Question

show that the following set of functions are subspaces of F(-infinity,infinity)
a) all everywhere differentiable function that satisfy f'+2f=0

Answer 1

Luck>

Answer 2

thats easy

Answer 3

all you need are closure under addition scalar multiplication and a zero vector

Answer 4

and those properties come straight from differenitation

Answer 5

\[\frac{d}{dx} ( \lambda f(x) ) = \lambda \frac{d}{dx} f(x)\]

Answer 6

i did not et the point please explain

Answer 7

\[(f+g)'(x) = f'(x)+g'(x)\]

Answer 8

and the zero vector (i.e. ) f=0 , obviously solves the equation

Answer 9

for a subspace you need three things:

1. closure under scalar multiplication

2. closure under vector addition

3. must include a zero vector

Answer 10

To prove 3, it is obvious , f=0 solves the equation

Answer 11

but you are not solving according to condition f'+2f=0

Answer 12

now lets prove 2.

consider multiplying both sides by \[\lambda\]

Answer 13

\[\lambda ( f' + 2f ) = \lambda (0) =0\]

Answer 14

so that \[(\lambda f )' +2 ( \lambda f ) =0 \]

Answer 15

which has the exact same form as the original equation , so if f solves the equation then any scalar multiple of f will solve the equation, so its closed under scalar multiplication

Answer 16

but how under addition

Answer 17

Sorry, i aint good with this.

Answer 18

then to prove part one

consider two functions f and g that satisify the equations , so that

f ' +2f =0 and

g' + 2g =0


add these two equations

f' +g' + 2f + 2g =0

and remember that the sum of the derivatives is the derivative of the sum, duifferentiation is a linear operation!.

group the LHS so:

(f+g) ' + 2 (f +g) =0

so if and f and g satisfy it, then their sum will satisfy it , so its closed under vector addition. There it is a subspace.

Answer 19

remember, you arent solving anything with these questions, its all basic algebra and remember basic things like linearity of differentiation and integration.

I did these last year , I can even get some slightly harder questions to put up if you really want , because this is the easiest they will ever be.

there is no maths you are just checking those three simple things.

Answer 20

@elecengineer please send some examples i got ur point thanks

Answer 21

The solution to the differential equation is of the form \[ce^{-2x}\]
Zero vector, c = 0
Additive closure, \[c_2e^{-2x} + c_1e^{-2x} = (c_1+c_2)e^{-2x}\]
Multiplicative closure, \[(kc)e^{-2x}\]

Answer 22

Damn , I thought I thought had some questions involving integrals but turns out I dont

Answer 23

\[(x+1)p ' (x) - 3p(x) =0 \]

show that is a subspace if you want, its pretty much the same ideas , using linear rules of differentiation

Answer 24

@ sir are you electrical engineer

Answer 25

or an even more trivial question would be \[p''(x) =0 \]

showing thats a subspace is done in 1 line