Proof of a dot product using sigma notation

In summary, you are trying to prove that the dot product of two vectors, ##(u_1, u_2, …, u_n, v_1, v_2, …, v_n)##, in ##\mathbb R^3## is equal to the sum of the vectors.
  • #1
happyparticle
400
20
Mentor note: Moved from a technical section, so is missing the homework template.
Hi,
I'm always not sure how to prove something in math and I'm wondering if this is enough.

##\vec r \cdot (\vec u + \vec v) ##

##\vec u + \vec v = (u_1+v_1, u_2+v_2,u_3+v_3) = \vec s##

##\vec r \cdot (\vec u + \vec v) = \vec r \cdot \vec s ##

## \sum_{i=1}^3 r_is_i = r_1s_1 + r_2s_2 + r_3s_3 = \vec r \cdot \vec s = \vec r \cdot (\vec u + \vec v)##
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
I'm a bit lost. What are you trying to prove? If it's that you can distribute the dot product, I don't think you've actually written down anything that shows that yet.
 
  • #3
Alright then I always don't know how where to begin. I have to prove I can distribute the dot product, but I have to use the sigma notation.
 
Last edited by a moderator:
  • #4
EpselonZero said:
I'm always not sure how to prove something in math and I'm wondering if this is enough.
EpselonZero said:
I have to prove I can distribute the dot product.
Start with ##\vec r \cdot (\vec u + \vec v) ##, and you should end up with ##\vec r \cdot \vec u + \vec r \cdot \vec v##. Each expression should be connected with '='.

You don't need ##\vec s##.

Is it given that these are vectors in ##\mathbb R^3##? Your summation seems to assume that.

Also, questions like this should be posted in the Homework & Coursework sections. I will move this thread.
 
  • #5
I would start by writing down the definition of ##r\cdot (u+v)## and ##r \cdot u + r \cdot v## using sigma notation, and then try to figure out how to get from one to the other.
 
  • #6
Thanks guys for the tips. Can you tell me if this is correct?

##\vec r \cdot (\vec u + \vec v ) = ##

##\sum_{n=1}^3 r_i(u_i+v_i)=##

##(r_1u_1,r_2u_2,r_3u_3) + (r_1v_1,r_2v_2r_3,v_3) = ##

## (r_1, r_2, r_3) \cdot [(u_1,u_2,u_3) + (v_1,v_2,v_3)] =##

##\vec r \cdot \vec u + \vec r \cdot \vec v##
 
  • #7
I think everything you wrote is a true statement, but doesn't seem to be particularly motivated as being obvious given what you are trying to prove. I think as a first question, to reiterate what mark asked, are these vectors in ##\mathbb{R}^3##?

I think you should try to never write down all the coordinates individually, given how the question is asked. What can you do with ##\sum r_i(u_i+v_i)##? Can you distribute that?
 
  • Like
Likes happyparticle
  • #8
These are vector in R³.
I don't see how I can do that if I don't have to write down the coordinates individually because I have to use
## \sum_{i=1}^3 u_iv_i##
else
## \sum_{i=1}^1 \vec r \cdot (\vec u + \vec v) ## , but that doesn't prove anything.
 
  • #9
So are you saying there are no mathematical operations you can do on ##\sum r_i (u_i+v_i)##? What about distribution over the real numbers?
 
  • #10
I'm sorry, but I'm not sure to understand.
The only thing I see is that
##\sum r_i(u_i + v_i) = r(u + v) = ru + rv##

However, in my mind that's not enough.
 
  • #11
What about something like
##\sum r_i(u_i +v_i) = \sum r_i u_i + r_i v_i##?

In some sense I feel like that is literally the only elementary algebraic operation you can do. What's the next step?
 
  • #12
EpselonZero said:
Thanks guys for the tips. Can you tell me if this is correct?
No, it's not correct.

##\vec r \cdot (\vec u + \vec v )## is a number whereas ##(r_1u_1,r_2u_2,r_3u_3) + (r_1v_1,r_2v_2r_3,v_3)## is a vector. They can't possibly be equal.
 
  • #13
EpselonZero said:
I'm sorry, but I'm not sure to understand.
The only thing I see is that
##\sum r_i(u_i + v_i) = r(u + v) = ru + rv##

However, in my mind that's not enough.
You're right that that's not enough. You're just asserting what you're being asked to prove.

One of the difficulties with these kinds of proofs is what you're supposed to prove seems obvious. You need to think like a mathematician. A mathematician would say ##r_i(u_i+v_i) = r_i u_i + r_i v_i## is axiomatically true because ##r_i##, ##u_i##, and ##v_i## are real numbers. ##\vec r \cdot (\vec u + \vec v)##, however, needs to be proved because you have new mathematical objects, namely, the vectors, and a different kind of addition and multiplication, vector addition and the dot product.

The idea here is to express the vector operation in terms of real numbers, according to its definition, working with the real quantities to massage the expression into a form where you can translate it back to vector notation, again according to the relevant definitions.
 
