Finding the infinitesimal order of a function as $n to infty$












0












$begingroup$


I have to find the infinitesimal order of a function $f(n)$ as $n to infty$,
this is what I did:



$$
begin{split}
f(n) &= sqrt{n+1} - sqrt{n} + frac{1}n \
&= frac{nsqrt{n+1} - nsqrt{n} + 1}{n} \
&= frac{sqrt{n^3(1+frac{1}{n^2})}-sqrt{n^3}+1}{n} \
&= frac{sqrt{n^3}-sqrt{n^3}+1}{n} \
&= frac{1}{n}
end{split}
$$
So the infinitesimal order is: 1



Is it correct?
Or should I have used some notable limit, taylor or asymptomatic approximation.






calculus functions infinitesimals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you have a typo or two? Or are you using "$=$" to mean something other than "equals"?
    $endgroup$
    – DanielWainfleet
    Jan 10 at 15:58












  • $begingroup$
    I had one, I already correct it, it was 1/n instead of 1/2, the = means "equals"
    $endgroup$
    – El Bryan
    Jan 10 at 15:59






  • 2




    $begingroup$
    @ElBryan : $left( 1 + frac{1}{n^2} right) neq 1$, so you are not using "$=$" to mean just "equals".
    $endgroup$
    – Eric Towers
    Jan 10 at 16:04












  • $begingroup$
    I see, then I'm wrong, I think I should have used "~"
    $endgroup$
    – El Bryan
    Jan 10 at 16:06










  • $begingroup$
    Is the infinitesimal order = 1?
    $endgroup$
    – El Bryan
    Jan 10 at 16:09
















0












$begingroup$


I have to find the infinitesimal order of a function $f(n)$ as $n to infty$,
this is what I did:



$$
begin{split}
f(n) &= sqrt{n+1} - sqrt{n} + frac{1}n \
&= frac{nsqrt{n+1} - nsqrt{n} + 1}{n} \
&= frac{sqrt{n^3(1+frac{1}{n^2})}-sqrt{n^3}+1}{n} \
&= frac{sqrt{n^3}-sqrt{n^3}+1}{n} \
&= frac{1}{n}
end{split}
$$
So the infinitesimal order is: 1



Is it correct?
Or should I have used some notable limit, taylor or asymptomatic approximation.






calculus functions infinitesimals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you have a typo or two? Or are you using "$=$" to mean something other than "equals"?
    $endgroup$
    – DanielWainfleet
    Jan 10 at 15:58












  • $begingroup$
    I had one, I already correct it, it was 1/n instead of 1/2, the = means "equals"
    $endgroup$
    – El Bryan
    Jan 10 at 15:59






  • 2




    $begingroup$
    @ElBryan : $left( 1 + frac{1}{n^2} right) neq 1$, so you are not using "$=$" to mean just "equals".
    $endgroup$
    – Eric Towers
    Jan 10 at 16:04












  • $begingroup$
    I see, then I'm wrong, I think I should have used "~"
    $endgroup$
    – El Bryan
    Jan 10 at 16:06










  • $begingroup$
    Is the infinitesimal order = 1?
    $endgroup$
    – El Bryan
    Jan 10 at 16:09














0












0








0





$begingroup$


I have to find the infinitesimal order of a function $f(n)$ as $n to infty$,
this is what I did:



$$
begin{split}
f(n) &= sqrt{n+1} - sqrt{n} + frac{1}n \
&= frac{nsqrt{n+1} - nsqrt{n} + 1}{n} \
&= frac{sqrt{n^3(1+frac{1}{n^2})}-sqrt{n^3}+1}{n} \
&= frac{sqrt{n^3}-sqrt{n^3}+1}{n} \
&= frac{1}{n}
end{split}
$$
So the infinitesimal order is: 1



Is it correct?
Or should I have used some notable limit, taylor or asymptomatic approximation.






calculus functions infinitesimals







share|cite|improve this question











$endgroup$




I have to find the infinitesimal order of a function $f(n)$ as $n to infty$,
this is what I did:



$$
begin{split}
f(n) &= sqrt{n+1} - sqrt{n} + frac{1}n \
&= frac{nsqrt{n+1} - nsqrt{n} + 1}{n} \
&= frac{sqrt{n^3(1+frac{1}{n^2})}-sqrt{n^3}+1}{n} \
&= frac{sqrt{n^3}-sqrt{n^3}+1}{n} \
&= frac{1}{n}
end{split}
$$
So the infinitesimal order is: 1



