#Linear independence and generating of a family of functions
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start with definition of linear independence
suppose a1 + bx + c*sinx = 0 for all x in R
in particular for x=0, conclude that a = 0
so now you're left with bx + csinx =0 for all x
if you take limit as x tends to + inf, the expression is unbounded given that b is nonzero
but the rhs is constantly 0, which is a contradiction
so b=0 must hold
and finally conclude that c=0 and you are done
@haughty ridge
Oh So you can use several x’s for the same scalars?
Thank you
of course
you have to keep in mind the vectors are functions in this case
equality of functions is defined pointwise

@little sigil sorry to bother but could you also just take x = 0 to prove a = 0 then x = pi to prove c = 0 then b will also be 0 for some other x?
yes
with x=0 we get a=0
then put x = pi, then bpi + csinpi = bpi = 0, hence b = 0
and finally c sin pi/2 = c = 0
@haughty ridge
you don't have to use the limit argument, but you can
Yeah that’s what I did. Thanks for explaining the limit regardless
@haughty ridge
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