If f is thrice continuously differentiatable ( f is in $C^{3}$) and its third derivative is bounded, then...
$begingroup$
|$frac{f(x+h)+f(x-h)-2f(x)}{h^{2}}-f''(x)$| is less than or equal to $h^{3}$ times some constant times the supremum of $|f''(y)|$ over y in R
Answer
$f'''(x)$ = $lim_{hrightarrow 0} frac{f(x+3h)-f(x+2h)}{h^{3}}-frac{2f(x+2h)-f(x+h)}{h^{3}}+frac{f(x+h)-f(x)}{h^{3}}$
$f'''(x)$= $lim_{hrightarrow 0} frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}$
not sure how to proceed
partial-derivative
asked Dec 18 '18 at 17:25
Tariro ManyikaTariro Manyika
619
$endgroup$
add a comment |
$begingroup$
|$frac{f(x+h)+f(x-h)-2f(x)}{h^{2}}-f''(x)$| is less than or equal to $h^{3}$ times some constant times the supremum of $|f''(y)|$ over y in R
Answer
$f'''(x)$ = $lim_{hrightarrow 0} frac{f(x+3h)-f(x+2h)}{h^{3}}-frac{2f(x+2h)-f(x+h)}{h^{3}}+frac{f(x+h)-f(x)}{h^{3}}$
$f'''(x)$= $lim_{hrightarrow 0} frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}$
not sure how to proceed
partial-derivative
asked Dec 18 '18 at 17:25
Tariro ManyikaTariro Manyika
619
$endgroup$
add a comment |
$begingroup$
|$frac{f(x+h)+f(x-h)-2f(x)}{h^{2}}-f''(x)$| is less than or equal to $h^{3}$ times some constant times the supremum of $|f''(y)|$ over y in R
Answer
$f'''(x)$ = $lim_{hrightarrow 0} frac{f(x+3h)-f(x+2h)}{h^{3}}-frac{2f(x+2h)-f(x+h)}{h^{3}}+frac{f(x+h)-f(x)}{h^{3}}$
$f'''(x)$= $lim_{hrightarrow 0} frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}$
not sure how to proceed
partial-derivative
asked Dec 18 '18 at 17:25
Tariro ManyikaTariro Manyika
619
$endgroup$
|$frac{f(x+h)+f(x-h)-2f(x)}{h^{2}}-f''(x)$| is less than or equal to $h^{3}$ times some constant times the supremum of $|f''(y)|$ over y in R
Answer
$f'''(x)$ = $lim_{hrightarrow 0} frac{f(x+3h)-f(x+2h)}{h^{3}}-frac{2f(x+2h)-f(x+h)}{h^{3}}+frac{f(x+h)-f(x)}{h^{3}}$
$f'''(x)$= $lim_{hrightarrow 0} frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}$
not sure how to proceed
partial-derivative
partial-derivative
asked Dec 18 '18 at 17:25
Tariro ManyikaTariro Manyika
619
asked Dec 18 '18 at 17:25
Tariro ManyikaTariro Manyika
619
asked Dec 18 '18 at 17:25
Tariro ManyikaTariro Manyika
619
asked Dec 18 '18 at 17:25
Tariro ManyikaTariro Manyika
619
asked Dec 18 '18 at 17:25
Tariro ManyikaTariro Manyika
619
619
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Hint: Taylor formulas yield that
$f(x pm h)=f(x) pm hf’(x)+frac{h^2}{2}int_0^1{(1-u)f’’(x pm hu)du}$.
answered Dec 18 '18 at 17:40
MindlackMindlack
4,910211
$endgroup$
$begingroup$
A bit vague , but thats for trying
$endgroup$
– Tariro Manyika
Dec 18 '18 at 21:55
$begingroup$
That is why I wrote it was a hint (you even asked how to proceed, not the full solution). You can figure out the solution from this, especially if you realize, say, that $f’’$ is Lipschitz continuous with constant $sup_y |f’’’|$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:47
$begingroup$
thanks a lot for the hint
$endgroup$
– Tariro Manyika
Dec 20 '18 at 9:19
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Taylor formulas yield that
$f(x pm h)=f(x) pm hf’(x)+frac{h^2}{2}int_0^1{(1-u)f’’(x pm hu)du}$.
