Are there continuous functions who are the same in an interval but differ in at least one other point?
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You are given a function $f: mathbb{R}rightarrow mathbb{R}$. Every derivative $frac{d^n}{dx^n}(f(x)), ,n >0$ of the function is continuous.
Is there a function $g: mathbb{R}rightarrow mathbb{R}$, for which every derivative $frac{d^n}{dx^n}(g(x)), ,n >0$ is also continuous, such that:
$$forall xin[a,b]: , g(x) = f(x)land , exists x notin [a,b]: f(x) neq g(x),, a neq b$$
Thanks!
real-analysis calculus
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add a comment |
$begingroup$
You are given a function $f: mathbb{R}rightarrow mathbb{R}$. Every derivative $frac{d^n}{dx^n}(f(x)), ,n >0$ of the function is continuous.
Is there a function $g: mathbb{R}rightarrow mathbb{R}$, for which every derivative $frac{d^n}{dx^n}(g(x)), ,n >0$ is also continuous, such that:
$$forall xin[a,b]: , g(x) = f(x)land , exists x notin [a,b]: f(x) neq g(x),, a neq b$$
Thanks!
real-analysis calculus
$endgroup$
add a comment |
$begingroup$
You are given a function $f: mathbb{R}rightarrow mathbb{R}$. Every derivative $frac{d^n}{dx^n}(f(x)), ,n >0$ of the function is continuous.
Is there a function $g: mathbb{R}rightarrow mathbb{R}$, for which every derivative $frac{d^n}{dx^n}(g(x)), ,n >0$ is also continuous, such that:
$$forall xin[a,b]: , g(x) = f(x)land , exists x notin [a,b]: f(x) neq g(x),, a neq b$$
Thanks!
real-analysis calculus
$endgroup$
You are given a function $f: mathbb{R}rightarrow mathbb{R}$. Every derivative $frac{d^n}{dx^n}(f(x)), ,n >0$ of the function is continuous.
Is there a function $g: mathbb{R}rightarrow mathbb{R}$, for which every derivative $frac{d^n}{dx^n}(g(x)), ,n >0$ is also continuous, such that:
$$forall xin[a,b]: , g(x) = f(x)land , exists x notin [a,b]: f(x) neq g(x),, a neq b$$
Thanks!
real-analysis calculus
real-analysis calculus
edited 1 hour ago
ZeroXLR
1,528519
1,528519
asked 3 hours ago
TVSuchtyTVSuchty
325
325
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Define the real functions $f$ and $g$ thus:
$$
f(x) = begin{cases} expBig(-frac{1}{(x - 1)^2}Big) &text{if } x > 1 \
0 &text{if } x in [-1, 1] \
expBig(-frac{1}{(x + 1)^2}Big) &text{if } x < -1
end{cases}
$$ and
$g(x) = 0$. $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.
Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by applying L'Hôpital's rule inductively. Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
begin{align*}
lim_{x to 1^+}frac{df(x)}{dx} &= limlimits_{x to 1^+}frac{2expbig(- frac{1}{(x - 1)^2}big)}{(x - 1)^3} \
&= 2lim_{x to 1^+}frac{frac{1}{(x - 1)^3}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{this limit is of the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 2 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-3}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 2 lim_{x to 1^+}frac{-3(x - 1)^{-4}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= 3lim_{x to 1^+}frac{(x - 1)^{-1}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{again, this has the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 3 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-1}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 3 lim_{x to 1^+}frac{-(x - 1)^{-2}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= frac{3}{2} lim_{x to 1^+}frac{x - 1}{expbig(frac{1}{(x - 1)^2}big)} \
&= frac{3}{2} lim_{x to 1^+} Big[(x - 1)expBig(-frac{1}{(x - 1)^2}Big)Big] \
&= frac{3}{2} Big[lim_{x to 1^+} (x - 1)Big] Big[lim_{x to 1^+} expBig(-frac{1}{(x - 1)^2}Big)Big] = frac{3}{2} times 0 times 0 = 0
end{align*} That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_{x to 1+}frac{d^nf}{dx^n} = 0$ for all $n in mathbb{Z}_+!$ At all other points i.e. on $(-infty, -1) cup (-1, 1) cup (1, infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.
Bonus Fact:
Both $frac{d^n f(x)}{dx^n}$ and $frac{d^n g(x)}{dx^n}$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!
