$f^{(k)}(z_0)=0$ for all $kgeq0$ implies that $f=0$












0












$begingroup$


I'm trying understand that given an analytic function in a domain, if the function and all its derivatives vanish at a point, then, the function itself is zero.



I was told that this is a direct consequence of principle of analytic continuation, but to my understanding, the principle of analytic continuation says that the zeros of an analytic function is isolated. I'm not sure how to connect the two pieces. Can anyone explain?






complex-analysis analyticity







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    0












    $begingroup$


    I'm trying understand that given an analytic function in a domain, if the function and all its derivatives vanish at a point, then, the function itself is zero.



    I was told that this is a direct consequence of principle of analytic continuation, but to my understanding, the principle of analytic continuation says that the zeros of an analytic function is isolated. I'm not sure how to connect the two pieces. Can anyone explain?






    complex-analysis analyticity







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I'm trying understand that given an analytic function in a domain, if the function and all its derivatives vanish at a point, then, the function itself is zero.



      I was told that this is a direct consequence of principle of analytic continuation, but to my understanding, the principle of analytic continuation says that the zeros of an analytic function is isolated. I'm not sure how to connect the two pieces. Can anyone explain?






      complex-analysis analyticity







      share|cite|improve this question









      $endgroup$




      I'm trying understand that given an analytic function in a domain, if the function and all its derivatives vanish at a point, then, the function itself is zero.



      I was told that this is a direct consequence of principle of analytic continuation, but to my understanding, the principle of analytic continuation says that the zeros of an analytic function is isolated. I'm not sure how to connect the two pieces. Can anyone explain?






      complex-analysis analyticity




      complex-analysis analyticity






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 1:58









      Ya GYa G

      541211




      541211






















          1 Answer
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          2












          $begingroup$

          Suppose the domain is connected. You can write $f(z)=sum a_nz^n$ in a open neighborhood $U$ of $0$ and $a_n$ is proportional to $f^{(n)}(0)$. This implies that $f$ vanishes on $U$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm trying to understand so please let me rephrase it in my understanding. So since the function is analytic, we can show it as a Taylor series centered at $z_0$, as $f(z)=f(z_0)+f'(z_0)z+frac{f''(z_0)}{2!}z^2+...$, and since each derivative is zero, the function itself is zero? is that what you are saying? So my question here is, what role does the open neighborhood $U$ play here in terms of representing the domain?
            $endgroup$
            – Ya G
            Jan 8 at 2:06










          • $begingroup$
            The rest of the argument is the point 2. of the answer of Jonatan Y. here math.stackexchange.com/questions/511718/…
            $endgroup$
            – Tsemo Aristide
            Jan 8 at 2:09










          • $begingroup$
            If $f$ all the derivative of $f$ at $x$ are zero vanishes $f$ vanishes on a neighborhood $U$. This implies the subset $C$ where all the derivatives vanish is open. The argument in the answer of Jonathan shows that this subset is also closed, so if the function is defined on a connected set, it is identically zero.
            $endgroup$
            – Tsemo Aristide
            Jan 8 at 2:15










          • $begingroup$
            @YaG The function $f(z)$ and the zero function coincide on $U$. Therefore by the analytic continuation they must coincide everywhere.
            $endgroup$
            – clark
            Jan 8 at 2:22












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          active

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          2












          $begingroup$

          Suppose the domain is connected. You can write $f(z)=sum a_nz^n$ in a open neighborhood $U$ of $0$ and $a_n$ is proportional to $f^{(n)}(0)$. This implies that $f$ vanishes on $U$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm trying to understand so please let me rephrase it in my understanding. So since the function is analytic, we can show it as a Taylor series centered at $z_0$, as $f(z)=f(z_0)+f'(z_0)z+frac{f''(z_0)}{2!}z^2+...$, and since each derivative is zero, the function itself is zero? is that what you are saying? So my question here is, what role does the open neighborhood $U$ play here in terms of representing the domain?
            $endgroup$
            – Ya G
            Jan 8 at 2:06










