Triangle Inequality: use to prove convergence

In summary: No but the only information is ##|w|>2R##, ##|z|<R## , I can see clearly that the aim is to express in multiplies of ##w## and I'm pretty sure I said I UNDERSTAND the first inequality so unsure why everyone is replying to that, but anyway, given this all we know is ##|w|>2R## , it could be infinity, that's all we know? how do we bound...We can bound ##|z|+|-2w|## as follows:##|z|+|-2w| \leq |z|+|w|##
  • #1
binbagsss
1,260
11

Homework Statement



Attached

psiconvergence.png


I understand the first bound but not the second.

I am fine with the rest of the derivation that follows after these bounds,

Homework Equations



I have this as the triangle inequality with a '+' sign enabling me to bound from above:

##|x+y| \leq |x|+|y| ## (1)
##|x-y| \geq |x|-|y| ## (2)

and this as the triangle inequality with a '-' sign enabling me to bound from below:

The Attempt at a Solution

So for the first bound we have:

##|z-w| \geq |z| - |w| ##

since we have a strict less than inequality for |z| and a strict greater than equality for |w| , both of these are consistent and we indeed loose the equality option in the triangle inequality to get ##|z-w| > -R ##

I am stuck on the second bound however.
1) I ionly have a upper bound for a subtraction and not a lower via the triangle inequalities. can i get a upper bound from (1) and (2)?

i.e. are you allowewd to do ##|z+(-2w)| \leq |z|+|-2w| ##?

(even if I am, unlike the lower bound, where it turns out the bound we have on ##z## and ##w## are consistent with the triangle inequality, (enabling us to loose the equality and get strictness) there is contrast between the inequalities in this case. ( both items on the right hand side would need lower bounds (or one upper and one equality) but z has a upper bound).

Many thanks .
 

Attachments

  • psiconvergence.png
    psiconvergence.png
    25.3 KB · Views: 915
Physics news on Phys.org
  • #2
binbagsss said:
i.e. are you allowewd to do ##|z+(-2w)| \leq |z|+|-2w| ##?
Yes that is a standard application of the triangle inequality.

(even if I am, unlike the lower bound, where it turns out the bound we have on ##z## and ##w## are consistent with the triangle inequality, (enabling us to loose the equality and get strictness) there is contrast between the inequalities in this case. ( both items on the right hand side would need lower bounds (or one upper and one equality) but z has a upper bound).
Note that the lower bounds are to be expressed as multiples of ##|\omega|##. It is trivial to get such a lower bound for the second term on the RHS. The two inequalities given in the problem enable us to get a lower bound in terms of ##|\omega|## for the first term.
 
  • #3
andrewkirk said:
Yes that is a standard application of the triangle inequality.
Yes that is a standard application of the triangle inequality.

Ok thanks so from similar manipulations with (2) I see that ##|x+y| ## and ##|x-y|## have the same upper and lower bounds from the triangle inequality which makes sense when I think about what the mod function does and possible combinations etc.

andrewkirk said:
The two inequalities given in the problem enable us to get a lower bound in terms of ##|\omega|## for the first term.

That is fine.

andrewkirk said:
Note that the lower bounds are to be expressed as multiples of ##|\omega|##. It is trivial to get such a lower bound for the second term on the RHS.

I still don't understand this. Why is it trivial?
As I said in my OP there is contrast between the inequalities in this case. ( both items on the right hand side would need lower bounds (or one upper and one equality) but z has a upper bound.
 
  • #4
binbagsss said:
I still don't understand this. Why is it trivial?
We are looking for upper bounds, not lower bounds.
We want to find an upper bound for ##|z|+|-2w|## that is no greater than ##\frac52|w|##.

The second term is ##|-2w|## and we want to find an upper bound for it that is a multiple of ##|w|##. Given that an equality is also a upper bound, can you express ##|-2w|## as a multiple of ##|w|## and thereby have an upper bound for ##|-2w|##?

