Mixing
write Taylor series for the function f(z)=1/z centered at z=i
$begingroup$
I know that I have to calculate $a_n$ first. $a_n=frac{1}{n!} f^{(n)}(z_0)$
so I arrived at this formula for $f^{(n)}(z)=(-1)^n.frac{n!}{z^{n+1}}$
then we have $a_n=frac{(-1)^n}{i^{n+1}}$
so from Taylor series we have $$f(z)=sum_{n=1}^{infty} a_n (z-z_0)^n $$
so for f(z)=1/z we have $$f(z)=frac{(-1)^n}{i^{n+1}} .(z-i)^n$$
but I can't confidentially say that this approach is true.If you see any problems with my solution please tell me.
complex-analysis taylor-expansion
asked Jan 2 at 17:41
mohamadrezamohamadreza
205
$endgroup$
add a comment |
$begingroup$
I know that I have to calculate $a_n$ first. $a_n=frac{1}{n!} f^{(n)}(z_0)$
so I arrived at this formula for $f^{(n)}(z)=(-1)^n.frac{n!}{z^{n+1}}$
then we have $a_n=frac{(-1)^n}{i^{n+1}}$
so from Taylor series we have $$f(z)=sum_{n=1}^{infty} a_n (z-z_0)^n $$
so for f(z)=1/z we have $$f(z)=frac{(-1)^n}{i^{n+1}} .(z-i)^n$$
but I can't confidentially say that this approach is true.If you see any problems with my solution please tell me.
complex-analysis taylor-expansion
asked Jan 2 at 17:41
mohamadrezamohamadreza
205
$endgroup$
$begingroup$
Your approach is fine. In the final result, you should have a series though. And, $-1 = i^2$ will simplify your result. The other approach would be to say $frac {1}{1-z} = sum z^n$ (the sum of a geometric series.) What do you need to do to algebraically manipulate your function to have the correct form?
$endgroup$
– Doug M
Jan 2 at 18:05
$begingroup$
please give me a hint
$endgroup$
– mohamadreza
Jan 2 at 18:53
add a comment |
$begingroup$
I know that I have to calculate $a_n$ first. $a_n=frac{1}{n!} f^{(n)}(z_0)$
so I arrived at this formula for $f^{(n)}(z)=(-1)^n.frac{n!}{z^{n+1}}$
then we have $a_n=frac{(-1)^n}{i^{n+1}}$
so from Taylor series we have $$f(z)=sum_{n=1}^{infty} a_n (z-z_0)^n $$
so for f(z)=1/z we have $$f(z)=frac{(-1)^n}{i^{n+1}} .(z-i)^n$$
but I can't confidentially say that this approach is true.If you see any problems with my solution please tell me.
complex-analysis taylor-expansion
asked Jan 2 at 17:41
mohamadrezamohamadreza
205
$endgroup$
I know that I have to calculate $a_n$ first. $a_n=frac{1}{n!} f^{(n)}(z_0)$
so I arrived at this formula for $f^{(n)}(z)=(-1)^n.frac{n!}{z^{n+1}}$
then we have $a_n=frac{(-1)^n}{i^{n+1}}$
so from Taylor series we have $$f(z)=sum_{n=1}^{infty} a_n (z-z_0)^n $$
so for f(z)=1/z we have $$f(z)=frac{(-1)^n}{i^{n+1}} .(z-i)^n$$
but I can't confidentially say that this approach is true.If you see any problems with my solution please tell me.
complex-analysis taylor-expansion
complex-analysis taylor-expansion
asked Jan 2 at 17:41
mohamadrezamohamadreza
205
asked Jan 2 at 17:41
mohamadrezamohamadreza
205
asked Jan 2 at 17:41
mohamadrezamohamadreza
205
asked Jan 2 at 17:41
mohamadrezamohamadreza
205
asked Jan 2 at 17:41
mohamadrezamohamadreza
205
205
$begingroup$
Your approach is fine. In the final result, you should have a series though. And, $-1 = i^2$ will simplify your result. The other approach would be to say $frac {1}{1-z} = sum z^n$ (the sum of a geometric series.) What do you need to do to algebraically manipulate your function to have the correct form?
$endgroup$
– Doug M
Jan 2 at 18:05
$begingroup$
please give me a hint
$endgroup$
– mohamadreza
Jan 2 at 18:53
add a comment |
$begingroup$
Your approach is fine. In the final result, you should have a series though. And, $-1 = i^2$ will simplify your result. The other approach would be to say $frac {1}{1-z} = sum z^n$ (the sum of a geometric series.) What do you need to do to algebraically manipulate your function to have the correct form?
$endgroup$
– Doug M
Jan 2 at 18:05
$begingroup$
please give me a hint
$endgroup$
– mohamadreza
Jan 2 at 18:53
$begingroup$
Your approach is fine. In the final result, you should have a series though. And, $-1 = i^2$ will simplify your result. The other approach would be to say $frac {1}{1-z} = sum z^n$ (the sum of a geometric series.) What do you need to do to algebraically manipulate your function to have the correct form?
$endgroup$
– Doug M
Jan 2 at 18:05
$begingroup$
Your approach is fine. In the final result, you should have a series though. And, $-1 = i^2$ will simplify your result. The other approach would be to say $frac {1}{1-z} = sum z^n$ (the sum of a geometric series.) What do you need to do to algebraically manipulate your function to have the correct form?