          • $begingroup$
            The rest of the argument is the point 2. of the answer of Jonatan Y. here math.stackexchange.com/questions/511718/…
            $endgroup$
            – Tsemo Aristide
            Jan 8 at 2:09










          • $begingroup$
            If $f$ all the derivative of $f$ at $x$ are zero vanishes $f$ vanishes on a neighborhood $U$. This implies the subset $C$ where all the derivatives vanish is open. The argument in the answer of Jonathan shows that this subset is also closed, so if the function is defined on a connected set, it is identically zero.
            $endgroup$
            – Tsemo Aristide
            Jan 8 at 2:15










          • $begingroup$
            @YaG The function $f(z)$ and the zero function coincide on $U$. Therefore by the analytic continuation they must coincide everywhere.
            $endgroup$
            – clark
            Jan 8 at 2:22
















          2












          $begingroup$

          Suppose the domain is connected. You can write $f(z)=sum a_nz^n$ in a open neighborhood $U$ of $0$ and $a_n$ is proportional to $f^{(n)}(0)$. This implies that $f$ vanishes on $U$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm trying to understand so please let me rephrase it in my understanding. So since the function is analytic, we can show it as a Taylor series centered at $z_0$, as $f(z)=f(z_0)+f'(z_0)z+frac{f''(z_0)}{2!}z^2+...$, and since each derivative is zero, the function itself is zero? is that what you are saying? So my question here is, what role does the open neighborhood $U$ play here in terms of representing the domain?
            $endgroup$
            – Ya G
            Jan 8 at 2:06










          • $begingroup$
            The rest of the argument is the point 2. of the answer of Jonatan Y. here math.stackexchange.com/questions/511718/…
            $endgroup$
            – Tsemo Aristide
            Jan 8 at 2:09










          • $begingroup$
            If $f$ all the derivative of $f$ at $x$ are zero vanishes $f$ vanishes on a neighborhood $U$. This implies the subset $C$ where all the derivatives vanish is open. The argument in the answer of Jonathan shows that this subset is also closed, so if the function is defined on a connected set, it is identically zero.
            $endgroup$
            – Tsemo Aristide
            Jan 8 at 2:15










          • $begingroup$
            @YaG The function $f(z)$ and the zero function coincide on $U$. Therefore by the analytic continuation they must coincide everywhere.
            $endgroup$
            – clark
            Jan 8 at 2:22














          2












          2








          2





          $begingroup$

          Suppose the domain is connected. You can write $f(z)=sum a_nz^n$ in a open neighborhood $U$ of $0$ and $a_n$ is proportional to $f^{(n)}(0)$. This implies that $f$ vanishes on $U$.






          share|cite|improve this answer











          $endgroup$



          Suppose the domain is connected. You can write $f(z)=sum a_nz^n$ in a open neighborhood $U$ of $0$ and $a_n$ is proportional to $f^{(n)}(0)$. This implies that $f$ vanishes on $U$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 2:07

























          answered Jan 8 at 2:01









          Tsemo AristideTsemo Aristide

          60.7k11446




          60.7k11446












          • $begingroup$
            I'm trying to understand so please let me rephrase it in my understanding. So since the function is analytic, we can show it as a Taylor series centered at $z_0$, as $f(z)=f(z_0)+f'(z_0)z+frac{f''(z_0)}{2!}z^2+...$, and since each derivative is zero, the function itself is zero? is that what you are saying? So my question here is, what role does the open neighborhood $U$ play here in terms of representing the domain?
            $endgroup$
            – Ya G
            Jan 8 at 2:06










          • $begingroup$
            The rest of the argument is the point 2. of the answer of Jonatan Y. here math.stackexchange.com/questions/511718/…
            $endgroup$
            – Tsemo Aristide
            Jan 8 at 2:09










          • $begingroup$
            If $f$ all the derivative of $f$ at $x$ are zero vanishes $f$ vanishes on a neighborhood $U$. This implies the subset $C$ where all the derivatives vanish is open. The argument in the answer of Jonathan shows that this subset is also closed, so if the function is defined on a connected set, it is identically zero.
            $endgroup$
            – Tsemo Aristide
            Jan 8 at 2:15










