Prove that the set of functions ${x, e^x, sin(x)}$ is linearly independent.
I am supposed to prove that ${x,e^x,sin(x)}$ is a linearly independent set.
I know that if ${x,e^x,sin(x)}$ is linearly independent, then we would have $ax+be^x+csin(x)=0$, for all $x in mathbb{R}$, and $a,b,c in mathbb{R}$, where a=b=c=0.
However, I'm skeptical how to proceed.
My gut tells me to begin taking derivatives, since I'm given a set of real function, and I may be able to to show that $a$, $b$, and $c$ must be equal to zero.
begin{align*}
(ax+be^x+csin(x))'&= a1+be^x+ccdot cos(x)=0\
(ax+be^x+csin(x))''&=a0+be^x-ccdot sin(x)=0\
(ax+be^x+csin(x))'''&=a0+be^x-ccdot cos(x)=0\
(ax+be^x+csin(x))''''&=a0+be^x+ccdot sin(x)=0\
end{align*}
Now subtracting $ccdot sin(x)$, we have
begin{equation*}
be^x=-c(sin x)
end{equation*}
From this I can't determine much, I know that $e^x >0$ for all $xinmathbb{R}$.
Can someone give me a push?
linear-algebra
add a comment |
I am supposed to prove that ${x,e^x,sin(x)}$ is a linearly independent set.
I know that if ${x,e^x,sin(x)}$ is linearly independent, then we would have $ax+be^x+csin(x)=0$, for all $x in mathbb{R}$, and $a,b,c in mathbb{R}$, where a=b=c=0.
However, I'm skeptical how to proceed.
My gut tells me to begin taking derivatives, since I'm given a set of real function, and I may be able to to show that $a$, $b$, and $c$ must be equal to zero.
begin{align*}
(ax+be^x+csin(x))'&= a1+be^x+ccdot cos(x)=0\
(ax+be^x+csin(x))''&=a0+be^x-ccdot sin(x)=0\
(ax+be^x+csin(x))'''&=a0+be^x-ccdot cos(x)=0\
(ax+be^x+csin(x))''''&=a0+be^x+ccdot sin(x)=0\
end{align*}
Now subtracting $ccdot sin(x)$, we have
begin{equation*}
be^x=-c(sin x)
end{equation*}
From this I can't determine much, I know that $e^x >0$ for all $xinmathbb{R}$.
Can someone give me a push?
linear-algebra
1
What you need to prove is that $$forall a,b,cin mathbb Rleft(forall xin mathbb Rleft(ax+be^x+csin(x)right)implies a=b=c=0right).$$ Take arbitrary real numbers $a,b,c$ and assume that $forall xin mathbb Rleft(ax+be^x+csin(x)right)$ holds. From here you can use a number of tricks, differentiation might help. Personally I'd just replace $x$ with suitable values (start with $x=0$ and see where it gets you, then try to annihilate $sin$ with something other than $0$).
– Git Gud
Oct 15 '15 at 22:06
1
Your approach is pretty good. Now notice you have $be^x -csin(x) = 0$ and $be^x+csin(x)=0$, so adding the two gives $be^x=0$. You can probably take it from there...
– Joey Zou
Oct 15 '15 at 22:09
On behalf of @Israel Barquín: I think Cramer method could've done very well with the system formed by your original sum, the second and third derivatives of it. The determinant of the system with the replaced column will be $0$ in each case.
– dantopa
Nov 28 at 21:13
add a comment |
I am supposed to prove that ${x,e^x,sin(x)}$ is a linearly independent set.
I know that if ${x,e^x,sin(x)}$ is linearly independent, then we would have $ax+be^x+csin(x)=0$, for all $x in mathbb{R}$, and $a,b,c in mathbb{R}$, where a=b=c=0.
However, I'm skeptical how to proceed.
