Derivative definition with double limit
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The following is an exercise from Calculus by Spivak.
Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$
My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:
Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}
This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.
real-analysis proof-verification proof-writing
asked Nov 20 at 18:04
高田航
1,283318
add a comment |
up vote
1
down vote
favorite
The following is an exercise from Calculus by Spivak.
Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$
My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:
Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}
This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.
real-analysis proof-verification proof-writing
asked Nov 20 at 18:04
高田航
1,283318
Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
– Federico
Nov 20 at 18:46
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
Nov 20 at 18:50
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The following is an exercise from Calculus by Spivak.
Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$
My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:
Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}
This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.
real-analysis proof-verification proof-writing
asked Nov 20 at 18:04
高田航
1,283318
The following is an exercise from Calculus by Spivak.
Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$
My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:
Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}
This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.
real-analysis proof-verification proof-writing
real-analysis proof-verification proof-writing
asked Nov 20 at 18:04
高田航
1,283318
asked Nov 20 at 18:04
高田航
1,283318
asked Nov 20 at 18:04
高田航
1,283318
asked Nov 20 at 18:04
高田航
1,283318
asked Nov 20 at 18:04
高田航
1,283318
1,283318
Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
– Federico
Nov 20 at 18:46
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
Nov 20 at 18:50
add a comment |
Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
– Federico
Nov 20 at 18:46
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
Nov 20 at 18:50
Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
– Federico
Nov 20 at 18:46
Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
– Federico
Nov 20 at 18:46
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
Nov 20 at 18:50
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
Nov 20 at 18:50
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$
answered Nov 20 at 18:54
Federico
2,910511
I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
– 高田航
Nov 21 at 7:47
1
$o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
– Federico
Nov 21 at 15:13
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$
answered Nov 20 at 18:54
Federico
2,910511
I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
– 高田航
Nov 21 at 7:47
1
$o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
– Federico
Nov 21 at 15:13
add a comment |
up vote
2
down vote
accepted
$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$
answered Nov 20 at 18:54
Federico
2,910511
I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
– 高田航
Nov 21 at 7:47
1
$o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
– Federico
Nov 21 at 15:13
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$
answered Nov 20 at 18:54
Federico
2,910511
$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$
answered Nov 20 at 18:54
Federico
2,910511
edited Nov 21 at 15:15
answered Nov 20 at 18:54
Federico
2,910511
answered Nov 20 at 18:54
Federico
2,910511
answered Nov 20 at 18:54
Federico
2,910511
2,910511
I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
– 高田航
Nov 21 at 7:47
1
$o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
– Federico
Nov 21 at 15:13
add a comment |
I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
– 高田航
Nov 21 at 7:47
1
$o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
– Federico
Nov 21 at 15:13
I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
– 高田航
Nov 21 at 7:47
I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
– 高田航
Nov 21 at 7:47
1
1
$o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
– Federico
Nov 21 at 15:13
$o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
– Federico
Nov 21 at 15:13
add a comment |
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Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
– Federico
Nov 20 at 18:46
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
Nov 20 at 18:50