Interesting limits.
$begingroup$
I recently came up with a problem which I think is quite interesting and here it is.
Let $$I(n)=lim_xto 0 left(frac 1(e^x-1)^n -frac 1left(frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!right)^nright) $$ Evaluate $I(n)$ in terms of $n$
Now I tried a lot of messing ups with this monster. Tried L'Hospital, Squeeze theorem, Stolz-Cesaro, Binomial theorem and Multinomial theorem, etc. But couldn't reach the general form. On writing out few terms using Wolfy I get $I(1)=frac -12$ , $I(2)=frac -13$, $I(3)=frac -18$,
$I(4)=frac -130$ and so on
On spending some time on these observations I could conjecture that
$$I(n)=frac -n(n+1)!$$
and tried this formula to check whether it was consistent with other values of $n$ and it indeed was. So Now I have the general form but no proof to it.
And something which is quite interesting in this question is that the first fraction has a denominator with a function raised to power $n$ while the denominator of second fraction is nothing but the Maclaurin expansion of that function with finite terms (here $n$ terms) raised to power of $n$. So I tried to test it with some other functions and the results impressed me a lot.
1) $$J(n)=lim_xto 0 left(frac 1(ln (1+x))^n -frac 1left(x-frac x^22+frac x^33-cdots +frac (-1)^n+1 x^nnright)^nright) =frac (-1)^n+1nn+1$$
2)$$G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright) $$
Now this limit is more interesting because it equals $0$ for all even natural $n$ and it doesn't exist for any odd natural $n$
Now what I actually want is a proper method to solve such problems. You can take any of top 2 limits ( i.e of $(e^x-1)$ or $ln (1+x)$) to demonstrate your method.
Thanks in advance for your attention.
calculus limits taylor-expansion
asked Aug 26 '18 at 11:46
DigammaDigamma
6,1621440
$endgroup$
add a comment |
$begingroup$
I recently came up with a problem which I think is quite interesting and here it is.
Let $$I(n)=lim_xto 0 left(frac 1(e^x-1)^n -frac 1left(frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!right)^nright) $$ Evaluate $I(n)$ in terms of $n$
Now I tried a lot of messing ups with this monster. Tried L'Hospital, Squeeze theorem, Stolz-Cesaro, Binomial theorem and Multinomial theorem, etc. But couldn't reach the general form. On writing out few terms using Wolfy I get $I(1)=frac -12$ , $I(2)=frac -13$, $I(3)=frac -18$,
$I(4)=frac -130$ and so on
On spending some time on these observations I could conjecture that
$$I(n)=frac -n(n+1)!$$
and tried this formula to check whether it was consistent with other values of $n$ and it indeed was. So Now I have the general form but no proof to it.
And something which is quite interesting in this question is that the first fraction has a denominator with a function raised to power $n$ while the denominator of second fraction is nothing but the Maclaurin expansion of that function with finite terms (here $n$ terms) raised to power of $n$. So I tried to test it with some other functions and the results impressed me a lot.
1) $$J(n)=lim_xto 0 left(frac 1(ln (1+x))^n -frac 1left(x-frac x^22+frac x^33-cdots +frac (-1)^n+1 x^nnright)^nright) =frac (-1)^n+1nn+1$$
2)$$G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright) $$
Now this limit is more interesting because it equals $0$ for all even natural $n$ and it doesn't exist for any odd natural $n$
Now what I actually want is a proper method to solve such problems. You can take any of top 2 limits ( i.e of $(e^x-1)$ or $ln (1+x)$) to demonstrate your method.
Thanks in advance for your attention.
calculus limits taylor-expansion
asked Aug 26 '18 at 11:46
DigammaDigamma
6,1621440
$endgroup$
$begingroup$
Very cool problem! Just as an aside: you can put a bracket under the expansion of the last formula to make it prettier $$G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)$$I didn't edit because it's not mandatory at all, if you want, here's the code G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)
$endgroup$
– Davide Morgante
Aug 26 '18 at 11:54
$begingroup$
@Davide Morgante Thanks for the code
$endgroup$
– Digamma
Aug 26 '18 at 12:03
add a comment |
$begingroup$
I recently came up with a problem which I think is quite interesting and here it is.
