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Computing the limit of $lambda y_1 - log( sum_i=1^N a_i e^lambda y_i)$ as $lambda$ goes to negative infinity
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I am wondering if anyone can give me a hint to evaluate the following limit:
$lim_lambda rightarrow -infty lambda y_1 - log( sum_i=1^N a_i e^lambda y_i)$.
The book I'm reading says that it's equal to $-log a_1$. Here, $a_i > 0$ however, I do not know the sign of the $y_i's$ -- it is not specified. Furthermore, it can be assumed that $y_1 < y_2 < dots y_N$.
Clearly, if $y_2 > 0$ (implying that the remaining $y_i$'s except for $y_1$ are positive), I recover the above result. Otherwise, what tricks can I use to evaluate this limit?
calculus limits
asked Aug 2 at 22:19
Tomas Jorovic
1,55721325
add a comment |Â
up vote
1
down vote
favorite
I am wondering if anyone can give me a hint to evaluate the following limit:
$lim_lambda rightarrow -infty lambda y_1 - log( sum_i=1^N a_i e^lambda y_i)$.
The book I'm reading says that it's equal to $-log a_1$. Here, $a_i > 0$ however, I do not know the sign of the $y_i's$ -- it is not specified. Furthermore, it can be assumed that $y_1 < y_2 < dots y_N$.
Clearly, if $y_2 > 0$ (implying that the remaining $y_i$'s except for $y_1$ are positive), I recover the above result. Otherwise, what tricks can I use to evaluate this limit?
calculus limits
asked Aug 2 at 22:19
Tomas Jorovic
1,55721325
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am wondering if anyone can give me a hint to evaluate the following limit:
$lim_lambda rightarrow -infty lambda y_1 - log( sum_i=1^N a_i e^lambda y_i)$.
The book I'm reading says that it's equal to $-log a_1$. Here, $a_i > 0$ however, I do not know the sign of the $y_i's$ -- it is not specified. Furthermore, it can be assumed that $y_1 < y_2 < dots y_N$.
Clearly, if $y_2 > 0$ (implying that the remaining $y_i$'s except for $y_1$ are positive), I recover the above result. Otherwise, what tricks can I use to evaluate this limit?
calculus limits
asked Aug 2 at 22:19
Tomas Jorovic
1,55721325
I am wondering if anyone can give me a hint to evaluate the following limit:
$lim_lambda rightarrow -infty lambda y_1 - log( sum_i=1^N a_i e^lambda y_i)$.
The book I'm reading says that it's equal to $-log a_1$. Here, $a_i > 0$ however, I do not know the sign of the $y_i's$ -- it is not specified. Furthermore, it can be assumed that $y_1 < y_2 < dots y_N$.
Clearly, if $y_2 > 0$ (implying that the remaining $y_i$'s except for $y_1$ are positive), I recover the above result. Otherwise, what tricks can I use to evaluate this limit?
calculus limits
asked Aug 2 at 22:19
Tomas Jorovic
1,55721325
edited Aug 3 at 0:13
asked Aug 2 at 22:19
Tomas Jorovic
1,55721325
asked Aug 2 at 22:19
Tomas Jorovic
1,55721325
asked Aug 2 at 22:19
Tomas Jorovic
1,55721325
1,55721325
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
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0
down vote
accepted
I am assuming you meant $y_1<y_2<dots <y_n$, as if $y_1=y_2$ this is not necessarily true. (You can check it for yourself with e.g. $N=2$, $a_1=a_2=1$ and $y_1=y_2 = 1$.)
Rewrite
$$
lambda y_1 - logsum_i=1^N a_i e^lambda y_i
= log e^lambda y_1 - logsum_i=1^N a_i e^lambda y_i
= log frace^lambda y_1sum_i=1^N a_i e^lambda y_i
= log frac1a_1+sum_i=2^N a_i e^lambda(y_i-y_1)tag1
$$
Since $y_i-y_1>0$ for every $2leq ileq N$, we have
$$
lim_lambdato-inftysum_i=2^N a_i e^lambda(y_i-y_1) = 0tag2
$$
and therefore, from (1),
$$
lim_lambdato-infty lambda y_1 - logsum_i=1^N a_i e^lambda y_i = log frac1a_1 + lim_lambdato-inftysum_i=2^N a_i e^lambda(y_i-y_1) = -log a_1,. tag3
$$
answered Aug 2 at 22:28
Clement C.
