Find the convergence radius of series
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0
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I need to find the convergence radius of the following series:
$$sum_{n=0}^infty cos in times z^n$$
Since
$$cos in = cosh n = {frac{e^{in} + e^{-in}}{2}}$$
therefore, using D'Alembert's property I find the limit as
$$lim {frac{a_{n+1}}{a_n}} = {frac{e^{i(n+1)} + e^{-i(n+1)}}{2}} times z^n times z times {frac{2}{(e^{in} + e^{-in})z^n}} = lim z times {frac{e^{i(n+1)} + e^{-i(n+1)}}{e^{in} + e^{-in}}} = z$$
Therefore the radius should be 1. But the answer on the book is ${frac{1}{e}}$.
What's wrong with my solution?
limits convergence complex-numbers
edited Nov 19 at 18:26
Bernard
117k637109
asked Nov 19 at 18:21
user3132457
856
|
show 5 more comments
up vote
0
down vote
favorite
I need to find the convergence radius of the following series:
$$sum_{n=0}^infty cos in times z^n$$
Since
$$cos in = cosh n = {frac{e^{in} + e^{-in}}{2}}$$
therefore, using D'Alembert's property I find the limit as
$$lim {frac{a_{n+1}}{a_n}} = {frac{e^{i(n+1)} + e^{-i(n+1)}}{2}} times z^n times z times {frac{2}{(e^{in} + e^{-in})z^n}} = lim z times {frac{e^{i(n+1)} + e^{-i(n+1)}}{e^{in} + e^{-in}}} = z$$
Therefore the radius should be 1. But the answer on the book is ${frac{1}{e}}$.
What's wrong with my solution?
limits convergence complex-numbers
edited Nov 19 at 18:26
Bernard
117k637109
asked Nov 19 at 18:21
user3132457
856
Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
– Clement C.
Nov 19 at 18:24
isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
– user3132457
Nov 19 at 18:26
(also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
– Clement C.
Nov 19 at 18:26
@OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
– Clement C.
Nov 19 at 18:27
1
So it should be $${frac{e^n + e^{-n}}{2}}$$?
– user3132457
Nov 19 at 18:28
|
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need to find the convergence radius of the following series:
$$sum_{n=0}^infty cos in times z^n$$
Since
$$cos in = cosh n = {frac{e^{in} + e^{-in}}{2}}$$
therefore, using D'Alembert's property I find the limit as
$$lim {frac{a_{n+1}}{a_n}} = {frac{e^{i(n+1)} + e^{-i(n+1)}}{2}} times z^n times z times {frac{2}{(e^{in} + e^{-in})z^n}} = lim z times {frac{e^{i(n+1)} + e^{-i(n+1)}}{e^{in} + e^{-in}}} = z$$
Therefore the radius should be 1. But the answer on the book is ${frac{1}{e}}$.
What's wrong with my solution?
limits convergence complex-numbers
edited Nov 19 at 18:26
Bernard
117k637109
asked Nov 19 at 18:21
user3132457
856
I need to find the convergence radius of the following series:
$$sum_{n=0}^infty cos in times z^n$$
Since
$$cos in = cosh n = {frac{e^{in} + e^{-in}}{2}}$$
therefore, using D'Alembert's property I find the limit as
$$lim {frac{a_{n+1}}{a_n}} = {frac{e^{i(n+1)} + e^{-i(n+1)}}{2}} times z^n times z times {frac{2}{(e^{in} + e^{-in})z^n}} = lim z times {frac{e^{i(n+1)} + e^{-i(n+1)}}{e^{in} + e^{-in}}} = z$$
Therefore the radius should be 1. But the answer on the book is ${frac{1}{e}}$.
What's wrong with my solution?
limits convergence complex-numbers
limits convergence complex-numbers
edited Nov 19 at 18:26
Bernard
117k637109
asked Nov 19 at 18:21
user3132457
856
edited Nov 19 at 18:26
Bernard
117k637109
asked Nov 19 at 18:21
user3132457
856
edited Nov 19 at 18:26
Bernard
117k637109
edited Nov 19 at 18:26
Bernard
117k637109
edited Nov 19 at 18:26
Bernard
117k637109
117k637109
asked Nov 19 at 18:21
user3132457
856
asked Nov 19 at 18:21
user3132457
856
asked Nov 19 at 18:21
user3132457
856
856
Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
– Clement C.
Nov 19 at 18:24
isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
– user3132457
Nov 19 at 18:26
(also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
– Clement C.
Nov 19 at 18:26
@OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
– Clement C.
