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What is the Taylor series of a square root?

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1

$begingroup$

I recently learned more about Taylor series, what I called infinite polynomials, and decided to find the Taylor series of $sqrt{x}$. Of course, because $frac{d}{dx}sqrt{x}$ at $x=0$ is undefined, I am actually asking about the Taylor series of $sqrt{x+1}$. I have found the Taylor series for this, kinda. It is $sum_{n=0}^∞left(left(prod_{m=1}^nleft(1.5-mright)right)cdotfrac{x^n}{n!}right)$. My real question is: Can this be made smaller?

calculus

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asked 9 hours ago

Infinite PlanesInfinite Planes

4277 bronze badges

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• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Taylor series of $sqrt{1+x}$ using sigma notation
  $endgroup$
  – YuiTo Cheng
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The OP is asking for the most "compact" closed form so I doubt the duplicate-target helps. Whether the inquiry is misguided is another issue.
  $endgroup$
  – Lee David Chung Lin
  8 secs ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I recently learned more about Taylor series, what I called infinite polynomials, and decided to find the Taylor series of $sqrt{x}$. Of course, because $frac{d}{dx}sqrt{x}$ at $x=0$ is undefined, I am actually asking about the Taylor series of $sqrt{x+1}$. I have found the Taylor series for this, kinda. It is $sum_{n=0}^∞left(left(prod_{m=1}^nleft(1.5-mright)right)cdotfrac{x^n}{n!}right)$. My real question is: Can this be made smaller?

calculus

share|cite|improve this question

asked 9 hours ago

Infinite PlanesInfinite Planes

4277 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Taylor series of $sqrt{1+x}$ using sigma notation
  $endgroup$
  – YuiTo Cheng
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The OP is asking for the most "compact" closed form so I doubt the duplicate-target helps. Whether the inquiry is misguided is another issue.
  $endgroup$
  – Lee David Chung Lin
  8 secs ago
  

  
  
  
  

add a comment | 

$begingroup$

I recently learned more about Taylor series, what I called infinite polynomials, and decided to find the Taylor series of $sqrt{x}$. Of course, because $frac{d}{dx}sqrt{x}$ at $x=0$ is undefined, I am actually asking about the Taylor series of $sqrt{x+1}$. I have found the Taylor series for this, kinda. It is $sum_{n=0}^∞left(left(prod_{m=1}^nleft(1.5-mright)right)cdotfrac{x^n}{n!}right)$. My real question is: Can this be made smaller?

calculus

share|cite|improve this question

asked 9 hours ago

Infinite PlanesInfinite Planes

4277 bronze badges

$endgroup$

share|cite|improve this question

asked 9 hours ago

Infinite PlanesInfinite Planes

4277 bronze badges

share|cite|improve this question

asked 9 hours ago

Infinite PlanesInfinite Planes

4277 bronze badges

asked 9 hours ago

Infinite PlanesInfinite Planes

4277 bronze badges

asked 9 hours ago

Infinite PlanesInfinite Planes

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3 Answers
3

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3

$begingroup$

Here is the "smallest" representation I know of.

$$sqrt{x+1} = sum_{nge 0} dbinom{tfrac{1}{2}}{n} x^n$$

This is not strictly a Taylor series. It is actually the Binomial Expansion. It is still an infinite series. Is this what you are looking for?

Edit: Actually, looking at the series you calculated, these two representations should be the same.

share|cite|improve this answer

answered 8 hours ago

InterstellarProbeInterstellarProbe

4,85399 silver badges3131 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It actually is a taylor series of $sqrt{x}$ at $x=1$, or a taylor series of $sqrt{x+1}$ at $x=0$
  $endgroup$
  – Calvin Khor
  8 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

There is a general formula for the expansion of a binomial: for any $alphainmathbf R$, one has
$$(1+x)^alpha=1+alpha x+frac{alpha(alpha-1)}{2!} x^2+frac{alpha(alpha-1)(alpha-2)}{3!} x^3+dots +binom{alpha}{n}x^n+dotsm,$$
where $binom{alpha}{n}$ is the generalised binomial coefficient:
$$binom{alpha}{n}=frac{alpha(alpha-1)dots(alpha -n+1)}{n!}.$$
This binomial series converges for all $|x|<1$.