My gut tells me to begin taking derivatives, since I'm given a set of real function, and I may be able to to show that $a$, $b$, and $c$ must be equal to zero.
begin{align*}
(ax+be^x+csin(x))'&= a1+be^x+ccdot cos(x)=0\
(ax+be^x+csin(x))''&=a0+be^x-ccdot sin(x)=0\
(ax+be^x+csin(x))'''&=a0+be^x-ccdot cos(x)=0\
(ax+be^x+csin(x))''''&=a0+be^x+ccdot sin(x)=0\
end{align*}
Now subtracting $ccdot sin(x)$, we have
begin{equation*}
be^x=-c(sin x)
end{equation*}
From this I can't determine much, I know that $e^x >0$ for all $xinmathbb{R}$.
Can someone give me a push?
linear-algebra
I am supposed to prove that ${x,e^x,sin(x)}$ is a linearly independent set.
I know that if ${x,e^x,sin(x)}$ is linearly independent, then we would have $ax+be^x+csin(x)=0$, for all $x in mathbb{R}$, and $a,b,c in mathbb{R}$, where a=b=c=0.
However, I'm skeptical how to proceed.
My gut tells me to begin taking derivatives, since I'm given a set of real function, and I may be able to to show that $a$, $b$, and $c$ must be equal to zero.
begin{align*}
(ax+be^x+csin(x))'&= a1+be^x+ccdot cos(x)=0\
(ax+be^x+csin(x))''&=a0+be^x-ccdot sin(x)=0\
(ax+be^x+csin(x))'''&=a0+be^x-ccdot cos(x)=0\
(ax+be^x+csin(x))''''&=a0+be^x+ccdot sin(x)=0\
end{align*}
Now subtracting $ccdot sin(x)$, we have
begin{equation*}
be^x=-c(sin x)
end{equation*}
From this I can't determine much, I know that $e^x >0$ for all $xinmathbb{R}$.
Can someone give me a push?
linear-algebra
linear-algebra
edited Oct 15 '15 at 22:44
asked Oct 15 '15 at 22:02
user225477
1
What you need to prove is that $$forall a,b,cin mathbb Rleft(forall xin mathbb Rleft(ax+be^x+csin(x)right)implies a=b=c=0right).$$ Take arbitrary real numbers $a,b,c$ and assume that $forall xin mathbb Rleft(ax+be^x+csin(x)right)$ holds. From here you can use a number of tricks, differentiation might help. Personally I'd just replace $x$ with suitable values (start with $x=0$ and see where it gets you, then try to annihilate $sin$ with something other than $0$).
– Git Gud
Oct 15 '15 at 22:06
1
Your approach is pretty good. Now notice you have $be^x -csin(x) = 0$ and $be^x+csin(x)=0$, so adding the two gives $be^x=0$. You can probably take it from there...
– Joey Zou
Oct 15 '15 at 22:09
On behalf of @Israel Barquín: I think Cramer method could've done very well with the system formed by your original sum, the second and third derivatives of it. The determinant of the system with the replaced column will be $0$ in each case.
– dantopa
Nov 28 at 21:13
add a comment |
1
What you need to prove is that $$forall a,b,cin mathbb Rleft(forall xin mathbb Rleft(ax+be^x+csin(x)right)implies a=b=c=0right).$$ Take arbitrary real numbers $a,b,c$ and assume that $forall xin mathbb Rleft(ax+be^x+csin(x)right)$ holds. From here you can use a number of tricks, differentiation might help. Personally I'd just replace $x$ with suitable values (start with $x=0$ and see where it gets you, then try to annihilate $sin$ with something other than $0$).
– Git Gud
Oct 15 '15 at 22:06
1
Your approach is pretty good. Now notice you have $be^x -csin(x) = 0$ and $be^x+csin(x)=0$, so adding the two gives $be^x=0$. You can probably take it from there...
– Joey Zou
Oct 15 '15 at 22:09
On behalf of @Israel Barquín: I think Cramer method could've done very well with the system formed by your original sum, the second and third derivatives of it. The determinant of the system with the replaced column will be $0$ in each case.
– dantopa
Nov 28 at 21:13
1
1
What you need to prove is that $$forall a,b,cin mathbb Rleft(forall xin mathbb Rleft(ax+be^x+csin(x)right)implies a=b=c=0right).$$ Take arbitrary real numbers $a,b,c$ and assume that $forall xin mathbb Rleft(ax+be^x+csin(x)right)$ holds. From here you can use a number of tricks, differentiation might help. Personally I'd just replace $x$ with suitable values (start with $x=0$ and see where it gets you, then try to annihilate $sin$ with something other than $0$).
