Mixing
Linear Independence/Dependence Contradiction
Clash Royale CLAN TAG#URR8PPP
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I am trying to determine if $u_1(x)=x^3, u_2(x)=|x|^3$ is linearly independent/dependent on $mathbbR$.
I first computed the Wronskian using two cases.
Case $1$ $(xgeq 0)$: $$W(x)=beginvmatrix
x^3 & x^3 \
3x^2 & 3x^2 \
notag
endvmatrix=0$$
Case $2$ $(x< 0)$: $$W(x)=beginvmatrix
x^3 & -x^3 \
3x^2 & -3x^2 \
notag
endvmatrix=0$$
I then used the definition of linear independence $(c_1,c_2inmathbbR)$:
$$(xge 0): c_1x^3+c_2x^3=0 (1) \ (x<0): c_1x^3-c_2x^3=0 (2)$$
$$(1)+(2)=2c_1x^3=0implies c_1=0 \ therefore c_2x^3=0implies c_2=0$$
So since $c_1=c_2=0$, both functions should be linearly independent.
What I don't understand is why can both functions be linearly constucted from the other, doesn't this show linear dependence?
e.g. if $$x<0implies |x|^3=-x^3 \ therefore -u_1(x)=u_2(x)$$
e.g. $$u_1(x)=sin(2x), u_2(x)=sin(x)cos(x)$$ are linearly dependent as they can be written as a linear combination of each other.
Linearly independent functions can not be written as a linear combination of said functions.
proof-verification independence
asked yesterday
Bell
54312
add a comment |Â
up vote
0
down vote
favorite
I am trying to determine if $u_1(x)=x^3, u_2(x)=|x|^3$ is linearly independent/dependent on $mathbbR$.
I first computed the Wronskian using two cases.
Case $1$ $(xgeq 0)$: $$W(x)=beginvmatrix
x^3 & x^3 \
3x^2 & 3x^2 \
notag
endvmatrix=0$$
Case $2$ $(x< 0)$: $$W(x)=beginvmatrix
x^3 & -x^3 \
3x^2 & -3x^2 \
notag
endvmatrix=0$$
I then used the definition of linear independence $(c_1,c_2inmathbbR)$:
$$(xge 0): c_1x^3+c_2x^3=0 (1) \ (x<0): c_1x^3-c_2x^3=0 (2)$$
$$(1)+(2)=2c_1x^3=0implies c_1=0 \ therefore c_2x^3=0implies c_2=0$$
So since $c_1=c_2=0$, both functions should be linearly independent.
What I don't understand is why can both functions be linearly constucted from the other, doesn't this show linear dependence?
e.g. if $$x<0implies |x|^3=-x^3 \ therefore -u_1(x)=u_2(x)$$
e.g. $$u_1(x)=sin(2x), u_2(x)=sin(x)cos(x)$$ are linearly dependent as they can be written as a linear combination of each other.
Linearly independent functions can not be written as a linear combination of said functions.
proof-verification independence
asked yesterday
Bell
54312
I can construct $sin$ using $cos$, by the formula $cos(pi/2 - x)=sin(x)$. However as you stated, it does not mean that $cos$ and $sin$ are linearly dependant, quite the contrary. Actually, the term linear really is important. It is not just any kind of dependancy, it is a very specific one.
– Suzet
yesterday
That may have been a bad example. I shall re-edit.
– Bell
yesterday
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to determine if $u_1(x)=x^3, u_2(x)=|x|^3$ is linearly independent/dependent on $mathbbR$.
I first computed the Wronskian using two cases.
