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1

$begingroup$

Here's the original example

To prove that when $y$ is real and numerically $y<1$.
$$log_e(1+y)=y-{1over2}y^2+{1over3}y^3-{1over4}y^4+...$$ ad inf.

Here are the first 2 steps of the solution. I understand the first one but not how 2nd one came.

$(1+y)^x=1+xlog_e(1+y)+{frac{x^2}{2!}}[log_e(1+y)]^2+...$ this comes from expansion of $a^x$ which I know.

But, since $y$ is real and numerically $<1$, we have
$$Rightarrow (1+y)^x=1+x.y+frac {x(x-1)}{1.2}y^2+frac{x(x-1)(x-2)}{1.2.3}y^3+...$$

How is this second line concluded? Please help me derive this statement not necessarily from first as it seems to independent derivation in the example.

sequences-and-series proof-explanation

share|cite|improve this question

asked Mar 15 at 9:40

Love InvariantsLove Invariants

89015

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know Newton's generalized binomial theorem ?
  $endgroup$
  – Zero
  Mar 15 at 9:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Power series of $(1+x)^{-1}$ could be integrated term by term to obtain this result.
  $endgroup$
  – Paras Khosla
  Mar 15 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zero- Yes ,Paras- I don't know power series. Just in the basics. Still don't understand what is the significance of $y<1$ except that the series converges.
  $endgroup$
  – Love Invariants
  Mar 15 at 9:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zero- My fault that I didn't see it was a regular binomial expansion.
  $endgroup$
  – Love Invariants
  Mar 15 at 10:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Quite unclear what you call the "second line", please specify.
  $endgroup$
  – Yves Daoust
  Mar 15 at 11:37
  
  
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

Here's the original example

To prove that when $y$ is real and numerically $y<1$.
$$log_e(1+y)=y-{1over2}y^2+{1over3}y^3-{1over4}y^4+...$$ ad inf.

Here are the first 2 steps of the solution. I understand the first one but not how 2nd one came.

$(1+y)^x=1+xlog_e(1+y)+{frac{x^2}{2!}}[log_e(1+y)]^2+...$ this comes from expansion of $a^x$ which I know.

But, since $y$ is real and numerically $<1$, we have
$$Rightarrow (1+y)^x=1+x.y+frac {x(x-1)}{1.2}y^2+frac{x(x-1)(x-2)}{1.2.3}y^3+...$$

How is this second line concluded? Please help me derive this statement not necessarily from first as it seems to independent derivation in the example.

sequences-and-series proof-explanation

share|cite|improve this question

asked Mar 15 at 9:40

Love InvariantsLove Invariants

89015

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know Newton's generalized binomial theorem ?
  $endgroup$
  – Zero
  Mar 15 at 9:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Power series of $(1+x)^{-1}$ could be integrated term by term to obtain this result.
  $endgroup$
  – Paras Khosla
  Mar 15 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zero- Yes ,Paras- I don't know power series. Just in the basics. Still don't understand what is the significance of $y<1$ except that the series converges.
  $endgroup$
  – Love Invariants
  Mar 15 at 9:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Zero- My fault that I didn't see it was a regular binomial expansion.
  $endgroup$
  – Love Invariants
  Mar 15 at 10:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Quite unclear what you call the "second line", please specify.
  $endgroup$
  – Yves Daoust
  Mar 15 at 11:37
  
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Here's the original example

To prove that when $y$ is real and numerically $y<1$.
$$log_e(1+y)=y-{1over2}y^2+{1over3}y^3-{1over4}y^4+...$$ ad inf.

Here are the first 2 steps of the solution. I understand the first one but not how 2nd one came.

$(1+y)^x=1+xlog_e(1+y)+{frac{x^2}{2!}}[log_e(1+y)]^2+...$ this comes from expansion of $a^x$ which I know.

But, since $y$ is real and numerically $<1$, we have
$$Rightarrow (1+y)^x=1+x.y+frac {x(x-1)}{1.2}y^2+frac{x(x-1)(x-2)}{1.2.3}y^3+...$$

How is this second line concluded? Please help me derive this statement not necessarily from first as it seems to independent derivation in the example.

sequences-and-series proof-explanation

share|cite|improve this question

asked Mar 15 at 9:40

Love InvariantsLove Invariants

89015

$endgroup$

share|cite|improve this question

asked Mar 15 at 9:40

Love InvariantsLove Invariants

89015

share|cite|improve this question

asked Mar 15 at 9:40

Love InvariantsLove Invariants

89015

asked Mar 15 at 9:40

Love InvariantsLove Invariants

89015

asked Mar 15 at 9:40

Love InvariantsLove Invariants

89015

2 Answers
2

active

oldest

votes

1

$begingroup$

The first line is the Taylor series expanded about $x=0.$ The second line is the Taylor series expanded about $y=0.$ It's just a change of independent variable.

share|cite|improve this answer

answered Mar 15 at 9:50

B. GoddardB. Goddard

19.8k21442

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thats not even in 2nd chapter. I know the series only. Not much in depth. This example is from chapter 1 of a trigonometry book which deals with complex plane. Still the basics are there.
  $endgroup$
  – Love Invariants
  Mar 15 at 9:54
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I'll be providing a different approach. Although this does not completely answer your question, it does provide you a relatively easier way to prove the result. Both ultimately somehow are connected as indicated by the coefficients of your infinte sum.

