Mixing
Expressing $c = 1 - expleft(lambda_1 p + lambda_3 qright)$ as a product of two terms
$begingroup$
This question is motivated by the response provided in this question
Considering the same equation which is shown below
$$c = 1 - expleft(lambda_1 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right) + lambda_3 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right)right)$$
It has been proven that it cannot be expressed as a sum. Still, I am wondering if I can express this as a product like $c = x * y$, where $x$ is in terms of $lambda_1$, and $y$ is in terms of $lambda_3 $? I will appreciate insights on this matter.
Pardon, I am asking a similar question, but as you can see, the dimensions are different.
exponential-function exponential-sum
asked Jan 12 at 3:17
AbdulhameedAbdulhameed
153115
$endgroup$
add a comment |
$begingroup$
This question is motivated by the response provided in this question
Considering the same equation which is shown below
$$c = 1 - expleft(lambda_1 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right) + lambda_3 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right)right)$$
It has been proven that it cannot be expressed as a sum. Still, I am wondering if I can express this as a product like $c = x * y$, where $x$ is in terms of $lambda_1$, and $y$ is in terms of $lambda_3 $? I will appreciate insights on this matter.
Pardon, I am asking a similar question, but as you can see, the dimensions are different.
exponential-function exponential-sum
asked Jan 12 at 3:17
AbdulhameedAbdulhameed
153115
$endgroup$
add a comment |
$begingroup$
This question is motivated by the response provided in this question
Considering the same equation which is shown below
$$c = 1 - expleft(lambda_1 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right) + lambda_3 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right)right)$$
It has been proven that it cannot be expressed as a sum. Still, I am wondering if I can express this as a product like $c = x * y$, where $x$ is in terms of $lambda_1$, and $y$ is in terms of $lambda_3 $? I will appreciate insights on this matter.
Pardon, I am asking a similar question, but as you can see, the dimensions are different.
exponential-function exponential-sum
asked Jan 12 at 3:17
AbdulhameedAbdulhameed
153115
$endgroup$
This question is motivated by the response provided in this question
Considering the same equation which is shown below
$$c = 1 - expleft(lambda_1 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right) + lambda_3 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right)right)$$
It has been proven that it cannot be expressed as a sum. Still, I am wondering if I can express this as a product like $c = x * y$, where $x$ is in terms of $lambda_1$, and $y$ is in terms of $lambda_3 $? I will appreciate insights on this matter.
Pardon, I am asking a similar question, but as you can see, the dimensions are different.
exponential-function exponential-sum
exponential-function exponential-sum
asked Jan 12 at 3:17
AbdulhameedAbdulhameed
153115
asked Jan 12 at 3:17
AbdulhameedAbdulhameed
153115
edited Jan 12 at 3:31
Abdulhameed
asked Jan 12 at 3:17
AbdulhameedAbdulhameed
153115
asked Jan 12 at 3:17
AbdulhameedAbdulhameed
153115
asked Jan 12 at 3:17
AbdulhameedAbdulhameed
153115
153115
add a comment |
add a comment |
1 Answer
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$begingroup$
Similarly as for the additive case, this is not possible unless $lambda_1=0$ or $lambda_3=0$ or $R=0$.
Suppose there exist two functions $x$ and $y$ such that
$$x(lambda_1)y(lambda_3)=1 - expleft(lambda_1 p + lambda_3 qright)$$
Then note that $x(0)y(0)=1-1=0$.
Also
$$ x(lambda_1)y(0)=1-e^{lambda_1p}$$
$$ x(0)y(lambda_3)=1-e^{lambda_3q}$$
Multiplying the two yields
$$ x(lambda_1)y(lambda_3)x(0)y(0)=(1-e^{lambda_1p})(1-e^{lambda_3q})$$
But since $x(0)y(0)=0$, that means
$$(1-e^{lambda_1p})(1-e^{lambda_3q})=0$$
which implies that $lambda_1=0$ or $lambda_3=0$ or $R=0$.
answered Jan 12 at 4:01
Stefan LafonStefan Lafon
1,87818
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add a comment |
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$begingroup$
Similarly as for the additive case, this is not possible unless $lambda_1=0$ or $lambda_3=0$ or $R=0$.
