How to evaluate $ lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n $ where $i=sqrt{-1}$?
up vote
0
down vote
favorite
How to evaluate the following limit?
$$
lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
$$
Here $i=sqrt{-1}$.
I got:
$$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
= lim_{nto infty} frac{(i-1)^n}{n(sqrt{2})^n}
$$
I know the lower part goes to infinity but what to do with the upper part? Is that usefull to use squeeze theorem or is there any simplier way?
sequences-and-series limits complex-numbers
edited Nov 24 at 23:16
user587192
1,295112
asked Nov 24 at 21:35
L. Spy
12
|
show 2 more comments
up vote
0
down vote
favorite
How to evaluate the following limit?
$$
lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
$$
Here $i=sqrt{-1}$.
I got:
$$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
= lim_{nto infty} frac{(i-1)^n}{n(sqrt{2})^n}
$$
I know the lower part goes to infinity but what to do with the upper part? Is that usefull to use squeeze theorem or is there any simplier way?
sequences-and-series limits complex-numbers
edited Nov 24 at 23:16
user587192
1,295112
asked Nov 24 at 21:35
L. Spy
12
1
It's very hard to understand what you wrote. Try mathjax
– DonAntonio
Nov 24 at 21:38
Express $(1+i)/sqrt{2}=exp(ipi/4)$.
– Diger
Nov 24 at 21:39
1
L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as$sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written$lim_{nto infty} f(x)$. Just a few pointers.
– amWhy
Nov 24 at 21:43
1
If your function $f(x)$ is a fraction, you can write it as follows$f(x) = frac{"numerator here"}{"denominator here"}$
– amWhy
Nov 24 at 21:46
Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
– Shubham Johri
Nov 24 at 21:57
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to evaluate the following limit?
$$
lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
$$
Here $i=sqrt{-1}$.
I got:
$$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
= lim_{nto infty} frac{(i-1)^n}{n(sqrt{2})^n}
$$
I know the lower part goes to infinity but what to do with the upper part? Is that usefull to use squeeze theorem or is there any simplier way?
sequences-and-series limits complex-numbers
edited Nov 24 at 23:16
user587192
1,295112
asked Nov 24 at 21:35
L. Spy
12
How to evaluate the following limit?
$$
lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
$$
Here $i=sqrt{-1}$.
I got:
$$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
= lim_{nto infty} frac{(i-1)^n}{n(sqrt{2})^n}
$$
I know the lower part goes to infinity but what to do with the upper part? Is that usefull to use squeeze theorem or is there any simplier way?
sequences-and-series limits complex-numbers
sequences-and-series limits complex-numbers
edited Nov 24 at 23:16
user587192
1,295112
asked Nov 24 at 21:35
L. Spy
12
edited Nov 24 at 23:16
user587192
1,295112
asked Nov 24 at 21:35
L. Spy
12
edited Nov 24 at 23:16
user587192
1,295112
edited Nov 24 at 23:16
user587192
1,295112
edited Nov 24 at 23:16
user587192
1,295112
1,295112
asked Nov 24 at 21:35
L. Spy
12
asked Nov 24 at 21:35
L. Spy
12
asked Nov 24 at 21:35
L. Spy
12
12
1
It's very hard to understand what you wrote. Try mathjax
– DonAntonio
Nov 24 at 21:38
Express $(1+i)/sqrt{2}=exp(ipi/4)$.
– Diger
Nov 24 at 21:39
1
L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as$sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written$lim_{nto infty} f(x)$. Just a few pointers.
– amWhy
Nov 24 at 21:43
1
If your function $f(x)$ is a fraction, you can write it as follows$f(x) = frac{"numerator here"}{"denominator here"}$
– amWhy
Nov 24 at 21:46
Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
– Shubham Johri
Nov 24 at 21:57
|
show 2 more comments
1
It's very hard to understand what you wrote. Try mathjax
– DonAntonio
Nov 24 at 21:38
Express $(1+i)/sqrt{2}=exp(ipi/4)$.
– Diger
Nov 24 at 21:39
1
L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as$sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written$lim_{nto infty} f(x)$. Just a few pointers.
– amWhy
Nov 24 at 21:43
1
If your function $f(x)$ is a fraction, you can write it as follows$f(x) = frac{"numerator here"}{"denominator here"}$
– amWhy
Nov 24 at 21:46
Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
– Shubham Johri
Nov 24 at 21:57
1
1
It's very hard to understand what you wrote. Try mathjax
– DonAntonio
Nov 24 at 21:38
It's very hard to understand what you wrote. Try mathjax
– DonAntonio
Nov 24 at 21:38
Express $(1+i)/sqrt{2}=exp(ipi/4)$.
– Diger
Nov 24 at 21:39
Express $(1+i)/sqrt{2}=exp(ipi/4)$.