46.9k33682
typo: $lambda_i - lambda_1$.
– Diger
Aug 2 at 22:32
@Diger Fixed, thanks.
– Clement C.
Aug 2 at 22:32
Thanks! I just forgot the technique that you can express $lambda y_1$ as $e^lambda y_1$! And yes, I made a mistake: the $y_i's$ are assumed distinct
– Tomas Jorovic
Aug 3 at 0:12
(With the log.) Yes, it's always useful to, weirdly, complicate something in order to simplify it.
– Clement C.
Aug 3 at 0:14
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I am assuming you meant $y_1<y_2<dots <y_n$, as if $y_1=y_2$ this is not necessarily true. (You can check it for yourself with e.g. $N=2$, $a_1=a_2=1$ and $y_1=y_2 = 1$.)
Rewrite
$$
lambda y_1 - logsum_i=1^N a_i e^lambda y_i
= log e^lambda y_1 - logsum_i=1^N a_i e^lambda y_i
= log frace^lambda y_1sum_i=1^N a_i e^lambda y_i
= log frac1a_1+sum_i=2^N a_i e^lambda(y_i-y_1)tag1
$$
Since $y_i-y_1>0$ for every $2leq ileq N$, we have
$$
lim_lambdato-inftysum_i=2^N a_i e^lambda(y_i-y_1) = 0tag2
$$
and therefore, from (1),
$$
lim_lambdato-infty lambda y_1 - logsum_i=1^N a_i e^lambda y_i = log frac1a_1 + lim_lambdato-inftysum_i=2^N a_i e^lambda(y_i-y_1) = -log a_1,. tag3
$$
answered Aug 2 at 22:28
Clement C.
46.9k33682
typo: $lambda_i - lambda_1$.
– Diger
Aug 2 at 22:32
@Diger Fixed, thanks.
– Clement C.
Aug 2 at 22:32
Thanks! I just forgot the technique that you can express $lambda y_1$ as $e^lambda y_1$! And yes, I made a mistake: the $y_i's$ are assumed distinct
– Tomas Jorovic
Aug 3 at 0:12
(With the log.) Yes, it's always useful to, weirdly, complicate something in order to simplify it.
– Clement C.
Aug 3 at 0:14
add a comment |Â
up vote
0
down vote
accepted
I am assuming you meant $y_1<y_2<dots <y_n$, as if $y_1=y_2$ this is not necessarily true. (You can check it for yourself with e.g. $N=2$, $a_1=a_2=1$ and $y_1=y_2 = 1$.)
Rewrite
$$
lambda y_1 - logsum_i=1^N a_i e^lambda y_i
= log e^lambda y_1 - logsum_i=1^N a_i e^lambda y_i
= log frace^lambda y_1sum_i=1^N a_i e^lambda y_i
= log frac1a_1+sum_i=2^N a_i e^lambda(y_i-y_1)tag1
$$
Since $y_i-y_1>0$ for every $2leq ileq N$, we have
$$
lim_lambdato-inftysum_i=2^N a_i e^lambda(y_i-y_1) = 0tag2
$$
and therefore, from (1),
$$
lim_lambdato-infty lambda y_1 - logsum_i=1^N a_i e^lambda y_i = log frac1a_1 + lim_lambdato-inftysum_i=2^N a_i e^lambda(y_i-y_1) = -log a_1,. tag3
$$
answered Aug 2 at 22:28
Clement C.
46.9k33682
typo: $lambda_i - lambda_1$.
– Diger
Aug 2 at 22:32
@Diger Fixed, thanks.
– Clement C.
Aug 2 at 22:32
Thanks! I just forgot the technique that you can express $lambda y_1$ as $e^lambda y_1$! And yes, I made a mistake: the $y_i's$ are assumed distinct
– Tomas Jorovic
Aug 3 at 0:12
(With the log.) Yes, it's always useful to, weirdly, complicate something in order to simplify it.
– Clement C.
Aug 3 at 0:14
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I am assuming you meant $y_1<y_2<dots <y_n$, as if $y_1=y_2$ this is not necessarily true. (You can check it for yourself with e.g. $N=2$, $a_1=a_2=1$ and $y_1=y_2 = 1$.)
