Find the power series about the originPower series and exponential excerciseFinding the Power Series of a Complex fuction.Let $∑_n=0^∞c_n z^n $ be a representation for the function $frac11-z-z^2 $. Find the coefficient $c_n$Find the Taylor Series expansion of the given analytic functionWhat is the radius of convergence of the power series expansion of $f(z)$ about $z=0$Find a power series centered at the origin that satisfies the BesselPower Series Extensions into the Complex PlaneFind power series of $frac1cosh z$Complex power series is arctan functionFind the Taylor series about $z=0$ and the Laurent series about $z=-3$
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Find the power series about the origin
Power series and exponential excerciseFinding the Power Series of a Complex fuction.Let $∑_n=0^∞c_n z^n $ be a representation for the function $frac11-z-z^2 $. Find the coefficient $c_n$Find the Taylor Series expansion of the given analytic functionWhat is the radius of convergence of the power series expansion of $f(z)$ about $z=0$Find a power series centered at the origin that satisfies the BesselPower Series Extensions into the Complex PlaneFind power series of $frac1cosh z$Complex power series is arctan functionFind the Taylor series about $z=0$ and the Laurent series about $z=-3$
$begingroup$
Find the power series about the origin for the given function $fracz^2(4-z)^2$ where |z|< 4. I am lost in the manipulation of the given function to bring it to one of the Taylor series that I know. I am given the following hint: $frac1(a-z)^2$ = $fracddz$ (a-z)^(-1). I would appreciate if someone could lead me!
complex-analysis
asked 2 days ago
Elvin JafaliElvin Jafali
214
$endgroup$
add a comment |
$begingroup$
Find the power series about the origin for the given function $fracz^2(4-z)^2$ where |z|< 4. I am lost in the manipulation of the given function to bring it to one of the Taylor series that I know. I am given the following hint: $frac1(a-z)^2$ = $fracddz$ (a-z)^(-1). I would appreciate if someone could lead me!
complex-analysis
asked 2 days ago
Elvin JafaliElvin Jafali
214
$endgroup$
2
$begingroup$
You want a power series centered at the origin or a power series which is valid for the origin (not necesary centered at $z=0$)??
$endgroup$
– JoseSquare
2 days ago
$begingroup$
centered about the origin
$endgroup$
– Elvin Jafali
2 days ago
1
$begingroup$
Hint. You should be able to write the series for $1/(4-z)$ starting from the geometric series for $1/(1-z)$. Then differentiate term by term and multiply by $z^2$.
$endgroup$
– Ethan Bolker
2 days ago
add a comment |
$begingroup$
Find the power series about the origin for the given function $fracz^2(4-z)^2$ where |z|< 4. I am lost in the manipulation of the given function to bring it to one of the Taylor series that I know. I am given the following hint: $frac1(a-z)^2$ = $fracddz$ (a-z)^(-1). I would appreciate if someone could lead me!
complex-analysis
asked 2 days ago
Elvin JafaliElvin Jafali
214
$endgroup$
Find the power series about the origin for the given function $fracz^2(4-z)^2$ where |z|< 4. I am lost in the manipulation of the given function to bring it to one of the Taylor series that I know. I am given the following hint: $frac1(a-z)^2$ = $fracddz$ (a-z)^(-1). I would appreciate if someone could lead me!
complex-analysis
complex-analysis
asked 2 days ago
Elvin JafaliElvin Jafali
214
asked 2 days ago
Elvin JafaliElvin Jafali
214
edited 2 days ago
Elvin Jafali
asked 2 days ago
Elvin JafaliElvin Jafali
214
asked 2 days ago
Elvin JafaliElvin Jafali
214
asked 2 days ago
Elvin JafaliElvin Jafali
214
214
2
$begingroup$
You want a power series centered at the origin or a power series which is valid for the origin (not necesary centered at $z=0$)??
$endgroup$
– JoseSquare
2 days ago
$begingroup$
centered about the origin
$endgroup$
– Elvin Jafali
2 days ago
1
$begingroup$
Hint. You should be able to write the series for $1/(4-z)$ starting from the geometric series for $1/(1-z)$. Then differentiate term by term and multiply by $z^2$.
$endgroup$
– Ethan Bolker
2 days ago
add a comment |
2
$begingroup$
You want a power series centered at the origin or a power series which is valid for the origin (not necesary centered at $z=0$)??
$endgroup$
– JoseSquare
2 days ago
$begingroup$
centered about the origin
$endgroup$
– Elvin Jafali
2 days ago
1
$begingroup$
Hint. You should be able to write the series for $1/(4-z)$ starting from the geometric series for $1/(1-z)$. Then differentiate term by term and multiply by $z^2$.
