Is my proof of this lemma about Taylor Expansion correct?

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I tried to prove the following lemma:

Lemma. Let $f,g:mathscr U_x_0to mathbb R$ differentiable $n$ times in $x_0$. Then
$$f(x)-g(x)=o((x-x_0)^n)iff f^(k)(x_0)=g^(k)(x_0), 0le kle n$$

Proof. (by induction)

The claim is true for $k=0$, let's suppose it is also true for $k=n$, and let's prove it for $k=n+1$:

"$(f(x)-g(x)=o((x-x_0)^n+1)implies f^(n+1)(x_0)=g^(n+1)(x_0))$"

$$lim_xto x_0fracf(x)-g(x)(x-x_0)^n+1= 0$$

Now my thought is using L'Hopital $n$ times to achieve

$$boxedlim_xto x_0fracf(x)-g(x)(x-x_0)^n+1= frac1(n+1)!lim_xto x_0fracf^(n)(x)-g^(n)(x)-overbrace(f^(n)(x_0)-g^(n)(x_0))^= 0text for induction x-x_0=0$$

"$(f^(n+1)(x_0)=g^(n+1)(x_0)implies f(x)-g(x)=o((x-x_0)^n+1))$"

$$lim_xto x_0fracf^(n)(x)-g^(n)(x)x-x_0=0$$
Applying L'Hopital the other way around (that is, going back to the primitive), I got

$$boxedlim_xto x_0fracf^(n)(x)-g^(n)(x)x-x_0=n!lim_xto x_0fracf(x)-g(x)(x-x_0)^n+1=0$$

I am not sure about the boxed equalities.

calculus real-analysis proof-verification taylor-expansion

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edited Jul 16 at 9:28

asked Jul 15 at 17:10

Lorenzo B.

1,5402418

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  Since you already have a fixed $n$ in the hypothesis, it would be better to refer to induction on $k$, for $kleq n$.... Have you tried induction on $k$ using the Reminder Formula for a power series ?
  – DanielWainfleet
  Jul 15 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your approach is fine except for the fact that you apply L'Hopital $(n+1)$ times. You can only apply it $n$ times.
  – Paramanand Singh
  Jul 16 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also you don't need induction. L'Hospital's Rule alone gives the implication in question.
  – Paramanand Singh
  Jul 16 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielWainfleet you are right, I should refer to $k$
  – Lorenzo B.
  Jul 16 at 9:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ParamanandSingh in fact I did apply it $n$ times but I wrote $n+1$, thank you for the correction
  – Lorenzo B.
  Jul 16 at 9:27
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

favorite

I tried to prove the following lemma:

Lemma. Let $f,g:mathscr U_x_0to mathbb R$ differentiable $n$ times in $x_0$. Then
$$f(x)-g(x)=o((x-x_0)^n)iff f^(k)(x_0)=g^(k)(x_0), 0le kle n$$

Proof. (by induction)

The claim is true for $k=0$, let's suppose it is also true for $k=n$, and let's prove it for $k=n+1$:

"$(f(x)-g(x)=o((x-x_0)^n+1)implies f^(n+1)(x_0)=g^(n+1)(x_0))$"

$$lim_xto x_0fracf(x)-g(x)(x-x_0)^n+1= 0$$

Now my thought is using L'Hopital $n$ times to achieve

$$boxedlim_xto x_0fracf(x)-g(x)(x-x_0)^n+1= frac1(n+1)!lim_xto x_0fracf^(n)(x)-g^(n)(x)-overbrace(f^(n)(x_0)-g^(n)(x_0))^= 0text for induction x-x_0=0$$

"$(f^(n+1)(x_0)=g^(n+1)(x_0)implies f(x)-g(x)=o((x-x_0)^n+1))$"

$$lim_xto x_0fracf^(n)(x)-g^(n)(x)x-x_0=0$$
Applying L'Hopital the other way around (that is, going back to the primitive), I got

$$boxedlim_xto x_0fracf^(n)(x)-g^(n)(x)x-x_0=n!lim_xto x_0fracf(x)-g(x)(x-x_0)^n+1=0$$

I am not sure about the boxed equalities.

calculus real-analysis proof-verification taylor-expansion

share|cite|improve this question

edited Jul 16 at 9:28

asked Jul 15 at 17:10

Lorenzo B.

