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} +...












0












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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?










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  • $begingroup$
    SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 5:39












  • $begingroup$
    @ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
    $endgroup$
    – drerD
    Dec 10 '18 at 6:17










  • $begingroup$
    @ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
    $endgroup$
    – drerD
    Dec 10 '18 at 8:41
















0












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 5:39












  • $begingroup$
    @ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
    $endgroup$
    – drerD
    Dec 10 '18 at 6:17










  • $begingroup$
    @ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
    $endgroup$
    – drerD
    Dec 10 '18 at 8:41














0












0








0





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






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edited Dec 10 '18 at 12:15









Harry Peter

5,47911439




5,47911439










asked Dec 10 '18 at 0:55









drerDdrerD

1609




1609












  • $begingroup$
    SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 5:39












  • $begingroup$
    @ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
    $endgroup$
    – drerD
    Dec 10 '18 at 6:17










  • $begingroup$
    @ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
    $endgroup$
    – drerD
    Dec 10 '18 at 8:41


















  • $begingroup$
    SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 5:39












  • $begingroup$
    @ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
    $endgroup$
    – drerD
    Dec 10 '18 at 6:17










  • $begingroup$
    @ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
    $endgroup$
    – drerD
    Dec 10 '18 at 8:41
















$begingroup$
SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
$endgroup$
– Claude Leibovici
Dec 10 '18 at 5:39






$begingroup$
SInce $x$ is a constant, let $e^{-x}=k$, $e^{-2x}=k^2$, $e^{-3x}=k^3$
$endgroup$
– Claude Leibovici
Dec 10 '18 at 5:39














$begingroup$
@ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
$endgroup$
– drerD
Dec 10 '18 at 6:17




$begingroup$
@ClaudeLeibovici interesting, I search about solving system of polynomials, and this problem seems to be more difficult than I thought.
$endgroup$
– drerD
Dec 10 '18 at 6:17












$begingroup$
@ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
$endgroup$
– drerD
Dec 10 '18 at 8:41




$begingroup$
@ClaudeLeibovici I tried substituting with $k^n$, do you mind checking if this is valid? math.stackexchange.com/q/3033628/235884
$endgroup$
– drerD
Dec 10 '18 at 8:41










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