# Showing a twice differentiable function is a vector space

• vcb003104
In summary: Thanks for the reply,Thus af(x) + bg(x)=a[sin(x)f"(x)+x^{2}f(x)] + b[sin(x)g"(x)+x^{2}g(X)] =a(0) +b(0)=0 Yes, that shows that applying the differential operator to both af and bg separately gives 0.But what you want to show is that sin(x)(af+ bg)''+ x^2(af+ bg)= 0Can you do that?
vcb003104

## Homework Statement

Show that the set of twice differentiable functions f: R→R satisfying the differential equation

sin(x)f"(x)+$x^{2}$f(x)=0is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f.

## Homework Equations

sin(x)f"(x)+$x^{2}$f(x)=0

## The Attempt at a Solution

I have no idea how to start this question.
I know that to prove it is a vector space, proving that it is a subspace is enough?
I flipped through the notes for my course and followed an example done in class. I got :

0(x) = sin(x)0"(x) + $x^{2}$0(x) = 0

vcb003104 said:

## Homework Statement

Show that the set of twice differentiable functions f: R→R satisfying the differential equation

sin(x)f"(x)+$x^{2}$f(x)=0is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f.

## Homework Equations

sin(x)f"(x)+$x^{2}$f(x)=0

## The Attempt at a Solution

I have no idea how to start this question.
I know that to prove it is a vector space, proving that it is a subspace is enough?
If your set was contained within an encompassing vector space. Can you establish that the space that your subset belongs to is a vector space? If you can't do that, you'll need to verify that all of the vector space axioms are satisfied.
vcb003104 said:
I flipped through the notes for my course and followed an example done in class. I got :

0(x) = sin(x)0"(x) + $x^{2}$0(x) = 0
This is pretty sketchy. What are you trying to say here?

Hi, thanks for the quick reply.

I copied that from another example that my lecturer did in class.

I know that I need to prove the axioms but I don't know how to?

For example, to prove u+v is in V, how do I do it? what's u and what's v?

Do I say something like:
sin(u)f"(u)+$u^{2}$f(u)=0
sin(v)f"(v)+$v^{2}$f(v)=0

since $u+v\in R$
sin(u+v)f"(u+v)+$(u+v)^{2}$f(u+v)=0thanks

Mark44 said:
If your set was contained within an encompassing vector space. Can you establish that the space that your subset belongs to is a vector space? If you can't do that, you'll need to verify that all of the vector space axioms are satisfied.

This is pretty sketchy. What are you trying to say here?
So I googled how to prove the u + v is an element of V thing.

In this context, if I do: sin(u+v)f"(u+v) + $(u+v)^{2}$f(u+v) = o
I want to get it to become something like

sin(u)f"(u) + $u^{2}$f(u) + sin(v)f"(v)+$v^{2}$f(v) = 0
so 0 + 0 = 0 right?

but how can I expand them? I mean with sin(u+v) I know I can at least get sin(u)cos(v) + sin(v)cos(u) but what about f(u+v) and f"(u+v)??

Thanks a bunch
and hope I make sense

Last edited:
No. The point is NOT to replace "x" with u+ v. You want to prove that the set of all solutions to this (linear, homogeneous) differential equation forms a vector space. So need to show that if f(x) and g(x) both satisfy the equation then so does af(x)+ bg(x) for any numbers a and b.

HallsofIvy said:
No. The point is NOT to replace "x" with u+ v. You want to prove that the set of all solutions to this (linear, homogeneous) differential equation forms a vector space. So need to show that if f(x) and g(x) both satisfy the equation then so does af(x)+ bg(x) for any numbers a and b.

