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Analyticity of holomorphic functions
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<h2 id="proof">Proof</h2> <p>The argument, first given by Cauchy, hinges on <a href="/facts/Cauchy%2527s_integral_formula/4B35nkvV">Cauchy's integral formula</a> and the power series expansion of the expression </p> 1 w − z . {\displaystyle {\frac {1}{w-z}}.} <p>Let D {\displaystyle D} be an open disk centered at a {\displaystyle a} and suppose f {\displaystyle f} is differentiable everywhere within an <a href="/facts/Open_neighborhood/XHKgrpkp">open neighborhood</a> containing the closure of D {\displaystyle D} . Let C {\displaystyle C} be the positively oriented (i.e., counterclockwise) circle which is the boundary of D {\displaystyle D} and let z {\displaystyle z} be a point in D {\displaystyle D} . Starting with Cauchy's integral formula, we have </p> f ( z ) = 1 2 π i ∫ C f ( w ) w − z d w = 1 2 π i ∫ C f ( w ) ( w − a ) − ( z − a ) d w = 1 2 π i ∫ C 1 w − a ⋅ 1 1 − z − a w − a f ( w ) d w = 1 2 π i ∫ C 1 w − a ⋅ ∑ n = 0 ∞ ( z − a w − a ) n f ( w ) d w = ∑ n = 0 ∞ 1 2 π i ∫ C ( z − a ) n ( w − a ) n + 1 f ( w ) d w . {\displaystyle {\begin{aligned}f(z)&{}={1 \over 2\pi i}\int _{C}{f(w) \over w-z}\,\mathrm {d} w\\[10pt]&{}={1 \over 2\pi i}\int _{C}{f(w) \over (w-a)-(z-a)}\,\mathrm {d} w\\[10pt]&{}={1 \over 2\pi i}\int _{C}{1 \over w-a}\cdot {1 \over 1-{z-a \over w-a}}f(w)\,\mathrm {d} w\\[10pt]&{}={1 \over 2\pi i}\int _{C}{1 \over w-a}\cdot {\sum _{n=0}^{\infty }\left({z-a \over w-a}\right)^{n}}f(w)\,\mathrm {d} w\\[10pt]&{}=\sum _{n=0}^{\infty }{1 \over 2\pi i}\int _{C}{(z-a)^{n} \over (w-a)^{n+1}}f(w)\,\mathrm {d} w.\end{aligned}}} <p>Interchange of the integral and infinite sum is justified by observing that f ( w ) / ( w − a ) {\displaystyle f(w)/(w-a)} is bounded on C {\displaystyle C} by some positive number M {\displaystyle M} , while for all w {\displaystyle w} in C {\displaystyle C} </p> | z − a w − a | ≤ r < 1 {\displaystyle \left|{\frac {z-a}{w-a}}\right|\leq r<1} <p>for some positive r {\displaystyle r} as well. We therefore have </p> | ( z − a ) n ( w − a ) n + 1 f ( w ) | ≤ M r n , {\displaystyle \left|{(z-a)^{n} \over (w-a)^{n+1}}f(w)\right|\leq Mr^{n},} <p>on C {\displaystyle C} , and as the <a href="/facts/Weierstrass_M-test/MzGjMAFR">Weierstrass M-test</a> shows the series <a href="/facts/Converges_uniformly/ecz9eNWn">converges uniformly</a> over C {\displaystyle C} , the sum and the integral may be interchanged. </p><p>As the factor ( z − a ) n {\displaystyle (z-a)^{n}} does not depend on the variable of integration w {\displaystyle w} , it may be factored out to yield </p> f ( z ) = ∑ n = 0 ∞ ( z − a ) n 1 2 π i ∫ C f ( w ) ( w − a ) n + 1 d w , {\displaystyle f(z)=\sum _{n=0}^{\infty }(z-a)^{n}{1 \over 2\pi i}\int _{C}{f(w) \over (w-a)^{n+1}}\,\mathrm {d} w,} <p>which has the desired form of a power series in z {\displaystyle z} : </p> f ( z ) = ∑ n = 0 ∞ c n ( z − a ) n {\displaystyle f(z)=\sum _{n=0}^{\infty }c_{n}(z-a)^{n}} <p>with coefficients </p> c n = 1 2 π i ∫ C f ( w ) ( w − a ) n + 1 d w . {\displaystyle c_{n}={1 \over 2\pi i}\int _{C}{f(w) \over (w-a)^{n+1}}\,\mathrm {d} w.} <h2 id="remarks">Remarks</h2> <ul><li>Since power series can be differentiated term-wise, applying the above argument in the reverse direction and the power series expression for 1 ( w − z ) n + 1 {\displaystyle {\frac {1}{(w-z)^{n+1}}}} gives f ( n ) ( a ) = n ! 2 π i ∫ C f ( w ) ( w − a ) n + 1 d w . {\displaystyle f^{(n)}(a)={n! \over 2\pi i}\int _{C}{f(w) \over (w-a)^{n+1}}\,dw.} This is a <a href="/facts/Cauchy%2527s_integral_formula/4B35nkvV">Cauchy integral formula</a> for derivatives. Therefore the power series obtained above is the <a href="/facts/Taylor_series/M0aTuftH">Taylor series</a> of f {\displaystyle f} .</li> <li>The argument works if z {\displaystyle z} is any point that is closer to the center a {\displaystyle a} than is any singularity of f {\displaystyle f} . Therefore, the radius of convergence of the Taylor series cannot be smaller than the distance from a {\displaystyle a} to the nearest singularity (nor can it be larger, since power series have no singularities in the interiors of their circles of convergence).</li> <li>A special case of the <a href="/facts/Identity_theorem/jEu4TXXn">identity theorem</a> follows from the preceding remark. If two holomorphic functions agree on a (possibly quite small) open neighborhood U {\displaystyle U} of a {\displaystyle a} , then they coincide on the open disk B d ( a ) {\displaystyle B_{d}(a)} , where d {\displaystyle d} is the distance from a {\displaystyle a} to the nearest singularity.</li></ul> <h2 id="external-links">External links</h2> <ul><li><a href="https://planetmath.org/ExistenceOfPowerSeries">"Existence of power series"</a>. <i><a href="/facts/PlanetMath/wphPCtLf">PlanetMath</a></i>.</li></ul>