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Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
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– TVSuchty
2 hours ago
1
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It basically says under certain conditions, $limlimits_{x to a}(f(x) / g(x)) = limlimits_{x to a}(frac{d f(x)}{dx} / frac{d g(x)}{dx})$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
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– ZeroXLR
2 hours ago
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I am stunned. Do you know of more complex solutions?
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– TVSuchty
2 hours ago
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@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
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– ZeroXLR
2 hours ago
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I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
|
show 2 more comments
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Taylors theorem implies that if two functions are the same in one interval, they must be the same everywhere. This is because when you look at one point in the interval, the nth derivatives of both will be equal. Thus, their Taylor series centered at that point will be equal. Then you can move away from the center and find the Taylor series of both centered around another point to get more information about the function, and they will still be equal. So anywhere you look, the two functions will be equal. (This applies for all analytic functions, not so much for piecewise functions)
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Define the real functions $f$ and $g$ thus:
$$
f(x) = begin{cases} expBig(-frac{1}{(x - 1)^2}Big) &text{if } x > 1 \
0 &text{if } x in [-1, 1] \
expBig(-frac{1}{(x + 1)^2}Big) &text{if } x < -1
end{cases}
$$ and
$g(x) = 0$. $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.
Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by applying L'Hôpital's rule inductively. Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
begin{align*}
lim_{x to 1^+}frac{df(x)}{dx} &= limlimits_{x to 1^+}frac{2expbig(- frac{1}{(x - 1)^2}big)}{(x - 1)^3} \
&= 2lim_{x to 1^+}frac{frac{1}{(x - 1)^3}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{this limit is of the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 2 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-3}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 2 lim_{x to 1^+}frac{-3(x - 1)^{-4}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= 3lim_{x to 1^+}frac{(x - 1)^{-1}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{again, this has the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 3 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-1}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 3 lim_{x to 1^+}frac{-(x - 1)^{-2}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= frac{3}{2} lim_{x to 1^+}frac{x - 1}{expbig(frac{1}{(x - 1)^2}big)} \
&= frac{3}{2} lim_{x to 1^+} Big[(x - 1)expBig(-frac{1}{(x - 1)^2}Big)Big] \
&= frac{3}{2} Big[lim_{x to 1^+} (x - 1)Big] Big[lim_{x to 1^+} expBig(-frac{1}{(x - 1)^2}Big)Big] = frac{3}{2} times 0 times 0 = 0
end{align*} That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_{x to 1+}frac{d^nf}{dx^n} = 0$ for all $n in mathbb{Z}_+!$ At all other points i.e. on $(-infty, -1) cup (-1, 1) cup (1, infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.
Bonus Fact:
Both $frac{d^n f(x)}{dx^n}$ and $frac{d^n g(x)}{dx^n}$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!
$endgroup$
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
2 hours ago
1
$begingroup$
It basically says under certain conditions, $limlimits_{x to a}(f(x) / g(x)) = limlimits_{x to a}(frac{d f(x)}{dx} / frac{d g(x)}{dx})$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I am stunned. Do you know of more complex solutions?
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
|
show 2 more comments
$begingroup$
Define the real functions $f$ and $g$ thus:
$$
f(x) = begin{cases} expBig(-frac{1}{(x - 1)^2}Big) &text{if } x > 1 \
0 &text{if } x in [-1, 1] \
expBig(-frac{1}{(x + 1)^2}Big) &text{if } x < -1
end{cases}
$$ and
$g(x) = 0$. $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.
Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by applying L'Hôpital's rule inductively. Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
begin{align*}
lim_{x to 1^+}frac{df(x)}{dx} &= limlimits_{x to 1^+}frac{2expbig(- frac{1}{(x - 1)^2}big)}{(x - 1)^3} \
&= 2lim_{x to 1^+}frac{frac{1}{(x - 1)^3}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{this limit is of the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 2 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-3}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 2 lim_{x to 1^+}frac{-3(x - 1)^{-4}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= 3lim_{x to 1^+}frac{(x - 1)^{-1}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{again, this has the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 3 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-1}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 3 lim_{x to 1^+}frac{-(x - 1)^{-2}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= frac{3}{2} lim_{x to 1^+}frac{x - 1}{expbig(frac{1}{(x - 1)^2}big)} \
&= frac{3}{2} lim_{x to 1^+} Big[(x - 1)expBig(-frac{1}{(x - 1)^2}Big)Big] \
&= frac{3}{2} Big[lim_{x to 1^+} (x - 1)Big] Big[lim_{x to 1^+} expBig(-frac{1}{(x - 1)^2}Big)Big] = frac{3}{2} times 0 times 0 = 0
end{align*} That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_{x to 1+}frac{d^nf}{dx^n} = 0$ for all $n in mathbb{Z}_+!$ At all other points i.e. on $(-infty, -1) cup (-1, 1) cup (1, infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.