Rewrite
$$
lambda y_1 - logsum_i=1^N a_i e^lambda y_i
= log e^lambda y_1 - logsum_i=1^N a_i e^lambda y_i
= log frace^lambda y_1sum_i=1^N a_i e^lambda y_i
= log frac1a_1+sum_i=2^N a_i e^lambda(y_i-y_1)tag1
$$
Since $y_i-y_1>0$ for every $2leq ileq N$, we have
$$
lim_lambdato-inftysum_i=2^N a_i e^lambda(y_i-y_1) = 0tag2
$$
and therefore, from (1),
$$
lim_lambdato-infty lambda y_1 - logsum_i=1^N a_i e^lambda y_i = log frac1a_1 + lim_lambdato-inftysum_i=2^N a_i e^lambda(y_i-y_1) = -log a_1,. tag3
$$
answered Aug 2 at 22:28
Clement C.
46.9k33682
I am assuming you meant $y_1<y_2<dots <y_n$, as if $y_1=y_2$ this is not necessarily true. (You can check it for yourself with e.g. $N=2$, $a_1=a_2=1$ and $y_1=y_2 = 1$.)
Rewrite
$$
lambda y_1 - logsum_i=1^N a_i e^lambda y_i
= log e^lambda y_1 - logsum_i=1^N a_i e^lambda y_i
= log frace^lambda y_1sum_i=1^N a_i e^lambda y_i
= log frac1a_1+sum_i=2^N a_i e^lambda(y_i-y_1)tag1
$$
Since $y_i-y_1>0$ for every $2leq ileq N$, we have
$$
lim_lambdato-inftysum_i=2^N a_i e^lambda(y_i-y_1) = 0tag2
$$
and therefore, from (1),
$$
lim_lambdato-infty lambda y_1 - logsum_i=1^N a_i e^lambda y_i = log frac1a_1 + lim_lambdato-inftysum_i=2^N a_i e^lambda(y_i-y_1) = -log a_1,. tag3
$$
answered Aug 2 at 22:28
Clement C.
46.9k33682
edited Aug 2 at 22:32
answered Aug 2 at 22:28
Clement C.
46.9k33682
answered Aug 2 at 22:28
Clement C.
46.9k33682
answered Aug 2 at 22:28
Clement C.
46.9k33682
46.9k33682
typo: $lambda_i - lambda_1$.
– Diger
Aug 2 at 22:32
@Diger Fixed, thanks.
– Clement C.
Aug 2 at 22:32
Thanks! I just forgot the technique that you can express $lambda y_1$ as $e^lambda y_1$! And yes, I made a mistake: the $y_i's$ are assumed distinct
– Tomas Jorovic
Aug 3 at 0:12
(With the log.) Yes, it's always useful to, weirdly, complicate something in order to simplify it.
– Clement C.
Aug 3 at 0:14
add a comment |Â
typo: $lambda_i - lambda_1$.
– Diger
Aug 2 at 22:32
@Diger Fixed, thanks.
– Clement C.
Aug 2 at 22:32
Thanks! I just forgot the technique that you can express $lambda y_1$ as $e^lambda y_1$! And yes, I made a mistake: the $y_i's$ are assumed distinct
– Tomas Jorovic
Aug 3 at 0:12
(With the log.) Yes, it's always useful to, weirdly, complicate something in order to simplify it.
– Clement C.
Aug 3 at 0:14
typo: $lambda_i - lambda_1$.
– Diger
Aug 2 at 22:32
typo: $lambda_i - lambda_1$.
– Diger
Aug 2 at 22:32
@Diger Fixed, thanks.
– Clement C.
Aug 2 at 22:32
@Diger Fixed, thanks.
– Clement C.
Aug 2 at 22:32
Thanks! I just forgot the technique that you can express $lambda y_1$ as $e^lambda y_1$! And yes, I made a mistake: the $y_i's$ are assumed distinct
– Tomas Jorovic
Aug 3 at 0:12
Thanks! I just forgot the technique that you can express $lambda y_1$ as $e^lambda y_1$! And yes, I made a mistake: the $y_i's$ are assumed distinct
– Tomas Jorovic
Aug 3 at 0:12
(With the log.) Yes, it's always useful to, weirdly, complicate something in order to simplify it.
– Clement C.
Aug 3 at 0:14
(With the log.) Yes, it's always useful to, weirdly, complicate something in order to simplify it.
– Clement C.
Aug 3 at 0:14
add a comment |Â
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