Case $1$ $(xgeq 0)$: $$W(x)=beginvmatrix
x^3 & x^3 \
3x^2 & 3x^2 \
notag
endvmatrix=0$$
Case $2$ $(x< 0)$: $$W(x)=beginvmatrix
x^3 & -x^3 \
3x^2 & -3x^2 \
notag
endvmatrix=0$$
I then used the definition of linear independence $(c_1,c_2inmathbbR)$:
$$(xge 0): c_1x^3+c_2x^3=0 (1) \ (x<0): c_1x^3-c_2x^3=0 (2)$$
$$(1)+(2)=2c_1x^3=0implies c_1=0 \ therefore c_2x^3=0implies c_2=0$$
So since $c_1=c_2=0$, both functions should be linearly independent.
What I don't understand is why can both functions be linearly constucted from the other, doesn't this show linear dependence?
e.g. if $$x<0implies |x|^3=-x^3 \ therefore -u_1(x)=u_2(x)$$
e.g. $$u_1(x)=sin(2x), u_2(x)=sin(x)cos(x)$$ are linearly dependent as they can be written as a linear combination of each other.
Linearly independent functions can not be written as a linear combination of said functions.
proof-verification independence
asked yesterday
Bell
54312
I am trying to determine if $u_1(x)=x^3, u_2(x)=|x|^3$ is linearly independent/dependent on $mathbbR$.
I first computed the Wronskian using two cases.
Case $1$ $(xgeq 0)$: $$W(x)=beginvmatrix
x^3 & x^3 \
3x^2 & 3x^2 \
notag
endvmatrix=0$$
Case $2$ $(x< 0)$: $$W(x)=beginvmatrix
x^3 & -x^3 \
3x^2 & -3x^2 \
notag
endvmatrix=0$$
I then used the definition of linear independence $(c_1,c_2inmathbbR)$:
$$(xge 0): c_1x^3+c_2x^3=0 (1) \ (x<0): c_1x^3-c_2x^3=0 (2)$$
$$(1)+(2)=2c_1x^3=0implies c_1=0 \ therefore c_2x^3=0implies c_2=0$$
So since $c_1=c_2=0$, both functions should be linearly independent.
What I don't understand is why can both functions be linearly constucted from the other, doesn't this show linear dependence?
e.g. if $$x<0implies |x|^3=-x^3 \ therefore -u_1(x)=u_2(x)$$
e.g. $$u_1(x)=sin(2x), u_2(x)=sin(x)cos(x)$$ are linearly dependent as they can be written as a linear combination of each other.
Linearly independent functions can not be written as a linear combination of said functions.
proof-verification independence
asked yesterday
Bell
54312
edited yesterday
asked yesterday
Bell
54312
asked yesterday
Bell
54312
asked yesterday
Bell
54312
54312
I can construct $sin$ using $cos$, by the formula $cos(pi/2 - x)=sin(x)$. However as you stated, it does not mean that $cos$ and $sin$ are linearly dependant, quite the contrary. Actually, the term linear really is important. It is not just any kind of dependancy, it is a very specific one.
– Suzet
yesterday
That may have been a bad example. I shall re-edit.
– Bell
yesterday
add a comment |Â
I can construct $sin$ using $cos$, by the formula $cos(pi/2 - x)=sin(x)$. However as you stated, it does not mean that $cos$ and $sin$ are linearly dependant, quite the contrary. Actually, the term linear really is important. It is not just any kind of dependancy, it is a very specific one.
– Suzet
yesterday
That may have been a bad example. I shall re-edit.
– Bell
yesterday
I can construct $sin$ using $cos$, by the formula $cos(pi/2 - x)=sin(x)$. However as you stated, it does not mean that $cos$ and $sin$ are linearly dependant, quite the contrary. Actually, the term linear really is important. It is not just any kind of dependancy, it is a very specific one.
– Suzet
yesterday
I can construct $sin$ using $cos$, by the formula $cos(pi/2 - x)=sin(x)$. However as you stated, it does not mean that $cos$ and $sin$ are linearly dependant, quite the contrary. Actually, the term linear really is important. It is not just any kind of dependancy, it is a very specific one.
– Suzet
yesterday
That may have been a bad example. I shall re-edit.
– Bell
yesterday
That may have been a bad example. I shall re-edit.
