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 12 hours ago
高田航
1,279318
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 12 hours ago
高田航
1,279318
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
11 hours ago
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
11 hours ago
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 12 hours ago
高田航
1,279318
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 12 hours ago
高田航
1,279318
asked 12 hours ago
高田航
1,279318
asked 12 hours ago
高田航
1,279318
asked 12 hours ago
高田航
1,279318
asked 12 hours ago
高田航
1,279318
1,279318
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
11 hours ago
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
11 hours ago
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
11 hours ago
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
11 hours ago
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
11 hours ago
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
11 hours ago
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
11 hours ago
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
11 hours ago
add a comment |
1 Answer
1
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1
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$$
frac{f(x+h)-f(x-k)}{h+k} = frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k}
= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} to f'(x) .
$$
answered 11 hours ago
Federico
1,98158
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$$
frac{f(x+h)-f(x-k)}{h+k} = frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k}
= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} to f'(x) .
$$
answered 11 hours ago
Federico
1,98158
add a comment |
up vote
1
down vote
$$
frac{f(x+h)-f(x-k)}{h+k} = frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k}
= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} to f'(x) .
$$
answered 11 hours ago
Federico
1,98158
add a comment |
up vote
1
down vote
up vote
1
down vote
$$
frac{f(x+h)-f(x-k)}{h+k} = frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k}
= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} to f'(x) .
$$
answered 11 hours ago
Federico
1,98158
$$
frac{f(x+h)-f(x-k)}{h+k} = frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k}
= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} to f'(x) .
$$
answered 11 hours ago
Federico
1,98158
answered 11 hours ago
Federico
1,98158
answered 11 hours ago
Federico
1,98158
answered 11 hours ago
Federico
1,98158
1,98158
add a comment |
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
11 hours ago
Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
11 hours ago