Roots of a real exponential sum
0
Suppose I had some exponential sum $ f(x) $ of the form:
$$f(x) = sum_{i=1}^{N} left( c_i e^{a_i x} right)$$
where:
$$c_i, a_i in R$$
$$a_i leq 0$$
Is there a quick way to find the roots, $ f(x) = 0 $?
real-analysis optimization numerical-optimization exponential-sum
asked Dec 2 '18 at 6:49
Michael Colomb
11
|
show 1 more comment
0
Suppose I had some exponential sum $ f(x) $ of the form:
$$f(x) = sum_{i=1}^{N} left( c_i e^{a_i x} right)$$
where:
$$c_i, a_i in R$$
$$a_i leq 0$$
Is there a quick way to find the roots, $ f(x) = 0 $?
real-analysis optimization numerical-optimization exponential-sum
asked Dec 2 '18 at 6:49
Michael Colomb
11
Consider the special case where $a_i = i-1$, so that $f(log x) = sum_{i=1}^{N} c_i x^{i-1}$ with $x>0$. This tells that the problem is at least as hard as determining positive zeros of arbitrary polynomial, which is already extremely hard and almost impossible to come up with a master answer.
– Sangchul Lee
Dec 2 '18 at 6:54
If I understand correctly, this actually isn't a special case, because $a_i leq 0$, right? But I suppose you could just define $a_i := 1 - i$. Hmm...
– Michael Colomb
Dec 2 '18 at 7:02
As you pointed out, we may still consider the case $a_i = 1-i$ and my point of opposition remains the same.
– Sangchul Lee
Dec 2 '18 at 7:27
If we had some known $lambda_i$ for each $i$, would it help to know that $sum_{i = 1}^N c_i / lambda_i = 1$?
– Michael Colomb
Dec 2 '18 at 8:09
and $lambda_i in [-1,1]$.
– Michael Colomb
Dec 2 '18 at 8:22
|
show 1 more comment
0
0
0
Suppose I had some exponential sum $ f(x) $ of the form:
$$f(x) = sum_{i=1}^{N} left( c_i e^{a_i x} right)$$
where:
$$c_i, a_i in R$$
$$a_i leq 0$$
Is there a quick way to find the roots, $ f(x) = 0 $?
real-analysis optimization numerical-optimization exponential-sum
asked Dec 2 '18 at 6:49
Michael Colomb
11
Suppose I had some exponential sum $ f(x) $ of the form:
$$f(x) = sum_{i=1}^{N} left( c_i e^{a_i x} right)$$
where:
$$c_i, a_i in R$$
$$a_i leq 0$$
Is there a quick way to find the roots, $ f(x) = 0 $?
real-analysis optimization numerical-optimization exponential-sum
real-analysis optimization numerical-optimization exponential-sum
asked Dec 2 '18 at 6:49
Michael Colomb
11
asked Dec 2 '18 at 6:49
Michael Colomb
11
edited Dec 2 '18 at 7:04
asked Dec 2 '18 at 6:49
Michael Colomb
11
asked Dec 2 '18 at 6:49
Michael Colomb
11
asked Dec 2 '18 at 6:49
Michael Colomb
11
11
Consider the special case where $a_i = i-1$, so that $f(log x) = sum_{i=1}^{N} c_i x^{i-1}$ with $x>0$. This tells that the problem is at least as hard as determining positive zeros of arbitrary polynomial, which is already extremely hard and almost impossible to come up with a master answer.
– Sangchul Lee
Dec 2 '18 at 6:54
If I understand correctly, this actually isn't a special case, because $a_i leq 0$, right? But I suppose you could just define $a_i := 1 - i$. Hmm...
– Michael Colomb
Dec 2 '18 at 7:02
As you pointed out, we may still consider the case $a_i = 1-i$ and my point of opposition remains the same.
– Sangchul Lee
Dec 2 '18 at 7:27
If we had some known $lambda_i$ for each $i$, would it help to know that $sum_{i = 1}^N c_i / lambda_i = 1$?
– Michael Colomb
Dec 2 '18 at 8:09
and $lambda_i in [-1,1]$.
