Question

are: x^3 and x^2*|x|, Linear independent?

Answer 1

\[(x^{3}) and( x^{2}|x|)\]

Answer 2

no (i don't think) because you can write
\[x^2|x| = x^3 \]or
\[x^2|x|=-x^3\]

Answer 3

yea, that is also what i assume, so I have to prove that there are no a and b so that(except the trivial case a=0 and b=0): \[a*(x3)+b*(x2|x|)=0\]

Answer 4

and a and b must be in \[\mathbb{R}\]

Answer 5

what space are you working in?

Answer 6

\[[-1,1]\rightarrow \mathbb{R}\]

Answer 7

because in fact it is possible to have
\[ax^3+bx^2|x|=0\] say x = -3, a = 1, b = -1

Answer 8

oh so space of continuous functions from
\[[-1,1]->R\]

Answer 9

those functions are in vectorspace of all functions, and to be shown is that they are linear independent if they are. I assumption was also all the time that they are independent But how would one go and prove it

Answer 10

yea [−1,1]−>R

Answer 11

i guess you have to work in cases.
\[ x^2|x| = \left\{
\begin{array}{lr}
x^3 & : x \geq 0\\
-x^3 & : x < 0
\end{array}
\right.\]

Answer 12

and the equality
\[ax^3+bx^2|x|=0\]
means
\[ax^3+bx^2|x|=0 \forall x\in [-1,1]\]

Answer 13

my thought was: \[x ^{2}|x|\] this is always positive, and because \[x^{3}\] can be also negative there is no any single a that could make \[a∗(x3)+b∗(x2|x|)=0\] true

Answer 14

a should be dependable about x (if <0 or if >0)

Answer 15

so since clearly you have positive values of x in [-1,1] you get
\[ax^3+bx^2|x|=0\] mean in particular that for all
\[x \in [0,1]\]
\[ax^3+bx^3=0\]

Answer 16

AND
\[ax^3-bx^3=0\]
\[\forall x \in [-1,0)\]

Answer 17

both have to be true, since it is true
\[\forall x \in [-1,1]\]

Answer 18

yes true and therefore I can't define any value a and b that would generally make it hold hold.

Answer 19

the first equality says
\[(a+b)x^3=0 \forall x \in [0,1]\] therefore
\[a=-b\]

Answer 20

the second says
\[(a-b)x^3=0 \forall x \in [-1,0)\] therefore
\[a=b\]

Answer 21

and the only way that
\[a=b\] and \[a=-b\] is if b = 0, hence so is a

Answer 22

But I understand that I should be after all be able to define a and b so that they would make it hold for every case -1 =< x = <1

Answer 23

right

Answer 24

otherwise they are linear independent

Answer 25

like the case here?

Answer 26

the logic goes like this:
assume
\[ax^3+bx^2|x|=0 \forall x \in [-1,1]\]

Answer 27

oh damn
i have just tried to show dependence and i showed independence!

Answer 28

i have showed that if
\[ax^3+bx^2|x|=0 \forall x \in [-1,1]\] then
\[a=b=0\]

Answer 29

so maybe my first assumption was wrong!

Answer 30

but that would be the case here i think since we cant define any specific a and b that would hold for -1 < x < 1

Answer 31

since there is a need to seperate 2 cases for x negative and positive

Answer 32

i always think of it as this: can you write one as a linear combination of the other. the formal definition says the same thing. so the question is can we write
\[x^2|x|\] as some linear combination of
\[x^3\] and i guess we cannot.

Answer 33

since one is a piecewise function and the other is not

Answer 34

i guess they are independent after all!

Answer 35

yes I think so too

Answer 36

whew! learn something new every day. mainly that i should not jump to a conclusion. since i thing the logic was sound and we have shown that if
\[ax^3+bx^2|x|=0 \forall x \in [-1,1]\] then
\[a=b=0\]

Answer 37

pretty sure actually, we would need a function in a place of a or b to make it work, but since it must be from \[R\] there is no way to make it work

Answer 38

then done. good work.

Answer 39

ok, thanks for your help