Solve for parameters of a system of non linear equations that have forms of $0 = e^x + A_1 e^x + A_2 e^{2x} +...
I'm self studying math and came across a problem that I have to solve for parameters of the following equations,
$$
0 = e^{-x} - A_1 - A_2 e^{-x} - A_3e^{-2x} \
0 = e^{-2x} - A_1e^{-x} - A_2 - A_3e^{-x} \
0 = e^{-3x} - A_1e^{-2x} - A_2e^{-x} - A_3 \
$$
The answer given is $A_1=e^{-x}, A_2=0, A_3 = 0$, but there's no method shown for finding these.
I was thinking I can put them in matrix form but I don't think that works, any suggestion of systemically finding the solution besides substituting individually?
exponential-function systems-of-equations
edited Dec 10 '18 at 12:15
Harry Peter
5,46111439
asked Dec 10 '18 at 0:55
drerD
1519
add a comment |
I'm self studying math and came across a problem that I have to solve for parameters of the following equations,
$$
0 = e^{-x} - A_1 - A_2 e^{-x} - A_3e^{-2x} \
0 = e^{-2x} - A_1e^{-x} - A_2 - A_3e^{-x} \
0 = e^{-3x} - A_1e^{-2x} - A_2e^{-x} - A_3 \
$$
The answer given is $A_1=e^{-x}, A_2=0, A_3 = 0$, but there's no method shown for finding these.
I was thinking I can put them in matrix form but I don't think that works, any suggestion of systemically finding the solution besides substituting individually?
exponential-function systems-of-equations
edited Dec 10 '18 at 12:15
Harry Peter
5,46111439
asked Dec 10 '18 at 0:55
drerD
1519
SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
– Claude Leibovici
Dec 10 '18 at 5:39
@ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
– drerD
Dec 10 '18 at 6:17
@ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
– drerD
Dec 10 '18 at 8:41
add a comment |
I'm self studying math and came across a problem that I have to solve for parameters of the following equations,
$$
0 = e^{-x} - A_1 - A_2 e^{-x} - A_3e^{-2x} \
0 = e^{-2x} - A_1e^{-x} - A_2 - A_3e^{-x} \
0 = e^{-3x} - A_1e^{-2x} - A_2e^{-x} - A_3 \
$$
The answer given is $A_1=e^{-x}, A_2=0, A_3 = 0$, but there's no method shown for finding these.
I was thinking I can put them in matrix form but I don't think that works, any suggestion of systemically finding the solution besides substituting individually?
exponential-function systems-of-equations
edited Dec 10 '18 at 12:15
Harry Peter
5,46111439
asked Dec 10 '18 at 0:55
drerD
1519
I'm self studying math and came across a problem that I have to solve for parameters of the following equations,
$$
0 = e^{-x} - A_1 - A_2 e^{-x} - A_3e^{-2x} \
0 = e^{-2x} - A_1e^{-x} - A_2 - A_3e^{-x} \
0 = e^{-3x} - A_1e^{-2x} - A_2e^{-x} - A_3 \
$$
The answer given is $A_1=e^{-x}, A_2=0, A_3 = 0$, but there's no method shown for finding these.
I was thinking I can put them in matrix form but I don't think that works, any suggestion of systemically finding the solution besides substituting individually?
exponential-function systems-of-equations
exponential-function systems-of-equations
edited Dec 10 '18 at 12:15
Harry Peter
5,46111439
asked Dec 10 '18 at 0:55
drerD
1519
edited Dec 10 '18 at 12:15
Harry Peter
5,46111439
asked Dec 10 '18 at 0:55
drerD
1519
edited Dec 10 '18 at 12:15
Harry Peter
5,46111439
edited Dec 10 '18 at 12:15
Harry Peter
5,46111439
edited Dec 10 '18 at 12:15
Harry Peter
5,46111439
5,46111439
asked Dec 10 '18 at 0:55
drerD
1519
asked Dec 10 '18 at 0:55
drerD
1519
asked Dec 10 '18 at 0:55
drerD
1519
1519
SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
– Claude Leibovici
Dec 10 '18 at 5:39
@ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
– drerD
Dec 10 '18 at 6:17
@ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
– drerD
Dec 10 '18 at 8:41
add a comment |
SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
– Claude Leibovici
Dec 10 '18 at 5:39
@ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
– drerD
Dec 10 '18 at 6:17
@ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
– drerD
Dec 10 '18 at 8:41
SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
– Claude Leibovici
Dec 10 '18 at 5:39
SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
– Claude Leibovici
Dec 10 '18 at 5:39
@ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
– drerD
Dec 10 '18 at 6:17
@ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
– drerD
Dec 10 '18 at 6:17
@ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
– drerD
Dec 10 '18 at 8:41
@ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
– drerD
Dec 10 '18 at 8:41
add a comment |
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SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
– Claude Leibovici
Dec 10 '18 at 5:39
@ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
– drerD
Dec 10 '18 at 6:17
@ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
– drerD
Dec 10 '18 at 8:41