Bonus Fact:
Both $frac{d^n f(x)}{dx^n}$ and $frac{d^n g(x)}{dx^n}$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!
$endgroup$
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
2 hours ago
1
$begingroup$
It basically says under certain conditions, $limlimits_{x to a}(f(x) / g(x)) = limlimits_{x to a}(frac{d f(x)}{dx} / frac{d g(x)}{dx})$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I am stunned. Do you know of more complex solutions?
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
|
show 2 more comments
$begingroup$
Define the real functions $f$ and $g$ thus:
$$
f(x) = begin{cases} expBig(-frac{1}{(x - 1)^2}Big) &text{if } x > 1 \
0 &text{if } x in [-1, 1] \
expBig(-frac{1}{(x + 1)^2}Big) &text{if } x < -1
end{cases}
$$ and
$g(x) = 0$. $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.
Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by applying L'Hôpital's rule inductively. Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
begin{align*}
lim_{x to 1^+}frac{df(x)}{dx} &= limlimits_{x to 1^+}frac{2expbig(- frac{1}{(x - 1)^2}big)}{(x - 1)^3} \
&= 2lim_{x to 1^+}frac{frac{1}{(x - 1)^3}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{this limit is of the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 2 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-3}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 2 lim_{x to 1^+}frac{-3(x - 1)^{-4}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= 3lim_{x to 1^+}frac{(x - 1)^{-1}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{again, this has the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 3 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-1}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 3 lim_{x to 1^+}frac{-(x - 1)^{-2}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= frac{3}{2} lim_{x to 1^+}frac{x - 1}{expbig(frac{1}{(x - 1)^2}big)} \
&= frac{3}{2} lim_{x to 1^+} Big[(x - 1)expBig(-frac{1}{(x - 1)^2}Big)Big] \
&= frac{3}{2} Big[lim_{x to 1^+} (x - 1)Big] Big[lim_{x to 1^+} expBig(-frac{1}{(x - 1)^2}Big)Big] = frac{3}{2} times 0 times 0 = 0
end{align*} That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_{x to 1+}frac{d^nf}{dx^n} = 0$ for all $n in mathbb{Z}_+!$ At all other points i.e. on $(-infty, -1) cup (-1, 1) cup (1, infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.
Bonus Fact:
Both $frac{d^n f(x)}{dx^n}$ and $frac{d^n g(x)}{dx^n}$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!
$endgroup$
Define the real functions $f$ and $g$ thus:
$$
f(x) = begin{cases} expBig(-frac{1}{(x - 1)^2}Big) &text{if } x > 1 \
0 &text{if } x in [-1, 1] \
expBig(-frac{1}{(x + 1)^2}Big) &text{if } x < -1
end{cases}
$$ and
$g(x) = 0$. $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.
Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by applying L'Hôpital's rule inductively. Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
begin{align*}
lim_{x to 1^+}frac{df(x)}{dx} &= limlimits_{x to 1^+}frac{2expbig(- frac{1}{(x - 1)^2}big)}{(x - 1)^3} \
&= 2lim_{x to 1^+}frac{frac{1}{(x - 1)^3}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{this limit is of the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 2 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-3}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 2 lim_{x to 1^+}frac{-3(x - 1)^{-4}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= 3lim_{x to 1^+}frac{(x - 1)^{-1}}{expbig(frac{1}{(x - 1)^2}big)} quadtext{again, this has the form } frac{infty}{infty} text{ so L'Hôpital applies} \
&= 3 lim_{x to 1^+}frac{frac{d}{dx}(x - 1)^{-1}}{frac{d}{dx}expbig(frac{1}{(x - 1)^2}big)} text{ by L'Hôpital} \
&= 3 lim_{x to 1^+}frac{-(x - 1)^{-2}}{-2expbig(frac{1}{(x - 1)^2}big)(x - 1)^{-3}} \
&= frac{3}{2} lim_{x to 1^+}frac{x - 1}{expbig(frac{1}{(x - 1)^2}big)} \
&= frac{3}{2} lim_{x to 1^+} Big[(x - 1)expBig(-frac{1}{(x - 1)^2}Big)Big] \
&= frac{3}{2} Big[lim_{x to 1^+} (x - 1)Big] Big[lim_{x to 1^+} expBig(-frac{1}{(x - 1)^2}Big)Big] = frac{3}{2} times 0 times 0 = 0
end{align*} That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_{x to 1+}frac{d^nf}{dx^n} = 0$ for all $n in mathbb{Z}_+!$ At all other points i.e. on $(-infty, -1) cup (-1, 1) cup (1, infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.