Is it correct?
Or should I have used some notable limit, taylor or asymptomatic approximation.






calculus functions infinitesimals




calculus functions infinitesimals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 16:06









gt6989b

35.6k22557




35.6k22557










asked Jan 10 at 15:46









El BryanEl Bryan

417




417












  • $begingroup$
    Do you have a typo or two? Or are you using "$=$" to mean something other than "equals"?
    $endgroup$
    – DanielWainfleet
    Jan 10 at 15:58












  • $begingroup$
    I had one, I already correct it, it was 1/n instead of 1/2, the = means "equals"
    $endgroup$
    – El Bryan
    Jan 10 at 15:59






  • 2




    $begingroup$
    @ElBryan : $left( 1 + frac{1}{n^2} right) neq 1$, so you are not using "$=$" to mean just "equals".
    $endgroup$
    – Eric Towers
    Jan 10 at 16:04












  • $begingroup$
    I see, then I'm wrong, I think I should have used "~"
    $endgroup$
    – El Bryan
    Jan 10 at 16:06










  • $begingroup$
    Is the infinitesimal order = 1?
    $endgroup$
    – El Bryan
    Jan 10 at 16:09


















  • $begingroup$
    Do you have a typo or two? Or are you using "$=$" to mean something other than "equals"?
    $endgroup$
    – DanielWainfleet
    Jan 10 at 15:58












  • $begingroup$
    I had one, I already correct it, it was 1/n instead of 1/2, the = means "equals"
    $endgroup$
    – El Bryan
    Jan 10 at 15:59






  • 2




    $begingroup$
    @ElBryan : $left( 1 + frac{1}{n^2} right) neq 1$, so you are not using "$=$" to mean just "equals".
    $endgroup$
    – Eric Towers
    Jan 10 at 16:04












  • $begingroup$
    I see, then I'm wrong, I think I should have used "~"
    $endgroup$
    – El Bryan
    Jan 10 at 16:06










  • $begingroup$
    Is the infinitesimal order = 1?
    $endgroup$
    – El Bryan
    Jan 10 at 16:09
















$begingroup$
Do you have a typo or two? Or are you using "$=$" to mean something other than "equals"?
$endgroup$
– DanielWainfleet
Jan 10 at 15:58






$begingroup$
Do you have a typo or two? Or are you using "$=$" to mean something other than "equals"?
$endgroup$
– DanielWainfleet
Jan 10 at 15:58














$begingroup$
I had one, I already correct it, it was 1/n instead of 1/2, the = means "equals"
$endgroup$
– El Bryan
Jan 10 at 15:59




$begingroup$
I had one, I already correct it, it was 1/n instead of 1/2, the = means "equals"
$endgroup$
– El Bryan
Jan 10 at 15:59




2




2




$begingroup$
@ElBryan : $left( 1 + frac{1}{n^2} right) neq 1$, so you are not using "$=$" to mean just "equals".
$endgroup$
– Eric Towers
Jan 10 at 16:04






$begingroup$
@ElBryan : $left( 1 + frac{1}{n^2} right) neq 1$, so you are not using "$=$" to mean just "equals".
$endgroup$
– Eric Towers
Jan 10 at 16:04














$begingroup$
I see, then I'm wrong, I think I should have used "~"
$endgroup$
– El Bryan
Jan 10 at 16:06




$begingroup$
I see, then I'm wrong, I think I should have used "~"
$endgroup$
– El Bryan
Jan 10 at 16:06












$begingroup$
Is the infinitesimal order = 1?
$endgroup$
– El Bryan
Jan 10 at 16:09




$begingroup$
Is the infinitesimal order = 1?
$endgroup$
– El Bryan
Jan 10 at 16:09