– Diger
Nov 24 at 21:39
1
1
L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as
$sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written $lim_{nto infty} f(x)$. Just a few pointers.– amWhy
Nov 24 at 21:43
L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as
$sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written $lim_{nto infty} f(x)$. Just a few pointers.– amWhy
Nov 24 at 21:43
1
1
If your function $f(x)$ is a fraction, you can write it as follows
$f(x) = frac{"numerator here"}{"denominator here"}$– amWhy
Nov 24 at 21:46
If your function $f(x)$ is a fraction, you can write it as follows
$f(x) = frac{"numerator here"}{"denominator here"}$– amWhy
Nov 24 at 21:46
Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
– Shubham Johri
Nov 24 at 21:57
Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
– Shubham Johri
Nov 24 at 21:57
|
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
We have
$$
left|frac{1+i}{sqrt{2}}right|=1
$$
Thus
$$
left|frac{i}{n}left(frac{1+i}{sqrt{2}}right)^{!n}right|=frac{1}{n}
$$
answered Nov 24 at 22:53
egreg
175k1383198
egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
– L. Spy
Nov 25 at 9:25
@L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
– egreg
Nov 25 at 9:40
add a comment |
up vote
1
down vote
$$lim_{nto infty} |frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n|=lim_{nto infty} {1over n}=0$$therefore $$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n=0$$
answered Nov 28 at 0:31
Mostafa Ayaz
13.3k3836
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
We have
$$
left|frac{1+i}{sqrt{2}}right|=1
$$
Thus
$$
left|frac{i}{n}left(frac{1+i}{sqrt{2}}right)^{!n}right|=frac{1}{n}
$$
answered Nov 24 at 22:53
egreg
175k1383198
egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
– L. Spy
Nov 25 at 9:25
@L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
– egreg
Nov 25 at 9:40
add a comment |
up vote
4
down vote
We have
$$
left|frac{1+i}{sqrt{2}}right|=1
$$
Thus
$$
left|frac{i}{n}left(frac{1+i}{sqrt{2}}right)^{!n}right|=frac{1}{n}
$$
answered Nov 24 at 22:53
egreg
175k1383198
egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
– L. Spy
Nov 25 at 9:25
@L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
– egreg
Nov 25 at 9:40
add a comment |
up vote
4
down vote
up vote
4
down vote
We have
$$
left|frac{1+i}{sqrt{2}}right|=1
$$
Thus
$$
left|frac{i}{n}left(frac{1+i}{sqrt{2}}right)^{!n}right|=frac{1}{n}
$$
answered Nov 24 at 22:53
egreg
175k1383198
We have
$$
left|frac{1+i}{sqrt{2}}right|=1
$$
Thus
$$
left|frac{i}{n}left(frac{1+i}{sqrt{2}}right)^{!n}right|=frac{1}{n}
$$
answered Nov 24 at 22:53
egreg
175k1383198
answered Nov 24 at 22:53
egreg
175k1383198
answered Nov 24 at 22:53
egreg
175k1383198
answered Nov 24 at 22:53
egreg
175k1383198
175k1383198
egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
– L. Spy
Nov 25 at 9:25
@L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
– egreg
Nov 25 at 9:40
add a comment |
egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
– L. Spy
Nov 25 at 9:25
@L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
– egreg
Nov 25 at 9:40
egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
– L. Spy
Nov 25 at 9:25
egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
– L. Spy
Nov 25 at 9:25
@L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
– egreg
Nov 25 at 9:40
@L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
– egreg
Nov 25 at 9:40
add a comment |
up vote
1
down vote
$$lim_{nto infty} |frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n|=lim_{nto infty} {1over n}=0$$therefore $$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n=0$$
answered Nov 28 at 0:31
Mostafa Ayaz
13.3k3836
add a comment |
up vote
1
down vote
$$lim_{nto infty} |frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n|=lim_{nto infty} {1over n}=0$$therefore $$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n=0$$
answered Nov 28 at 0:31
Mostafa Ayaz
13.3k3836
add a comment |
up vote
1
down vote
up vote
1
down vote
$$lim_{nto infty} |frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n|=lim_{nto infty} {1over n}=0$$therefore $$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n=0$$
answered Nov 28 at 0:31
Mostafa Ayaz
13.3k3836
$$lim_{nto infty} |frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n|=lim_{nto infty} {1over n}=0$$therefore $$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n=0$$
answered Nov 28 at 0:31
Mostafa Ayaz
13.3k3836
answered Nov 28 at 0:31
Mostafa Ayaz
13.3k3836
answered Nov 28 at 0:31
Mostafa Ayaz
13.3k3836
answered Nov 28 at 0:31
Mostafa Ayaz
13.3k3836
13.3k3836
add a comment |
add a comment |
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1
It's very hard to understand what you wrote. Try mathjax
– DonAntonio
Nov 24 at 21:38
Express $(1+i)/sqrt{2}=exp(ipi/4)$.
– Diger
Nov 24 at 21:39
1
L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as
$sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written$lim_{nto infty} f(x)$. Just a few pointers.– amWhy
Nov 24 at 21:43
1
If your function $f(x)$ is a fraction, you can write it as follows
$f(x) = frac{"numerator here"}{"denominator here"}$– amWhy
Nov 24 at 21:46
Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
– Shubham Johri
Nov 24 at 21:57