Limit of a series of difference quotients minus derivatives of functions
$begingroup$
Let $a,b in mathbb{R}_+$ be real positive numbers with and $frac{1}{2}<a<1$ and let $I=[0,b]$ be a closed real interval
Let $forall n in mathbb{N}: f_n(x) : I to mathbb{R}$ be the sequence of functions with
$$
f_n(x)
=
{
dfrac{n}{(2n-1)^{a+x}}
-
dfrac{n}{(2n)^{a+x}}
}
$$
and let $forall n in mathbb{N}: g_n(x) : I to mathbb{R}$ be the sequence of functions with
$$
g_n(x)
=
dfrac
{d}{dx}
f_n(x)
=
dfrac{n cdot ln(2n)}{(2n)^{a+x}}
-
dfrac{n cdot ln(2n-1)}{(2n-1)^{a+x}}
$$
Let $h : I to mathbb{R}$ be the funtion
$$
h(x)=
sum_{n=1}^infty
left|
dfrac{f_n(x)-f_n(0)}{x}
-
g_n(0)
right|^2
$$
My question is if it is true that
$$
lim_{x to 0^+}
h(x)=0
$$
Thanks.
real-analysis sequences-and-series limits
asked Dec 14 '18 at 10:29
Matey MathMatey Math
846514
$endgroup$
add a comment |
$begingroup$
Let $a,b in mathbb{R}_+$ be real positive numbers with and $frac{1}{2}<a<1$ and let $I=[0,b]$ be a closed real interval
Let $forall n in mathbb{N}: f_n(x) : I to mathbb{R}$ be the sequence of functions with
$$
f_n(x)
=
{
dfrac{n}{(2n-1)^{a+x}}
-
dfrac{n}{(2n)^{a+x}}
}
$$
and let $forall n in mathbb{N}: g_n(x) : I to mathbb{R}$ be the sequence of functions with
$$
g_n(x)
=
dfrac
{d}{dx}
f_n(x)
=
dfrac{n cdot ln(2n)}{(2n)^{a+x}}
-
dfrac{n cdot ln(2n-1)}{(2n-1)^{a+x}}
$$
Let $h : I to mathbb{R}$ be the funtion
$$
h(x)=
sum_{n=1}^infty
left|
dfrac{f_n(x)-f_n(0)}{x}
-
g_n(0)
right|^2
$$
My question is if it is true that
$$
lim_{x to 0^+}
h(x)=0
$$
Thanks.
real-analysis sequences-and-series limits
asked Dec 14 '18 at 10:29
Matey MathMatey Math
846514
$endgroup$
add a comment |
$begingroup$
Let $a,b in mathbb{R}_+$ be real positive numbers with and $frac{1}{2}<a<1$ and let $I=[0,b]$ be a closed real interval
Let $forall n in mathbb{N}: f_n(x) : I to mathbb{R}$ be the sequence of functions with
$$
f_n(x)
=
{
dfrac{n}{(2n-1)^{a+x}}
-
dfrac{n}{(2n)^{a+x}}
}
$$
and let $forall n in mathbb{N}: g_n(x) : I to mathbb{R}$ be the sequence of functions with
$$
g_n(x)
=
dfrac
{d}{dx}
f_n(x)
=
dfrac{n cdot ln(2n)}{(2n)^{a+x}}
-
dfrac{n cdot ln(2n-1)}{(2n-1)^{a+x}}
$$
Let $h : I to mathbb{R}$ be the funtion
$$
h(x)=
sum_{n=1}^infty
left|
dfrac{f_n(x)-f_n(0)}{x}
-
g_n(0)
right|^2
$$
My question is if it is true that
$$
lim_{x to 0^+}
h(x)=0
$$
Thanks.