1,5402418

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since you already have a fixed $n$ in the hypothesis, it would be better to refer to induction on $k$, for $kleq n$.... Have you tried induction on $k$ using the Reminder Formula for a power series ?
  – DanielWainfleet
  Jul 15 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your approach is fine except for the fact that you apply L'Hopital $(n+1)$ times. You can only apply it $n$ times.
  – Paramanand Singh
  Jul 16 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also you don't need induction. L'Hospital's Rule alone gives the implication in question.
  – Paramanand Singh
  Jul 16 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielWainfleet you are right, I should refer to $k$
  – Lorenzo B.
  Jul 16 at 9:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ParamanandSingh in fact I did apply it $n$ times but I wrote $n+1$, thank you for the correction
  – Lorenzo B.
  Jul 16 at 9:27
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I tried to prove the following lemma:

Lemma. Let $f,g:mathscr U_x_0to mathbb R$ differentiable $n$ times in $x_0$. Then
$$f(x)-g(x)=o((x-x_0)^n)iff f^(k)(x_0)=g^(k)(x_0), 0le kle n$$

Proof. (by induction)

The claim is true for $k=0$, let's suppose it is also true for $k=n$, and let's prove it for $k=n+1$:

"$(f(x)-g(x)=o((x-x_0)^n+1)implies f^(n+1)(x_0)=g^(n+1)(x_0))$"

$$lim_xto x_0fracf(x)-g(x)(x-x_0)^n+1= 0$$

Now my thought is using L'Hopital $n$ times to achieve

$$boxedlim_xto x_0fracf(x)-g(x)(x-x_0)^n+1= frac1(n+1)!lim_xto x_0fracf^(n)(x)-g^(n)(x)-overbrace(f^(n)(x_0)-g^(n)(x_0))^= 0text for induction x-x_0=0$$

"$(f^(n+1)(x_0)=g^(n+1)(x_0)implies f(x)-g(x)=o((x-x_0)^n+1))$"

$$lim_xto x_0fracf^(n)(x)-g^(n)(x)x-x_0=0$$
Applying L'Hopital the other way around (that is, going back to the primitive), I got

$$boxedlim_xto x_0fracf^(n)(x)-g^(n)(x)x-x_0=n!lim_xto x_0fracf(x)-g(x)(x-x_0)^n+1=0$$

I am not sure about the boxed equalities.

calculus real-analysis proof-verification taylor-expansion

share|cite|improve this question

edited Jul 16 at 9:28

asked Jul 15 at 17:10

Lorenzo B.

1,5402418

I tried to prove the following lemma:

Lemma. Let $f,g:mathscr U_x_0to mathbb R$ differentiable $n$ times in $x_0$. Then
$$f(x)-g(x)=o((x-x_0)^n)iff f^(k)(x_0)=g^(k)(x_0), 0le kle n$$

Proof. (by induction)

The claim is true for $k=0$, let's suppose it is also true for $k=n$, and let's prove it for $k=n+1$:

"$(f(x)-g(x)=o((x-x_0)^n+1)implies f^(n+1)(x_0)=g^(n+1)(x_0))$"

$$lim_xto x_0fracf(x)-g(x)(x-x_0)^n+1= 0$$

Now my thought is using L'Hopital $n$ times to achieve

$$boxedlim_xto x_0fracf(x)-g(x)(x-x_0)^n+1= frac1(n+1)!lim_xto x_0fracf^(n)(x)-g^(n)(x)-overbrace(f^(n)(x_0)-g^(n)(x_0))^= 0text for induction x-x_0=0$$

"$(f^(n+1)(x_0)=g^(n+1)(x_0)implies f(x)-g(x)=o((x-x_0)^n+1))$"

$$lim_xto x_0fracf^(n)(x)-g^(n)(x)x-x_0=0$$
Applying L'Hopital the other way around (that is, going back to the primitive), I got

$$boxedlim_xto x_0fracf^(n)(x)-g^(n)(x)x-x_0=n!lim_xto x_0fracf(x)-g(x)(x-x_0)^n+1=0$$

I am not sure about the boxed equalities.

calculus real-analysis proof-verification taylor-expansion

share|cite|improve this question

edited Jul 16 at 9:28

asked Jul 15 at 17:10

Lorenzo B.

