1078: For 2 Not-Both-0 Complex Numbers and Non-Negative Real Number p, p-th Power of Absolute Value of Sum Is Smaller Than (p-th Power of 2) Times Sum of p-th Powers of Absolute Values

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description/proof of that for $2$ not-both-$0$ complex numbers and non-negative real number $p$, $p$-th power of absolute value of sum is smaller than ($p$-th power of $2$) times sum of $p$-th powers of absolute values

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note 1
• 3: Proof
• 4: Note 2

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Starting Context

• The reader knows a definition of complex numbers set.

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Target Context

• The reader will have a description and a proof of the proposition that for any 2 not-both-0 complex numbers and any non-negative real number, p, the p-th power of the absolute value of the sum of the numbers is smaller than (the p-th power of 2) times the sum of the p-th powers of the absolute values of the numbers.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$c_1$: $\in \mathbb{C}$
$c_2$: $\in \mathbb{C}$
$p$: $\in \mathbb{R}$ such that $0\le p$
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Statements:
$|c_1|\ne 0\vee |c_2|\ne 0$
$⟹$
$|c_1+c_2{|}^p<2^p(|c_1{|}^p+|c_2{|}^p)$
//

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2: Note 1

We are not saying that this is the best possible evaluation: for example, for $p=1$, $|c_1+c_2|\le |c_1|+|c_2|$, and for $p=2$, $|c_1+c_2{|}^2\le 2(|c_1{|}^2+|c_2{|}^2)$.

But in many cases, our concern is not having the best possible $c$ for $|c_1+c_2{|}^p<c(|c_1{|}^p+|c_2{|}^p)$ but to see that there is such any constant $c$ that does not depend on $c_1,c_2$ (but can depend on $p$).

This proposition is about getting a $c$ by the arguably easiest way.

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3: Proof

Whole Strategy: Step 1: see that $|c_1+c_2|\le |c_1|+|c_2|$; Step 2: suppose that $|c_2|\le |c_1|$ and let $|c_2|=r|c_1|$ for an $r\in \mathbb{R}$ such that $0\le r\le 1$; Step 3: evaluate $(|c_1|+|c_2|{)}^p=(|c_1|+r|c_1|{)}^p$; Step 4: see that supposing that $|c_1|<|c_2|$ does not cause any problem.

Step 1:

$|c_1+c_2|\le |c_1|+|c_2|$, which is a well-known fact, but to explain it succinctly, $c_j$ can be regarded to be a vector on ${\mathbb{R}}^2$ and $|c_1+c_2|$ is the length of the sum of the vectors and $|c_1|$ and $|c_2|$ are the lengths of $c_1$ and $c_2$, and $|c_1+c_2|\le |c_1|+|c_2|$ holds by the Euclidean geometry.

Step 2:

Let us suppose that $|c_2|\le |c_1|$.

Then, $|c_2|=r|c_1|$ for an $r\in \mathbb{R}$ such that $0\le r\le 1$.

Step 3:

$|c_1+c_2{|}^p\le (|c_1|+|c_2|{)}^p=(|c_1|+r|c_1|{)}^p=(|c_1|(1+r){)}^p=(1+r{)}^p|c_1{|}^p$.

Let us suppose that $|c_2|<|c_1|$, which means that $r<1$.

Then, $<2^p|c_1{|}^p\le 2^p(|c_1{|}^p+|c_2{|}^p)$.

Let us suppose that $|c_2|=|c_1|$.

Then, $|c_1+c_2{|}^p\le (|c_1|+|c_2|{)}^p=(2|c_1|{)}^p=2^p|c_1{|}^p<2^p(|c_1{|}^p+|c_2{|}^p)$: $0<|c_2|$, because otherwise, $|c_2|=|c_1|=0$, a contradiction against $|c_1|\ne 0\vee |c_2|\ne 0$.

So, anyway, $|c_1+c_2{|}^p<2^p(|c_1{|}^p+|c_2{|}^p)$.

Step 4:

Let us suppose that $|c_1|<|c_2|$.

Then, $|c_1+c_2{|}^p=|c_2+c_1{|}^p$, and by the previous steps, $|c_2+c_1{|}^p<2^p(|c_2{|}^p+|c_1{|}^p)=2^p(|c_1{|}^p+|c_2{|}^p)$.

So, $|c_1+c_2{|}^p<2^p(|c_1{|}^p+|c_2{|}^p)$.

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4: Note 2

Let us see a proof of that for $p=2$, $|c_1+c_2{|}^p\le 2(|c_1{|}^2+|c_2{|}^2)$.

$|c_1+c_2{|}^2\le (|c_1|+|c_2|{)}^2=|c_1{|}^2+|c_2{|}^2+2|c_1||c_2|$.

But $0\le (|c_1|-|c_2|{)}^2=|c_1{|}^2+|c_2{|}^2-2|c_1||c_2|$, which implies that $2|c_1||c_2|\le |c_1{|}^2+|c_2{|}^2$.

So, $|c_1+c_2{|}^2\le |c_1{|}^2+|c_2{|}^2+|c_1{|}^2+|c_2{|}^2=2(|c_1{|}^2+|c_2{|}^2)$.

That is the best possible evaluation, because $|1+1{|}^2=2^2=4=2(1+1)=2(1^2+1^2)$.

In order to make $\le$ $<$, we can just take any $2<c$.

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References

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