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Splitting a function in $mathbb{C}$ into even and uneven parts

How do I divide a function into even and odd sections?Proof of an odd function plus an even functionExistence of a function from $f : mathbb Z^2 rightarrow mathbb Z$Derivative of a distributionif f is a function with domain $mathbb{R}$ that can be written $f = E + O$ where E is an even function…Suppose $Ain mathbb R$ and the product $(-3+8i) (-3+Ai)$ is also a real number …Need a help in understanding a solution of a forth problem in Israel Gohberg.Finding an injective function $f:mathbb{Z}timesmathbb{Z}to mathbb{N}$ - need help with understandingGiven a function a to b, its inverse relation will be a function iff the function is bijectiveIf $B$ is a basis of $V$ and $Usubseteq V$ is linearly independent then there exists a $Csubseteq B$ such that$ tilde{B}:= Udotcup C$ is a basis

0

$begingroup$

Let $X$ be a subset of $mathbb{C}$ auch that if $xin X$ then also $-xin X$. A function $f:Xrightarrow mathbb{C}$ is called even if $f(x)=f(-x)$, a function is called uneven if $-f(x)=f(-x)$. Show that for every function $varphi:Xrightarrow mathbb{C}$ there exists unique $e$ and $u$ such that $varphi=e+u$ where $e$ is even and $u$ is uneven.

I did not understand the solution, can somebody explain me the chain of thoughts so I could derrive to the solution myself and also there are specific parts of the uniqueness proof that I could not understand, I hope somebody can explain them to me.

The existence solution is given without further comment that $e(x)=frac{1}{2}(varphi(x)+varphi(-x))$ and $u(x)=frac{1}{2}(varphi(x)-varphi(-x))$

I have already checked that they already fullfill every condition but I have no clue how the solution was built.

The uniquenes proof is:

The uniqueness of the decomposition $varphi = e + u$ results from the fact that from an analogous presentation $0=e+u$ necessarily $e=u=0$ follows. Latter follows from the both identities $e(z)+u(z)=0$ and $e(z)-u(z)=e(-z)+u(-z)=0$.

I did not understand anything about the uniqueness proof. I hope that the Translation of the solution text was done correctly.

What is the analogos presentation ? Analogous to what? And what identites does he mean? And how do we know that those identities are true?

functions complex-numbers proof-explanation

share|cite|improve this question

asked Mar 12 at 14:47

New2MathNew2Math

10413

$endgroup$

add a comment | 

0

$begingroup$

Let $X$ be a subset of $mathbb{C}$ auch that if $xin X$ then also $-xin X$. A function $f:Xrightarrow mathbb{C}$ is called even if $f(x)=f(-x)$, a function is called uneven if $-f(x)=f(-x)$. Show that for every function $varphi:Xrightarrow mathbb{C}$ there exists unique $e$ and $u$ such that $varphi=e+u$ where $e$ is even and $u$ is uneven.

I did not understand the solution, can somebody explain me the chain of thoughts so I could derrive to the solution myself and also there are specific parts of the uniqueness proof that I could not understand, I hope somebody can explain them to me.

The existence solution is given without further comment that $e(x)=frac{1}{2}(varphi(x)+varphi(-x))$ and $u(x)=frac{1}{2}(varphi(x)-varphi(-x))$

I have already checked that they already fullfill every condition but I have no clue how the solution was built.

The uniquenes proof is:

The uniqueness of the decomposition $varphi = e + u$ results from the fact that from an analogous presentation $0=e+u$ necessarily $e=u=0$ follows. Latter follows from the both identities $e(z)+u(z)=0$ and $e(z)-u(z)=e(-z)+u(-z)=0$.

I did not understand anything about the uniqueness proof. I hope that the Translation of the solution text was done correctly.