real-analysis sequences-and-series limits
asked Dec 14 '18 at 10:29
Matey MathMatey Math
846514
$endgroup$
Let $a,b in mathbb{R}_+$ be real positive numbers with and $frac{1}{2}<a<1$ and let $I=[0,b]$ be a closed real interval
Let $forall n in mathbb{N}: f_n(x) : I to mathbb{R}$ be the sequence of functions with
$$
f_n(x)
=
{
dfrac{n}{(2n-1)^{a+x}}
-
dfrac{n}{(2n)^{a+x}}
}
$$
and let $forall n in mathbb{N}: g_n(x) : I to mathbb{R}$ be the sequence of functions with
$$
g_n(x)
=
dfrac
{d}{dx}
f_n(x)
=
dfrac{n cdot ln(2n)}{(2n)^{a+x}}
-
dfrac{n cdot ln(2n-1)}{(2n-1)^{a+x}}
$$
Let $h : I to mathbb{R}$ be the funtion
$$
h(x)=
sum_{n=1}^infty
left|
dfrac{f_n(x)-f_n(0)}{x}
-
g_n(0)
right|^2
$$
My question is if it is true that
$$
lim_{x to 0^+}
h(x)=0
$$
Thanks.
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
asked Dec 14 '18 at 10:29
Matey MathMatey Math
846514
asked Dec 14 '18 at 10:29
Matey MathMatey Math
846514
edited Dec 14 '18 at 12:59
Matey Math
asked Dec 14 '18 at 10:29
Matey MathMatey Math
846514
asked Dec 14 '18 at 10:29
Matey MathMatey Math
846514
asked Dec 14 '18 at 10:29
Matey MathMatey Math
846514
846514
add a comment |
add a comment |
1 Answer
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$begingroup$
Basically, what you are asking is that we are to sum the squares of the difference between the actual rate of how much a function rises minus the first derivative rate for n ranging $(1,infty)$. Since the difference between $x=0^+$ and $x=0$ does not depend upon $n$, I'd say that the limit is $0$. But if you give some dependence of $h=0^+-0$ on $n$, the answer may change.
By taylor expansion:$$f_n(h)=f_n(0)+frac h{1!}g_n(0)+frac{h^2}{2!}f_n^{''}(0)+frac{h^3}{3!}f_n^{'''}(0)+...infty$$
or
$$frac{f_n(h)-f_n(0)}{h}-g_n(0)=frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...infty$$
$$sum_{n=1}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=Bigg|-frac{h}{2!}frac{ln^2(2)}{2^{a+h}}+frac{h^2}{3!}frac{ln^3(2)}{2^{a+h}}...inftyBigg|^2+sum_{n=2}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=h^2times i(n)$$
where $i(n)$ is some function depending on the magnitude of $n$ and may be however big but limit ends up at $0$ because $h$ can be made independently smaller than anything.
answered Dec 14 '18 at 17:08
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
Basically, what you are asking is that we are to sum the squares of the difference between the actual rate of how much a function rises minus the first derivative rate for n ranging $(1,infty)$. Since the difference between $x=0^+$ and $x=0$ does not depend upon $n$, I'd say that the limit is $0$. But if you give some dependence of $h=0^+-0$ on $n$, the answer may change.
By taylor expansion:$$f_n(h)=f_n(0)+frac h{1!}g_n(0)+frac{h^2}{2!}f_n^{''}(0)+frac{h^3}{3!}f_n^{'''}(0)+...infty$$
or
$$frac{f_n(h)-f_n(0)}{h}-g_n(0)=frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...infty$$
$$sum_{n=1}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=Bigg|-frac{h}{2!}frac{ln^2(2)}{2^{a+h}}+frac{h^2}{3!}frac{ln^3(2)}{2^{a+h}}...inftyBigg|^2+sum_{n=2}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=h^2times i(n)$$
where $i(n)$ is some function depending on the magnitude of $n$ and may be however big but limit ends up at $0$ because $h$ can be made independently smaller than anything.
answered Dec 14 '18 at 17:08
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
$begingroup$
Basically, what you are asking is that we are to sum the squares of the difference between the actual rate of how much a function rises minus the first derivative rate for n ranging $(1,infty)$. Since the difference between $x=0^+$ and $x=0$ does not depend upon $n$, I'd say that the limit is $0$. But if you give some dependence of $h=0^+-0$ on $n$, the answer may change.