– Michael Colomb
Dec 2 '18 at 8:22
|
show 1 more comment
Consider the special case where $a_i = i-1$, so that $f(log x) = sum_{i=1}^{N} c_i x^{i-1}$ with $x>0$. This tells that the problem is at least as hard as determining positive zeros of arbitrary polynomial, which is already extremely hard and almost impossible to come up with a master answer.
– Sangchul Lee
Dec 2 '18 at 6:54
If I understand correctly, this actually isn't a special case, because $a_i leq 0$, right? But I suppose you could just define $a_i := 1 - i$. Hmm...
– Michael Colomb
Dec 2 '18 at 7:02
As you pointed out, we may still consider the case $a_i = 1-i$ and my point of opposition remains the same.
– Sangchul Lee
Dec 2 '18 at 7:27
If we had some known $lambda_i$ for each $i$, would it help to know that $sum_{i = 1}^N c_i / lambda_i = 1$?
– Michael Colomb
Dec 2 '18 at 8:09
and $lambda_i in [-1,1]$.
– Michael Colomb
Dec 2 '18 at 8:22
Consider the special case where $a_i = i-1$, so that $f(log x) = sum_{i=1}^{N} c_i x^{i-1}$ with $x>0$. This tells that the problem is at least as hard as determining positive zeros of arbitrary polynomial, which is already extremely hard and almost impossible to come up with a master answer.
– Sangchul Lee
Dec 2 '18 at 6:54
Consider the special case where $a_i = i-1$, so that $f(log x) = sum_{i=1}^{N} c_i x^{i-1}$ with $x>0$. This tells that the problem is at least as hard as determining positive zeros of arbitrary polynomial, which is already extremely hard and almost impossible to come up with a master answer.
– Sangchul Lee
Dec 2 '18 at 6:54
If I understand correctly, this actually isn't a special case, because $a_i leq 0$, right? But I suppose you could just define $a_i := 1 - i$. Hmm...
– Michael Colomb
Dec 2 '18 at 7:02
If I understand correctly, this actually isn't a special case, because $a_i leq 0$, right? But I suppose you could just define $a_i := 1 - i$. Hmm...
– Michael Colomb
Dec 2 '18 at 7:02
As you pointed out, we may still consider the case $a_i = 1-i$ and my point of opposition remains the same.
– Sangchul Lee
Dec 2 '18 at 7:27
As you pointed out, we may still consider the case $a_i = 1-i$ and my point of opposition remains the same.
– Sangchul Lee
Dec 2 '18 at 7:27
If we had some known $lambda_i$ for each $i$, would it help to know that $sum_{i = 1}^N c_i / lambda_i = 1$?
– Michael Colomb
Dec 2 '18 at 8:09
If we had some known $lambda_i$ for each $i$, would it help to know that $sum_{i = 1}^N c_i / lambda_i = 1$?
– Michael Colomb
Dec 2 '18 at 8:09
and $lambda_i in [-1,1]$.
– Michael Colomb
Dec 2 '18 at 8:22
and $lambda_i in [-1,1]$.
– Michael Colomb
Dec 2 '18 at 8:22
|
show 1 more comment
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Consider the special case where $a_i = i-1$, so that $f(log x) = sum_{i=1}^{N} c_i x^{i-1}$ with $x>0$. This tells that the problem is at least as hard as determining positive zeros of arbitrary polynomial, which is already extremely hard and almost impossible to come up with a master answer.
– Sangchul Lee
Dec 2 '18 at 6:54
If I understand correctly, this actually isn't a special case, because $a_i leq 0$, right? But I suppose you could just define $a_i := 1 - i$. Hmm...
– Michael Colomb
Dec 2 '18 at 7:02
As you pointed out, we may still consider the case $a_i = 1-i$ and my point of opposition remains the same.
– Sangchul Lee
Dec 2 '18 at 7:27
If we had some known $lambda_i$ for each $i$, would it help to know that $sum_{i = 1}^N c_i / lambda_i = 1$?
– Michael Colomb
Dec 2 '18 at 8:09
and $lambda_i in [-1,1]$.
– Michael Colomb
Dec 2 '18 at 8:22