What is the analogos presentation ? Analogous to what? And what identites does he mean? And how do we know that those identities are true?

functions complex-numbers proof-explanation

share|cite|improve this question

asked Mar 12 at 14:47

New2MathNew2Math

10413

$endgroup$

add a comment | 

$begingroup$

Let $X$ be a subset of $mathbb{C}$ auch that if $xin X$ then also $-xin X$. A function $f:Xrightarrow mathbb{C}$ is called even if $f(x)=f(-x)$, a function is called uneven if $-f(x)=f(-x)$. Show that for every function $varphi:Xrightarrow mathbb{C}$ there exists unique $e$ and $u$ such that $varphi=e+u$ where $e$ is even and $u$ is uneven.

I did not understand the solution, can somebody explain me the chain of thoughts so I could derrive to the solution myself and also there are specific parts of the uniqueness proof that I could not understand, I hope somebody can explain them to me.

The existence solution is given without further comment that $e(x)=frac{1}{2}(varphi(x)+varphi(-x))$ and $u(x)=frac{1}{2}(varphi(x)-varphi(-x))$

I have already checked that they already fullfill every condition but I have no clue how the solution was built.

The uniquenes proof is:

The uniqueness of the decomposition $varphi = e + u$ results from the fact that from an analogous presentation $0=e+u$ necessarily $e=u=0$ follows. Latter follows from the both identities $e(z)+u(z)=0$ and $e(z)-u(z)=e(-z)+u(-z)=0$.

I did not understand anything about the uniqueness proof. I hope that the Translation of the solution text was done correctly.

What is the analogos presentation ? Analogous to what? And what identites does he mean? And how do we know that those identities are true?

functions complex-numbers proof-explanation

share|cite|improve this question

asked Mar 12 at 14:47

New2MathNew2Math

10413

$endgroup$

share|cite|improve this question

asked Mar 12 at 14:47

New2MathNew2Math

10413

share|cite|improve this question

asked Mar 12 at 14:47

New2MathNew2Math

10413

asked Mar 12 at 14:47

New2MathNew2Math

10413

asked Mar 12 at 14:47

New2MathNew2Math

10413

1 Answer
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active

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1

$begingroup$

If $e_1$ and $e_2$ are even functions, and $u_1$ and $u_2$ are uneven, let us assume that $e_1 + u_1 = e_2 + u_2$. Then $e_1 - e_2 = u_2 - u_1$. $e_1 - e_2$ is an even function, and $u_2 - u_1$ is an uneven function. But we know that he only function that is both even and uneven is the constant $0$. So $e_1 = e_2$ and $u_1 = u_2$.

EDIT: Lemma: If a function $f$ is both even and uneven, then it is the constant $0$.

Proof: Let $x in mathbb{R}$. Then $f(-x) = f(x)$ because $f$ is even, and then $f(-x) = -f(x)$ because $f$ is uneven. So $f(x) = -f(x)$, and so $f(x) = 0$.

share|cite|improve this answer

edited Mar 14 at 12:54

answered Mar 12 at 14:55

PlopPlop

465216

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I would avoid "uneven" and use the standard term "odd". "Uneven" suggests "not even", which is not the same thing as "odd".
  $endgroup$
  – MPW
  Mar 12 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was using the same term as the person who asked the question. But thanks for the information !
  $endgroup$
  – Plop
  Mar 12 at 15:19
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I see, not sure how I missed that. Then this should have been directed at OP instead of you.
  $endgroup$
  – MPW
  Mar 12 at 15:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do we know that the only function that is both even and uneven is the constant $0$?
  $endgroup$
  – New2Math
  Mar 12 at 17:48
  

  
  
  
  

add a comment | 

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1

$begingroup$

If $e_1$ and $e_2$ are even functions, and $u_1$ and $u_2$ are uneven, let us assume that $e_1 + u_1 = e_2 + u_2$. Then $e_1 - e_2 = u_2 - u_1$. $e_1 - e_2$ is an even function, and $u_2 - u_1$ is an uneven function. But we know that he only function that is both even and uneven is the constant $0$. So $e_1 = e_2$ and $u_1 = u_2$.