$endgroup$
– Ethan Bolker
2 days ago
2
2
$begingroup$
You want a power series centered at the origin or a power series which is valid for the origin (not necesary centered at $z=0$)??
$endgroup$
– JoseSquare
2 days ago
$begingroup$
You want a power series centered at the origin or a power series which is valid for the origin (not necesary centered at $z=0$)??
$endgroup$
– JoseSquare
2 days ago
$begingroup$
centered about the origin
$endgroup$
– Elvin Jafali
2 days ago
$begingroup$
centered about the origin
$endgroup$
– Elvin Jafali
2 days ago
1
1
$begingroup$
Hint. You should be able to write the series for $1/(4-z)$ starting from the geometric series for $1/(1-z)$. Then differentiate term by term and multiply by $z^2$.
$endgroup$
– Ethan Bolker
2 days ago
$begingroup$
Hint. You should be able to write the series for $1/(4-z)$ starting from the geometric series for $1/(1-z)$. Then differentiate term by term and multiply by $z^2$.
$endgroup$
– Ethan Bolker
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The expression given, is :
$$fracz^2(4-z)^2 = z^2 fracmathrmdmathrmdz(4-z)^-1= z^2fracmathrmdmathrmdzleft(frac14-zright)$$
Now, recalling the general geometric series $(1-z)^-1 = sum_k=0^infty z^k$ for $|z| <1$, you can reach the expression :
$$frac14-z = sum_k=0^infty 4^-(k+1)z^k, quad |z| < 4$$
I'll leave this short part up to you.
Finally, this yields us :
$$fracz^2(4-z)^2 = z^2fracmathrmdmathrmdzleft(sum_k=0^infty 4^-(k+1)z^kright)= z^2 sum_k=1^infty 4^-(k+1)kz^k-1 = boxedsum_k=1^infty 4^-(k+1)kz^k+1$$
answered 2 days ago
RebellosRebellos
15.3k31250
$endgroup$
$begingroup$
@ElvinJafali This is equivalent to what I reached to. By the way, the index should be $i=1$ due to differentiating.
$endgroup$
– Rebellos
2 days ago
$begingroup$
@ElvinJafali Check the boxed expression.
$endgroup$
– Rebellos
2 days ago
$begingroup$
Thank you, I will try to understand your step-by-step solution and thank you!
$endgroup$
– Elvin Jafali
2 days ago
add a comment |
Your Answer
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1 Answer
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$begingroup$
The expression given, is :
$$fracz^2(4-z)^2 = z^2 fracmathrmdmathrmdz(4-z)^-1= z^2fracmathrmdmathrmdzleft(frac14-zright)$$
Now, recalling the general geometric series $(1-z)^-1 = sum_k=0^infty z^k$ for $|z| <1$, you can reach the expression :
$$frac14-z = sum_k=0^infty 4^-(k+1)z^k, quad |z| < 4$$
I'll leave this short part up to you.
Finally, this yields us :
$$fracz^2(4-z)^2 = z^2fracmathrmdmathrmdzleft(sum_k=0^infty 4^-(k+1)z^kright)= z^2 sum_k=1^infty 4^-(k+1)kz^k-1 = boxedsum_k=1^infty 4^-(k+1)kz^k+1$$
answered 2 days ago
RebellosRebellos
15.3k31250
$endgroup$
$begingroup$
@ElvinJafali This is equivalent to what I reached to. By the way, the index should be $i=1$ due to differentiating.
$endgroup$
– Rebellos
2 days ago
$begingroup$
@ElvinJafali Check the boxed expression.
$endgroup$
– Rebellos
2 days ago
$begingroup$
Thank you, I will try to understand your step-by-step solution and thank you!
$endgroup$
– Elvin Jafali
2 days ago
add a comment |
$begingroup$
The expression given, is :
$$fracz^2(4-z)^2 = z^2 fracmathrmdmathrmdz(4-z)^-1= z^2fracmathrmdmathrmdzleft(frac14-zright)$$
Now, recalling the general geometric series $(1-z)^-1 = sum_k=0^infty z^k$ for $|z| <1$, you can reach the expression :
$$frac14-z = sum_k=0^infty 4^-(k+1)z^k, quad |z| < 4$$
I'll leave this short part up to you.
Finally, this yields us :
$$fracz^2(4-z)^2 = z^2fracmathrmdmathrmdzleft(sum_k=0^infty 4^-(k+1)z^kright)= z^2 sum_k=1^infty 4^-(k+1)kz^k-1 = boxedsum_k=1^infty 4^-(k+1)kz^k+1$$
answered 2 days ago
RebellosRebellos
15.3k31250
$endgroup$
$begingroup$
@ElvinJafali This is equivalent to what I reached to. By the way, the index should be $i=1$ due to differentiating.