Let $$I(n)=lim_xto 0 left(frac 1(e^x-1)^n -frac 1left(frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!right)^nright) $$ Evaluate $I(n)$ in terms of $n$
Now I tried a lot of messing ups with this monster. Tried L'Hospital, Squeeze theorem, Stolz-Cesaro, Binomial theorem and Multinomial theorem, etc. But couldn't reach the general form. On writing out few terms using Wolfy I get $I(1)=frac -12$ , $I(2)=frac -13$, $I(3)=frac -18$,
$I(4)=frac -130$ and so on
On spending some time on these observations I could conjecture that
$$I(n)=frac -n(n+1)!$$
and tried this formula to check whether it was consistent with other values of $n$ and it indeed was. So Now I have the general form but no proof to it.
And something which is quite interesting in this question is that the first fraction has a denominator with a function raised to power $n$ while the denominator of second fraction is nothing but the Maclaurin expansion of that function with finite terms (here $n$ terms) raised to power of $n$. So I tried to test it with some other functions and the results impressed me a lot.
1) $$J(n)=lim_xto 0 left(frac 1(ln (1+x))^n -frac 1left(x-frac x^22+frac x^33-cdots +frac (-1)^n+1 x^nnright)^nright) =frac (-1)^n+1nn+1$$
2)$$G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright) $$
Now this limit is more interesting because it equals $0$ for all even natural $n$ and it doesn't exist for any odd natural $n$
Now what I actually want is a proper method to solve such problems. You can take any of top 2 limits ( i.e of $(e^x-1)$ or $ln (1+x)$) to demonstrate your method.
Thanks in advance for your attention.
calculus limits taylor-expansion
asked Aug 26 '18 at 11:46
DigammaDigamma
6,1621440
$endgroup$
I recently came up with a problem which I think is quite interesting and here it is.
Let $$I(n)=lim_xto 0 left(frac 1(e^x-1)^n -frac 1left(frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!right)^nright) $$ Evaluate $I(n)$ in terms of $n$
Now I tried a lot of messing ups with this monster. Tried L'Hospital, Squeeze theorem, Stolz-Cesaro, Binomial theorem and Multinomial theorem, etc. But couldn't reach the general form. On writing out few terms using Wolfy I get $I(1)=frac -12$ , $I(2)=frac -13$, $I(3)=frac -18$,
$I(4)=frac -130$ and so on
On spending some time on these observations I could conjecture that
$$I(n)=frac -n(n+1)!$$
and tried this formula to check whether it was consistent with other values of $n$ and it indeed was. So Now I have the general form but no proof to it.
And something which is quite interesting in this question is that the first fraction has a denominator with a function raised to power $n$ while the denominator of second fraction is nothing but the Maclaurin expansion of that function with finite terms (here $n$ terms) raised to power of $n$. So I tried to test it with some other functions and the results impressed me a lot.
1) $$J(n)=lim_xto 0 left(frac 1(ln (1+x))^n -frac 1left(x-frac x^22+frac x^33-cdots +frac (-1)^n+1 x^nnright)^nright) =frac (-1)^n+1nn+1$$
2)$$G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright) $$
Now this limit is more interesting because it equals $0$ for all even natural $n$ and it doesn't exist for any odd natural $n$
Now what I actually want is a proper method to solve such problems. You can take any of top 2 limits ( i.e of $(e^x-1)$ or $ln (1+x)$) to demonstrate your method.
Thanks in advance for your attention.