Suppose there exist two functions $x$ and $y$ such that
$$x(lambda_1)y(lambda_3)=1 - expleft(lambda_1 p + lambda_3 qright)$$
Then note that $x(0)y(0)=1-1=0$.
Also
$$ x(lambda_1)y(0)=1-e^{lambda_1p}$$
$$ x(0)y(lambda_3)=1-e^{lambda_3q}$$
Multiplying the two yields
$$ x(lambda_1)y(lambda_3)x(0)y(0)=(1-e^{lambda_1p})(1-e^{lambda_3q})$$
But since $x(0)y(0)=0$, that means
$$(1-e^{lambda_1p})(1-e^{lambda_3q})=0$$
which implies that $lambda_1=0$ or $lambda_3=0$ or $R=0$.
answered Jan 12 at 4:01
Stefan LafonStefan Lafon
1,87818
$endgroup$
add a comment |
$begingroup$
Similarly as for the additive case, this is not possible unless $lambda_1=0$ or $lambda_3=0$ or $R=0$.
Suppose there exist two functions $x$ and $y$ such that
$$x(lambda_1)y(lambda_3)=1 - expleft(lambda_1 p + lambda_3 qright)$$
Then note that $x(0)y(0)=1-1=0$.
Also
$$ x(lambda_1)y(0)=1-e^{lambda_1p}$$
$$ x(0)y(lambda_3)=1-e^{lambda_3q}$$
Multiplying the two yields
$$ x(lambda_1)y(lambda_3)x(0)y(0)=(1-e^{lambda_1p})(1-e^{lambda_3q})$$
But since $x(0)y(0)=0$, that means
$$(1-e^{lambda_1p})(1-e^{lambda_3q})=0$$
which implies that $lambda_1=0$ or $lambda_3=0$ or $R=0$.
answered Jan 12 at 4:01
Stefan LafonStefan Lafon
1,87818
$endgroup$
add a comment |
$begingroup$
Similarly as for the additive case, this is not possible unless $lambda_1=0$ or $lambda_3=0$ or $R=0$.
Suppose there exist two functions $x$ and $y$ such that
$$x(lambda_1)y(lambda_3)=1 - expleft(lambda_1 p + lambda_3 qright)$$
Then note that $x(0)y(0)=1-1=0$.
Also
$$ x(lambda_1)y(0)=1-e^{lambda_1p}$$
$$ x(0)y(lambda_3)=1-e^{lambda_3q}$$
Multiplying the two yields
$$ x(lambda_1)y(lambda_3)x(0)y(0)=(1-e^{lambda_1p})(1-e^{lambda_3q})$$
But since $x(0)y(0)=0$, that means
$$(1-e^{lambda_1p})(1-e^{lambda_3q})=0$$
which implies that $lambda_1=0$ or $lambda_3=0$ or $R=0$.
answered Jan 12 at 4:01
Stefan LafonStefan Lafon
1,87818
$endgroup$
Similarly as for the additive case, this is not possible unless $lambda_1=0$ or $lambda_3=0$ or $R=0$.
Suppose there exist two functions $x$ and $y$ such that
$$x(lambda_1)y(lambda_3)=1 - expleft(lambda_1 p + lambda_3 qright)$$
Then note that $x(0)y(0)=1-1=0$.
Also
$$ x(lambda_1)y(0)=1-e^{lambda_1p}$$
$$ x(0)y(lambda_3)=1-e^{lambda_3q}$$
Multiplying the two yields
$$ x(lambda_1)y(lambda_3)x(0)y(0)=(1-e^{lambda_1p})(1-e^{lambda_3q})$$
But since $x(0)y(0)=0$, that means
$$(1-e^{lambda_1p})(1-e^{lambda_3q})=0$$
which implies that $lambda_1=0$ or $lambda_3=0$ or $R=0$.
answered Jan 12 at 4:01
Stefan LafonStefan Lafon
1,87818
answered Jan 12 at 4:01
Stefan LafonStefan Lafon
1,87818
answered Jan 12 at 4:01
Stefan LafonStefan Lafon
1,87818
answered Jan 12 at 4:01
Stefan LafonStefan Lafon
1,87818
1,87818
add a comment |
add a comment |
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