\int_\alpha\left(T^k f\right)(x)\left(T^{-k} g\right)(\zeta-x) d x$$

$$\left(T^N-I\right)^K(T-I) h(\zeta)=0$$

MATLAB代写

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