$endgroup$
– Rebellos
2 days ago
$begingroup$
@ElvinJafali Check the boxed expression.
$endgroup$
– Rebellos
2 days ago
$begingroup$
Thank you, I will try to understand your step-by-step solution and thank you!
$endgroup$
– Elvin Jafali
2 days ago
add a comment |
$begingroup$
The expression given, is :
$$fracz^2(4-z)^2 = z^2 fracmathrmdmathrmdz(4-z)^-1= z^2fracmathrmdmathrmdzleft(frac14-zright)$$
Now, recalling the general geometric series $(1-z)^-1 = sum_k=0^infty z^k$ for $|z| <1$, you can reach the expression :
$$frac14-z = sum_k=0^infty 4^-(k+1)z^k, quad |z| < 4$$
I'll leave this short part up to you.
Finally, this yields us :
$$fracz^2(4-z)^2 = z^2fracmathrmdmathrmdzleft(sum_k=0^infty 4^-(k+1)z^kright)= z^2 sum_k=1^infty 4^-(k+1)kz^k-1 = boxedsum_k=1^infty 4^-(k+1)kz^k+1$$
answered 2 days ago
RebellosRebellos
15.3k31250
$endgroup$
The expression given, is :
$$fracz^2(4-z)^2 = z^2 fracmathrmdmathrmdz(4-z)^-1= z^2fracmathrmdmathrmdzleft(frac14-zright)$$
Now, recalling the general geometric series $(1-z)^-1 = sum_k=0^infty z^k$ for $|z| <1$, you can reach the expression :
$$frac14-z = sum_k=0^infty 4^-(k+1)z^k, quad |z| < 4$$
I'll leave this short part up to you.
Finally, this yields us :
$$fracz^2(4-z)^2 = z^2fracmathrmdmathrmdzleft(sum_k=0^infty 4^-(k+1)z^kright)= z^2 sum_k=1^infty 4^-(k+1)kz^k-1 = boxedsum_k=1^infty 4^-(k+1)kz^k+1$$
answered 2 days ago
RebellosRebellos
15.3k31250
edited 2 days ago
answered 2 days ago
RebellosRebellos
15.3k31250
answered 2 days ago
RebellosRebellos
15.3k31250
answered 2 days ago
RebellosRebellos
15.3k31250
15.3k31250
$begingroup$
@ElvinJafali This is equivalent to what I reached to. By the way, the index should be $i=1$ due to differentiating.
$endgroup$
– Rebellos
2 days ago
$begingroup$
@ElvinJafali Check the boxed expression.
$endgroup$
– Rebellos
2 days ago
$begingroup$
Thank you, I will try to understand your step-by-step solution and thank you!
$endgroup$
– Elvin Jafali
2 days ago
add a comment |
$begingroup$
@ElvinJafali This is equivalent to what I reached to. By the way, the index should be $i=1$ due to differentiating.
$endgroup$
– Rebellos
2 days ago
$begingroup$
@ElvinJafali Check the boxed expression.
$endgroup$
– Rebellos
2 days ago
$begingroup$
Thank you, I will try to understand your step-by-step solution and thank you!
$endgroup$
– Elvin Jafali
2 days ago
$begingroup$
@ElvinJafali This is equivalent to what I reached to. By the way, the index should be $i=1$ due to differentiating.
$endgroup$
– Rebellos
2 days ago
$begingroup$
@ElvinJafali This is equivalent to what I reached to. By the way, the index should be $i=1$ due to differentiating.
$endgroup$
– Rebellos
2 days ago
$begingroup$
@ElvinJafali Check the boxed expression.
$endgroup$
– Rebellos
2 days ago
$begingroup$
@ElvinJafali Check the boxed expression.
$endgroup$
– Rebellos
2 days ago
$begingroup$
Thank you, I will try to understand your step-by-step solution and thank you!
$endgroup$
– Elvin Jafali
2 days ago
$begingroup$
Thank you, I will try to understand your step-by-step solution and thank you!
$endgroup$
– Elvin Jafali
2 days ago
add a comment |
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2
$begingroup$
You want a power series centered at the origin or a power series which is valid for the origin (not necesary centered at $z=0$)??
$endgroup$
– JoseSquare
2 days ago
$begingroup$
centered about the origin
$endgroup$
– Elvin Jafali
2 days ago
1
$begingroup$
Hint. You should be able to write the series for $1/(4-z)$ starting from the geometric series for $1/(1-z)$. Then differentiate term by term and multiply by $z^2$.
$endgroup$
– Ethan Bolker
2 days ago