Expand $frac1z^2$ as a series on $|z - 2| < 2$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Series Expansion of $frac11-e^int$Complex power series expansion for $f(z) = frace^za-z$Taylor Series of $frac11-cos x$notation for first and second derivatives of a power seriesIntegrating the Fourier series to find the Fourier series of $frac12x^2$Understanding ProofWiki's proof of differentiation of sine power seriesPower series representation of $fracx^527x^3 + 1$How can I express the natural logarithm of a power series in terms of another power series?Complex power series: radius of convergence and derivativechallenging power series expansion
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Expand $frac1z^2$ as a series on $|z - 2|
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Series Expansion of $frac11-e^int$Complex power series expansion for $f(z) = frace^za-z$Taylor Series of $frac11-cos x$notation for first and second derivatives of a power seriesIntegrating the Fourier series to find the Fourier series of $frac12x^2$Understanding ProofWiki's proof of differentiation of sine power seriesPower series representation of $fracx^527x^3 + 1$How can I express the natural logarithm of a power series in terms of another power series?Complex power series: radius of convergence and derivativechallenging power series expansion
$begingroup$
I am asked to show that
$$frac1z^2 = frac14 + frac14sum_n = 1^infty(-1)^n+1(n+1)left(fracz-22right)^n$$
on $|z - 2| < 2$.
My plan is to consider a series expansion of $frac-1z$ in $|z - 2| < 2$ and differentiate this series term by term to yield the desired series (This is valid from theorems we have proved in class).
Indeed,
$$frac-1z = frac-12-2+z = frac-12cdotfrac11+left(fracz-22right)$$
$$ = frac-12sum_n = 0^infty(-1)^nleft(fracz-22right)^n $$
$$ = frac12sum_n = 0^infty(-1)^n+1left(fracz-22right)^n $$
Differentiating term by term gives:
$$ frac1z^2 = frac12sum_n = 1^infty(-1)^n+1nleft(fracz-22right)^n-1cdotfrac12 $$
$$ = frac14sum_n = 1^infty(-1)^n+1nleft(fracz-22right)^n-1$$
Now, if we let $k = n - 1$, we can re-index the series as:
$$ = frac14sum_k = 0^infty(-1)^k+2(k+1)left(fracz-22right)^k$$
$$ = frac14 + frac14sum_k = 1^infty(-1)^k(k+1)left(fracz-22right)^k$$
Unfortunately, the power of $(-1)$ present in my series does not match what the prompt reads. I am off by one. Did I make a mistake or is the prompt incorrect?
proof-verification power-series
asked Apr 8 at 20:29
Nicholas RobertsNicholas Roberts
149113
$endgroup$
add a comment |
$begingroup$
I am asked to show that
$$frac1z^2 = frac14 + frac14sum_n = 1^infty(-1)^n+1(n+1)left(fracz-22right)^n$$
on $|z - 2| < 2$.
My plan is to consider a series expansion of $frac-1z$ in $|z - 2| < 2$ and differentiate this series term by term to yield the desired series (This is valid from theorems we have proved in class).
Indeed,
$$frac-1z = frac-12-2+z = frac-12cdotfrac11+left(fracz-22right)$$
$$ = frac-12sum_n = 0^infty(-1)^nleft(fracz-22right)^n $$
$$ = frac12sum_n = 0^infty(-1)^n+1left(fracz-22right)^n $$
Differentiating term by term gives:
$$ frac1z^2 = frac12sum_n = 1^infty(-1)^n+1nleft(fracz-22right)^n-1cdotfrac12 $$
$$ = frac14sum_n = 1^infty(-1)^n+1nleft(fracz-22right)^n-1$$
Now, if we let $k = n - 1$, we can re-index the series as:
$$ = frac14sum_k = 0^infty(-1)^k+2(k+1)left(fracz-22right)^k$$
$$ = frac14 + frac14sum_k = 1^infty(-1)^k(k+1)left(fracz-22right)^k$$
Unfortunately, the power of $(-1)$ present in my series does not match what the prompt reads. I am off by one. Did I make a mistake or is the prompt incorrect?
proof-verification power-series
asked Apr 8 at 20:29
Nicholas RobertsNicholas Roberts
149113
$endgroup$
$begingroup$
Your solution looks correct.
$endgroup$
– Umberto P.
Apr 8 at 20:50
add a comment |
$begingroup$
I am asked to show that
$$frac1z^2 = frac14 + frac14sum_n = 1^infty(-1)^n+1(n+1)left(fracz-22right)^n$$
on $|z - 2| < 2$.
My plan is to consider a series expansion of $frac-1z$ in $|z - 2| < 2$ and differentiate this series term by term to yield the desired series (This is valid from theorems we have proved in class).
Indeed,
$$frac-1z = frac-12-2+z = frac-12cdotfrac11+left(fracz-22right)$$
$$ = frac-12sum_n = 0^infty(-1)^nleft(fracz-22right)^n $$
$$ = frac12sum_n = 0^infty(-1)^n+1left(fracz-22right)^n $$
Differentiating term by term gives:
$$ frac1z^2 = frac12sum_n = 1^infty(-1)^n+1nleft(fracz-22right)^n-1cdotfrac12 $$
$$ = frac14sum_n = 1^infty(-1)^n+1nleft(fracz-22right)^n-1$$
Now, if we let $k = n - 1$, we can re-index the series as:
$$ = frac14sum_k = 0^infty(-1)^k+2(k+1)left(fracz-22right)^k$$
$$ = frac14 + frac14sum_k = 1^infty(-1)^k(k+1)left(fracz-22right)^k$$
Unfortunately, the power of $(-1)$ present in my series does not match what the prompt reads. I am off by one. Did I make a mistake or is the prompt incorrect?
proof-verification power-series
asked Apr 8 at 20:29
Nicholas RobertsNicholas Roberts
149113
$endgroup$
I am asked to show that
$$frac1z^2 = frac14 + frac14sum_n = 1^infty(-1)^n+1(n+1)left(fracz-22right)^n$$
on $|z - 2| < 2$.