– Bell
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The Theorem about Wronskian and linear independent is a statement about functions which are solutions to linear differential equations not arbitrary functions such as $x^3$ and $|x|^3$
What you have discovered is that these two functions are not solutions to a lineare differential equation on the whole real line.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
I understand that the linear Independence/dependence of said arbitrary functions cannot be determined by their Wronskian alone, that is why I have resorted to the definition of linear independence. My question is, why does the formal definition confirm linear independence, but then in the case when, for example, $x<0$, does $-u_1=u_2$, as I mentioned. If they were linearly independent, this should not be possible, right?
– Bell
yesterday
Your functions are linearly dependent on $R^>0$ and on $R ^<0$ but they are not linearly dependent on $R$
– Mohammad Riazi-Kermani
yesterday
I see. Is this why we must test this using the definition of linear independence, as this considers the whole real line? My example that I mentioned $(-u_1=u_2)$ only takes into account $mathbbR^<0$
– Bell
yesterday
Yes, that is correct.
– Mohammad Riazi-Kermani
yesterday
1
That is correct.
– Mohammad Riazi-Kermani
yesterday
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The Theorem about Wronskian and linear independent is a statement about functions which are solutions to linear differential equations not arbitrary functions such as $x^3$ and $|x|^3$
What you have discovered is that these two functions are not solutions to a lineare differential equation on the whole real line.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
I understand that the linear Independence/dependence of said arbitrary functions cannot be determined by their Wronskian alone, that is why I have resorted to the definition of linear independence. My question is, why does the formal definition confirm linear independence, but then in the case when, for example, $x<0$, does $-u_1=u_2$, as I mentioned. If they were linearly independent, this should not be possible, right?
– Bell
yesterday
Your functions are linearly dependent on $R^>0$ and on $R ^<0$ but they are not linearly dependent on $R$
– Mohammad Riazi-Kermani
yesterday
I see. Is this why we must test this using the definition of linear independence, as this considers the whole real line? My example that I mentioned $(-u_1=u_2)$ only takes into account $mathbbR^<0$
– Bell
yesterday
Yes, that is correct.
– Mohammad Riazi-Kermani
yesterday
1
That is correct.
– Mohammad Riazi-Kermani
yesterday
 |Â
show 5 more comments
up vote
1
down vote
accepted
The Theorem about Wronskian and linear independent is a statement about functions which are solutions to linear differential equations not arbitrary functions such as $x^3$ and $|x|^3$
What you have discovered is that these two functions are not solutions to a lineare differential equation on the whole real line.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
I understand that the linear Independence/dependence of said arbitrary functions cannot be determined by their Wronskian alone, that is why I have resorted to the definition of linear independence. My question is, why does the formal definition confirm linear independence, but then in the case when, for example, $x<0$, does $-u_1=u_2$, as I mentioned. If they were linearly independent, this should not be possible, right?
– Bell
yesterday
Your functions are linearly dependent on $R^>0$ and on $R ^<0$ but they are not linearly dependent on $R$
– Mohammad Riazi-Kermani
yesterday
I see. Is this why we must test this using the definition of linear independence, as this considers the whole real line? My example that I mentioned $(-u_1=u_2)$ only takes into account $mathbbR^<0$
– Bell
yesterday
Yes, that is correct.
– Mohammad Riazi-Kermani
yesterday
1
That is correct.
– Mohammad Riazi-Kermani
yesterday
 |Â
show 5 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The Theorem about Wronskian and linear independent is a statement about functions which are solutions to linear differential equations not arbitrary functions such as $x^3$ and $|x|^3$
What you have discovered is that these two functions are not solutions to a lineare differential equation on the whole real line.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
The Theorem about Wronskian and linear independent is a statement about functions which are solutions to linear differential equations not arbitrary functions such as $x^3$ and $|x|^3$
What you have discovered is that these two functions are not solutions to a lineare differential equation on the whole real line.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
26.9k41849
I understand that the linear Independence/dependence of said arbitrary functions cannot be determined by their Wronskian alone, that is why I have resorted to the definition of linear independence. My question is, why does the formal definition confirm linear independence, but then in the case when, for example, $x<0$, does $-u_1=u_2$, as I mentioned. If they were linearly independent, this should not be possible, right?
