Show That $e^x$ and $e^-x$ are Linearly Independent

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To test for linear independence, let $$c_1e^x+c_2e^-x=0$$
Where $c_1$ and $c_2$ must equal $0$ for the two functions to be demonstrated to be linearly independent.

So for $x=0$: $$c_1=-c_2$$
And for $x=ln(1)$: $$c_1 = c_2$$
For $c_1$ to equal both $c_2$ and $-c_2$ requires $c_1$ to equal $0$, thus $c_2$ also equals $0$.

Is this sufficient to demonstrate linear independence?

edited Aug 22 at 23:49

asked Aug 19 at 2:59

Alessandro Verniani

406

Your attempt makes no sense. $ln(1) = 0$...
â€“Â user21820
Aug 23 at 14:35

add a commentÂ |Â

up vote
-3
down vote

favorite

To test for linear independence, let $$c_1e^x+c_2e^-x=0$$
Where $c_1$ and $c_2$ must equal $0$ for the two functions to be demonstrated to be linearly independent.

So for $x=0$: $$c_1=-c_2$$
And for $x=ln(1)$: $$c_1 = c_2$$
For $c_1$ to equal both $c_2$ and $-c_2$ requires $c_1$ to equal $0$, thus $c_2$ also equals $0$.

Is this sufficient to demonstrate linear independence?

edited Aug 22 at 23:49

asked Aug 19 at 2:59

Alessandro Verniani

406

Your attempt makes no sense. $ln(1) = 0$...
â€“Â user21820
Aug 23 at 14:35

add a commentÂ |Â

up vote
-3
down vote

favorite

To test for linear independence, let $$c_1e^x+c_2e^-x=0$$
Where $c_1$ and $c_2$ must equal $0$ for the two functions to be demonstrated to be linearly independent.

So for $x=0$: $$c_1=-c_2$$
And for $x=ln(1)$: $$c_1 = c_2$$
For $c_1$ to equal both $c_2$ and $-c_2$ requires $c_1$ to equal $0$, thus $c_2$ also equals $0$.

Is this sufficient to demonstrate linear independence?

edited Aug 22 at 23:49

asked Aug 19 at 2:59

Alessandro Verniani

406

To test for linear independence, let $$c_1e^x+c_2e^-x=0$$
Where $c_1$ and $c_2$ must equal $0$ for the two functions to be demonstrated to be linearly independent.

So for $x=0$: $$c_1=-c_2$$
And for $x=ln(1)$: $$c_1 = c_2$$
For $c_1$ to equal both $c_2$ and $-c_2$ requires $c_1$ to equal $0$, thus $c_2$ also equals $0$.

Is this sufficient to demonstrate linear independence?

edited Aug 22 at 23:49

asked Aug 19 at 2:59

Alessandro Verniani

406

edited Aug 22 at 23:49

asked Aug 19 at 2:59

Alessandro Verniani

406

asked Aug 19 at 2:59

Alessandro Verniani

406

asked Aug 19 at 2:59

Alessandro Verniani

406

Your attempt makes no sense. $ln(1) = 0$...
â€“Â user21820
Aug 23 at 14:35

add a commentÂ |Â

Your attempt makes no sense. $ln(1) = 0$...
â€“Â user21820
Aug 23 at 14:35

Your attempt makes no sense. $ln(1) = 0$...
â€“Â user21820
Aug 23 at 14:35

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
4
down vote

accepted

Let $x$ be another value. For example, let $x=1$ and you will have $$c_1e+c_2e^-1=0$$

with that you can solve for $c_1$ and $c_2$ uniquely and they are zero or show the determinant of the corresponding matrix is non-zero.

$$beginbmatrixe & e^-1 \ 1 & 1 endbmatrix beginbmatrix c_1 \ c_2endbmatrix= beginbmatrix 0 \ 0endbmatrix$$

answered Aug 19 at 3:03

Siong Thye Goh

79.7k145299

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up vote
2
down vote

For $x=Ln(2)$, $2c_1+1over 2c_2=0$, with $c_1+c_2=0$ you deduce the result.

answered Aug 19 at 3:03

Tsemo Aristide

51.8k11244

Do you deduce the result by plugging in $c_1=-c_2$ such that $2c_1-0.5c_1=0$, meaning $1.5c_1=0$ and thus that $c_1=0=c_2$?
â€“Â Alessandro Verniani
Aug 19 at 3:14

add a commentÂ |Â

up vote
2
down vote

Wronskian of $e^x$ and $e^-x$:

$$beginvmatrix
e^x&e^-x\
e^x&-e^-x\endvmatrix=-2neq 0$$so.....?

answered Aug 19 at 3:17

Chinnapparaj R

1,859317

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
4
down vote

accepted

Let $x$ be another value. For example, let $x=1$ and you will have $$c_1e+c_2e^-1=0$$

with that you can solve for $c_1$ and $c_2$ uniquely and they are zero or show the determinant of the corresponding matrix is non-zero.

$$beginbmatrixe & e^-1 \ 1 & 1 endbmatrix beginbmatrix c_1 \ c_2endbmatrix= beginbmatrix 0 \ 0endbmatrix$$

answered Aug 19 at 3:03

Siong Thye Goh

79.7k145299

add a commentÂ |Â

up vote
4
down vote

accepted

Let $x$ be another value. For example, let $x=1$ and you will have $$c_1e+c_2e^-1=0$$

with that you can solve for $c_1$ and $c_2$ uniquely and they are zero or show the determinant of the corresponding matrix is non-zero.

