Nth derivative in zero The Next CEO of Stack OverflowThe nth derivative of a trigonometric functionNth derivative can be expressed like that?Formula for nth derivative of $arcsin^k(x/2)$Strange claim by WA involving nth derivativeNth Derivative of a fucntionFinding the derivative to nth orderFind the nth derivativeweak derivative of a function on intervallsIs the derivative of a nonconstant Lipschitz functions non-zero almost everywhere?Derivative at $0$?
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Nth derivative in zero
The Next CEO of Stack OverflowThe nth derivative of a trigonometric functionNth derivative can be expressed like that?Formula for nth derivative of $arcsin^k(x/2)$Strange claim by WA involving nth derivativeNth Derivative of a fucntionFinding the derivative to nth orderFind the nth derivativeweak derivative of a function on intervallsIs the derivative of a nonconstant Lipschitz functions non-zero almost everywhere?Derivative at $0$?
$begingroup$
For $f:mathbbRrightarrowmathbbR$ where $f(x)=frac12+3x^2$ how can we find out, that $f^(1001)(0)=0$?
real-analysis derivatives
edited Mar 18 at 10:57
Bernard
123k741117
asked Mar 18 at 7:11
avan1235avan1235
3688
$endgroup$
add a comment |
$begingroup$
For $f:mathbbRrightarrowmathbbR$ where $f(x)=frac12+3x^2$ how can we find out, that $f^(1001)(0)=0$?
real-analysis derivatives
edited Mar 18 at 10:57
Bernard
123k741117
asked Mar 18 at 7:11
avan1235avan1235
3688
$endgroup$
add a comment |
$begingroup$
For $f:mathbbRrightarrowmathbbR$ where $f(x)=frac12+3x^2$ how can we find out, that $f^(1001)(0)=0$?
real-analysis derivatives
edited Mar 18 at 10:57
Bernard
123k741117
asked Mar 18 at 7:11
avan1235avan1235
3688
$endgroup$
For $f:mathbbRrightarrowmathbbR$ where $f(x)=frac12+3x^2$ how can we find out, that $f^(1001)(0)=0$?
real-analysis derivatives
real-analysis derivatives
edited Mar 18 at 10:57
Bernard
123k741117
asked Mar 18 at 7:11
avan1235avan1235
3688
edited Mar 18 at 10:57
Bernard
123k741117
asked Mar 18 at 7:11
avan1235avan1235
3688
edited Mar 18 at 10:57
Bernard
123k741117
edited Mar 18 at 10:57
Bernard
123k741117
edited Mar 18 at 10:57
Bernard
123k741117
123k741117
asked Mar 18 at 7:11
avan1235avan1235
3688
asked Mar 18 at 7:11
avan1235avan1235
3688
asked Mar 18 at 7:11
avan1235avan1235
3688
3688
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
We have $f(x)=frac12cdot frac11+frac32x^2.$ Now use the geometric series to get $f(x)= sum_n=0^inftya_nx^n.$.
Then we have $ f^(n)(0)= a_n n!$ for $n in mathbb N_0.$
answered Mar 18 at 7:18
FredFred
48.8k11849
$endgroup$
add a comment |
$begingroup$
$f(x)=frac 1 2 frac 1 1+3x^2/2= frac 1 2sumlimits_k=0^infty (-3x^2/2)^k$ for $|x| <2/3$. You can find all derivatives at $0$ from this! All the odd order derivatives are $0$.
answered Mar 18 at 7:21
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
$endgroup$
add a comment |
$begingroup$
Calculations-free argument:
$$fhbox evenimplies f'hbox oddimplies
f''hbox evenimpliescdotsimplies
f^(1001)hbox oddimplies f^(1001)(0) = 0.$$
answered Mar 18 at 9:06
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
1
$begingroup$
(+1) I was just about to post a very similar answer.
