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Re-writing pi notation using exponentials

Proving the uniform convergence of $sin(fracxn) e^-x^2$Evaluating integral using Riemann sumsShow that it is LipschitzInfinite exponentialsUpper bounding using negative exponentialsIntegration and little o notationIs the proof approach for $sqrt[n]Pi_j=1^n x_j leq frac1nsum_j=1^n x_j$ futile?n vs. N in notation (proof writing)Multiplication proof by induction

-1

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How might one show that the following is true by re-arranging the term on the left: $$Pi_r=0^n-1 e^frac2ripin=(-1)^n-1$$

real-analysis analysis

share|cite|improve this question

asked Mar 21 at 19:11

Ali LodhiAli Lodhi

383

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Remember in general $prodlimits _re^a_r = e^sumlimits_r a_r$.
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 19:30
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

How might one show that the following is true by re-arranging the term on the left: $$Pi_r=0^n-1 e^frac2ripin=(-1)^n-1$$

real-analysis analysis

share|cite|improve this question

asked Mar 21 at 19:11

Ali LodhiAli Lodhi

383

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Remember in general $prodlimits _re^a_r = e^sumlimits_r a_r$.
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 19:30
  
  

  
  
  
  

add a comment | 

$begingroup$

How might one show that the following is true by re-arranging the term on the left: $$Pi_r=0^n-1 e^frac2ripin=(-1)^n-1$$

real-analysis analysis

share|cite|improve this question

asked Mar 21 at 19:11

Ali LodhiAli Lodhi

383

$endgroup$

share|cite|improve this question

asked Mar 21 at 19:11

Ali LodhiAli Lodhi

383

share|cite|improve this question

asked Mar 21 at 19:11

Ali LodhiAli Lodhi

383

asked Mar 21 at 19:11

Ali LodhiAli Lodhi

383

asked Mar 21 at 19:11

Ali LodhiAli Lodhi

383

3 Answers
3

active

oldest

votes

2

$begingroup$

Let $omega_r:=e^frac2pi irn$. Then the $omega_r$'s, $r=0,ldots,n-1$ are the $n$ $n^th$ roots of unity. That is, they are the solutions of
$$z^n-1=0.$$

Observe that the constant term of the polynomial
$z^n-1=(z-omega_0)cdots (z-omega_n-1)$
is equal to $-1$ which should be from the product of the constant terms of the linear factors. Thus,
$$(-1)^nprod_r=0^n-1e^frac2pi i rn=prod_r=0^n-1(-omega_r)=-1$$
or dividing both sides by $(-1)^n$, gives
$$prod_r=0^n-1e^frac2pi i rn=-1/(-1)^n=(-1)^1-n=(-1)^n-1.$$

share|cite|improve this answer

edited Mar 21 at 19:32

answered Mar 21 at 19:26

chhrochhro

1,442311

$endgroup$

add a comment | 

1

$begingroup$

Change the pi notation to a sigma in the exponent. Use the fact that the sum of $r$ from $1$ to $(n-1)$ is $n(n-1)/2$. Cancel out the $2/n$ in the exponent and the answer follows from there. Idk if this is the method you were looking for.

share|cite|improve this answer

edited Mar 21 at 19:25

Shubham Johri

5,500818

answered Mar 21 at 19:16

uhhhhidkuhhhhidk

1266

$endgroup$

add a comment | 

0

$begingroup$

Re-writing pi notation using exponentials
beginalign
prod_r=0^n-1 e^2ripi/n &= expleft(sum_r=0^n-1frac2pi i rnright)
=expleft(frac2pi insum_r=0^n-1rright)
\
&=expleft(frac2pi in;fracn(n-1)2right)
=e^pi i(n-1)
=big(e^pi ibig)^n-1 = (-1)^n-1
endalign

share|cite|improve this answer

edited Mar 21 at 19:33

answered Mar 21 at 19:32

GEdgarGEdgar

63.4k269174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is base e ^frac2piirn removed?
  $endgroup$
  – Ali Lodhi
  Mar 21 at 19:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We use Euler's formula $e^pi i = -1$. Special case of $e^ti = cos t + i sin t$.
  $endgroup$
  – GEdgar
  Mar 21 at 19:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Apologies, edited. I meant the base e in the first term which is not in the second term.
  $endgroup$
  – Ali Lodhi
  Mar 21 at 19:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Notation $exp(x)$ means $e^x$. We use it when $x$ is complicated, to avoid writing things so small they are illegible.
  $endgroup$
  – GEdgar
  Mar 21 at 19:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean the base e in e^(2piri/n). Is this because the conversion from Pi to Sigma notation involves a ln operator?
  $endgroup$
  – Ali Lodhi
  Mar 21 at 19:44
  