$$beginbmatrixe & e^-1 \ 1 & 1 endbmatrix beginbmatrix c_1 \ c_2endbmatrix= beginbmatrix 0 \ 0endbmatrix$$

answered Aug 19 at 3:03

Siong Thye Goh

79.7k145299

add a commentÂ |Â

up vote
4
down vote

accepted

Let $x$ be another value. For example, let $x=1$ and you will have $$c_1e+c_2e^-1=0$$

with that you can solve for $c_1$ and $c_2$ uniquely and they are zero or show the determinant of the corresponding matrix is non-zero.

$$beginbmatrixe & e^-1 \ 1 & 1 endbmatrix beginbmatrix c_1 \ c_2endbmatrix= beginbmatrix 0 \ 0endbmatrix$$

answered Aug 19 at 3:03

Siong Thye Goh

79.7k145299

Let $x$ be another value. For example, let $x=1$ and you will have $$c_1e+c_2e^-1=0$$

with that you can solve for $c_1$ and $c_2$ uniquely and they are zero or show the determinant of the corresponding matrix is non-zero.

$$beginbmatrixe & e^-1 \ 1 & 1 endbmatrix beginbmatrix c_1 \ c_2endbmatrix= beginbmatrix 0 \ 0endbmatrix$$

answered Aug 19 at 3:03

Siong Thye Goh

79.7k145299

answered Aug 19 at 3:03

Siong Thye Goh

79.7k145299

answered Aug 19 at 3:03

Siong Thye Goh

79.7k145299

answered Aug 19 at 3:03

Siong Thye Goh

79.7k145299

add a commentÂ |Â

up vote
2
down vote

For $x=Ln(2)$, $2c_1+1over 2c_2=0$, with $c_1+c_2=0$ you deduce the result.

answered Aug 19 at 3:03

Tsemo Aristide

51.8k11244

Do you deduce the result by plugging in $c_1=-c_2$ such that $2c_1-0.5c_1=0$, meaning $1.5c_1=0$ and thus that $c_1=0=c_2$?
â€“Â Alessandro Verniani
Aug 19 at 3:14

add a commentÂ |Â

up vote
2
down vote

For $x=Ln(2)$, $2c_1+1over 2c_2=0$, with $c_1+c_2=0$ you deduce the result.

answered Aug 19 at 3:03

Tsemo Aristide

51.8k11244

Do you deduce the result by plugging in $c_1=-c_2$ such that $2c_1-0.5c_1=0$, meaning $1.5c_1=0$ and thus that $c_1=0=c_2$?
â€“Â Alessandro Verniani
Aug 19 at 3:14

add a commentÂ |Â

up vote
2
down vote

For $x=Ln(2)$, $2c_1+1over 2c_2=0$, with $c_1+c_2=0$ you deduce the result.

answered Aug 19 at 3:03

Tsemo Aristide

51.8k11244

For $x=Ln(2)$, $2c_1+1over 2c_2=0$, with $c_1+c_2=0$ you deduce the result.

answered Aug 19 at 3:03

Tsemo Aristide

51.8k11244

answered Aug 19 at 3:03

Tsemo Aristide

51.8k11244

answered Aug 19 at 3:03

Tsemo Aristide

51.8k11244

answered Aug 19 at 3:03

Tsemo Aristide

51.8k11244

Do you deduce the result by plugging in $c_1=-c_2$ such that $2c_1-0.5c_1=0$, meaning $1.5c_1=0$ and thus that $c_1=0=c_2$?
â€“Â Alessandro Verniani
Aug 19 at 3:14

add a commentÂ |Â

Do you deduce the result by plugging in $c_1=-c_2$ such that $2c_1-0.5c_1=0$, meaning $1.5c_1=0$ and thus that $c_1=0=c_2$?
â€“Â Alessandro Verniani
Aug 19 at 3:14

Do you deduce the result by plugging in $c_1=-c_2$ such that $2c_1-0.5c_1=0$, meaning $1.5c_1=0$ and thus that $c_1=0=c_2$?
â€“Â Alessandro Verniani
Aug 19 at 3:14

add a commentÂ |Â

up vote
2
down vote

Wronskian of $e^x$ and $e^-x$:

$$beginvmatrix
e^x&e^-x\
e^x&-e^-x\endvmatrix=-2neq 0$$so.....?

answered Aug 19 at 3:17

Chinnapparaj R

1,859317

add a commentÂ |Â

up vote
2
down vote

Wronskian of $e^x$ and $e^-x$:

$$beginvmatrix
e^x&e^-x\
e^x&-e^-x\endvmatrix=-2neq 0$$so.....?

answered Aug 19 at 3:17

Chinnapparaj R

1,859317

add a commentÂ |Â

up vote
2
down vote

Wronskian of $e^x$ and $e^-x$:

$$beginvmatrix
e^x&e^-x\
e^x&-e^-x\endvmatrix=-2neq 0$$so.....?

answered Aug 19 at 3:17

Chinnapparaj R

1,859317

Wronskian of $e^x$ and $e^-x$:

$$beginvmatrix
e^x&e^-x\
e^x&-e^-x\endvmatrix=-2neq 0$$so.....?

answered Aug 19 at 3:17

Chinnapparaj R

1,859317

answered Aug 19 at 3:17

Chinnapparaj R

1,859317

answered Aug 19 at 3:17

Chinnapparaj R

1,859317

answered Aug 19 at 3:17

Chinnapparaj R

1,859317

add a commentÂ |Â

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