– Git Gud
Oct 15 '15 at 22:06
What you need to prove is that $$forall a,b,cin mathbb Rleft(forall xin mathbb Rleft(ax+be^x+csin(x)right)implies a=b=c=0right).$$ Take arbitrary real numbers $a,b,c$ and assume that $forall xin mathbb Rleft(ax+be^x+csin(x)right)$ holds. From here you can use a number of tricks, differentiation might help. Personally I'd just replace $x$ with suitable values (start with $x=0$ and see where it gets you, then try to annihilate $sin$ with something other than $0$).
– Git Gud
Oct 15 '15 at 22:06
1
1
Your approach is pretty good. Now notice you have $be^x -csin(x) = 0$ and $be^x+csin(x)=0$, so adding the two gives $be^x=0$. You can probably take it from there...
– Joey Zou
Oct 15 '15 at 22:09
Your approach is pretty good. Now notice you have $be^x -csin(x) = 0$ and $be^x+csin(x)=0$, so adding the two gives $be^x=0$. You can probably take it from there...
– Joey Zou
Oct 15 '15 at 22:09
On behalf of @Israel Barquín: I think Cramer method could've done very well with the system formed by your original sum, the second and third derivatives of it. The determinant of the system with the replaced column will be $0$ in each case.
– dantopa
Nov 28 at 21:13
On behalf of @Israel Barquín: I think Cramer method could've done very well with the system formed by your original sum, the second and third derivatives of it. The determinant of the system with the replaced column will be $0$ in each case.
– dantopa
Nov 28 at 21:13
add a comment |
3 Answers
3
active
oldest
votes
$ax +be^x +csin(x) =0$, $x=0$ implies $b=0$, $ax+csin(x) = 0$, $x=pi$, $a=0$ thus $c=0$
answered Oct 15 '15 at 22:11
Tsemo Aristide
55.4k11444
add a comment |
Since $x$, $e^x$, and $sin x$ are analytic, then their non-vanishing Wronskian is sufficient to conclude that the set is linearly independent.
The Wronskian is equal to the determinant $W$ of the matrix given by
$$begin{bmatrix}
x & e^x & sin x\
1 & e^x & cos x\
0 & e^x & -sin x\
end{bmatrix}$$
Therefore, we have
$$W=
begin{vmatrix}
x & e^x & sin x\
1 & e^x & cos x\
0 & e^x & -sin x\
end{vmatrix}
=-xe^x(sin x+cos x)+2e^x sin x$$
Inasmuch as $Wne 0$ for all $xin mathscr{R}$ (e.g., take $x=pi/2$), then $x$, $e^x$ and $sin x$ are linearly independent.
answered Oct 15 '15 at 22:31
Mark Viola
130k1273170
add a comment |
Suppose
$ax +be^x +csin(x) =0
$.
If
$b ne 0$,
then,
for large enough $x$,
$b e^x$
is much larger than
the other two terms,
which is a contradiction.
Therefore
$b = 0$,
so we have
$ax+csin(x)
= 0
$.
Again,
if $a ne 0$,
we can choose $x$
large enough so that
$|ax| > |c|$.
Since
$|sin(x)| le 1$,
this can not hold.
Therefore
$a = 0$,
so that
$c sin(x) = 0$.
This obviously implies that
$c=0$.
And we are done.