Bonus Fact:
Both $frac{d^n f(x)}{dx^n}$ and $frac{d^n g(x)}{dx^n}$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!
edited 1 hour ago
answered 2 hours ago
ZeroXLRZeroXLR
1,528519
1,528519
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
2 hours ago
1
$begingroup$
It basically says under certain conditions, $limlimits_{x to a}(f(x) / g(x)) = limlimits_{x to a}(frac{d f(x)}{dx} / frac{d g(x)}{dx})$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I am stunned. Do you know of more complex solutions?
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
|
show 2 more comments
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
2 hours ago
1
$begingroup$
It basically says under certain conditions, $limlimits_{x to a}(f(x) / g(x)) = limlimits_{x to a}(frac{d f(x)}{dx} / frac{d g(x)}{dx})$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I am stunned. Do you know of more complex solutions?
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
2 hours ago
1
1
$begingroup$
It basically says under certain conditions, $limlimits_{x to a}(f(x) / g(x)) = limlimits_{x to a}(frac{d f(x)}{dx} / frac{d g(x)}{dx})$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
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– ZeroXLR
2 hours ago
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It basically says under certain conditions, $limlimits_{x to a}(f(x) / g(x)) = limlimits_{x to a}(frac{d f(x)}{dx} / frac{d g(x)}{dx})$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
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– ZeroXLR
2 hours ago
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I am stunned. Do you know of more complex solutions?
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– TVSuchty
2 hours ago
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I am stunned. Do you know of more complex solutions?
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– TVSuchty
2 hours ago
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@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
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– ZeroXLR
2 hours ago
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@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
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– ZeroXLR
2 hours ago
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I look forward to. Thank you for your assistance.
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– TVSuchty
2 hours ago
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I look forward to. Thank you for your assistance.
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– TVSuchty
2 hours ago
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show 2 more comments
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Taylors theorem implies that if two functions are the same in one interval, they must be the same everywhere. This is because when you look at one point in the interval, the nth derivatives of both will be equal. Thus, their Taylor series centered at that point will be equal. Then you can move away from the center and find the Taylor series of both centered around another point to get more information about the function, and they will still be equal. So anywhere you look, the two functions will be equal. (This applies for all analytic functions, not so much for piecewise functions)
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add a comment |
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Taylors theorem implies that if two functions are the same in one interval, they must be the same everywhere. This is because when you look at one point in the interval, the nth derivatives of both will be equal. Thus, their Taylor series centered at that point will be equal. Then you can move away from the center and find the Taylor series of both centered around another point to get more information about the function, and they will still be equal. So anywhere you look, the two functions will be equal. (This applies for all analytic functions, not so much for piecewise functions)
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add a comment |
$begingroup$
Taylors theorem implies that if two functions are the same in one interval, they must be the same everywhere. This is because when you look at one point in the interval, the nth derivatives of both will be equal. Thus, their Taylor series centered at that point will be equal. Then you can move away from the center and find the Taylor series of both centered around another point to get more information about the function, and they will still be equal. So anywhere you look, the two functions will be equal. (This applies for all analytic functions, not so much for piecewise functions)
$endgroup$
Taylors theorem implies that if two functions are the same in one interval, they must be the same everywhere. This is because when you look at one point in the interval, the nth derivatives of both will be equal. Thus, their Taylor series centered at that point will be equal. Then you can move away from the center and find the Taylor series of both centered around another point to get more information about the function, and they will still be equal. So anywhere you look, the two functions will be equal. (This applies for all analytic functions, not so much for piecewise functions)
answered 54 mins ago
uhhhhidkuhhhhidk
1266
1266
add a comment |
add a comment |
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