1,5402418

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 9:28

edited Jul 16 at 9:28

edited Jul 16 at 9:28

asked Jul 15 at 17:10

Lorenzo B.

1,5402418

asked Jul 15 at 17:10

Lorenzo B.

1,5402418

asked Jul 15 at 17:10

Lorenzo B.

1,5402418

1,5402418

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since you already have a fixed $n$ in the hypothesis, it would be better to refer to induction on $k$, for $kleq n$.... Have you tried induction on $k$ using the Reminder Formula for a power series ?
  – DanielWainfleet
  Jul 15 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your approach is fine except for the fact that you apply L'Hopital $(n+1)$ times. You can only apply it $n$ times.
  – Paramanand Singh
  Jul 16 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also you don't need induction. L'Hospital's Rule alone gives the implication in question.
  – Paramanand Singh
  Jul 16 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielWainfleet you are right, I should refer to $k$
  – Lorenzo B.
  Jul 16 at 9:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ParamanandSingh in fact I did apply it $n$ times but I wrote $n+1$, thank you for the correction
  – Lorenzo B.
  Jul 16 at 9:27
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Since you already have a fixed $n$ in the hypothesis, it would be better to refer to induction on $k$, for $kleq n$.... Have you tried induction on $k$ using the Reminder Formula for a power series ?
  – DanielWainfleet
  Jul 15 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your approach is fine except for the fact that you apply L'Hopital $(n+1)$ times. You can only apply it $n$ times.
  – Paramanand Singh
  Jul 16 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also you don't need induction. L'Hospital's Rule alone gives the implication in question.
  – Paramanand Singh
  Jul 16 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DanielWainfleet you are right, I should refer to $k$
  – Lorenzo B.
  Jul 16 at 9:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ParamanandSingh in fact I did apply it $n$ times but I wrote $n+1$, thank you for the correction
  – Lorenzo B.
  Jul 16 at 9:27
  

  
  
  
  

1

1

Since you already have a fixed $n$ in the hypothesis, it would be better to refer to induction on $k$, for $kleq n$.... Have you tried induction on $k$ using the Reminder Formula for a power series ?
– DanielWainfleet
Jul 15 at 21:57

Since you already have a fixed $n$ in the hypothesis, it would be better to refer to induction on $k$, for $kleq n$.... Have you tried induction on $k$ using the Reminder Formula for a power series ?
– DanielWainfleet
Jul 15 at 21:57

Your approach is fine except for the fact that you apply L'Hopital $(n+1)$ times. You can only apply it $n$ times.
– Paramanand Singh
Jul 16 at 8:42

Your approach is fine except for the fact that you apply L'Hopital $(n+1)$ times. You can only apply it $n$ times.
– Paramanand Singh
Jul 16 at 8:42

Also you don't need induction. L'Hospital's Rule alone gives the implication in question.
– Paramanand Singh
Jul 16 at 8:51

Also you don't need induction. L'Hospital's Rule alone gives the implication in question.
– Paramanand Singh
Jul 16 at 8:51

@DanielWainfleet you are right, I should refer to $k$
– Lorenzo B.
Jul 16 at 9:26

@DanielWainfleet you are right, I should refer to $k$
– Lorenzo B.
Jul 16 at 9:26

@ParamanandSingh in fact I did apply it $n$ times but I wrote $n+1$, thank you for the correction
– Lorenzo B.
Jul 16 at 9:27

@ParamanandSingh in fact I did apply it $n$ times but I wrote $n+1$, thank you for the correction
– Lorenzo B.
Jul 16 at 9:27

 | 
show 2 more comments

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