Changing of index in sum notation

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The following is the derivative of the power-series expansion of $e^x$. I can't seem to understand why the starting point changes to $1$ after the second $text“=text''.$ Would this surely not be missing out a term?
$$
sum_n=0^infty frac d dx , fracx^nn! = sum_n=0^infty fracnx^n-1n! = sum_n=1^infty fracx^n-1(n-1)! = sum_n=0^infty fracx^nn! = e^x.
$$

calculus summation power-series

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edited Jul 18 at 19:42

Michael Hardy

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asked Jul 18 at 19:28

ojd

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  $fracnn! = frac1(n-1)!$, and if $n=0$ then this first term would just disappear. So we can really just start at $n=1$
  – gd1035
  Jul 18 at 19:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Derivative of a constant term is zero
  – kingW3
  Jul 18 at 19:43
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

The following is the derivative of the power-series expansion of $e^x$. I can't seem to understand why the starting point changes to $1$ after the second $text“=text''.$ Would this surely not be missing out a term?
$$
sum_n=0^infty frac d dx , fracx^nn! = sum_n=0^infty fracnx^n-1n! = sum_n=1^infty fracx^n-1(n-1)! = sum_n=0^infty fracx^nn! = e^x.
$$

calculus summation power-series

share|cite|improve this question

edited Jul 18 at 19:42

Michael Hardy

204k23186462

asked Jul 18 at 19:28

ojd

424

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $fracnn! = frac1(n-1)!$, and if $n=0$ then this first term would just disappear. So we can really just start at $n=1$
  – gd1035
  Jul 18 at 19:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Derivative of a constant term is zero
  – kingW3
  Jul 18 at 19:43
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

The following is the derivative of the power-series expansion of $e^x$. I can't seem to understand why the starting point changes to $1$ after the second $text“=text''.$ Would this surely not be missing out a term?
$$
sum_n=0^infty frac d dx , fracx^nn! = sum_n=0^infty fracnx^n-1n! = sum_n=1^infty fracx^n-1(n-1)! = sum_n=0^infty fracx^nn! = e^x.
$$

calculus summation power-series

share|cite|improve this question

edited Jul 18 at 19:42

Michael Hardy

204k23186462

asked Jul 18 at 19:28

ojd

424

The following is the derivative of the power-series expansion of $e^x$. I can't seem to understand why the starting point changes to $1$ after the second $text“=text''.$ Would this surely not be missing out a term?
$$
sum_n=0^infty frac d dx , fracx^nn! = sum_n=0^infty fracnx^n-1n! = sum_n=1^infty fracx^n-1(n-1)! = sum_n=0^infty fracx^nn! = e^x.
$$

calculus summation power-series

share|cite|improve this question

edited Jul 18 at 19:42

Michael Hardy

204k23186462

asked Jul 18 at 19:28

ojd

424

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 19:42

Michael Hardy

204k23186462

edited Jul 18 at 19:42

Michael Hardy

204k23186462

edited Jul 18 at 19:42

Michael Hardy

204k23186462

204k23186462

asked Jul 18 at 19:28

ojd

424

asked Jul 18 at 19:28

ojd

424

asked Jul 18 at 19:28

ojd

424

424

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $fracnn! = frac1(n-1)!$, and if $n=0$ then this first term would just disappear. So we can really just start at $n=1$
  – gd1035
  Jul 18 at 19:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Derivative of a constant term is zero
  – kingW3
  Jul 18 at 19:43
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $fracnn! = frac1(n-1)!$, and if $n=0$ then this first term would just disappear. So we can really just start at $n=1$
  – gd1035
  Jul 18 at 19:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Derivative of a constant term is zero
  – kingW3
  Jul 18 at 19:43
  

  
  
  
  

3

3

$fracnn! = frac1(n-1)!$, and if $n=0$ then this first term would just disappear. So we can really just start at $n=1$
– gd1035
Jul 18 at 19:32

$fracnn! = frac1(n-1)!$, and if $n=0$ then this first term would just disappear. So we can really just start at $n=1$
– gd1035
Jul 18 at 19:32

Derivative of a constant term is zero
– kingW3
Jul 18 at 19:43

Derivative of a constant term is zero
– kingW3
Jul 18 at 19:43

add a comment | 

3 Answers
3

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oldest

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up vote
7
down vote



accepted

$dfracnx^n-1n! = 0$ when $n=0,$ so the $n=0$ term of the sum $displaystylesum_n=0^infty fracnx^n-1n!$ can be discarded, yielding $displaystyle sum_n=1^infty$ instead of $displaystylesum_n=0^infty.$

share|cite|improve this answer

answered Jul 18 at 19:33

Michael Hardy

204k23186462

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's a bit sloppy to do that and cancel the $n$ in the same step, IMHO. If they would've done it as two separate steps, OP might not have been as confused.
  – Kevin
  Jul 18 at 21:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kevin True, but proofs are sometimes like spaghetti code - it looks so obvious when laying it down that surely no one would be confused. Upon rereading some time later, the same person might need help to decipher their own logic. :)
  – Lawrence
  Jul 19 at 3:33
  

