How can I use modulus and inequalities to simplify my equations?

In summary, by using the triangle inequality and properties of absolute values, it is possible to show that |f(t)|^2 is less than or equal to 1 based on information about |f_n(t)|^2 and |f_n(t) - f(t)|^2. This can be useful in proving properties related to uniform convergence in metric spaces.
  • #1
Somefantastik
230
0
Hello,

I've got

[tex] |a|^{2} = |a - b + b|^{2} [/tex]

What can I do with this guy? Usually when the square isn't there I use the triangle inequality and things fall out pretty quick, for example,

[tex] |a| = |a - b + b| \leq |a-b| + |b| [/tex]

Is there something like that I can do with the orginial guy?
 
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  • #2
Somefantastik said:
Hello,

I've got

[tex] |a|^{2} = |a - b + b|^{2} [/tex]

What can I do with this guy? Usually when the square isn't there I use the triangle inequality and things fall out pretty quick, for example,

[tex] |a| = |a - b + b| \leq |a-b| + |b| [/tex]

Is there something like that I can do with the orginial guy?
Well, |a - b + b|2 = |a|2
 
  • #3
What are you looking for?
 
  • #4
mathman said:
What are you looking for?

I know a property concerning |a-b|2 (it's less than some epsilon) and I know what |b|2 is, so, if I can break this down into an inequality, I can say something about |a|2. Does that make sense? this is part of a bigger problem concerning metric spaces.
 
  • #5
Are you looking for something like this?

[tex]\left|a\right|^2=\left|a^2\right|=\left|a^2-b^2+b^2\right|\leq\left|a^2-b^2\right|+\left|b^2\right|[/tex]

Now, if [itex]\left|a^2-b^2\right|\leq\epsilon[/itex] and [itex]\left|b^2\right|\leq\epsilon[/itex], then [itex]\left|a\right|^2\leq 2\epsilon[/itex]
 
  • #6
No, my a is a function in C[0,1] so that statement may not be true. I'm looking for a way to massage out that to which I responded to mathman.
 
  • #7
Well, my guess is that you are working with the sup norm, and want to prove something related to uniform convergence, but I think you must give more details. After all, you stated that you had information about |a2-b2|, |b2| and needed to relate these to |a2|. Given only this, I can't add anything else.
 
  • #8
I have

[tex]|f_{n}(t)-f(t)|^{2} \leq\epsilon [/tex]

and [tex] |f_{n}(t)|^{2} \leq 1 [/tex]

I wish to show

[tex] |f(t)|^{2} \leq 1 [/tex]

as well.

As much as I'd like to, I cannot say

[tex] |f(t)|^{2} = |f(t) - f_{n} (t) + f_{n} (t)|^{2} \leq |f_{n}(t) - f(t)|^{2} + |f_{n}(t)|^{2} [/tex]
 
  • #9
No, but you can say:

[tex]
|f(t)|^{2} = |f(t) - f_{n} (t) + f_{n} (t)|^{2} \leq |f_{n}(t) - f(t)|^{2} +2|f_{n} (t)||f_{n}(t) - f(t)|+ |f_{n}(t)|^{2}
[/tex]

And you know how to estimate all terms on the right; just keep in mind that [itex]\epsilon[/itex] is arbitrarily small (or at least I think it is).
 
  • #10
I see you just expanded it...I didn't know you could do that inside absolute value.
 
  • #11
It's just one of those Analysis tricks. After doing a fair number of these proofs, you start to spot them more rapidly.:smile:
 
  • #12
Well thank you so much! And yes, you are right, it did begin with the sup norm.
 
  • #13
Glad to help.:wink:
 
  • #14
Somefantastik said:
I see you just expanded it...I didn't know you could do that inside absolute value.
When you expand the square of the sum or difference of two terms, you get two squares and a product. Taking absolute values for each term can only make the sum of the three terms bigger.
 
  • #15
or

[tex] |f_n(t) - f(t)|^2 \leq \epsilon \implies [/tex]
[tex] |f_n(t) - f(t)| \leq \epsilon' \implies [/tex] by reverse triangle inequality
[tex] |f(t)| - |f_n(t)| \leq \epsilon' \implies [/tex]
[tex] |f(t)| \leq 1 + \epsilon' \implies [/tex]
[tex] |f(t)|^2 \leq 1 [/tex]

edit: this is true for one n if the OP's statements still hold. as in if [itex] |f_n(t) - f(t) | \leq \epsilon [/itex] for all epsilon without incrementing n.
 
Last edited:

Related to How can I use modulus and inequalities to simplify my equations?

1. What is modulus and how is it used in inequalities?

Modulus, also known as absolute value, refers to the distance of a number from zero on a number line. Inequalities involving modulus typically have two parts: an expression involving a modulus and a numerical value. The solution to the inequality is any values that make the expression equal to or greater than the given numerical value.

2. How do you solve modulus inequalities?

To solve modulus inequalities, you can follow these steps:
1. Rewrite the inequality as two separate inequalities, one with the positive value of the modulus and one with the negative value of the modulus.
2. Solve each inequality separately.
3. Combine the solutions to find the final solution set.
Note: The solution set may include both positive and negative numbers depending on the given inequality.

3. Can modulus inequalities have more than one solution?

Yes, modulus inequalities can have more than one solution. This is because the absolute value of a number can be the same for both positive and negative values. For example, the inequality |x| < 3 has two solutions: x < 3 and x > -3.

4. How do you graph modulus inequalities?

To graph modulus inequalities, you can follow these steps:
1. Rewrite the inequality as two separate inequalities, one with the positive value of the modulus and one with the negative value of the modulus.
2. Graph each inequality on a number line.
3. Shade the regions that satisfy both inequalities, which will be the solution set.
Note: The solution set may include both positive and negative numbers depending on the given inequality.

5. What are some real-life applications of modulus inequalities?

Modulus inequalities are commonly used in physics, engineering, and economics to model real-world situations. For example, in physics, modulus inequalities are used to determine the range of possible values for a physical quantity. In economics, they are used to analyze profit and loss in business. In engineering, modulus inequalities are used to design safe and efficient structures.

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