– Bell
yesterday
Your functions are linearly dependent on $R^>0$ and on $R ^<0$ but they are not linearly dependent on $R$
– Mohammad Riazi-Kermani
yesterday
I see. Is this why we must test this using the definition of linear independence, as this considers the whole real line? My example that I mentioned $(-u_1=u_2)$ only takes into account $mathbbR^<0$
– Bell
yesterday
Yes, that is correct.
– Mohammad Riazi-Kermani
yesterday
1
That is correct.
– Mohammad Riazi-Kermani
yesterday
 |Â
show 5 more comments
I understand that the linear Independence/dependence of said arbitrary functions cannot be determined by their Wronskian alone, that is why I have resorted to the definition of linear independence. My question is, why does the formal definition confirm linear independence, but then in the case when, for example, $x<0$, does $-u_1=u_2$, as I mentioned. If they were linearly independent, this should not be possible, right?
– Bell
yesterday
Your functions are linearly dependent on $R^>0$ and on $R ^<0$ but they are not linearly dependent on $R$
– Mohammad Riazi-Kermani
yesterday
I see. Is this why we must test this using the definition of linear independence, as this considers the whole real line? My example that I mentioned $(-u_1=u_2)$ only takes into account $mathbbR^<0$
– Bell
yesterday
Yes, that is correct.
– Mohammad Riazi-Kermani
yesterday
1
That is correct.
– Mohammad Riazi-Kermani
yesterday
I understand that the linear Independence/dependence of said arbitrary functions cannot be determined by their Wronskian alone, that is why I have resorted to the definition of linear independence. My question is, why does the formal definition confirm linear independence, but then in the case when, for example, $x<0$, does $-u_1=u_2$, as I mentioned. If they were linearly independent, this should not be possible, right?
– Bell
yesterday
I understand that the linear Independence/dependence of said arbitrary functions cannot be determined by their Wronskian alone, that is why I have resorted to the definition of linear independence. My question is, why does the formal definition confirm linear independence, but then in the case when, for example, $x<0$, does $-u_1=u_2$, as I mentioned. If they were linearly independent, this should not be possible, right?
– Bell
yesterday
Your functions are linearly dependent on $R^>0$ and on $R ^<0$ but they are not linearly dependent on $R$
– Mohammad Riazi-Kermani
yesterday
Your functions are linearly dependent on $R^>0$ and on $R ^<0$ but they are not linearly dependent on $R$
– Mohammad Riazi-Kermani
yesterday
I see. Is this why we must test this using the definition of linear independence, as this considers the whole real line? My example that I mentioned $(-u_1=u_2)$ only takes into account $mathbbR^<0$
– Bell
yesterday
I see. Is this why we must test this using the definition of linear independence, as this considers the whole real line? My example that I mentioned $(-u_1=u_2)$ only takes into account $mathbbR^<0$
– Bell
yesterday
Yes, that is correct.
– Mohammad Riazi-Kermani
yesterday
Yes, that is correct.
– Mohammad Riazi-Kermani
yesterday
1
1
That is correct.
– Mohammad Riazi-Kermani
yesterday
That is correct.
– Mohammad Riazi-Kermani
yesterday
 |Â
show 5 more comments
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I can construct $sin$ using $cos$, by the formula $cos(pi/2 - x)=sin(x)$. However as you stated, it does not mean that $cos$ and $sin$ are linearly dependant, quite the contrary. Actually, the term linear really is important. It is not just any kind of dependancy, it is a very specific one.
– Suzet
yesterday
That may have been a bad example. I shall re-edit.
– Bell
yesterday