2 Answers
2






active

oldest

votes


















1












$begingroup$

Generally it's easier to note that
$$
sqrt{n+1}=sqrt{n}sqrt{1+1/n}sim sqrt{n}left(1+frac{1}{2n}right)=sqrt{n}+frac{1}{2sqrt{n}},
$$
using the Taylor series for $sqrt{1+x}$. Then
$$
sqrt{n+1}-sqrt{n}+frac{1}{n}simfrac{1}{2sqrt{n}}+frac{1}{n},
$$
with omitted terms $O(n^{-3/2})$. The error in your original calculation is that $nsqrt{n+1}=sqrt{n^3+n^2}=sqrt{n^3(1+1/n)}$, not $sqrt{n^3(1+1/n^2)}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your answer is -3/2, the other user answer is -1/2, which one is the correct?
    $endgroup$
    – El Bryan
    Jan 10 at 16:21










  • $begingroup$
    No, both answers agree: $f(n)$ is asymptotic to a term of order $n^{-1/2}$ plus additional smaller terms.
    $endgroup$
    – mjqxxxx
    Jan 10 at 20:15



















1












$begingroup$

I would consider the following instead:
$$
begin{split}
sqrt{n+1} - sqrt{n}
&= left(sqrt{n+1} - sqrt{n}right) times
frac{sqrt{n+1} + sqrt{n}}{sqrt{n+1} + sqrt{n}}\
&= frac{n+1-n}{sqrt{n+1} + sqrt{n}} \
&= frac{1}{sqrt{n+1} + sqrt{n}} \
&= Thetaleft(n^{-1/2}right)
end{split}
$$
therefore,
$$
begin{split}
f(n)
&= sqrt{n+1} - sqrt{n} + frac1n \
&= frac{1}{sqrt{n+1} + sqrt{n}} + frac1n \
&= Thetaleft(n^{-1/2} + n^{-1}right) \
&= Thetaleft(n^{-1/2}right)
end{split}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So my answer is wrong?, it was that easy....
    $endgroup$
    – El Bryan
    Jan 10 at 16:12










  • $begingroup$
    @ElBryan your approach was incorrect, the error you ignored in your approximation was too big
    $endgroup$
    – gt6989b
    Jan 10 at 16:13












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Generally it's easier to note that
$$
sqrt{n+1}=sqrt{n}sqrt{1+1/n}sim sqrt{n}left(1+frac{1}{2n}right)=sqrt{n}+frac{1}{2sqrt{n}},
$$
using the Taylor series for $sqrt{1+x}$. Then
$$
sqrt{n+1}-sqrt{n}+frac{1}{n}simfrac{1}{2sqrt{n}}+frac{1}{n},
$$
with omitted terms $O(n^{-3/2})$. The error in your original calculation is that $nsqrt{n+1}=sqrt{n^3+n^2}=sqrt{n^3(1+1/n)}$, not $sqrt{n^3(1+1/n^2)}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your answer is -3/2, the other user answer is -1/2, which one is the correct?
    $endgroup$
    – El Bryan
    Jan 10 at 16:21










  • $begingroup$
    No, both answers agree: $f(n)$ is asymptotic to a term of order $n^{-1/2}$ plus additional smaller terms.
    $endgroup$
    – mjqxxxx
    Jan 10 at 20:15
















1












$begingroup$

Generally it's easier to note that
$$
sqrt{n+1}=sqrt{n}sqrt{1+1/n}sim sqrt{n}left(1+frac{1}{2n}right)=sqrt{n}+frac{1}{2sqrt{n}},
$$
using the Taylor series for $sqrt{1+x}$. Then
$$
sqrt{n+1}-sqrt{n}+frac{1}{n}simfrac{1}{2sqrt{n}}+frac{1}{n},
$$
with omitted terms $O(n^{-3/2})$. The error in your original calculation is that $nsqrt{n+1}=sqrt{n^3+n^2}=sqrt{n^3(1+1/n)}$, not $sqrt{n^3(1+1/n^2)}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your answer is -3/2, the other user answer is -1/2, which one is the correct?
    $endgroup$
    – El Bryan
    Jan 10 at 16:21










  • $begingroup$
    No, both answers agree: $f(n)$ is asymptotic to a term of order $n^{-1/2}$ plus additional smaller terms.
    $endgroup$
    – mjqxxxx
    Jan 10 at 20:15