$endgroup$
– Doug M
Jan 2 at 18:05
$begingroup$
please give me a hint
$endgroup$
– mohamadreza
Jan 2 at 18:53
$begingroup$
please give me a hint
$endgroup$
– mohamadreza
Jan 2 at 18:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can either say.
$frac {f^{(n)}(z)}{n!} = (-1)^n z^{-n-1}$
evaluated at $z=i$
$a_n = (-1)^n (i)^{-n-1}\
-1 = i^2\
a_n = (i)^{n-1}\
f(z) = sum_limits{0}^{infty} i^{n-1} {(z-i)^n}$
Alternatively, you could use the formula for a geometric series.
$sum_limits{n=0}^infty z^n = frac {1}{1-z}$ and manipulate your function to fit this form.
$f(z) = frac {1}{z} = frac {1}{i+(z-i)} = frac {i}{-1+(z+i)i}= -isum i^n(z-i)^n = sum i^{n-1}(z-i)^n$
answered Jan 2 at 18:02
Doug MDoug M
44.3k31854
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
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active
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votes
$begingroup$
You can either say.
$frac {f^{(n)}(z)}{n!} = (-1)^n z^{-n-1}$
evaluated at $z=i$
$a_n = (-1)^n (i)^{-n-1}\
-1 = i^2\
a_n = (i)^{n-1}\
f(z) = sum_limits{0}^{infty} i^{n-1} {(z-i)^n}$
Alternatively, you could use the formula for a geometric series.
$sum_limits{n=0}^infty z^n = frac {1}{1-z}$ and manipulate your function to fit this form.
$f(z) = frac {1}{z} = frac {1}{i+(z-i)} = frac {i}{-1+(z+i)i}= -isum i^n(z-i)^n = sum i^{n-1}(z-i)^n$
answered Jan 2 at 18:02
Doug MDoug M
44.3k31854
$endgroup$
add a comment |
$begingroup$
You can either say.
$frac {f^{(n)}(z)}{n!} = (-1)^n z^{-n-1}$
evaluated at $z=i$
$a_n = (-1)^n (i)^{-n-1}\
-1 = i^2\
a_n = (i)^{n-1}\
f(z) = sum_limits{0}^{infty} i^{n-1} {(z-i)^n}$
Alternatively, you could use the formula for a geometric series.
$sum_limits{n=0}^infty z^n = frac {1}{1-z}$ and manipulate your function to fit this form.
$f(z) = frac {1}{z} = frac {1}{i+(z-i)} = frac {i}{-1+(z+i)i}= -isum i^n(z-i)^n = sum i^{n-1}(z-i)^n$
answered Jan 2 at 18:02
Doug MDoug M
44.3k31854
$endgroup$
add a comment |
$begingroup$
You can either say.
$frac {f^{(n)}(z)}{n!} = (-1)^n z^{-n-1}$
evaluated at $z=i$
$a_n = (-1)^n (i)^{-n-1}\
-1 = i^2\
a_n = (i)^{n-1}\
f(z) = sum_limits{0}^{infty} i^{n-1} {(z-i)^n}$
Alternatively, you could use the formula for a geometric series.
$sum_limits{n=0}^infty z^n = frac {1}{1-z}$ and manipulate your function to fit this form.
$f(z) = frac {1}{z} = frac {1}{i+(z-i)} = frac {i}{-1+(z+i)i}= -isum i^n(z-i)^n = sum i^{n-1}(z-i)^n$
answered Jan 2 at 18:02
Doug MDoug M
44.3k31854
$endgroup$
You can either say.
$frac {f^{(n)}(z)}{n!} = (-1)^n z^{-n-1}$
evaluated at $z=i$
$a_n = (-1)^n (i)^{-n-1}\
-1 = i^2\
a_n = (i)^{n-1}\
f(z) = sum_limits{0}^{infty} i^{n-1} {(z-i)^n}$
Alternatively, you could use the formula for a geometric series.
$sum_limits{n=0}^infty z^n = frac {1}{1-z}$ and manipulate your function to fit this form.
$f(z) = frac {1}{z} = frac {1}{i+(z-i)} = frac {i}{-1+(z+i)i}= -isum i^n(z-i)^n = sum i^{n-1}(z-i)^n$
answered Jan 2 at 18:02
Doug MDoug M
44.3k31854
edited Jan 2 at 18:56
answered Jan 2 at 18:02
Doug MDoug M
44.3k31854
answered Jan 2 at 18:02
Doug MDoug M
44.3k31854
answered Jan 2 at 18:02
Doug MDoug M
44.3k31854
44.3k31854
add a comment |
add a comment |
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$begingroup$
Your approach is fine. In the final result, you should have a series though. And, $-1 = i^2$ will simplify your result. The other approach would be to say $frac {1}{1-z} = sum z^n$ (the sum of a geometric series.) What do you need to do to algebraically manipulate your function to have the correct form?
$endgroup$
– Doug M
Jan 2 at 18:05
$begingroup$
please give me a hint
$endgroup$
– mohamadreza
Jan 2 at 18:53