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

33.9k22455




33.9k22455










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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068790%2ffinding-the-infinitesimal-order-of-a-function-as-n-to-infty%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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

31.3k24086




31.3k24086












  • $begingroup$
    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

33.9k22455




33.9k22455












  • $begingroup$
    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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068790%2ffinding-the-infinitesimal-order-of-a-function-as-n-to-infty%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

Antonio Litta Visconti Arese

Image
Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
Read more