1












1








1





$begingroup$

Generally it's easier to note that
$$
sqrt{n+1}=sqrt{n}sqrt{1+1/n}sim sqrt{n}left(1+frac{1}{2n}right)=sqrt{n}+frac{1}{2sqrt{n}},
$$
using the Taylor series for $sqrt{1+x}$. Then
$$
sqrt{n+1}-sqrt{n}+frac{1}{n}simfrac{1}{2sqrt{n}}+frac{1}{n},
$$
with omitted terms $O(n^{-3/2})$. The error in your original calculation is that $nsqrt{n+1}=sqrt{n^3+n^2}=sqrt{n^3(1+1/n)}$, not $sqrt{n^3(1+1/n^2)}$.






share|cite|improve this answer









$endgroup$



Generally it's easier to note that
$$
sqrt{n+1}=sqrt{n}sqrt{1+1/n}sim sqrt{n}left(1+frac{1}{2n}right)=sqrt{n}+frac{1}{2sqrt{n}},
$$
using the Taylor series for $sqrt{1+x}$. Then
$$
sqrt{n+1}-sqrt{n}+frac{1}{n}simfrac{1}{2sqrt{n}}+frac{1}{n},
$$
with omitted terms $O(n^{-3/2})$. The error in your original calculation is that $nsqrt{n+1}=sqrt{n^3+n^2}=sqrt{n^3(1+1/n)}$, not $sqrt{n^3(1+1/n^2)}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 16:12









mjqxxxxmjqxxxx

31.6k24186




31.6k24186












  • $begingroup$
    Your answer is -3/2, the other user answer is -1/2, which one is the correct?
    $endgroup$
    – El Bryan
    Jan 10 at 16:21










  • $begingroup$
    No, both answers agree: $f(n)$ is asymptotic to a term of order $n^{-1/2}$ plus additional smaller terms.
    $endgroup$
    – mjqxxxx
    Jan 10 at 20:15


















  • $begingroup$
    Your answer is -3/2, the other user answer is -1/2, which one is the correct?
    $endgroup$
    – El Bryan
    Jan 10 at 16:21










  • $begingroup$
    No, both answers agree: $f(n)$ is asymptotic to a term of order $n^{-1/2}$ plus additional smaller terms.
    $endgroup$
    – mjqxxxx
    Jan 10 at 20:15
















$begingroup$
Your answer is -3/2, the other user answer is -1/2, which one is the correct?
$endgroup$
– El Bryan
Jan 10 at 16:21




$begingroup$
Your answer is -3/2, the other user answer is -1/2, which one is the correct?
$endgroup$
– El Bryan
Jan 10 at 16:21












$begingroup$
No, both answers agree: $f(n)$ is asymptotic to a term of order $n^{-1/2}$ plus additional smaller terms.
$endgroup$
– mjqxxxx
Jan 10 at 20:15




$begingroup$
No, both answers agree: $f(n)$ is asymptotic to a term of order $n^{-1/2}$ plus additional smaller terms.
$endgroup$
– mjqxxxx
Jan 10 at 20:15











1












$begingroup$

I would consider the following instead:
$$
begin{split}
sqrt{n+1} - sqrt{n}
&= left(sqrt{n+1} - sqrt{n}right) times
frac{sqrt{n+1} + sqrt{n}}{sqrt{n+1} + sqrt{n}}\
&= frac{n+1-n}{sqrt{n+1} + sqrt{n}} \
&= frac{1}{sqrt{n+1} + sqrt{n}} \
&= Thetaleft(n^{-1/2}right)
end{split}
$$
therefore,
$$
begin{split}
f(n)
&= sqrt{n+1} - sqrt{n} + frac1n \
&= frac{1}{sqrt{n+1} + sqrt{n}} + frac1n \
&= Thetaleft(n^{-1/2} + n^{-1}right) \
&= Thetaleft(n^{-1/2}right)
end{split}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So my answer is wrong?, it was that easy....
    $endgroup$
    – El Bryan
    Jan 10 at 16:12










  • $begingroup$
    @ElBryan your approach was incorrect, the error you ignored in your approximation was too big
    $endgroup$
    – gt6989b
    Jan 10 at 16:13
