  
  
  
  

 | 
show 1 more comment

Your Answer

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2

$begingroup$

Let $omega_r:=e^frac2pi irn$. Then the $omega_r$'s, $r=0,ldots,n-1$ are the $n$ $n^th$ roots of unity. That is, they are the solutions of
$$z^n-1=0.$$

Observe that the constant term of the polynomial
$z^n-1=(z-omega_0)cdots (z-omega_n-1)$
is equal to $-1$ which should be from the product of the constant terms of the linear factors. Thus,
$$(-1)^nprod_r=0^n-1e^frac2pi i rn=prod_r=0^n-1(-omega_r)=-1$$
or dividing both sides by $(-1)^n$, gives
$$prod_r=0^n-1e^frac2pi i rn=-1/(-1)^n=(-1)^1-n=(-1)^n-1.$$

share|cite|improve this answer

edited Mar 21 at 19:32

answered Mar 21 at 19:26

chhrochhro

1,442311

$endgroup$

add a comment | 

2

$begingroup$

Let $omega_r:=e^frac2pi irn$. Then the $omega_r$'s, $r=0,ldots,n-1$ are the $n$ $n^th$ roots of unity. That is, they are the solutions of
$$z^n-1=0.$$

Observe that the constant term of the polynomial
$z^n-1=(z-omega_0)cdots (z-omega_n-1)$
is equal to $-1$ which should be from the product of the constant terms of the linear factors. Thus,
$$(-1)^nprod_r=0^n-1e^frac2pi i rn=prod_r=0^n-1(-omega_r)=-1$$
or dividing both sides by $(-1)^n$, gives
$$prod_r=0^n-1e^frac2pi i rn=-1/(-1)^n=(-1)^1-n=(-1)^n-1.$$

share|cite|improve this answer

edited Mar 21 at 19:32

answered Mar 21 at 19:26

chhrochhro

1,442311

$endgroup$

add a comment | 

$begingroup$

Let $omega_r:=e^frac2pi irn$. Then the $omega_r$'s, $r=0,ldots,n-1$ are the $n$ $n^th$ roots of unity. That is, they are the solutions of
$$z^n-1=0.$$

Observe that the constant term of the polynomial
$z^n-1=(z-omega_0)cdots (z-omega_n-1)$
is equal to $-1$ which should be from the product of the constant terms of the linear factors. Thus,
$$(-1)^nprod_r=0^n-1e^frac2pi i rn=prod_r=0^n-1(-omega_r)=-1$$
or dividing both sides by $(-1)^n$, gives
$$prod_r=0^n-1e^frac2pi i rn=-1/(-1)^n=(-1)^1-n=(-1)^n-1.$$

share|cite|improve this answer

edited Mar 21 at 19:32

answered Mar 21 at 19:26

chhrochhro

1,442311

$endgroup$

share|cite|improve this answer

edited Mar 21 at 19:32

answered Mar 21 at 19:26

chhrochhro

1,442311

answered Mar 21 at 19:26

chhrochhro

1,442311

answered Mar 21 at 19:26

chhrochhro

1,442311

1

$begingroup$

Change the pi notation to a sigma in the exponent. Use the fact that the sum of $r$ from $1$ to $(n-1)$ is $n(n-1)/2$. Cancel out the $2/n$ in the exponent and the answer follows from there. Idk if this is the method you were looking for.

share|cite|improve this answer

edited Mar 21 at 19:25

Shubham Johri

5,500818

answered Mar 21 at 19:16

uhhhhidkuhhhhidk

1266

$endgroup$

add a comment | 

1

$begingroup$

Change the pi notation to a sigma in the exponent. Use the fact that the sum of $r$ from $1$ to $(n-1)$ is $n(n-1)/2$. Cancel out the $2/n$ in the exponent and the answer follows from there. Idk if this is the method you were looking for.

share|cite|improve this answer

edited Mar 21 at 19:25

Shubham Johri

5,500818

answered Mar 21 at 19:16

uhhhhidkuhhhhidk

1266

$endgroup$

add a comment | 

$begingroup$

Change the pi notation to a sigma in the exponent. Use the fact that the sum of $r$ from $1$ to $(n-1)$ is $n(n-1)/2$. Cancel out the $2/n$ in the exponent and the answer follows from there. Idk if this is the method you were looking for.