For $sqrt{1+x}=(1+x)^{1/2}$, it begins with
$$sqrt{1+x}=1+frac x2 -frac18x^2+frac1{16}x^3-frac5{128} x^4+dotsm$$

share|cite|improve this answer

answered 8 hours ago

BernardBernard

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add a comment | 

2

$begingroup$

$$
begin{align}
sqrt{1+x}
&=sum_{n=0}^inftybinom{1/2}{n}x^n\
&=1+sum_{n=0}^inftyfrac{(-1)^nbinom{2n}{n},x^{n+1}}{(n+1),2^{2n+1}}\
&=1+sum_{n=0}^inftyfrac{(-1)^n}{2^{2n+1}},C_n,x^{n+1}
end{align}
$$
Where $C_n$ are the Catalan Numbers.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

robjohn♦robjohn

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add a comment | 

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3

$begingroup$

Here is the "smallest" representation I know of.

$$sqrt{x+1} = sum_{nge 0} dbinom{tfrac{1}{2}}{n} x^n$$

This is not strictly a Taylor series. It is actually the Binomial Expansion. It is still an infinite series. Is this what you are looking for?

Edit: Actually, looking at the series you calculated, these two representations should be the same.

share|cite|improve this answer

answered 8 hours ago

InterstellarProbeInterstellarProbe

4,85399 silver badges3131 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It actually is a taylor series of $sqrt{x}$ at $x=1$, or a taylor series of $sqrt{x+1}$ at $x=0$
  $endgroup$
  – Calvin Khor
  8 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Here is the "smallest" representation I know of.

$$sqrt{x+1} = sum_{nge 0} dbinom{tfrac{1}{2}}{n} x^n$$

This is not strictly a Taylor series. It is actually the Binomial Expansion. It is still an infinite series. Is this what you are looking for?

Edit: Actually, looking at the series you calculated, these two representations should be the same.

share|cite|improve this answer

answered 8 hours ago

InterstellarProbeInterstellarProbe

4,85399 silver badges3131 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It actually is a taylor series of $sqrt{x}$ at $x=1$, or a taylor series of $sqrt{x+1}$ at $x=0$
  $endgroup$
  – Calvin Khor
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Here is the "smallest" representation I know of.

$$sqrt{x+1} = sum_{nge 0} dbinom{tfrac{1}{2}}{n} x^n$$

This is not strictly a Taylor series. It is actually the Binomial Expansion. It is still an infinite series. Is this what you are looking for?

Edit: Actually, looking at the series you calculated, these two representations should be the same.

share|cite|improve this answer

answered 8 hours ago

InterstellarProbeInterstellarProbe

4,85399 silver badges3131 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

InterstellarProbeInterstellarProbe

4,85399 silver badges3131 bronze badges

answered 8 hours ago

InterstellarProbeInterstellarProbe

4,85399 silver badges3131 bronze badges

answered 8 hours ago

InterstellarProbeInterstellarProbe

4,85399 silver badges3131 bronze badges

2

$begingroup$

There is a general formula for the expansion of a binomial: for any $alphainmathbf R$, one has
$$(1+x)^alpha=1+alpha x+frac{alpha(alpha-1)}{2!} x^2+frac{alpha(alpha-1)(alpha-2)}{3!} x^3+dots +binom{alpha}{n}x^n+dotsm,$$
where $binom{alpha}{n}$ is the generalised binomial coefficient:
$$binom{alpha}{n}=frac{alpha(alpha-1)dots(alpha -n+1)}{n!}.$$
This binomial series converges for all $|x|<1$.

