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How do i show that $f_{1}(x)=1$, $f_{2}(x)=e^{x}$ and $f_{3}(x)=\sin{x}$ are linearly independent, as elements of the vector space, of continuous functions $\mathcal{C}[0,1]$.

So for showing these elements are linearly independent, one needs to show that if $$ a_{1} \cdot 1 + a_{2} \cdot e^{x} + a_{3} \cdot \sin{x}=0$$ then from this we should conclude that $a_{1}=a_{2}=a_{3}=0$. But i am not being able to deduce this.

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You don't mean linearly independent over C[0,1], you mean linearly independent in C[0,1]. The functions are dependent over C[0,1]. Let a1 = e^x, a2 = 1, a3 = 0. –  Carl Mummert Dec 3 '10 at 14:32
    
@carl: thanks carl –  anonymous Dec 3 '10 at 14:38
    
Perhaps you can use the taylor expansions of the functions to make this simpler. –  AnonymousCoward Dec 3 '10 at 18:08

6 Answers 6

up vote 14 down vote accepted

If you differentiate your equation with respect to $x$ four times, you get that $a_2 e^x+a_3 \sin x = 0$, from which it follows that $a_1=0$. Now, letting $x=0$, you see that $a_2 e^x=0$, so $a_2=0$, and it follows readily that $a_3=0$.

With problems like this, always remember that you are dealing with functions. That means that you can treat them as functions as well: you can differentiate them and set their values equal to something.

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Nice! Also it's enough to differentiate twice for the same argument to work. –  user1119 Dec 4 '10 at 21:10

You may calculate the Wronskian of $f_1(x)$, $f_2(x)$, $f_3(x)$ and check that it doesn't vanish on $[0,1]$. This implies that the functions are linearly independent.

A caveat. If the Wronskian of a set of functions is equal to zero identically, it doesn't follow in general that the functions are linearly dependent.

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Following up on Fredrik Meyer's post, you can just pick points to evaluate the functions at and show that the resulting equations require $a_1, a_2$, and $a_3$ are 0. So at 0 you have $a_1+a_2=0$. At 1 you have $a_1+a_2\exp (1) +a_3\sin (1)=0$. At $\frac{\pi}{4}$ you have $a_1+a_2\exp(\frac{\pi}{4}) + a_3\frac{\sqrt{2}}{2}=0$. These will show that $a_1, a_2$, and $a_3$ are 0.

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HINT $\rm\ \ y(0) = a+b,\ \ y'(0) = b+c,\ \ y''(0) = b\ $ and they must all be $\:0\:$ if $\rm\ y(x) = 0 $

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You have $a + b \sum_{k\geq 0} \frac{x^k}{k!} + c \sum_{k\geq 0} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!} = 0$

All the coefficients must be 0 above, so, for the constant term we have $a+b = 0$ and comparing coefficients of $x$ we have $b+c=0$, comparing coefficient of $x^2$ we have $b=0$, from which it follows $a=b=c=0.$

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You need to show that there are no $\alpha$, $\beta$ such that $\alpha e^x + \beta \sin x = \mathrm{const} \neq 0$ on $[0, 1]$. To do that, observe that $e^x$ is strictly convex and $\sin x$ is strictly concave on $[0, 1]$ and both are strictly increasing. It is easy to show that convexness and concaveness are invariant in respect to multiplication by a positive number, and they swap under multiplication by a negative number, so if $\alpha > 0$ and $\beta > 0$ or both are less than zero, then the linear combination is not constant (otherwise the summands would be either both strictly convex or both strictly concave). If $\alpha > 0$ and $\beta < 0$, then one summand is strictly increasing while the other is strictly decreasing, which raises the same problem. The last case to check is when either $\alpha$ or $\beta$ is zero, but it is then obvious.

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