  
  
  
  

add a comment | 

up vote
5
down vote

Note that $$sum_n=0^inftya_n = a_0 + sum_n=1^inftya_n.$$ Therefore, $$sum_n=0^inftyfracn x^n-1n! = 0+sum_n=1^inftyfracn x^n-1n!.$$

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answered Jul 18 at 19:33

Math Lover

12.4k21232

add a comment | 

up vote
0
down vote

The correct form should have been $$frac d dx e^x =frac d dx sum_n=0^infty fracx^nn!=
sum_n=0^infty frac d dx , fracx^nn! = sum_n=1^infty fracx^n-1(n-1)! = sum_n=0^infty fracx^nn! = e^x$$

$$frac ddx ( 1+x+x^2/2 + x^3/6+cdots)= 1+x+x^2/2 + x^3/6+cdots$$

share|cite|improve this answer

edited Jul 18 at 23:52

Michael Hardy

204k23186462

answered Jul 18 at 19:47

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote



accepted

$dfracnx^n-1n! = 0$ when $n=0,$ so the $n=0$ term of the sum $displaystylesum_n=0^infty fracnx^n-1n!$ can be discarded, yielding $displaystyle sum_n=1^infty$ instead of $displaystylesum_n=0^infty.$

share|cite|improve this answer

answered Jul 18 at 19:33

Michael Hardy

204k23186462

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's a bit sloppy to do that and cancel the $n$ in the same step, IMHO. If they would've done it as two separate steps, OP might not have been as confused.
  – Kevin
  Jul 18 at 21:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kevin True, but proofs are sometimes like spaghetti code - it looks so obvious when laying it down that surely no one would be confused. Upon rereading some time later, the same person might need help to decipher their own logic. :)
  – Lawrence
  Jul 19 at 3:33
  

  
  
  
  

add a comment | 

up vote
7
down vote



accepted

$dfracnx^n-1n! = 0$ when $n=0,$ so the $n=0$ term of the sum $displaystylesum_n=0^infty fracnx^n-1n!$ can be discarded, yielding $displaystyle sum_n=1^infty$ instead of $displaystylesum_n=0^infty.$

share|cite|improve this answer

answered Jul 18 at 19:33

Michael Hardy

204k23186462

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's a bit sloppy to do that and cancel the $n$ in the same step, IMHO. If they would've done it as two separate steps, OP might not have been as confused.
  – Kevin
  Jul 18 at 21:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kevin True, but proofs are sometimes like spaghetti code - it looks so obvious when laying it down that surely no one would be confused. Upon rereading some time later, the same person might need help to decipher their own logic. :)
  – Lawrence
  Jul 19 at 3:33
  

  
  
  
  

add a comment | 

up vote
7
down vote



accepted

up vote
7
down vote



accepted

$dfracnx^n-1n! = 0$ when $n=0,$ so the $n=0$ term of the sum $displaystylesum_n=0^infty fracnx^n-1n!$ can be discarded, yielding $displaystyle sum_n=1^infty$ instead of $displaystylesum_n=0^infty.$

share|cite|improve this answer

answered Jul 18 at 19:33

Michael Hardy

204k23186462

$dfracnx^n-1n! = 0$ when $n=0,$ so the $n=0$ term of the sum $displaystylesum_n=0^infty fracnx^n-1n!$ can be discarded, yielding $displaystyle sum_n=1^infty$ instead of $displaystylesum_n=0^infty.$

share|cite|improve this answer

answered Jul 18 at 19:33

Michael Hardy

204k23186462

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 19:33

Michael Hardy

204k23186462

answered Jul 18 at 19:33

Michael Hardy

204k23186462

answered Jul 18 at 19:33

Michael Hardy

204k23186462

204k23186462

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's a bit sloppy to do that and cancel the $n$ in the same step, IMHO. If they would've done it as two separate steps, OP might not have been as confused.
  – Kevin
  Jul 18 at 21:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kevin True, but proofs are sometimes like spaghetti code - it looks so obvious when laying it down that surely no one would be confused. Upon rereading some time later, the same person might need help to decipher their own logic. :)
  – Lawrence
  Jul 19 at 3:33
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's a bit sloppy to do that and cancel the $n$ in the same step, IMHO. If they would've done it as two separate steps, OP might not have been as confused.
  – Kevin
  Jul 18 at 21:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kevin True, but proofs are sometimes like spaghetti code - it looks so obvious when laying it down that surely no one would be confused. Upon rereading some time later, the same person might need help to decipher their own logic. :)
  – Lawrence
  Jul 19 at 3:33
  

  
  
  
  

1

1

It's a bit sloppy to do that and cancel the $n$ in the same step, IMHO. If they would've done it as two separate steps, OP might not have been as confused.
– Kevin
Jul 18 at 21:32

It's a bit sloppy to do that and cancel the $n$ in the same step, IMHO. If they would've done it as two separate steps, OP might not have been as confused.
– Kevin
Jul 18 at 21:32