Hi there,

So to show f(x) and g(x) both satisfy the equation, do I just write

sin(x)f"(x)+$x^{2}$f(x)=0
sin(x)g"(x)+$x^{2}$g(X)=0

Thus af(x) + bg(x)
=a[sin(x)f"(x)+$x^{2}$f(x)] + b[sin(x)g"(x)+$x^{2}$g(X)]
=a(0) +b(0)
=0

Sorry I'm a bit slow at this

vcb003104 said:
Hi there,

So to show f(x) and g(x) both satisfy the equation, do I just write

sin(x)f"(x)+$x^{2}$f(x)=0
sin(x)g"(x)+$x^{2}$g(X)=0Thus af(x) + bg(x)
=a[sin(x)f"(x)+$x^{2}$f(x)] + b[sin(x)g"(x)+$x^{2}$g(X)]
=a(0) +b(0)
=0
Yes, that shows that applying the differential operator to both af and bg separately gives 0.
But what you want to show is that
$sin(x)(af+ bg)''+ x^2(af+ bg)= 0$

Can you do that? (It is important that this is a "linear, homogeneous" differential equation and that differentiation is a "linear" operation.)

Sorry I'm a bit slow at this

1 person
HallsofIvy said:
Yes, that shows that applying the differential operator to both af and bg separately gives 0.
But what you want to show is that
$sin(x)(af+ bg)''+ x^2(af+ bg)= 0$

Can you do that? (It is important that this is a "linear, homogeneous" differential equation and that differentiation is a "linear" operation.)

So I can just expand it?

Can I say, showing that this is a vector space via the addition of functions:
$sin(x)(af+ bg)''+ x^2(af+ bg)= 0$

LHS = sin(x)(af+ bg)''+ $x^{2}$(af+ bg)
= sin(x)(af)"+sin(x)(bg)" + $x^{2}$ (af) + $x^{2}$ (bg)
= sin(x)af"(x) + $x^{2}$af(x) + sin(x) bg"(x) + $x^{2}$ bg(x)
= a[sin(x)f"(x) + $x^{2}$f(x)] +b[sin(x) g"(x) + $x^{2}$ g(x)]
= 0 + 0
= 0
= RHS

vcb003104 said:
So I can just expand it?

Can I say, showing that this is a vector space via the addition of functions:
$sin(x)(af+ bg)''+ x^2(af+ bg)= 0$

LHS = sin(x)(af+ bg)''+ $x^{2}$(af+ bg)
= sin(x)(af)"+sin(x)(bg)" + $x^{2}$ (af) + $x^{2}$ (bg)
= sin(x)af"(x) + $x^{2}$af(x) + sin(x) bg"(x) + $x^{2}$ bg(x)
= a[sin(x)f"(x) + $x^{2}$f(x)] +b[sin(x) g"(x) + $x^{2}$ g(x)]
= 0 + 0
= 0
= RHS

Yes, that's it.

So that is the additional of functions right?

How about multiplication? is it included in the equation as well?

vcb003104 said:
So that is the additional of functions right?

How about multiplication? is it included in the equation as well?

Yes. If you take the case a=1 and b=1, it tests addition on f+g. If you take the case g=0 it tests multiplication on af. Your test does both at once.

## 1. What does it mean for a function to be twice differentiable?

A function is said to be twice differentiable if its first and second derivatives exist and are continuous over its entire domain.

## 2. How is a twice differentiable function different from a once differentiable function?

A twice differentiable function has a second derivative, while a once differentiable function only has a first derivative. This means that a twice differentiable function is smoother and has a more gradual change in slope compared to a once differentiable function.

## 3. Why is it important to show that a twice differentiable function is a vector space?

Showing that a twice differentiable function is a vector space means that it satisfies all the properties of a vector space, such as closure under addition and scalar multiplication. This allows for the use of vector space techniques and theorems in analyzing and solving problems involving the function.

## 4. How do you prove that a twice differentiable function is a vector space?

To prove that a twice differentiable function is a vector space, one must show that it satisfies the 10 axioms of a vector space. These include properties such as closure under addition and scalar multiplication, existence of an additive identity and inverse, and associativity and distributivity.

## 5. Can any twice differentiable function be considered a vector space?

No, not all twice differentiable functions can be considered vector spaces. Only those that satisfy all the axioms of a vector space can be considered as such. A function that does not satisfy even one of the axioms cannot be considered a vector space.

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