calculus limits taylor-expansion
calculus limits taylor-expansion
asked Aug 26 '18 at 11:46
DigammaDigamma
6,1621440
asked Aug 26 '18 at 11:46
DigammaDigamma
6,1621440
edited Aug 26 '18 at 12:02
Digamma
asked Aug 26 '18 at 11:46
DigammaDigamma
6,1621440
asked Aug 26 '18 at 11:46
DigammaDigamma
6,1621440
asked Aug 26 '18 at 11:46
DigammaDigamma
6,1621440
6,1621440
$begingroup$
Very cool problem! Just as an aside: you can put a bracket under the expansion of the last formula to make it prettier $$G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)$$I didn't edit because it's not mandatory at all, if you want, here's the code G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)
$endgroup$
– Davide Morgante
Aug 26 '18 at 11:54
$begingroup$
@Davide Morgante Thanks for the code
$endgroup$
– Digamma
Aug 26 '18 at 12:03
add a comment |
$begingroup$
Very cool problem! Just as an aside: you can put a bracket under the expansion of the last formula to make it prettier $$G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)$$I didn't edit because it's not mandatory at all, if you want, here's the code G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)
$endgroup$
– Davide Morgante
Aug 26 '18 at 11:54
$begingroup$
@Davide Morgante Thanks for the code
$endgroup$
– Digamma
Aug 26 '18 at 12:03
$begingroup$
Very cool problem! Just as an aside: you can put a bracket under the expansion of the last formula to make it prettier $$G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)$$I didn't edit because it's not mandatory at all, if you want, here's the code G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)
$endgroup$
– Davide Morgante
Aug 26 '18 at 11:54
$begingroup$
Very cool problem! Just as an aside: you can put a bracket under the expansion of the last formula to make it prettier $$G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)$$I didn't edit because it's not mandatory at all, if you want, here's the code G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)
$endgroup$
– Davide Morgante
Aug 26 '18 at 11:54
$begingroup$
@Davide Morgante Thanks for the code
$endgroup$
– Digamma
Aug 26 '18 at 12:03
$begingroup$
@Davide Morgante Thanks for the code
$endgroup$
– Digamma
Aug 26 '18 at 12:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $a=e^x-1$ and $b=frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!$, so is it easy to see: $asim x$ , $bsim x$ and $b-asim -fracx^n+1(n+1)!$ as $xto 0.$
$$frac 1(e^x-1)^n -frac 1left(frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!right)^n$$
$$=fracb^n-a^na^nb^n=frac(b-a)(b^n-1+b^n-2a+cdots+a^n-1)a^n b^n$$
$$simfracleft(-fracx^n+1(n+1)!right)(nx^n-1)x^2ntofrac -n(n+1)!,$$
as $xto 0$.
answered Aug 26 '18 at 12:21
RiemannRiemann
3,4551322
$endgroup$
$begingroup$
Did you solve this problem somewhere else before because when I asked my professor about this problem a few minutes before he too hinted towards the same method.
$endgroup$
– Digamma
Aug 26 '18 at 12:47
2
$begingroup$
I think you need to do more exercise, and then you will do this problem very quickly.
$endgroup$
– Riemann
Aug 26 '18 at 12:54
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $a=e^x-1$ and $b=frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!$, so is it easy to see: $asim x$ , $bsim x$ and $b-asim -fracx^n+1(n+1)!$ as $xto 0.$
$$frac 1(e^x-1)^n -frac 1left(frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!right)^n$$
$$=fracb^n-a^na^nb^n=frac(b-a)(b^n-1+b^n-2a+cdots+a^n-1)a^n b^n$$
$$simfracleft(-fracx^n+1(n+1)!right)(nx^n-1)x^2ntofrac -n(n+1)!,$$
as $xto 0$.
answered Aug 26 '18 at 12:21
RiemannRiemann
3,4551322
$endgroup$
$begingroup$
Did you solve this problem somewhere else before because when I asked my professor about this problem a few minutes before he too hinted towards the same method.
$endgroup$
– Digamma
Aug 26 '18 at 12:47
2
$begingroup$
I think you need to do more exercise, and then you will do this problem very quickly.
$endgroup$
– Riemann
Aug 26 '18 at 12:54
add a comment |
$begingroup$
Let $a=e^x-1$ and $b=frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!$, so is it easy to see: $asim x$ , $bsim x$ and $b-asim -fracx^n+1(n+1)!$ as $xto 0.$
$$frac 1(e^x-1)^n -frac 1left(frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!right)^n$$
$$=fracb^n-a^na^nb^n=frac(b-a)(b^n-1+b^n-2a+cdots+a^n-1)a^n b^n$$
$$simfracleft(-fracx^n+1(n+1)!right)(nx^n-1)x^2ntofrac -n(n+1)!,$$
as $xto 0$.
answered Aug 26 '18 at 12:21
RiemannRiemann
3,4551322
$endgroup$
$begingroup$
Did you solve this problem somewhere else before because when I asked my professor about this problem a few minutes before he too hinted towards the same method.