share|cite|improve this answer

edited Mar 21 at 19:25

Shubham Johri

5,500818

answered Mar 21 at 19:16

uhhhhidkuhhhhidk

1266

$endgroup$

share|cite|improve this answer

edited Mar 21 at 19:25

Shubham Johri

5,500818

answered Mar 21 at 19:16

uhhhhidkuhhhhidk

1266

edited Mar 21 at 19:25

Shubham Johri

5,500818

edited Mar 21 at 19:25

Shubham Johri

5,500818

answered Mar 21 at 19:16

uhhhhidkuhhhhidk

1266

answered Mar 21 at 19:16

uhhhhidkuhhhhidk

1266

0

$begingroup$

Re-writing pi notation using exponentials
beginalign
prod_r=0^n-1 e^2ripi/n &= expleft(sum_r=0^n-1frac2pi i rnright)
=expleft(frac2pi insum_r=0^n-1rright)
\
&=expleft(frac2pi in;fracn(n-1)2right)
=e^pi i(n-1)
=big(e^pi ibig)^n-1 = (-1)^n-1
endalign

share|cite|improve this answer

edited Mar 21 at 19:33

answered Mar 21 at 19:32

GEdgarGEdgar

63.4k269174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is base e ^frac2piirn removed?
  $endgroup$
  – Ali Lodhi
  Mar 21 at 19:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We use Euler's formula $e^pi i = -1$. Special case of $e^ti = cos t + i sin t$.
  $endgroup$
  – GEdgar
  Mar 21 at 19:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Apologies, edited. I meant the base e in the first term which is not in the second term.
  $endgroup$
  – Ali Lodhi
  Mar 21 at 19:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Notation $exp(x)$ means $e^x$. We use it when $x$ is complicated, to avoid writing things so small they are illegible.
  $endgroup$
  – GEdgar
  Mar 21 at 19:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean the base e in e^(2piri/n). Is this because the conversion from Pi to Sigma notation involves a ln operator?
  $endgroup$
  – Ali Lodhi
  Mar 21 at 19:44
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Re-writing pi notation using exponentials
beginalign
prod_r=0^n-1 e^2ripi/n &= expleft(sum_r=0^n-1frac2pi i rnright)
=expleft(frac2pi insum_r=0^n-1rright)
\
&=expleft(frac2pi in;fracn(n-1)2right)
=e^pi i(n-1)
=big(e^pi ibig)^n-1 = (-1)^n-1
endalign

share|cite|improve this answer

edited Mar 21 at 19:33

answered Mar 21 at 19:32

GEdgarGEdgar

63.4k269174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is base e ^frac2piirn removed?
  $endgroup$
  – Ali Lodhi
  Mar 21 at 19:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We use Euler's formula $e^pi i = -1$. Special case of $e^ti = cos t + i sin t$.
  $endgroup$
  – GEdgar
  Mar 21 at 19:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Apologies, edited. I meant the base e in the first term which is not in the second term.
  $endgroup$
  – Ali Lodhi
  Mar 21 at 19:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Notation $exp(x)$ means $e^x$. We use it when $x$ is complicated, to avoid writing things so small they are illegible.
  $endgroup$
  – GEdgar
  Mar 21 at 19:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean the base e in e^(2piri/n). Is this because the conversion from Pi to Sigma notation involves a ln operator?
  $endgroup$
  – Ali Lodhi
  Mar 21 at 19:44
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Re-writing pi notation using exponentials
beginalign
prod_r=0^n-1 e^2ripi/n &= expleft(sum_r=0^n-1frac2pi i rnright)
=expleft(frac2pi insum_r=0^n-1rright)
\
&=expleft(frac2pi in;fracn(n-1)2right)
=e^pi i(n-1)
=big(e^pi ibig)^n-1 = (-1)^n-1
endalign

share|cite|improve this answer

edited Mar 21 at 19:33

answered Mar 21 at 19:32

GEdgarGEdgar

63.4k269174

$endgroup$

share|cite|improve this answer

edited Mar 21 at 19:33

answered Mar 21 at 19:32

GEdgarGEdgar

63.4k269174

answered Mar 21 at 19:32

GEdgarGEdgar

63.4k269174

answered Mar 21 at 19:32

GEdgarGEdgar

63.4k269174