By Waclaw Sierpinski

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**Extra info for 250 problems in elementary number theory**

**Sample text**

I) / " ( x ) = 0. Thus, f{x ) is of the form f{x ) = ax + 6. (ii) f { x + 1) = / ( x ) . Thus, / ( x ) = b. (iii) / ( 5 ) = Thus, /( x ) = -ilo g (2 7 r). First, we show (i).

For example, in the case of the Lfunction of automorphic forms on G L(n), the fact that no zero exists in Re(s) > 1 has been proved, only by the method using the Eisen stein series for GL(n + 1 ). (See Chapter 11. ) (c ) R a n k in -S elb erg m e th o d . As an application of E{s^z), there is a method of obtaining analytic continuation of C functions of modular forms. 3 we proved the analytic continuation and functional equation of L(s, A ) by using the Foruier transform (Mellin transform) of the automorphy of A.

MODULAR FORMS (calculation by Euler). Indeed, in the formula / \ ^i 1 '^W - i d u ‘ ‘"’ “ si, “ H - 2 (“ + ; ) ) “ we substitute v = |г¿, and we obtain Thus, in the formula we substitute ^ = fj;, and we obtain e x p (-(»+ ^ ))»4 ^ = -{y/ T zK i{z)). Prom this we see that y/^K i{ z) = Ce~^, C = J^ e - % - i d r = r ( i ) = V^. Therefore, we obtain ^ ^ ( 0 , z) =\ogy - - 4 ^ ^(y^)^-27rmy cos( 27rmx). 771=1 On the other hand, we have log(j/®|A(2:)|) = 61ogj/ + log|A(z)| = 61ogy + Re(log A (z )) OO OO = 6 logy - 2wy - 2 4 R e ( ^ n = l 771=1 OO = 6 logy — 27ry — 2 4 ^ “ OO ^ 77=1 771=1 OO / - — g-27r77my cos( 27rmnx) m \ ^ '^) ^-27rmy cos( 27Tmx), = 6 logy — 2Try — 24 771=1 and the formula in question follows.