  • Like
Likes PeroK
  • #14
Let me add that when you wrote
$$\vec r \cdot (\vec u + \vec v) = \sum_{i=1}^3 r_i(u_i+v_i),$$ you actually glossed over a step. It really should be
$$\vec r \cdot (\vec u + \vec v) = \sum_{i=1}^3 r_i(\vec u+\vec v)_i = \sum_{i=1}^3 r_i(u_i+v_i),$$ where the first equality comes from the definition of the dot product and the second equality is justified by the definition of vector addition. (Whether you need to be this nit-picky though depends on what your instructor requires.)

The next step would be what the Shredder suggested
$$\sum_{i=1}^3 r_i(u_i+v_i) = \sum_{i=1}^3 (r_i u_i+r_i v_i),$$ which you can justify as being true because of the distributive property of real numbers.
 
  • Like
Likes happyparticle
  • #15
As you said, one of my mistake was about the vectors and the real numbers.

## \vec r \cdot (\vec v + \vec u) = ##
## \sum_{i=1}^3 r_i(\vec v + \vec u)_i =##
## \sum_{i=1}^3 r_i( v_i + u_i) =##
##\sum_{i=1}^3 (r_iv_i + r_iu_i) =##
##(r_1v_1 + r_1u_1) + (r_2v_2 + r_2u_2) + (r_3v_3 + r_3u_3) =##
##(r_1v_1 + r_2v_2 + r_3v_3 ) + (r_1u_1 + r_2u_2 + r_3u_3) =##
## \vec r \cdot \vec v + \vec r \cdot \vec u##

I'm not sure ##(r_1 + r_2 + r_3 ) = \vec r ##
 
  • #16
EpselonZero said:
As you said, one of my mistake was about the vectors and the real numbers.

## \vec r \cdot (\vec v + \vec u) = ##
## \sum_{i=1}^3 r_i(\vec v + \vec u)_i =##
## \sum_{i=1}^3 r_i( v_i + u_i) =##
##\sum_{i=1}^3 (r_iv_i + r_iu_i) =##
You don't need the two lines below. The above is equal to
##\sum_{i=1}^3 r_iv_i + \sum_{i=1}^3 r_iu_i##
The above can be rewritten as
## \vec r \cdot \vec v + \vec r \cdot \vec u##, which is what you are intended to show.
EpselonZero said:
##(r_1v_1 + r_1u_1) + (r_2v_2 + r_2u_2) + (r_3v_3 + r_3u_3) =##
##(r_1v_1 + r_2v_2 + r_3v_3 ) + (r_1u_1 + r_2u_2 + r_3u_3) =##
## \vec r \cdot \vec v + \vec r \cdot \vec u##

I'm not sure ##(r_1 + r_2 + r_3 ) = \vec r ##
They are not equal. The left side is a real number; the right side is a vector. A vector isn't equal to the sum of its components.
 
  • Like
Likes happyparticle
  • #17
Alright, thanks guys for your patience.
one more thing.

##r_1v_1 + r_2v_2 + r_3v_3 = \vec r \cdot \vec v## ?
 
  • Like
Likes PeroK
  • #18
EpselonZero said:
Alright, thanks guys for your patience.
one more thing.

##r_1v_1 + r_2v_2 + r_3v_3 = \vec r \cdot \vec v## ?
Yes, because that's generally the definition of the dot product.
 
  • Like
Likes happyparticle
  • #19
EpselonZero said:
##r_1v_1 + r_2v_2 + r_3v_3 = \vec r \cdot \vec v## ?
PeroK said:
Yes, because that's generally the definition of the dot product.
Keep in mind that the above is the coordinate definition of the dot product (of vectors in ##\mathbb R^3##).
The coordinate-free definition uses the magnitudes of the vectors and the cosine of the angle between them.
## \vec u \cdot \vec v = |\vec u | |\vec v | \cos(\theta)##
 
  • #20
Thanks alot!
 

1. What is a dot product?

A dot product is a mathematical operation that takes two vectors and produces a scalar quantity. It is also known as the scalar product or inner product.

2. What is sigma notation?

Sigma notation is a shorthand way of writing mathematical series or sums. It uses the Greek letter sigma (∑) to represent the sum of a sequence of terms.

3. How is a dot product expressed using sigma notation?

A dot product can be expressed using sigma notation as follows: ∑i=1n ai * bi, where ai and bi are the corresponding components of the two vectors and n is the number of components.

4. What is the purpose of using sigma notation in proving a dot product?

Sigma notation makes it easier to express and manipulate mathematical series or sums, making the proof of a dot product more concise and easier to understand.

5. Can you provide an example of a proof of a dot product using sigma notation?

Yes, here is an example: Let vector a = [1, 2, 3] and vector b = [4, 5, 6]. The dot product of a and b can be expressed as ∑i=13 ai * bi = (1*4) + (2*5) + (3*6) = 32. This can also be written as aTb = [1, 2, 3] * [4, 5, 6] = 32, where aT is the transpose of vector a.

Similar threads

Replies
3
Views
1K
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
814
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Special and General Relativity
Replies
1
Views
541
  • Linear and Abstract Algebra
Replies
5
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
821
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
Back
Top