EDIT: Lemma: If a function $f$ is both even and uneven, then it is the constant $0$.

Proof: Let $x in mathbb{R}$. Then $f(-x) = f(x)$ because $f$ is even, and then $f(-x) = -f(x)$ because $f$ is uneven. So $f(x) = -f(x)$, and so $f(x) = 0$.

share|cite|improve this answer

edited Mar 14 at 12:54

answered Mar 12 at 14:55

PlopPlop

465216

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I would avoid "uneven" and use the standard term "odd". "Uneven" suggests "not even", which is not the same thing as "odd".
  $endgroup$
  – MPW
  Mar 12 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was using the same term as the person who asked the question. But thanks for the information !
  $endgroup$
  – Plop
  Mar 12 at 15:19
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I see, not sure how I missed that. Then this should have been directed at OP instead of you.
  $endgroup$
  – MPW
  Mar 12 at 15:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do we know that the only function that is both even and uneven is the constant $0$?
  $endgroup$
  – New2Math
  Mar 12 at 17:48
  

  
  
  
  

add a comment | 

1

$begingroup$

If $e_1$ and $e_2$ are even functions, and $u_1$ and $u_2$ are uneven, let us assume that $e_1 + u_1 = e_2 + u_2$. Then $e_1 - e_2 = u_2 - u_1$. $e_1 - e_2$ is an even function, and $u_2 - u_1$ is an uneven function. But we know that he only function that is both even and uneven is the constant $0$. So $e_1 = e_2$ and $u_1 = u_2$.

EDIT: Lemma: If a function $f$ is both even and uneven, then it is the constant $0$.

Proof: Let $x in mathbb{R}$. Then $f(-x) = f(x)$ because $f$ is even, and then $f(-x) = -f(x)$ because $f$ is uneven. So $f(x) = -f(x)$, and so $f(x) = 0$.

share|cite|improve this answer

edited Mar 14 at 12:54

answered Mar 12 at 14:55

PlopPlop

465216

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I would avoid "uneven" and use the standard term "odd". "Uneven" suggests "not even", which is not the same thing as "odd".
  $endgroup$
  – MPW
  Mar 12 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was using the same term as the person who asked the question. But thanks for the information !
  $endgroup$
  – Plop
  Mar 12 at 15:19
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I see, not sure how I missed that. Then this should have been directed at OP instead of you.
  $endgroup$
  – MPW
  Mar 12 at 15:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do we know that the only function that is both even and uneven is the constant $0$?
  $endgroup$
  – New2Math
  Mar 12 at 17:48
  

  
  
  
  

add a comment | 

$begingroup$

If $e_1$ and $e_2$ are even functions, and $u_1$ and $u_2$ are uneven, let us assume that $e_1 + u_1 = e_2 + u_2$. Then $e_1 - e_2 = u_2 - u_1$. $e_1 - e_2$ is an even function, and $u_2 - u_1$ is an uneven function. But we know that he only function that is both even and uneven is the constant $0$. So $e_1 = e_2$ and $u_1 = u_2$.

EDIT: Lemma: If a function $f$ is both even and uneven, then it is the constant $0$.

Proof: Let $x in mathbb{R}$. Then $f(-x) = f(x)$ because $f$ is even, and then $f(-x) = -f(x)$ because $f$ is uneven. So $f(x) = -f(x)$, and so $f(x) = 0$.

share|cite|improve this answer

edited Mar 14 at 12:54

answered Mar 12 at 14:55

PlopPlop

465216

$endgroup$

share|cite|improve this answer

edited Mar 14 at 12:54

answered Mar 12 at 14:55

PlopPlop

465216

answered Mar 12 at 14:55

PlopPlop

465216

answered Mar 12 at 14:55

PlopPlop

465216