1












$begingroup$

I would consider the following instead:
$$
begin{split}
sqrt{n+1} - sqrt{n}
&= left(sqrt{n+1} - sqrt{n}right) times
frac{sqrt{n+1} + sqrt{n}}{sqrt{n+1} + sqrt{n}}\
&= frac{n+1-n}{sqrt{n+1} + sqrt{n}} \
&= frac{1}{sqrt{n+1} + sqrt{n}} \
&= Thetaleft(n^{-1/2}right)
end{split}
$$
therefore,
$$
begin{split}
f(n)
&= sqrt{n+1} - sqrt{n} + frac1n \
&= frac{1}{sqrt{n+1} + sqrt{n}} + frac1n \
&= Thetaleft(n^{-1/2} + n^{-1}right) \
&= Thetaleft(n^{-1/2}right)
end{split}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So my answer is wrong?, it was that easy....
    $endgroup$
    – El Bryan
    Jan 10 at 16:12










  • $begingroup$
    @ElBryan your approach was incorrect, the error you ignored in your approximation was too big
    $endgroup$
    – gt6989b
    Jan 10 at 16:13














1












1








1





$begingroup$

I would consider the following instead:
$$
begin{split}
sqrt{n+1} - sqrt{n}
&= left(sqrt{n+1} - sqrt{n}right) times
frac{sqrt{n+1} + sqrt{n}}{sqrt{n+1} + sqrt{n}}\
&= frac{n+1-n}{sqrt{n+1} + sqrt{n}} \
&= frac{1}{sqrt{n+1} + sqrt{n}} \
&= Thetaleft(n^{-1/2}right)
end{split}
$$
therefore,
$$
begin{split}
f(n)
&= sqrt{n+1} - sqrt{n} + frac1n \
&= frac{1}{sqrt{n+1} + sqrt{n}} + frac1n \
&= Thetaleft(n^{-1/2} + n^{-1}right) \
&= Thetaleft(n^{-1/2}right)
end{split}
$$






share|cite|improve this answer











$endgroup$



I would consider the following instead:
$$
begin{split}
sqrt{n+1} - sqrt{n}
&= left(sqrt{n+1} - sqrt{n}right) times
frac{sqrt{n+1} + sqrt{n}}{sqrt{n+1} + sqrt{n}}\
&= frac{n+1-n}{sqrt{n+1} + sqrt{n}} \
&= frac{1}{sqrt{n+1} + sqrt{n}} \
&= Thetaleft(n^{-1/2}right)
end{split}
$$
therefore,
$$
begin{split}
f(n)
&= sqrt{n+1} - sqrt{n} + frac1n \
&= frac{1}{sqrt{n+1} + sqrt{n}} + frac1n \
&= Thetaleft(n^{-1/2} + n^{-1}right) \
&= Thetaleft(n^{-1/2}right)
end{split}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 16:12

























answered Jan 10 at 16:10









gt6989bgt6989b

35.6k22557




35.6k22557












  • $begingroup$
    So my answer is wrong?, it was that easy....
    $endgroup$
    – El Bryan
    Jan 10 at 16:12










  • $begingroup$
    @ElBryan your approach was incorrect, the error you ignored in your approximation was too big
    $endgroup$
    – gt6989b
    Jan 10 at 16:13


















  • $begingroup$
    So my answer is wrong?, it was that easy....
    $endgroup$
    – El Bryan
    Jan 10 at 16:12










  • $begingroup$
    @ElBryan your approach was incorrect, the error you ignored in your approximation was too big
    $endgroup$
    – gt6989b
    Jan 10 at 16:13
















$begingroup$
So my answer is wrong?, it was that easy....
$endgroup$
– El Bryan
Jan 10 at 16:12




$begingroup$
So my answer is wrong?, it was that easy....
$endgroup$
– El Bryan
Jan 10 at 16:12












$begingroup$
@ElBryan your approach was incorrect, the error you ignored in your approximation was too big
$endgroup$
– gt6989b
Jan 10 at 16:13




$begingroup$
@ElBryan your approach was incorrect, the error you ignored in your approximation was too big
$endgroup$
– gt6989b
Jan 10 at 16:13


















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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Gyllenstierna

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