Verify linear independence of function

Question

The prompt is to to verify linear independence among the following functions $\{1, \sin{x}, e^{x^{2}}\}$

The way I went on solving this problem was by multiplying them with random variables, like $$C_11 + C_2\sin{x} + C_3 e^{x^2}$$ and try to prove that $C_1 = C_2 = C_3 = 0$ which I haven't been able to do assuming $x = \frac{\pi}{2}$

Answer 1

Take $x=0$, $x=\frac\pi2$, and $x=\pi$. You will get $3$ linearly independent vectors.

Answer 2

To check linear independence you need to calculate the Wronskian.$$W(f_1,f_2,f_3)=|W|$$where$$W=\begin{pmatrix} 1&\sin x &e^{x^2}\\0&\cos x&2xe^{x^2}\\0 &-\sin x&e^{x^2}(2+4x^2)\end{pmatrix}$$therefore by calculating the determinant we have:$$W(f_1,f_2,f_3)=e^{x^2}([2+4x^2]\cos x+2x\sin x)$$which is zero only in the roots of $[2+4x^2]\cos x+2x\sin x=0$ and obviously$$W(0)=2\\W(\dfrac{\pi}{2})=W(-\dfrac{\pi}{2})=\pi e^{\dfrac{\pi^2}{4}}$$therefore the functions are linearly independent.

Answer 3

With $x=0$ and $x=\pi$ (roots of the sine),

$$\begin{cases}C_1+C_3=0,\\C_1+ C_3\lambda=0,\end{cases}$$ where $\lambda\ne1$ (no need to evaluate it). Hence $C_1=C_3=0$.

Now only $C_2\sin x=0$ remains and you can conclude.

Answer 4

To put up yet another answer, which comes from the comment I wrote to the OP...

Substituting $x=0, x=\pi/2, x=-\pi/2$ you get:

$$\begin{array}{rlrlrlr}C_1&&&+&C_3&=&0\\C_1&+&C_2&+&KC_3&=&0\\C_1&-&C_2&+&KC_3&=&0\end{array}$$

where $K=e^{(\pi/2)^2}$. This system can be solved in $C_1, C_2, C_3$ and the unique solution is $C_1=C_2=C_3=0$.

(Actually you have a lot of freedom in choosing the three values for $x$, I just find those particular values $x=0, x=\pi/2, x=-\pi/2$ to make for the simplest system of equations I can think of.)