answered Oct 16 '15 at 0:27
marty cohen
72.2k549127
add a comment |
Your Answer
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$ax +be^x +csin(x) =0$, $x=0$ implies $b=0$, $ax+csin(x) = 0$, $x=pi$, $a=0$ thus $c=0$
answered Oct 15 '15 at 22:11
Tsemo Aristide
55.4k11444
add a comment |
$ax +be^x +csin(x) =0$, $x=0$ implies $b=0$, $ax+csin(x) = 0$, $x=pi$, $a=0$ thus $c=0$
answered Oct 15 '15 at 22:11
Tsemo Aristide
55.4k11444
add a comment |
$ax +be^x +csin(x) =0$, $x=0$ implies $b=0$, $ax+csin(x) = 0$, $x=pi$, $a=0$ thus $c=0$
answered Oct 15 '15 at 22:11
Tsemo Aristide
55.4k11444
$ax +be^x +csin(x) =0$, $x=0$ implies $b=0$, $ax+csin(x) = 0$, $x=pi$, $a=0$ thus $c=0$
answered Oct 15 '15 at 22:11
Tsemo Aristide
55.4k11444
edited Oct 15 '15 at 23:23
user223391
answered Oct 15 '15 at 22:11
Tsemo Aristide
55.4k11444
answered Oct 15 '15 at 22:11
Tsemo Aristide
55.4k11444
answered Oct 15 '15 at 22:11
Tsemo Aristide
55.4k11444
55.4k11444
add a comment |
add a comment |
Since $x$, $e^x$, and $sin x$ are analytic, then their non-vanishing Wronskian is sufficient to conclude that the set is linearly independent.
The Wronskian is equal to the determinant $W$ of the matrix given by
$$begin{bmatrix}
x & e^x & sin x\
1 & e^x & cos x\
0 & e^x & -sin x\
end{bmatrix}$$
Therefore, we have
$$W=
begin{vmatrix}
x & e^x & sin x\
1 & e^x & cos x\
0 & e^x & -sin x\
end{vmatrix}
=-xe^x(sin x+cos x)+2e^x sin x$$
Inasmuch as $Wne 0$ for all $xin mathscr{R}$ (e.g., take $x=pi/2$), then $x$, $e^x$ and $sin x$ are linearly independent.
answered Oct 15 '15 at 22:31
Mark Viola
130k1273170
add a comment |
Since $x$, $e^x$, and $sin x$ are analytic, then their non-vanishing Wronskian is sufficient to conclude that the set is linearly independent.
The Wronskian is equal to the determinant $W$ of the matrix given by
$$begin{bmatrix}
x & e^x & sin x\
1 & e^x & cos x\
0 & e^x & -sin x\
end{bmatrix}$$
Therefore, we have
$$W=
begin{vmatrix}
x & e^x & sin x\
1 & e^x & cos x\
0 & e^x & -sin x\
end{vmatrix}
=-xe^x(sin x+cos x)+2e^x sin x$$
Inasmuch as $Wne 0$ for all $xin mathscr{R}$ (e.g., take $x=pi/2$), then $x$, $e^x$ and $sin x$ are linearly independent.
answered Oct 15 '15 at 22:31
Mark Viola
130k1273170
add a comment |
Since $x$, $e^x$, and $sin x$ are analytic, then their non-vanishing Wronskian is sufficient to conclude that the set is linearly independent.
The Wronskian is equal to the determinant $W$ of the matrix given by
$$begin{bmatrix}
x & e^x & sin x\
1 & e^x & cos x\
0 & e^x & -sin x\
end{bmatrix}$$
Therefore, we have
$$W=
begin{vmatrix}
x & e^x & sin x\
1 & e^x & cos x\
0 & e^x & -sin x\
end{vmatrix}
=-xe^x(sin x+cos x)+2e^x sin x$$
Inasmuch as $Wne 0$ for all $xin mathscr{R}$ (e.g., take $x=pi/2$), then $x$, $e^x$ and $sin x$ are linearly independent.
answered Oct 15 '15 at 22:31
Mark Viola
130k1273170
Since $x$, $e^x$, and $sin x$ are analytic, then their non-vanishing Wronskian is sufficient to conclude that the set is linearly independent.
The Wronskian is equal to the determinant $W$ of the matrix given by
$$begin{bmatrix}
x & e^x & sin x\
1 & e^x & cos x\
0 & e^x & -sin x\
end{bmatrix}$$
Therefore, we have
$$W=
begin{vmatrix}
x & e^x & sin x\
1 & e^x & cos x\
0 & e^x & -sin x\
end{vmatrix}
=-xe^x(sin x+cos x)+2e^x sin x$$
Inasmuch as $Wne 0$ for all $xin mathscr{R}$ (e.g., take $x=pi/2$), then $x$, $e^x$ and $sin x$ are linearly independent.
answered Oct 15 '15 at 22:31
Mark Viola
130k1273170
edited Oct 15 '15 at 22:37
answered Oct 15 '15 at 22:31
Mark Viola
130k1273170
answered Oct 15 '15 at 22:31
Mark Viola
130k1273170
answered Oct 15 '15 at 22:31
Mark Viola
130k1273170
130k1273170
add a comment |
add a comment |
Suppose
$ax +be^x +csin(x) =0
$.
