Angle Between 1 and x in C[0,1] Using Inner Product (3)

In summary, In C[0,1], with inner product defined by (3), consider the vectors 1 and x. Find the angle theta between 1 and x.
  • #1
Dustinsfl
2,281
5
In C[0,1], with inner product defined by (3), consider the vectors 1 and x.

Find the angle theta between 1 and x.

(3)[tex]\int_{0}^{1}f(x)g(x)dx[/tex]

Find the angle theta between 1 and x

I don't know what to do with polynomial inner product vector space
 
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  • #2
Can't you compute the inner product of 1 and x given the definition?
 
  • #3
Yes but how is that going to find the angle?
 
  • #4
How would you find the angle between two ordinary vectors? Wouldn't you use something like this?

[itex]u \cdot v = |u| |v| cos(\theta)[/itex]

This idea can be generalized to any inner product space.
 
  • #5
After taking the integral, I obtain 1/2. How does that help me obtain the angle of pi/6?
 
  • #6
Dustinsfl said:
After taking the integral, I obtain 1/2. How does that help me obtain the angle of pi/6?

Didn't you see Mark44's suggestion? Set u=1 and v=x and then figure out u.v, |u|, |v| and then what cos(theta) is. Remember |v|=sqrt(v.v). It's a number, not a function.
 
  • #7
By definition, <1,1> is the integral from 0 to 1 of 1^2 which is 1/2.
And <x,x> is the integral from 0 to 1 of x^2 which is 1/3.

Now the equation is 1/2=1/6 cos theta so theta is arccos 3 which isn't pi/6.
 
  • #8
Dustinsfl said:
By definition, <1,1> is the integral from 0 to 1 of 1^2 which is 1/2.
And <x,x> is the integral from 0 to 1 of x^2 which is 1/3.

Now the equation is 1/2=1/6 cos theta so theta is arccos 3 which isn't pi/6.

The integral of 1*1 from 0 to 1 is 1/2? And I already warned you that |x|=sqrt(<x,x>).
 
  • #9
1 sorry. The definition of inner product on this space is (3)

<u, v>=[tex]\int_{0}^{1}f(x)g(x)[/tex] which implies <x, x>=[tex]\int_{0}^{1}x^{2}=1/3[/tex].


So now we have 1/2=1/3 cos theta and now 3/2=cos theta which is greater than 1 so it doesn't exist.
 
  • #10
Dustinsfl said:
1 sorry. The definition of inner product on this space is (3)

<u, v>=[tex]\int_{0}^{1}f(x)g(x)[/tex] which implies <x, x>=[tex]\int_{0}^{1}x^{2}=1/3[/tex].


So now we have 1/2=1/3 cos theta and now 3/2=cos theta which is greater than 1 so it doesn't exist.

<x,x> is 1/3. That doesn't mean |x|=1/3.
 
  • #11
Ok I understand now.
 

1. What is the definition of the inner product in this context?

The inner product in this context refers to the mathematical operation that measures the angle between two vectors in the vector space C[0,1]. It is calculated by taking the dot product of the two vectors and dividing it by the product of their magnitudes.

2. How is the angle between 1 and x in C[0,1] calculated using the inner product?

The angle between 1 and x in C[0,1] is calculated by first finding the inner product between the two vectors, and then using the formula: cosθ = (u · v)/(|u|*|v|) where u is the vector 1 and v is the vector x. The resulting angle θ is the measure of the angle between the two vectors.

3. Can the angle between 1 and x in C[0,1] be negative?

No, the angle between 1 and x in C[0,1] cannot be negative. The inner product always returns a positive value, and the formula for calculating the angle only uses the magnitude of the vectors, making the resulting angle always positive.

4. How does the angle between 1 and x relate to the shape of the functions in C[0,1]?

The angle between 1 and x in C[0,1] represents the similarity or dissimilarity between the two functions. If the angle is close to 0, it means that the two functions are very similar, while a larger angle indicates a greater difference between the two functions.

5. Can the angle between 1 and x in C[0,1] be used to compare functions in other vector spaces?

Yes, the concept of the angle between two vectors using the inner product can be applied to any vector space. However, the specific formula for calculating the angle may differ depending on the properties of the vector space.

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