answered Dec 18 '18 at 17:40
MindlackMindlack
4,910211
$endgroup$
$begingroup$
A bit vague , but thats for trying
$endgroup$
– Tariro Manyika
Dec 18 '18 at 21:55
$begingroup$
That is why I wrote it was a hint (you even asked how to proceed, not the full solution). You can figure out the solution from this, especially if you realize, say, that $f’’$ is Lipschitz continuous with constant $sup_y |f’’’|$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:47
$begingroup$
thanks a lot for the hint
$endgroup$
– Tariro Manyika
Dec 20 '18 at 9:19
add a comment |
$begingroup$
Hint: Taylor formulas yield that
$f(x pm h)=f(x) pm hf’(x)+frac{h^2}{2}int_0^1{(1-u)f’’(x pm hu)du}$.
answered Dec 18 '18 at 17:40
MindlackMindlack
4,910211
$endgroup$
$begingroup$
A bit vague , but thats for trying
$endgroup$
– Tariro Manyika
Dec 18 '18 at 21:55
$begingroup$
That is why I wrote it was a hint (you even asked how to proceed, not the full solution). You can figure out the solution from this, especially if you realize, say, that $f’’$ is Lipschitz continuous with constant $sup_y |f’’’|$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:47
$begingroup$
thanks a lot for the hint
$endgroup$
– Tariro Manyika
Dec 20 '18 at 9:19
add a comment |
$begingroup$
Hint: Taylor formulas yield that
$f(x pm h)=f(x) pm hf’(x)+frac{h^2}{2}int_0^1{(1-u)f’’(x pm hu)du}$.
answered Dec 18 '18 at 17:40
MindlackMindlack
4,910211
$endgroup$
Hint: Taylor formulas yield that
$f(x pm h)=f(x) pm hf’(x)+frac{h^2}{2}int_0^1{(1-u)f’’(x pm hu)du}$.
answered Dec 18 '18 at 17:40
MindlackMindlack
4,910211
answered Dec 18 '18 at 17:40
MindlackMindlack
4,910211
answered Dec 18 '18 at 17:40
MindlackMindlack
4,910211
answered Dec 18 '18 at 17:40
MindlackMindlack
4,910211
4,910211
$begingroup$
A bit vague , but thats for trying
$endgroup$
– Tariro Manyika
Dec 18 '18 at 21:55
$begingroup$
That is why I wrote it was a hint (you even asked how to proceed, not the full solution). You can figure out the solution from this, especially if you realize, say, that $f’’$ is Lipschitz continuous with constant $sup_y |f’’’|$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:47
$begingroup$
thanks a lot for the hint
$endgroup$
– Tariro Manyika
Dec 20 '18 at 9:19
add a comment |
$begingroup$
A bit vague , but thats for trying
$endgroup$
– Tariro Manyika
Dec 18 '18 at 21:55
$begingroup$
That is why I wrote it was a hint (you even asked how to proceed, not the full solution). You can figure out the solution from this, especially if you realize, say, that $f’’$ is Lipschitz continuous with constant $sup_y |f’’’|$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:47
$begingroup$
thanks a lot for the hint
$endgroup$
– Tariro Manyika
Dec 20 '18 at 9:19
$begingroup$
A bit vague , but thats for trying
$endgroup$
– Tariro Manyika
Dec 18 '18 at 21:55
$begingroup$
A bit vague , but thats for trying
$endgroup$
– Tariro Manyika
Dec 18 '18 at 21:55
$begingroup$
That is why I wrote it was a hint (you even asked how to proceed, not the full solution). You can figure out the solution from this, especially if you realize, say, that $f’’$ is Lipschitz continuous with constant $sup_y |f’’’|$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:47
$begingroup$
That is why I wrote it was a hint (you even asked how to proceed, not the full solution). You can figure out the solution from this, especially if you realize, say, that $f’’$ is Lipschitz continuous with constant $sup_y |f’’’|$.
$endgroup$
– Mindlack
Dec 18 '18 at 23:47
$begingroup$
thanks a lot for the hint
$endgroup$
– Tariro Manyika
Dec 20 '18 at 9:19
$begingroup$
thanks a lot for the hint
$endgroup$
– Tariro Manyika
Dec 20 '18 at 9:19
add a comment |
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