General term of Taylor Series of $sin(2z-1)$ centered at $dfrac{pi}{4}$.











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I started with $sin(z)$ expansion but couldn't move to powers of $z-frac{pi}{4}$ from $(2z-1)^{2n+1} ..$ is there another way to reach it?






taylor-expansion







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  • Please see math.meta.stackexchange.com/questions/5020/…
    – Lord Shark the Unknown
    Nov 17 at 7:16










  • Let $f(z)=sin(2z-1)$. identify the function $g(w)=f(pi/4+w)$.
    – Lord Shark the Unknown
    Nov 17 at 7:18















up vote
0
down vote

favorite












I started with $sin(z)$ expansion but couldn't move to powers of $z-frac{pi}{4}$ from $(2z-1)^{2n+1} ..$ is there another way to reach it?






taylor-expansion







share|cite|improve this question
























  • Please see math.meta.stackexchange.com/questions/5020/…
    – Lord Shark the Unknown
    Nov 17 at 7:16










  • Let $f(z)=sin(2z-1)$. identify the function $g(w)=f(pi/4+w)$.
    – Lord Shark the Unknown
    Nov 17 at 7:18













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I started with $sin(z)$ expansion but couldn't move to powers of $z-frac{pi}{4}$ from $(2z-1)^{2n+1} ..$ is there another way to reach it?






taylor-expansion







share|cite|improve this question















I started with $sin(z)$ expansion but couldn't move to powers of $z-frac{pi}{4}$ from $(2z-1)^{2n+1} ..$ is there another way to reach it?






taylor-expansion




taylor-expansion






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edited Nov 17 at 7:23









Yadati Kiran

1,288317




1,288317










asked Nov 17 at 7:15









Aya

11




11












  • Please see math.meta.stackexchange.com/questions/5020/…
    – Lord Shark the Unknown
    Nov 17 at 7:16










  • Let $f(z)=sin(2z-1)$. identify the function $g(w)=f(pi/4+w)$.
    – Lord Shark the Unknown
    Nov 17 at 7:18


















  • Please see math.meta.stackexchange.com/questions/5020/…
    – Lord Shark the Unknown
    Nov 17 at 7:16










  • Let $f(z)=sin(2z-1)$. identify the function $g(w)=f(pi/4+w)$.
    – Lord Shark the Unknown
    Nov 17 at 7:18
















Please see math.meta.stackexchange.com/questions/5020/…
– Lord Shark the Unknown
Nov 17 at 7:16




Please see math.meta.stackexchange.com/questions/5020/…
– Lord Shark the Unknown
Nov 17 at 7:16












Let $f(z)=sin(2z-1)$. identify the function $g(w)=f(pi/4+w)$.
– Lord Shark the Unknown
Nov 17 at 7:18




Let $f(z)=sin(2z-1)$. identify the function $g(w)=f(pi/4+w)$.
– Lord Shark the Unknown
Nov 17 at 7:18










1 Answer
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To make life simpler, as Lord Shark the Unknown commented, start with $2z=t+frac pi 2$.



So
$$sin(2z-1)=cos (1-t)=cos (1) cos (t)-sin (1) sin (t)$$ Now, use the series expansions of $sin(t)$ and $cos(t)$ around $t=0$ and, when finished, make $t=2z-frac pi 2$






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    1 Answer
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    active

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    1 Answer
    1






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    up vote
    0
    down vote













    To make life simpler, as Lord Shark the Unknown commented, start with $2z=t+frac pi 2$.



    So
    $$sin(2z-1)=cos (1-t)=cos (1) cos (t)-sin (1) sin (t)$$ Now, use the series expansions of $sin(t)$ and $cos(t)$ around $t=0$ and, when finished, make $t=2z-frac pi 2$






    share|cite|improve this answer

























      up vote
      0
      down vote













      To make life simpler, as Lord Shark the Unknown commented, start with $2z=t+frac pi 2$.



      So
      $$sin(2z-1)=cos (1-t)=cos (1) cos (t)-sin (1) sin (t)$$ Now, use the series expansions of $sin(t)$ and $cos(t)$ around $t=0$ and, when finished, make $t=2z-frac pi 2$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        To make life simpler, as Lord Shark the Unknown commented, start with $2z=t+frac pi 2$.



        So
        $$sin(2z-1)=cos (1-t)=cos (1) cos (t)-sin (1) sin (t)$$ Now, use the series expansions of $sin(t)$ and $cos(t)$ around $t=0$ and, when finished, make $t=2z-frac pi 2$






        share|cite|improve this answer












        To make life simpler, as Lord Shark the Unknown commented, start with $2z=t+frac pi 2$.



        So
        $$sin(2z-1)=cos (1-t)=cos (1) cos (t)-sin (1) sin (t)$$ Now, use the series expansions of $sin(t)$ and $cos(t)$ around $t=0$ and, when finished, make $t=2z-frac pi 2$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 17 at 9:01









        Claude Leibovici

        116k1156131




        116k1156131






























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