Mixing
Convergence of an exponent function
$begingroup$
Following the answer to this MSE question, it is claimed that we easily show the following convergence:
$$ text{exp}(-sqrt{lambda}it + lambda (e^{it/sqrt{lambda}}-1)) rightarrow_{lambda rightarrow infty} expleft(-frac{t^2}{2}right)$$
Which is proved by noting that the following expression holds:
$$ expleft(frac{it}{sqrt{lambda}} right) -1 = frac{it}{sqrt{lambda}} - frac{t^2}{2lambda} + o(lambda^{-1})$$
While I understand how the former was derived, my question is how does this prove the convergence?
Since in this case,
$$lambda (e^{it/sqrt{lambda}}-1) =lambdafrac{it}{sqrt{lambda}} - frac{t^2}{2} + o(lambda^{-1/2}) $$
And I can't see why the $sqrt{lambda} it$ member disappears at the limit here. Does it have to do anything with the imaginary part?
What am I missing?
calculus probability-theory convergence uniform-convergence
asked Jan 19 at 22:01
NutleNutle
320110
$endgroup$
add a comment |
$begingroup$
Following the answer to this MSE question, it is claimed that we easily show the following convergence:
$$ text{exp}(-sqrt{lambda}it + lambda (e^{it/sqrt{lambda}}-1)) rightarrow_{lambda rightarrow infty} expleft(-frac{t^2}{2}right)$$
Which is proved by noting that the following expression holds:
$$ expleft(frac{it}{sqrt{lambda}} right) -1 = frac{it}{sqrt{lambda}} - frac{t^2}{2lambda} + o(lambda^{-1})$$
While I understand how the former was derived, my question is how does this prove the convergence?
Since in this case,
$$lambda (e^{it/sqrt{lambda}}-1) =lambdafrac{it}{sqrt{lambda}} - frac{t^2}{2} + o(lambda^{-1/2}) $$
And I can't see why the $sqrt{lambda} it$ member disappears at the limit here. Does it have to do anything with the imaginary part?
What am I missing?
calculus probability-theory convergence uniform-convergence
asked Jan 19 at 22:01
NutleNutle
320110
$endgroup$
add a comment |
$begingroup$
Following the answer to this MSE question, it is claimed that we easily show the following convergence:
$$ text{exp}(-sqrt{lambda}it + lambda (e^{it/sqrt{lambda}}-1)) rightarrow_{lambda rightarrow infty} expleft(-frac{t^2}{2}right)$$
Which is proved by noting that the following expression holds:
$$ expleft(frac{it}{sqrt{lambda}} right) -1 = frac{it}{sqrt{lambda}} - frac{t^2}{2lambda} + o(lambda^{-1})$$
While I understand how the former was derived, my question is how does this prove the convergence?
Since in this case,
$$lambda (e^{it/sqrt{lambda}}-1) =lambdafrac{it}{sqrt{lambda}} - frac{t^2}{2} + o(lambda^{-1/2}) $$
And I can't see why the $sqrt{lambda} it$ member disappears at the limit here. Does it have to do anything with the imaginary part?
What am I missing?
calculus probability-theory convergence uniform-convergence
asked Jan 19 at 22:01
NutleNutle
320110
$endgroup$
Following the answer to this MSE question, it is claimed that we easily show the following convergence:
$$ text{exp}(-sqrt{lambda}it + lambda (e^{it/sqrt{lambda}}-1)) rightarrow_{lambda rightarrow infty} expleft(-frac{t^2}{2}right)$$
Which is proved by noting that the following expression holds:
$$ expleft(frac{it}{sqrt{lambda}} right) -1 = frac{it}{sqrt{lambda}} - frac{t^2}{2lambda} + o(lambda^{-1})$$
While I understand how the former was derived, my question is how does this prove the convergence?
Since in this case,
$$lambda (e^{it/sqrt{lambda}}-1) =lambdafrac{it}{sqrt{lambda}} - frac{t^2}{2} + o(lambda^{-1/2}) $$
And I can't see why the $sqrt{lambda} it$ member disappears at the limit here. Does it have to do anything with the imaginary part?
What am I missing?
calculus probability-theory convergence uniform-convergence
calculus probability-theory convergence uniform-convergence
asked Jan 19 at 22:01
NutleNutle
320110
asked Jan 19 at 22:01
NutleNutle
320110
asked Jan 19 at 22:01
NutleNutle
320110
asked Jan 19 at 22:01
NutleNutle
320110
asked Jan 19 at 22:01
NutleNutle
320110
320110
add a comment |
add a comment |
1 Answer
1
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$begingroup$
It follows that
$$
exp(-sqrt{lambda}it + lambda (e^{it/sqrt{lambda}}-1))=exp(-sqrt{lambda}it+sqrt{lambda}it-frac{t^2}{2}+lambda o(lambda^{-1}))=exp(-frac{t^2}{2}+lambda o(lambda^{-1})).
$$
We note that
$$
lambda o(lambda^{-1})stackrel{lambdatoinfty}{to}0 implies exp(-frac{t^2}{2}+lambda o(lambda^{-1}))stackrel{lambdatoinfty}{to}exp(-t^2/2)
$$
by continuity of the exponential function.
answered Jan 19 at 22:18
Foobaz JohnFoobaz John
22.3k41452
$endgroup$
$begingroup$
Oh, that's right, the other $-sqrt{lambda}it$ was right in front of me.. Discarded it too soon. Thanks!