          • $begingroup$
            @YaG The function $f(z)$ and the zero function coincide on $U$. Therefore by the analytic continuation they must coincide everywhere.
            $endgroup$
            – clark
            Jan 8 at 2:22


















          • $begingroup$
            I'm trying to understand so please let me rephrase it in my understanding. So since the function is analytic, we can show it as a Taylor series centered at $z_0$, as $f(z)=f(z_0)+f'(z_0)z+frac{f''(z_0)}{2!}z^2+...$, and since each derivative is zero, the function itself is zero? is that what you are saying? So my question here is, what role does the open neighborhood $U$ play here in terms of representing the domain?
            $endgroup$
            – Ya G
            Jan 8 at 2:06










          • $begingroup$
            The rest of the argument is the point 2. of the answer of Jonatan Y. here math.stackexchange.com/questions/511718/…
            $endgroup$
            – Tsemo Aristide
            Jan 8 at 2:09










          • $begingroup$
            If $f$ all the derivative of $f$ at $x$ are zero vanishes $f$ vanishes on a neighborhood $U$. This implies the subset $C$ where all the derivatives vanish is open. The argument in the answer of Jonathan shows that this subset is also closed, so if the function is defined on a connected set, it is identically zero.
            $endgroup$
            – Tsemo Aristide
            Jan 8 at 2:15










          • $begingroup$
            @YaG The function $f(z)$ and the zero function coincide on $U$. Therefore by the analytic continuation they must coincide everywhere.
            $endgroup$
            – clark
            Jan 8 at 2:22
















          $begingroup$
          I'm trying to understand so please let me rephrase it in my understanding. So since the function is analytic, we can show it as a Taylor series centered at $z_0$, as $f(z)=f(z_0)+f'(z_0)z+frac{f''(z_0)}{2!}z^2+...$, and since each derivative is zero, the function itself is zero? is that what you are saying? So my question here is, what role does the open neighborhood $U$ play here in terms of representing the domain?
          $endgroup$
          – Ya G
          Jan 8 at 2:06




          $begingroup$
          I'm trying to understand so please let me rephrase it in my understanding. So since the function is analytic, we can show it as a Taylor series centered at $z_0$, as $f(z)=f(z_0)+f'(z_0)z+frac{f''(z_0)}{2!}z^2+...$, and since each derivative is zero, the function itself is zero? is that what you are saying? So my question here is, what role does the open neighborhood $U$ play here in terms of representing the domain?
          $endgroup$
          – Ya G
          Jan 8 at 2:06












          $begingroup$
          The rest of the argument is the point 2. of the answer of Jonatan Y. here math.stackexchange.com/questions/511718/…
          $endgroup$
          – Tsemo Aristide
          Jan 8 at 2:09




          $begingroup$
          The rest of the argument is the point 2. of the answer of Jonatan Y. here math.stackexchange.com/questions/511718/…
          $endgroup$
          – Tsemo Aristide
          Jan 8 at 2:09












          $begingroup$
          If $f$ all the derivative of $f$ at $x$ are zero vanishes $f$ vanishes on a neighborhood $U$. This implies the subset $C$ where all the derivatives vanish is open. The argument in the answer of Jonathan shows that this subset is also closed, so if the function is defined on a connected set, it is identically zero.
          $endgroup$
          – Tsemo Aristide
          Jan 8 at 2:15




          $begingroup$
          If $f$ all the derivative of $f$ at $x$ are zero vanishes $f$ vanishes on a neighborhood $U$. This implies the subset $C$ where all the derivatives vanish is open. The argument in the answer of Jonathan shows that this subset is also closed, so if the function is defined on a connected set, it is identically zero.
          $endgroup$
          – Tsemo Aristide
          Jan 8 at 2:15












          $begingroup$
          @YaG The function $f(z)$ and the zero function coincide on $U$. Therefore by the analytic continuation they must coincide everywhere.
          $endgroup$
          – clark
          Jan 8 at 2:22




          $begingroup$
          @YaG The function $f(z)$ and the zero function coincide on $U$. Therefore by the analytic continuation they must coincide everywhere.
          $endgroup$
          – clark
          Jan 8 at 2:22


















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