Nov 19 at 18:27
1
So it should be $${frac{e^n + e^{-n}}{2}}$$?
– user3132457
Nov 19 at 18:28
|
show 5 more comments
Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
– Clement C.
Nov 19 at 18:24
isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
– user3132457
Nov 19 at 18:26
(also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
– Clement C.
Nov 19 at 18:26
@OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
– Clement C.
Nov 19 at 18:27
1
So it should be $${frac{e^n + e^{-n}}{2}}$$?
– user3132457
Nov 19 at 18:28
Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
– Clement C.
Nov 19 at 18:24
Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
– Clement C.
Nov 19 at 18:24
isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
– user3132457
Nov 19 at 18:26
isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
– user3132457
Nov 19 at 18:26
(also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
– Clement C.
Nov 19 at 18:26
(also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
– Clement C.
Nov 19 at 18:26
@OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
– Clement C.
Nov 19 at 18:27
@OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
– Clement C.
Nov 19 at 18:27
1
1
So it should be $${frac{e^n + e^{-n}}{2}}$$?
– user3132457
Nov 19 at 18:28
So it should be $${frac{e^n + e^{-n}}{2}}$$?
– user3132457
Nov 19 at 18:28
|
show 5 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
You have two errors here. The first is that
$$
cosh n = frac{e^n + e^{-n}}{2}
$$
not what you wrote (which is $cos n$. To remember: $cosh$ is unbounded on $mathbb{R}$, so you have "true" exponentials, while $cos$ is bounded, so you have $e^{i x}$ which is bounded). So you need to compute
$$lim_{ntoinfty} frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} tag{1}$$ and not $lim_{ntoinfty} frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).
So let's compute (1):
$$
frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}}
= frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} =
= frac{e+e^{-(2n+1)}}{1+e^{-2n}} xrightarrow[ntoinfty]{} frac{e+0}{1+0} = e,.
$$
Therefore,
$$
lim_{ntoinfty} frac{a_{n+1}}{a_n} = zcdot e tag{2}
$$
which explains why the radius is $frac{1}{e}$.
answered Nov 19 at 18:36
Clement C.
49.2k33785
add a comment |
up vote
1
down vote
Hint:
We can use https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
far more easily,so that we need $$|ez|<1$$ and $$left|dfrac zeright|<1$$
answered Nov 19 at 18:51
lab bhattacharjee
222k15155273
The OP is using the ratio test.
– Clement C.
Nov 19 at 18:52
@ClementC, Please find updated answer
– lab bhattacharjee
Nov 19 at 19:05
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have two errors here. The first is that
$$
cosh n = frac{e^n + e^{-n}}{2}
$$
not what you wrote (which is $cos n$. To remember: $cosh$ is unbounded on $mathbb{R}$, so you have "true" exponentials, while $cos$ is bounded, so you have $e^{i x}$ which is bounded). So you need to compute
$$lim_{ntoinfty} frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} tag{1}$$ and not $lim_{ntoinfty} frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).
So let's compute (1):
$$
frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}}
= frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} =
= frac{e+e^{-(2n+1)}}{1+e^{-2n}} xrightarrow[ntoinfty]{} frac{e+0}{1+0} = e,.
$$
Therefore,
$$
lim_{ntoinfty} frac{a_{n+1}}{a_n} = zcdot e tag{2}
$$
which explains why the radius is $frac{1}{e}$.
answered Nov 19 at 18:36
Clement C.
49.2k33785
add a comment |
up vote
1
down vote
accepted
You have two errors here. The first is that
$$
cosh n = frac{e^n + e^{-n}}{2}
$$
not what you wrote (which is $cos n$. To remember: $cosh$ is unbounded on $mathbb{R}$, so you have "true" exponentials, while $cos$ is bounded, so you have $e^{i x}$ which is bounded). So you need to compute
$$lim_{ntoinfty} frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} tag{1}$$ and not $lim_{ntoinfty} frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).
So let's compute (1):
$$
frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}}
= frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} =
= frac{e+e^{-(2n+1)}}{1+e^{-2n}} xrightarrow[ntoinfty]{} frac{e+0}{1+0} = e,.
$$
Therefore,
$$
lim_{ntoinfty} frac{a_{n+1}}{a_n} = zcdot e tag{2}
$$
which explains why the radius is $frac{1}{e}$.
answered Nov 19 at 18:36
Clement C.