$endgroup$
– Nutle
Jan 19 at 22:30
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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$begingroup$
It follows that
$$
exp(-sqrt{lambda}it + lambda (e^{it/sqrt{lambda}}-1))=exp(-sqrt{lambda}it+sqrt{lambda}it-frac{t^2}{2}+lambda o(lambda^{-1}))=exp(-frac{t^2}{2}+lambda o(lambda^{-1})).
$$
We note that
$$
lambda o(lambda^{-1})stackrel{lambdatoinfty}{to}0 implies exp(-frac{t^2}{2}+lambda o(lambda^{-1}))stackrel{lambdatoinfty}{to}exp(-t^2/2)
$$
by continuity of the exponential function.
answered Jan 19 at 22:18
Foobaz JohnFoobaz John
22.3k41452
$endgroup$
$begingroup$
Oh, that's right, the other $-sqrt{lambda}it$ was right in front of me.. Discarded it too soon. Thanks!
$endgroup$
– Nutle
Jan 19 at 22:30
add a comment |
$begingroup$
It follows that
$$
exp(-sqrt{lambda}it + lambda (e^{it/sqrt{lambda}}-1))=exp(-sqrt{lambda}it+sqrt{lambda}it-frac{t^2}{2}+lambda o(lambda^{-1}))=exp(-frac{t^2}{2}+lambda o(lambda^{-1})).
$$
We note that
$$
lambda o(lambda^{-1})stackrel{lambdatoinfty}{to}0 implies exp(-frac{t^2}{2}+lambda o(lambda^{-1}))stackrel{lambdatoinfty}{to}exp(-t^2/2)
$$
by continuity of the exponential function.
answered Jan 19 at 22:18
Foobaz JohnFoobaz John
22.3k41452
$endgroup$
$begingroup$
Oh, that's right, the other $-sqrt{lambda}it$ was right in front of me.. Discarded it too soon. Thanks!
$endgroup$
– Nutle
Jan 19 at 22:30
add a comment |
$begingroup$
It follows that
$$
exp(-sqrt{lambda}it + lambda (e^{it/sqrt{lambda}}-1))=exp(-sqrt{lambda}it+sqrt{lambda}it-frac{t^2}{2}+lambda o(lambda^{-1}))=exp(-frac{t^2}{2}+lambda o(lambda^{-1})).
$$
We note that
$$
lambda o(lambda^{-1})stackrel{lambdatoinfty}{to}0 implies exp(-frac{t^2}{2}+lambda o(lambda^{-1}))stackrel{lambdatoinfty}{to}exp(-t^2/2)
$$
by continuity of the exponential function.
answered Jan 19 at 22:18
Foobaz JohnFoobaz John
22.3k41452
$endgroup$
It follows that
$$
exp(-sqrt{lambda}it + lambda (e^{it/sqrt{lambda}}-1))=exp(-sqrt{lambda}it+sqrt{lambda}it-frac{t^2}{2}+lambda o(lambda^{-1}))=exp(-frac{t^2}{2}+lambda o(lambda^{-1})).
$$
We note that
$$
lambda o(lambda^{-1})stackrel{lambdatoinfty}{to}0 implies exp(-frac{t^2}{2}+lambda o(lambda^{-1}))stackrel{lambdatoinfty}{to}exp(-t^2/2)
$$
by continuity of the exponential function.
answered Jan 19 at 22:18
Foobaz JohnFoobaz John
22.3k41452
answered Jan 19 at 22:18
Foobaz JohnFoobaz John
22.3k41452
answered Jan 19 at 22:18
Foobaz JohnFoobaz John
22.3k41452
answered Jan 19 at 22:18
Foobaz JohnFoobaz John
22.3k41452
22.3k41452
$begingroup$
Oh, that's right, the other $-sqrt{lambda}it$ was right in front of me.. Discarded it too soon. Thanks!
$endgroup$
– Nutle
Jan 19 at 22:30
add a comment |
$begingroup$
Oh, that's right, the other $-sqrt{lambda}it$ was right in front of me.. Discarded it too soon. Thanks!
$endgroup$
– Nutle
Jan 19 at 22:30
$begingroup$
Oh, that's right, the other $-sqrt{lambda}it$ was right in front of me.. Discarded it too soon. Thanks!
$endgroup$
– Nutle
Jan 19 at 22:30
$begingroup$
Oh, that's right, the other $-sqrt{lambda}it$ was right in front of me.. Discarded it too soon. Thanks!
$endgroup$
– Nutle
Jan 19 at 22:30
add a comment |
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