For the first term ##|z|##, the two inequalities you've been given are ##|z|<R## and ##|w|>2R##. How can you use those to get an upper bound for ##|z|## that is a multiple of ##|w|##?
 
  • #5
andrewkirk said:
We are looking for upper bounds, not lower bounds.
We want to find an upper bound for ##|z|+|-2w|## that is no greater than ##\frac52|w|##.

The second term is ##|-2w|## and we want to find an upper bound for it that is a multiple of ##|w|##. Given that an equality is also a upper bound, can you express ##|-2w|## as a multiple of ##|w|## and thereby have an upper bound for ##|-2w|##?

For the first term ##|z|##, the two inequalities you've been given are ##|z|<R## and ##|w|>2R##. How can you use those to get an upper bound for ##|z|## that is a multiple of ##|w|##?

No but the only information is ##|w|>2R##, ##|z|<R## , I can see clearly that the aim is to express in multiplies of ##w## and I'm pretty sure I said I UNDERSTAND the first inequality so unsure why everyone is replying to that, but anyway, given this all we know is ##|w|>2R## , it could be infinity, that's all we know? how do we bound that?
 
  • #6
binbagsss said:
No but the only information is ##|w|>2R##, ##|z|<R##
Put those two together to get an inequality relation between |z| and |w|. Then use that to get an upper bound on |z|+|-2w| (first converting the second term of that into a multiple of |w|).
 
  • #7
andrewkirk said:
Put those two together to get an inequality relation between |z| and |w|. Then use that to get an upper bound on |z|+|-2w| (first converting the second term of that into a multiple of |w|).

I don't believe I'd have any issue with going from a bound between |z| and |w| to one of |z|+|-2w| . My issue in my previous still stands whether I am upper bounding - I assume this is what you meant by an inequality - |z|+|-2w| or |z|+|w| - as far as I can see z can be as large as it likes and so I can't see how we can upper bound anything involving an addition of z. Ta.
 
  • #8
binbagsss said:
I don't believe I'd have any issue with going from a bound between |z| and |w| to one of |z|+|-2w| . My issue in my previous still stands whether I am upper bounding - I assume this is what you meant by an inequality - |z|+|-2w| or |z|+|w| - as far as I can see z can be as large as it likes and so I can't see how we can upper bound anything involving an addition of z. Ta.
?
 
  • #9
$$|z| < R < 2R < |\omega| \text{ ?}$$
 
  • #10
LCKurtz said:
$$|z| < R < 2R < |\omega| \text{ ?}$$

Apologies typo, \omega can be as large as it wants
 

Related to Triangle Inequality: use to prove convergence

What is the Triangle Inequality?

The Triangle Inequality is a mathematical theorem that states that the sum of any two sides of a triangle must be greater than the third side. In other words, the shortest distance between two points is a straight line.

How is the Triangle Inequality used to prove convergence?

The Triangle Inequality is used in mathematical proofs to show that a sequence or series of numbers is convergent. By using the Triangle Inequality, we can bound the terms of the sequence or series and show that they approach a specific limit.

What is the significance of the Triangle Inequality in mathematics?

The Triangle Inequality is an important tool in mathematics because it allows us to prove many theorems and properties, especially in geometry and analysis. It is also used in many real-world applications, such as in computer graphics and optimization problems.

Can the Triangle Inequality be extended to higher dimensions?

Yes, the Triangle Inequality can be extended to higher dimensions, such as in three-dimensional space. In this case, the sum of the lengths of any two sides of a three-dimensional triangle must be greater than the length of the third side.

Are there any variations of the Triangle Inequality?

Yes, there are several variations of the Triangle Inequality, such as the Reverse Triangle Inequality, which states that the absolute value of the difference between two sides of a triangle must be less than or equal to the third side. There are also variations for different types of triangles, such as acute, obtuse, and right triangles.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
340
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
3K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
246
  • Calculus and Beyond Homework Help
Replies
2
Views
823
  • General Math
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
19
Views
2K
Back
Top