49.2k33785
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have two errors here. The first is that
$$
cosh n = frac{e^n + e^{-n}}{2}
$$
not what you wrote (which is $cos n$. To remember: $cosh$ is unbounded on $mathbb{R}$, so you have "true" exponentials, while $cos$ is bounded, so you have $e^{i x}$ which is bounded). So you need to compute
$$lim_{ntoinfty} frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} tag{1}$$ and not $lim_{ntoinfty} frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).
So let's compute (1):
$$
frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}}
= frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} =
= frac{e+e^{-(2n+1)}}{1+e^{-2n}} xrightarrow[ntoinfty]{} frac{e+0}{1+0} = e,.
$$
Therefore,
$$
lim_{ntoinfty} frac{a_{n+1}}{a_n} = zcdot e tag{2}
$$
which explains why the radius is $frac{1}{e}$.
answered Nov 19 at 18:36
Clement C.
49.2k33785
You have two errors here. The first is that
$$
cosh n = frac{e^n + e^{-n}}{2}
$$
not what you wrote (which is $cos n$. To remember: $cosh$ is unbounded on $mathbb{R}$, so you have "true" exponentials, while $cos$ is bounded, so you have $e^{i x}$ which is bounded). So you need to compute
$$lim_{ntoinfty} frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} tag{1}$$ and not $lim_{ntoinfty} frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).
So let's compute (1):
$$
frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}}
= frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} =
= frac{e+e^{-(2n+1)}}{1+e^{-2n}} xrightarrow[ntoinfty]{} frac{e+0}{1+0} = e,.
$$
Therefore,
$$
lim_{ntoinfty} frac{a_{n+1}}{a_n} = zcdot e tag{2}
$$
which explains why the radius is $frac{1}{e}$.
answered Nov 19 at 18:36
Clement C.
49.2k33785
answered Nov 19 at 18:36
Clement C.
49.2k33785
answered Nov 19 at 18:36
Clement C.
49.2k33785
answered Nov 19 at 18:36
Clement C.
49.2k33785
49.2k33785
add a comment |
add a comment |
up vote
1
down vote
Hint:
We can use https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
far more easily,so that we need $$|ez|<1$$ and $$left|dfrac zeright|<1$$
answered Nov 19 at 18:51
lab bhattacharjee
222k15155273
The OP is using the ratio test.
– Clement C.
Nov 19 at 18:52
@ClementC, Please find updated answer
– lab bhattacharjee
Nov 19 at 19:05
add a comment |
up vote
1
down vote
Hint:
We can use https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
far more easily,so that we need $$|ez|<1$$ and $$left|dfrac zeright|<1$$
answered Nov 19 at 18:51
lab bhattacharjee
222k15155273
The OP is using the ratio test.
– Clement C.
Nov 19 at 18:52
@ClementC, Please find updated answer
– lab bhattacharjee
Nov 19 at 19:05
add a comment |
up vote
1
down vote
up vote
1
down vote
Hint:
We can use https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
far more easily,so that we need $$|ez|<1$$ and $$left|dfrac zeright|<1$$
answered Nov 19 at 18:51
lab bhattacharjee
222k15155273
Hint:
We can use https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence
far more easily,so that we need $$|ez|<1$$ and $$left|dfrac zeright|<1$$
answered Nov 19 at 18:51
lab bhattacharjee
222k15155273
edited Nov 19 at 19:04
answered Nov 19 at 18:51
lab bhattacharjee
222k15155273
answered Nov 19 at 18:51
lab bhattacharjee
222k15155273
answered Nov 19 at 18:51
lab bhattacharjee
222k15155273
222k15155273
The OP is using the ratio test.
– Clement C.
Nov 19 at 18:52
@ClementC, Please find updated answer
– lab bhattacharjee
Nov 19 at 19:05
add a comment |
The OP is using the ratio test.
– Clement C.
Nov 19 at 18:52
@ClementC, Please find updated answer
– lab bhattacharjee
Nov 19 at 19:05
The OP is using the ratio test.
– Clement C.
Nov 19 at 18:52
The OP is using the ratio test.
– Clement C.
Nov 19 at 18:52
@ClementC, Please find updated answer
– lab bhattacharjee
Nov 19 at 19:05
@ClementC, Please find updated answer
– lab bhattacharjee
Nov 19 at 19:05
add a comment |
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Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
– Clement C.
Nov 19 at 18:24
isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
– user3132457
Nov 19 at 18:26
(also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
– Clement C.
Nov 19 at 18:26
@OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
– Clement C.
Nov 19 at 18:27
1
So it should be $${frac{e^n + e^{-n}}{2}}$$?
– user3132457
Nov 19 at 18:28