For $sqrt{1+x}=(1+x)^{1/2}$, it begins with
$$sqrt{1+x}=1+frac x2 -frac18x^2+frac1{16}x^3-frac5{128} x^4+dotsm$$

share|cite|improve this answer

answered 8 hours ago

BernardBernard

129k77 gold badges4343 silver badges122122 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

There is a general formula for the expansion of a binomial: for any $alphainmathbf R$, one has
$$(1+x)^alpha=1+alpha x+frac{alpha(alpha-1)}{2!} x^2+frac{alpha(alpha-1)(alpha-2)}{3!} x^3+dots +binom{alpha}{n}x^n+dotsm,$$
where $binom{alpha}{n}$ is the generalised binomial coefficient:
$$binom{alpha}{n}=frac{alpha(alpha-1)dots(alpha -n+1)}{n!}.$$
This binomial series converges for all $|x|<1$.

For $sqrt{1+x}=(1+x)^{1/2}$, it begins with
$$sqrt{1+x}=1+frac x2 -frac18x^2+frac1{16}x^3-frac5{128} x^4+dotsm$$

share|cite|improve this answer

answered 8 hours ago

BernardBernard

129k77 gold badges4343 silver badges122122 bronze badges

$endgroup$

add a comment | 

$begingroup$

There is a general formula for the expansion of a binomial: for any $alphainmathbf R$, one has
$$(1+x)^alpha=1+alpha x+frac{alpha(alpha-1)}{2!} x^2+frac{alpha(alpha-1)(alpha-2)}{3!} x^3+dots +binom{alpha}{n}x^n+dotsm,$$
where $binom{alpha}{n}$ is the generalised binomial coefficient:
$$binom{alpha}{n}=frac{alpha(alpha-1)dots(alpha -n+1)}{n!}.$$
This binomial series converges for all $|x|<1$.

For $sqrt{1+x}=(1+x)^{1/2}$, it begins with
$$sqrt{1+x}=1+frac x2 -frac18x^2+frac1{16}x^3-frac5{128} x^4+dotsm$$

share|cite|improve this answer

answered 8 hours ago

BernardBernard

129k77 gold badges4343 silver badges122122 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

BernardBernard

129k77 gold badges4343 silver badges122122 bronze badges

answered 8 hours ago

BernardBernard

129k77 gold badges4343 silver badges122122 bronze badges

answered 8 hours ago

BernardBernard

129k77 gold badges4343 silver badges122122 bronze badges

2

$begingroup$

$$
begin{align}
sqrt{1+x}
&=sum_{n=0}^inftybinom{1/2}{n}x^n\
&=1+sum_{n=0}^inftyfrac{(-1)^nbinom{2n}{n},x^{n+1}}{(n+1),2^{2n+1}}\
&=1+sum_{n=0}^inftyfrac{(-1)^n}{2^{2n+1}},C_n,x^{n+1}
end{align}
$$
Where $C_n$ are the Catalan Numbers.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

robjohn♦robjohn

275k2828 gold badges324324 silver badges655655 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

$$
begin{align}
sqrt{1+x}
&=sum_{n=0}^inftybinom{1/2}{n}x^n\
&=1+sum_{n=0}^inftyfrac{(-1)^nbinom{2n}{n},x^{n+1}}{(n+1),2^{2n+1}}\
&=1+sum_{n=0}^inftyfrac{(-1)^n}{2^{2n+1}},C_n,x^{n+1}
end{align}
$$
Where $C_n$ are the Catalan Numbers.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

robjohn♦robjohn

275k2828 gold badges324324 silver badges655655 bronze badges

$endgroup$

add a comment | 

$begingroup$

$$
begin{align}
sqrt{1+x}
&=sum_{n=0}^inftybinom{1/2}{n}x^n\
&=1+sum_{n=0}^inftyfrac{(-1)^nbinom{2n}{n},x^{n+1}}{(n+1),2^{2n+1}}\
&=1+sum_{n=0}^inftyfrac{(-1)^n}{2^{2n+1}},C_n,x^{n+1}
end{align}
$$
Where $C_n$ are the Catalan Numbers.

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

robjohn♦robjohn

275k2828 gold badges324324 silver badges655655 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

robjohn♦robjohn

275k2828 gold badges324324 silver badges655655 bronze badges

answered 7 hours ago

robjohn♦robjohn

275k2828 gold badges324324 silver badges655655 bronze badges

answered 7 hours ago

robjohn♦robjohn

275k2828 gold badges324324 silver badges655655 bronze badges