By taylor expansion:$$f_n(h)=f_n(0)+frac h{1!}g_n(0)+frac{h^2}{2!}f_n^{''}(0)+frac{h^3}{3!}f_n^{'''}(0)+...infty$$
or
$$frac{f_n(h)-f_n(0)}{h}-g_n(0)=frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...infty$$
$$sum_{n=1}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=Bigg|-frac{h}{2!}frac{ln^2(2)}{2^{a+h}}+frac{h^2}{3!}frac{ln^3(2)}{2^{a+h}}...inftyBigg|^2+sum_{n=2}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=h^2times i(n)$$
where $i(n)$ is some function depending on the magnitude of $n$ and may be however big but limit ends up at $0$ because $h$ can be made independently smaller than anything.
answered Dec 14 '18 at 17:08
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
$begingroup$
Basically, what you are asking is that we are to sum the squares of the difference between the actual rate of how much a function rises minus the first derivative rate for n ranging $(1,infty)$. Since the difference between $x=0^+$ and $x=0$ does not depend upon $n$, I'd say that the limit is $0$. But if you give some dependence of $h=0^+-0$ on $n$, the answer may change.
By taylor expansion:$$f_n(h)=f_n(0)+frac h{1!}g_n(0)+frac{h^2}{2!}f_n^{''}(0)+frac{h^3}{3!}f_n^{'''}(0)+...infty$$
or
$$frac{f_n(h)-f_n(0)}{h}-g_n(0)=frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...infty$$
$$sum_{n=1}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=Bigg|-frac{h}{2!}frac{ln^2(2)}{2^{a+h}}+frac{h^2}{3!}frac{ln^3(2)}{2^{a+h}}...inftyBigg|^2+sum_{n=2}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=h^2times i(n)$$
where $i(n)$ is some function depending on the magnitude of $n$ and may be however big but limit ends up at $0$ because $h$ can be made independently smaller than anything.
answered Dec 14 '18 at 17:08
Sameer BahetiSameer Baheti
5168
$endgroup$
Basically, what you are asking is that we are to sum the squares of the difference between the actual rate of how much a function rises minus the first derivative rate for n ranging $(1,infty)$. Since the difference between $x=0^+$ and $x=0$ does not depend upon $n$, I'd say that the limit is $0$. But if you give some dependence of $h=0^+-0$ on $n$, the answer may change.
By taylor expansion:$$f_n(h)=f_n(0)+frac h{1!}g_n(0)+frac{h^2}{2!}f_n^{''}(0)+frac{h^3}{3!}f_n^{'''}(0)+...infty$$
or
$$frac{f_n(h)-f_n(0)}{h}-g_n(0)=frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...infty$$
$$sum_{n=1}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=Bigg|-frac{h}{2!}frac{ln^2(2)}{2^{a+h}}+frac{h^2}{3!}frac{ln^3(2)}{2^{a+h}}...inftyBigg|^2+sum_{n=2}^inftylim_{hto 0}Bigg|frac{h}{2!}f_n^{''}(0)+frac{h^2}{3!}f_n^{'''}(0)+...inftyBigg|^2=h^2times i(n)$$
where $i(n)$ is some function depending on the magnitude of $n$ and may be however big but limit ends up at $0$ because $h$ can be made independently smaller than anything.
answered Dec 14 '18 at 17:08
Sameer BahetiSameer Baheti
5168
answered Dec 14 '18 at 17:08
Sameer BahetiSameer Baheti
5168
answered Dec 14 '18 at 17:08
Sameer BahetiSameer Baheti
5168
answered Dec 14 '18 at 17:08
Sameer BahetiSameer Baheti
5168
5168
add a comment |
add a comment |
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