Let y_1=sin(x) and y_2=cos(x). Which of the following functions are linearly independent from y_1, y_2? A. sin^2(x) B. 2cos(x) C. sin(x)+cos(x) D. 0 E. sin(2x)
B. 2 * y_2 (linearly dependent) C. y_1 + y_2 (linearly dependent) D. 0 * y1_1 (linearly dependent) E. 2 * y_1 * y_2 (linearly dependent)
A. y_1 ^ 2 (NOT linearly dependent)
so a*sin(x)+b*cos(x)=0 where a and b are not both 0 if a =0, then b*cos(x)=0 and b can be any number since cos does equal 0 sometimes ---- a=1 and b=1 so sin(x)+cos(x)=0 happens so this combination works (1,1) ---- sin(2x)=2sin(x)cos(x)=sin(x)cos(x)+sin(x)cos(x)=0 so I'm thinking any of the above that can be written as a*sin(x)+b*cos(x) should work where a and b i guess can vary as it did in the sin(2x) --- so i think two of the sets are linear dependent --- I could be wrong though
a*sin(x)+b*cos(x)=0 where a and b are not all 0 means {sin(x),cos(x)} is linearly independent
but I'm thinking a and b don't exactly have to be constant numbers as shown in my sin(2x)
just thinking
oops got my definitions turned around
a*sin(x)+b*cos(x)=0 where a and b are not all 0 means {sin(x),cos(x)} is linearly dependent so i think two of the sets are linear independent *
expressions not sets*
I am not sure of my answers.
I think all of your answers are good I'm just iffy on the 0 there
I was curious about the 0 also..
@vmarroquin1002 Are you allowed to pick more than one answer?
Two vectors u and v are linearly independent if the only numbers x and y satisfying xu+yv=0 are x=y=0. If we let
Yes I am. Sorry I forgot to clarify that.
this was copied from http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/lindep/lindep.html
so according to that definition all of the things can be zero for the definition linearly independent to apply
so @vmarroquin1002 are you cool? like do you feel you need any further explanation on anything?
actually i'm still thinking about these answers
It was A and E
What is the definition for linear independence given in your book/notes?
a*sin(x)+b*cos(x) I guess that means a and b have to be constants and can all be zero?
Yeah, that is what it looks like.
using my logic from earlier i could have said sin^2(x)=sin(x)*sin(x)+0*cos(x) lol
the only one that wouldn't work was the 0 since a and b couldn't both be 0
from my logic earlier
I wasn't even thinking vectors. I was trying to express each choice as a combination of y_1 and y_2 without raising either one to a power (other than 1).
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