@Kevin True, but proofs are sometimes like spaghetti code - it looks so obvious when laying it down that surely no one would be confused. Upon rereading some time later, the same person might need help to decipher their own logic. :)
– Lawrence
Jul 19 at 3:33

@Kevin True, but proofs are sometimes like spaghetti code - it looks so obvious when laying it down that surely no one would be confused. Upon rereading some time later, the same person might need help to decipher their own logic. :)
– Lawrence
Jul 19 at 3:33

add a comment | 

up vote
5
down vote

Note that $$sum_n=0^inftya_n = a_0 + sum_n=1^inftya_n.$$ Therefore, $$sum_n=0^inftyfracn x^n-1n! = 0+sum_n=1^inftyfracn x^n-1n!.$$

share|cite|improve this answer

answered Jul 18 at 19:33

Math Lover

12.4k21232

add a comment | 

up vote
5
down vote

Note that $$sum_n=0^inftya_n = a_0 + sum_n=1^inftya_n.$$ Therefore, $$sum_n=0^inftyfracn x^n-1n! = 0+sum_n=1^inftyfracn x^n-1n!.$$

share|cite|improve this answer

answered Jul 18 at 19:33

Math Lover

12.4k21232

add a comment | 

up vote
5
down vote

up vote
5
down vote

Note that $$sum_n=0^inftya_n = a_0 + sum_n=1^inftya_n.$$ Therefore, $$sum_n=0^inftyfracn x^n-1n! = 0+sum_n=1^inftyfracn x^n-1n!.$$

share|cite|improve this answer

answered Jul 18 at 19:33

Math Lover

12.4k21232

Note that $$sum_n=0^inftya_n = a_0 + sum_n=1^inftya_n.$$ Therefore, $$sum_n=0^inftyfracn x^n-1n! = 0+sum_n=1^inftyfracn x^n-1n!.$$

share|cite|improve this answer

answered Jul 18 at 19:33

Math Lover

12.4k21232

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 19:33

Math Lover

12.4k21232

answered Jul 18 at 19:33

Math Lover

12.4k21232

answered Jul 18 at 19:33

Math Lover

12.4k21232

12.4k21232

add a comment | 

add a comment | 

up vote
0
down vote

The correct form should have been $$frac d dx e^x =frac d dx sum_n=0^infty fracx^nn!=
sum_n=0^infty frac d dx , fracx^nn! = sum_n=1^infty fracx^n-1(n-1)! = sum_n=0^infty fracx^nn! = e^x$$

$$frac ddx ( 1+x+x^2/2 + x^3/6+cdots)= 1+x+x^2/2 + x^3/6+cdots$$

share|cite|improve this answer

edited Jul 18 at 23:52

Michael Hardy

204k23186462

answered Jul 18 at 19:47

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
0
down vote

The correct form should have been $$frac d dx e^x =frac d dx sum_n=0^infty fracx^nn!=
sum_n=0^infty frac d dx , fracx^nn! = sum_n=1^infty fracx^n-1(n-1)! = sum_n=0^infty fracx^nn! = e^x$$

$$frac ddx ( 1+x+x^2/2 + x^3/6+cdots)= 1+x+x^2/2 + x^3/6+cdots$$

share|cite|improve this answer

edited Jul 18 at 23:52

Michael Hardy

204k23186462

answered Jul 18 at 19:47

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
0
down vote

up vote
0
down vote

The correct form should have been $$frac d dx e^x =frac d dx sum_n=0^infty fracx^nn!=
sum_n=0^infty frac d dx , fracx^nn! = sum_n=1^infty fracx^n-1(n-1)! = sum_n=0^infty fracx^nn! = e^x$$

$$frac ddx ( 1+x+x^2/2 + x^3/6+cdots)= 1+x+x^2/2 + x^3/6+cdots$$

share|cite|improve this answer

edited Jul 18 at 23:52

Michael Hardy

204k23186462

answered Jul 18 at 19:47

Mohammad Riazi-Kermani

27.5k41852

The correct form should have been $$frac d dx e^x =frac d dx sum_n=0^infty fracx^nn!=
sum_n=0^infty frac d dx , fracx^nn! = sum_n=1^infty fracx^n-1(n-1)! = sum_n=0^infty fracx^nn! = e^x$$

$$frac ddx ( 1+x+x^2/2 + x^3/6+cdots)= 1+x+x^2/2 + x^3/6+cdots$$

share|cite|improve this answer

edited Jul 18 at 23:52

Michael Hardy

204k23186462

answered Jul 18 at 19:47

Mohammad Riazi-Kermani

27.5k41852

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 23:52

Michael Hardy

204k23186462

edited Jul 18 at 23:52

Michael Hardy

204k23186462

edited Jul 18 at 23:52

Michael Hardy

204k23186462

204k23186462

answered Jul 18 at 19:47

Mohammad Riazi-Kermani

27.5k41852

answered Jul 18 at 19:47

Mohammad Riazi-Kermani

27.5k41852

answered Jul 18 at 19:47

Mohammad Riazi-Kermani

27.5k41852

27.5k41852

add a comment | 

add a comment | 

 

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