If
$b ne 0$,
then,
for large enough $x$,
$b e^x$
is much larger than
the other two terms,
which is a contradiction.
Therefore
$b = 0$,
so we have
$ax+csin(x)
= 0
$.
Again,
if $a ne 0$,
we can choose $x$
large enough so that
$|ax| > |c|$.
Since
$|sin(x)| le 1$,
this can not hold.
Therefore
$a = 0$,
so that
$c sin(x) = 0$.
This obviously implies that
$c=0$.
And we are done.
answered Oct 16 '15 at 0:27
marty cohen
72.2k549127
add a comment |
Suppose
$ax +be^x +csin(x) =0
$.
If
$b ne 0$,
then,
for large enough $x$,
$b e^x$
is much larger than
the other two terms,
which is a contradiction.
Therefore
$b = 0$,
so we have
$ax+csin(x)
= 0
$.
Again,
if $a ne 0$,
we can choose $x$
large enough so that
$|ax| > |c|$.
Since
$|sin(x)| le 1$,
this can not hold.
Therefore
$a = 0$,
so that
$c sin(x) = 0$.
This obviously implies that
$c=0$.
And we are done.
answered Oct 16 '15 at 0:27
marty cohen
72.2k549127
add a comment |
Suppose
$ax +be^x +csin(x) =0
$.
If
$b ne 0$,
then,
for large enough $x$,
$b e^x$
is much larger than
the other two terms,
which is a contradiction.
Therefore
$b = 0$,
so we have
$ax+csin(x)
= 0
$.
Again,
if $a ne 0$,
we can choose $x$
large enough so that
$|ax| > |c|$.
Since
$|sin(x)| le 1$,
this can not hold.
Therefore
$a = 0$,
so that
$c sin(x) = 0$.
This obviously implies that
$c=0$.
And we are done.
answered Oct 16 '15 at 0:27
marty cohen
72.2k549127
Suppose
$ax +be^x +csin(x) =0
$.
If
$b ne 0$,
then,
for large enough $x$,
$b e^x$
is much larger than
the other two terms,
which is a contradiction.
Therefore
$b = 0$,
so we have
$ax+csin(x)
= 0
$.
Again,
if $a ne 0$,
we can choose $x$
large enough so that
$|ax| > |c|$.
Since
$|sin(x)| le 1$,
this can not hold.
Therefore
$a = 0$,
so that
$c sin(x) = 0$.
This obviously implies that
$c=0$.
And we are done.
answered Oct 16 '15 at 0:27
marty cohen
72.2k549127
answered Oct 16 '15 at 0:27
marty cohen
72.2k549127
answered Oct 16 '15 at 0:27
marty cohen
72.2k549127
answered Oct 16 '15 at 0:27
marty cohen
72.2k549127
72.2k549127
add a comment |
add a comment |
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1
What you need to prove is that $$forall a,b,cin mathbb Rleft(forall xin mathbb Rleft(ax+be^x+csin(x)right)implies a=b=c=0right).$$ Take arbitrary real numbers $a,b,c$ and assume that $forall xin mathbb Rleft(ax+be^x+csin(x)right)$ holds. From here you can use a number of tricks, differentiation might help. Personally I'd just replace $x$ with suitable values (start with $x=0$ and see where it gets you, then try to annihilate $sin$ with something other than $0$).
– Git Gud
Oct 15 '15 at 22:06
1
Your approach is pretty good. Now notice you have $be^x -csin(x) = 0$ and $be^x+csin(x)=0$, so adding the two gives $be^x=0$. You can probably take it from there...
– Joey Zou
Oct 15 '15 at 22:09
On behalf of @Israel Barquín: I think Cramer method could've done very well with the system formed by your original sum, the second and third derivatives of it. The determinant of the system with the replaced column will be $0$ in each case.
– dantopa
Nov 28 at 21:13