My plan is to consider a series expansion of $frac-1z$ in $|z - 2| < 2$ and differentiate this series term by term to yield the desired series (This is valid from theorems we have proved in class).
Indeed,
$$frac-1z = frac-12-2+z = frac-12cdotfrac11+left(fracz-22right)$$
$$ = frac-12sum_n = 0^infty(-1)^nleft(fracz-22right)^n $$
$$ = frac12sum_n = 0^infty(-1)^n+1left(fracz-22right)^n $$
Differentiating term by term gives:
$$ frac1z^2 = frac12sum_n = 1^infty(-1)^n+1nleft(fracz-22right)^n-1cdotfrac12 $$
$$ = frac14sum_n = 1^infty(-1)^n+1nleft(fracz-22right)^n-1$$
Now, if we let $k = n - 1$, we can re-index the series as:
$$ = frac14sum_k = 0^infty(-1)^k+2(k+1)left(fracz-22right)^k$$
$$ = frac14 + frac14sum_k = 1^infty(-1)^k(k+1)left(fracz-22right)^k$$
Unfortunately, the power of $(-1)$ present in my series does not match what the prompt reads. I am off by one. Did I make a mistake or is the prompt incorrect?
proof-verification power-series
proof-verification power-series
asked Apr 8 at 20:29
Nicholas RobertsNicholas Roberts
149113
asked Apr 8 at 20:29
Nicholas RobertsNicholas Roberts
149113
edited Apr 8 at 20:48
Nicholas Roberts
asked Apr 8 at 20:29
Nicholas RobertsNicholas Roberts
149113
asked Apr 8 at 20:29
Nicholas RobertsNicholas Roberts
149113
asked Apr 8 at 20:29
Nicholas RobertsNicholas Roberts
149113
149113
$begingroup$
Your solution looks correct.
$endgroup$
– Umberto P.
Apr 8 at 20:50
add a comment |
$begingroup$
Your solution looks correct.
$endgroup$
– Umberto P.
Apr 8 at 20:50
$begingroup$
Your solution looks correct.
$endgroup$
– Umberto P.
Apr 8 at 20:50
$begingroup$
Your solution looks correct.
$endgroup$
– Umberto P.
Apr 8 at 20:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Perhaps a simpler approach would be to write
$$
1 over z^2 = 1 over 1 - left(1 - z^2right).
$$
Letting $a = 1 - z^2$, we see that the above expression has (at least formally) the geometric series expansion:
$$
1 over 1 - left(1 - z^2right) = 1 over 1 - a = sum_n geq 0 a^n.
$$
Now expand each power
$$
a^n = left(1 - z^2right)^n
$$
using the Binomial Theorem.
answered Apr 8 at 20:46
avsavs
4,197515
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Perhaps a simpler approach would be to write
$$
1 over z^2 = 1 over 1 - left(1 - z^2right).
$$
Letting $a = 1 - z^2$, we see that the above expression has (at least formally) the geometric series expansion:
$$
1 over 1 - left(1 - z^2right) = 1 over 1 - a = sum_n geq 0 a^n.
$$
Now expand each power
$$
a^n = left(1 - z^2right)^n
$$
using the Binomial Theorem.
answered Apr 8 at 20:46
avsavs
4,197515
$endgroup$
add a comment |
$begingroup$
Perhaps a simpler approach would be to write
$$
1 over z^2 = 1 over 1 - left(1 - z^2right).
$$
Letting $a = 1 - z^2$, we see that the above expression has (at least formally) the geometric series expansion:
$$
1 over 1 - left(1 - z^2right) = 1 over 1 - a = sum_n geq 0 a^n.
$$
Now expand each power
$$
a^n = left(1 - z^2right)^n
$$
using the Binomial Theorem.
answered Apr 8 at 20:46
avsavs
4,197515
$endgroup$
add a comment |
$begingroup$
Perhaps a simpler approach would be to write
$$
1 over z^2 = 1 over 1 - left(1 - z^2right).
$$
Letting $a = 1 - z^2$, we see that the above expression has (at least formally) the geometric series expansion:
$$
1 over 1 - left(1 - z^2right) = 1 over 1 - a = sum_n geq 0 a^n.
$$
Now expand each power
$$
a^n = left(1 - z^2right)^n
$$
using the Binomial Theorem.
answered Apr 8 at 20:46
avsavs
4,197515
$endgroup$
Perhaps a simpler approach would be to write
$$
1 over z^2 = 1 over 1 - left(1 - z^2right).
$$
Letting $a = 1 - z^2$, we see that the above expression has (at least formally) the geometric series expansion:
$$
1 over 1 - left(1 - z^2right) = 1 over 1 - a = sum_n geq 0 a^n.
$$
Now expand each power
$$
a^n = left(1 - z^2right)^n
$$
using the Binomial Theorem.
answered Apr 8 at 20:46
avsavs
4,197515
answered Apr 8 at 20:46
avsavs
4,197515
answered Apr 8 at 20:46
avsavs
4,197515
answered Apr 8 at 20:46
avsavs
4,197515
4,197515
add a comment |
add a comment |
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$begingroup$
Your solution looks correct.
$endgroup$
– Umberto P.
Apr 8 at 20:50