[SOUND] The role of the tailor expansion in our start is explained by the following proposition In Jacobi form, I'm working Jacobi form weight K index M Is determined Uniquely By its first. Taylor coefficient, more exactly, by f0(tau) and so on til f2m(tau). Those are Taylor coefficients, t the coefficient 2m, Determine the Jacobi form uniquely. The proof is very simple. This again an implication of the theorem involves a number of zeroes. Suppose, that we have two functions, phi and psi with the same Taylor coefficient from F0 till F2M. Then the order of zero, of Z equal to zero, of pie minus sa, is greater or equal to M plus one. And the index of this function is m, therefore this function vanishes identically. Through using this proposition, we would like to estimate the dimension of the space of Jacobi form, using only some number of the tela coefficient. Let us analyze a Taylor expansion. As usual in this section, ka is Jacobi form of weight K and index M. Let's take the following [SOUND] Matrix in Sl2Z. Jacobi form is modular with respect to this. Writing therefore, we have the following useful property. Tao to minus z is equal to minus 1 to the power k f to z. Therefore, our Jacobi form is invariant with respect to the multiplication by minus 1 if the weight is even. And until bearing if the weight is off. It follows that in the Taylor expansion of the Jacobi form, Phi, tau, Z. We have only Taylor coefficient, With index d, congruent to the weight, modular 2 D is not equal. Therefore, in the Taylor expansion we have only even In this is D if K is a weight is even and only odd in this is D if K is odd. Coming back to the property to the previous proposition. We see that the number of Taylor coefficient would have draw alive. And this proposition is certainly small for 2 M. The Taylor co-efficient as they told you, they're not the tele co-efficient, they're not modular form, but the first non vanish teller co-efficient, is always a modular form. We can prove the following proposition. R is Jacobi form of k, and index M.. The first Taylor coefficient, the 0 of Taylor coefficient is a modular form of weight K. That if, This function chooses identity, then, fd0 the first vanishing Taylor coefficient is a cusp form of weight k + d0. Proof. To prove this serum we have to analyze the modular property of Jacobi form. And let us calculate for any M equal to A,B,C,D. And S two Zed, we have to calculate the tailor expansion on both sides of the modular equation e to the power of 2 pi i c z to the square over C tau + d. phi (tau, z). On the left hand side, We get The following teller experience, M D M tau Z is upon D over C tau plus D To the power d. There is a small problem here, we have two letter d but this d, two d are different, is equal. And now, we have to calculate the right of this identity. I'm sorry, in this identity we have to add the weight K. We get C Tau + d to the power k, then e to the power 2 pi i m c z to the square over C Tau + d. And then the Taylor expansion of the function, r. Now, we can compare the tail coefficients in the Left and right part. To do this we have to add the tail of expansion of this exponential function, we get a rather complicated formula. But if d0 = the order of 0 Of Jacobi form 0. We can calculate, The relation for this minimal degree, z to the power d0. And we get [SOUND] I repeat that if d is greater than the 0, then we have a rather complicated relations because we have to add some powers of this holomorphic factor. But the first coefficient, we have very simple equation. Which tells us that, f d0 in Tau is a modular form of weight K plus D zero, moreover according to the result about the free expansion of this function of non-zero tell of the coefficient. We see that, This function is a cusp form. If d0 is strictly positive. In this case, this function has no constant term [SOUND] [MUSIC]