$endgroup$
– Digamma
Aug 26 '18 at 12:47
2
$begingroup$
I think you need to do more exercise, and then you will do this problem very quickly.
$endgroup$
– Riemann
Aug 26 '18 at 12:54
add a comment |
$begingroup$
Let $a=e^x-1$ and $b=frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!$, so is it easy to see: $asim x$ , $bsim x$ and $b-asim -fracx^n+1(n+1)!$ as $xto 0.$
$$frac 1(e^x-1)^n -frac 1left(frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!right)^n$$
$$=fracb^n-a^na^nb^n=frac(b-a)(b^n-1+b^n-2a+cdots+a^n-1)a^n b^n$$
$$simfracleft(-fracx^n+1(n+1)!right)(nx^n-1)x^2ntofrac -n(n+1)!,$$
as $xto 0$.
answered Aug 26 '18 at 12:21
RiemannRiemann
3,4551322
$endgroup$
Let $a=e^x-1$ and $b=frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!$, so is it easy to see: $asim x$ , $bsim x$ and $b-asim -fracx^n+1(n+1)!$ as $xto 0.$
$$frac 1(e^x-1)^n -frac 1left(frac x1!+frac x^22!+frac x^33!+cdots +frac x^nn!right)^n$$
$$=fracb^n-a^na^nb^n=frac(b-a)(b^n-1+b^n-2a+cdots+a^n-1)a^n b^n$$
$$simfracleft(-fracx^n+1(n+1)!right)(nx^n-1)x^2ntofrac -n(n+1)!,$$
as $xto 0$.
answered Aug 26 '18 at 12:21
RiemannRiemann
3,4551322
answered Aug 26 '18 at 12:21
RiemannRiemann
3,4551322
answered Aug 26 '18 at 12:21
RiemannRiemann
3,4551322
answered Aug 26 '18 at 12:21
RiemannRiemann
3,4551322
3,4551322
$begingroup$
Did you solve this problem somewhere else before because when I asked my professor about this problem a few minutes before he too hinted towards the same method.
$endgroup$
– Digamma
Aug 26 '18 at 12:47
2
$begingroup$
I think you need to do more exercise, and then you will do this problem very quickly.
$endgroup$
– Riemann
Aug 26 '18 at 12:54
add a comment |
$begingroup$
Did you solve this problem somewhere else before because when I asked my professor about this problem a few minutes before he too hinted towards the same method.
$endgroup$
– Digamma
Aug 26 '18 at 12:47
2
$begingroup$
I think you need to do more exercise, and then you will do this problem very quickly.
$endgroup$
– Riemann
Aug 26 '18 at 12:54
$begingroup$
Did you solve this problem somewhere else before because when I asked my professor about this problem a few minutes before he too hinted towards the same method.
$endgroup$
– Digamma
Aug 26 '18 at 12:47
$begingroup$
Did you solve this problem somewhere else before because when I asked my professor about this problem a few minutes before he too hinted towards the same method.
$endgroup$
– Digamma
Aug 26 '18 at 12:47
2
2
$begingroup$
I think you need to do more exercise, and then you will do this problem very quickly.
$endgroup$
– Riemann
Aug 26 '18 at 12:54
$begingroup$
I think you need to do more exercise, and then you will do this problem very quickly.
$endgroup$
– Riemann
Aug 26 '18 at 12:54
add a comment |
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$begingroup$
Very cool problem! Just as an aside: you can put a bracket under the expansion of the last formula to make it prettier $$G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)$$I didn't edit because it's not mandatory at all, if you want, here's the code G(n)=lim_xto 0 left(frac 1(arccos x-frac pi2)^n -frac 1left(underbracefrac x1+frac x^36+frac 3x^540+frac 5x^7112+cdots_text n termsright)^nright)
$endgroup$
– Davide Morgante
Aug 26 '18 at 11:54
$begingroup$
@Davide Morgante Thanks for the code
$endgroup$
– Digamma
Aug 26 '18 at 12:03