From the Formula for Sum of a Geometric Series we have:

$$dfrac{1}{1+x}=sum_{k=0}^{infty}(-1)^k x^k$$
Now, Integrating term by term gives us $ln(1+x)$ on the left hand side as follows: $$ln(1+x)=intsum_{k=0}^{infty}(-1)^kx^kmathrm dx=sum_{k=0}^{infty}(-1)^kint x^kmathrm dx=sum_{k=0}^{infty}(-1)^kdfrac{x^{k+1}}{k+1}=x-dfrac{x^2}{2}+-cdots$$

share|cite|improve this answer

edited Mar 15 at 11:33

answered Mar 15 at 9:55

Paras KhoslaParas Khosla

2,676323

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks you for the help. It wasn't my real question but still pretty. :D
  $endgroup$
  – Love Invariants
  Mar 15 at 9:59
  
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

The first line is the Taylor series expanded about $x=0.$ The second line is the Taylor series expanded about $y=0.$ It's just a change of independent variable.

share|cite|improve this answer

answered Mar 15 at 9:50

B. GoddardB. Goddard

19.8k21442

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thats not even in 2nd chapter. I know the series only. Not much in depth. This example is from chapter 1 of a trigonometry book which deals with complex plane. Still the basics are there.
  $endgroup$
  – Love Invariants
  Mar 15 at 9:54
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

The first line is the Taylor series expanded about $x=0.$ The second line is the Taylor series expanded about $y=0.$ It's just a change of independent variable.

share|cite|improve this answer

answered Mar 15 at 9:50

B. GoddardB. Goddard

19.8k21442

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thats not even in 2nd chapter. I know the series only. Not much in depth. This example is from chapter 1 of a trigonometry book which deals with complex plane. Still the basics are there.
  $endgroup$
  – Love Invariants
  Mar 15 at 9:54
  
  
  

  
  
  
  

add a comment | 

$begingroup$

The first line is the Taylor series expanded about $x=0.$ The second line is the Taylor series expanded about $y=0.$ It's just a change of independent variable.

share|cite|improve this answer

answered Mar 15 at 9:50

B. GoddardB. Goddard

19.8k21442

$endgroup$

share|cite|improve this answer

answered Mar 15 at 9:50

B. GoddardB. Goddard

19.8k21442

answered Mar 15 at 9:50

B. GoddardB. Goddard

19.8k21442

answered Mar 15 at 9:50

B. GoddardB. Goddard

19.8k21442

1

$begingroup$

I'll be providing a different approach. Although this does not completely answer your question, it does provide you a relatively easier way to prove the result. Both ultimately somehow are connected as indicated by the coefficients of your infinte sum.

From the Formula for Sum of a Geometric Series we have:

$$dfrac{1}{1+x}=sum_{k=0}^{infty}(-1)^k x^k$$
Now, Integrating term by term gives us $ln(1+x)$ on the left hand side as follows: $$ln(1+x)=intsum_{k=0}^{infty}(-1)^kx^kmathrm dx=sum_{k=0}^{infty}(-1)^kint x^kmathrm dx=sum_{k=0}^{infty}(-1)^kdfrac{x^{k+1}}{k+1}=x-dfrac{x^2}{2}+-cdots$$

share|cite|improve this answer

edited Mar 15 at 11:33

answered Mar 15 at 9:55

Paras KhoslaParas Khosla

2,676323

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks you for the help. It wasn't my real question but still pretty. :D
  $endgroup$
  – Love Invariants
  Mar 15 at 9:59
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I'll be providing a different approach. Although this does not completely answer your question, it does provide you a relatively easier way to prove the result. Both ultimately somehow are connected as indicated by the coefficients of your infinte sum.

From the Formula for Sum of a Geometric Series we have:

$$dfrac{1}{1+x}=sum_{k=0}^{infty}(-1)^k x^k$$
Now, Integrating term by term gives us $ln(1+x)$ on the left hand side as follows: $$ln(1+x)=intsum_{k=0}^{infty}(-1)^kx^kmathrm dx=sum_{k=0}^{infty}(-1)^kint x^kmathrm dx=sum_{k=0}^{infty}(-1)^kdfrac{x^{k+1}}{k+1}=x-dfrac{x^2}{2}+-cdots$$

share|cite|improve this answer

edited Mar 15 at 11:33

answered Mar 15 at 9:55

Paras KhoslaParas Khosla

2,676323

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks you for the help. It wasn't my real question but still pretty. :D
  $endgroup$
  – Love Invariants
  Mar 15 at 9:59
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I'll be providing a different approach. Although this does not completely answer your question, it does provide you a relatively easier way to prove the result. Both ultimately somehow are connected as indicated by the coefficients of your infinte sum.

From the Formula for Sum of a Geometric Series we have:

$$dfrac{1}{1+x}=sum_{k=0}^{infty}(-1)^k x^k$$
Now, Integrating term by term gives us $ln(1+x)$ on the left hand side as follows: $$ln(1+x)=intsum_{k=0}^{infty}(-1)^kx^kmathrm dx=sum_{k=0}^{infty}(-1)^kint x^kmathrm dx=sum_{k=0}^{infty}(-1)^kdfrac{x^{k+1}}{k+1}=x-dfrac{x^2}{2}+-cdots$$

share|cite|improve this answer

edited Mar 15 at 11:33

answered Mar 15 at 9:55

Paras KhoslaParas Khosla

2,676323

$endgroup$

share|cite|improve this answer

edited Mar 15 at 11:33

answered Mar 15 at 9:55

Paras KhoslaParas Khosla

2,676323

answered Mar 15 at 9:55

Paras KhoslaParas Khosla

2,676323

answered Mar 15 at 9:55

Paras KhoslaParas Khosla

2,676323