$endgroup$
– robjohn♦
Mar 18 at 9:09
add a comment |
$begingroup$
Alternatively:
$$beginalignf(x)&=frac12+3x^2=frac12sqrt2left(frac1sqrt2-sqrt3xi+frac1sqrt2+sqrt3xiright);\
f'(x)&=frac12sqrt2left(frac(-1)(-sqrt3i)(sqrt2-sqrt3xi)^2+frac(-1)(sqrt3i)(sqrt2+sqrt3xi)^2right); f'(0)=0;\
f''(x)&=frac12sqrt2left(frac(-1)(-2)(-sqrt3i)^2(sqrt2-sqrt3xi)^3+frac(-1)(-2)(sqrt3i)^2(sqrt2+sqrt3xi)^3right); f''(0)=-frac32;\
vdots\
f^(2n+1)(x)&=frac12sqrt2left(frac(-1)^2n+1(2n+1)!(-sqrt3i)^2n+1(sqrt2-sqrt3xi)^2n+2+frac(-1)^2n+1(2n+1)!(sqrt3i)^2n+1(sqrt2+sqrt3xi)^2n+2right); \
f^(2n+1)(0)&=0.endalign$$
answered Mar 18 at 11:33
farruhotafarruhota
21.6k2842
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have $f(x)=frac12cdot frac11+frac32x^2.$ Now use the geometric series to get $f(x)= sum_n=0^inftya_nx^n.$.
Then we have $ f^(n)(0)= a_n n!$ for $n in mathbb N_0.$
answered Mar 18 at 7:18
FredFred
48.8k11849
$endgroup$
add a comment |
$begingroup$
We have $f(x)=frac12cdot frac11+frac32x^2.$ Now use the geometric series to get $f(x)= sum_n=0^inftya_nx^n.$.
Then we have $ f^(n)(0)= a_n n!$ for $n in mathbb N_0.$
answered Mar 18 at 7:18
FredFred
48.8k11849
$endgroup$
add a comment |
$begingroup$
We have $f(x)=frac12cdot frac11+frac32x^2.$ Now use the geometric series to get $f(x)= sum_n=0^inftya_nx^n.$.
Then we have $ f^(n)(0)= a_n n!$ for $n in mathbb N_0.$
answered Mar 18 at 7:18
FredFred
48.8k11849
$endgroup$
We have $f(x)=frac12cdot frac11+frac32x^2.$ Now use the geometric series to get $f(x)= sum_n=0^inftya_nx^n.$.
Then we have $ f^(n)(0)= a_n n!$ for $n in mathbb N_0.$
answered Mar 18 at 7:18
FredFred
48.8k11849
edited Mar 18 at 8:38
answered Mar 18 at 7:18
FredFred
48.8k11849
answered Mar 18 at 7:18
FredFred
48.8k11849
answered Mar 18 at 7:18
FredFred
48.8k11849
48.8k11849
add a comment |
add a comment |
$begingroup$
$f(x)=frac 1 2 frac 1 1+3x^2/2= frac 1 2sumlimits_k=0^infty (-3x^2/2)^k$ for $|x| <2/3$. You can find all derivatives at $0$ from this! All the odd order derivatives are $0$.
answered Mar 18 at 7:21
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
$endgroup$
add a comment |
$begingroup$
$f(x)=frac 1 2 frac 1 1+3x^2/2= frac 1 2sumlimits_k=0^infty (-3x^2/2)^k$ for $|x| <2/3$. You can find all derivatives at $0$ from this! All the odd order derivatives are $0$.
answered Mar 18 at 7:21
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
$endgroup$
add a comment |
$begingroup$
$f(x)=frac 1 2 frac 1 1+3x^2/2= frac 1 2sumlimits_k=0^infty (-3x^2/2)^k$ for $|x| <2/3$. You can find all derivatives at $0$ from this! All the odd order derivatives are $0$.
answered Mar 18 at 7:21
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
$endgroup$
$f(x)=frac 1 2 frac 1 1+3x^2/2= frac 1 2sumlimits_k=0^infty (-3x^2/2)^k$ for $|x| <2/3$. You can find all derivatives at $0$ from this! All the odd order derivatives are $0$.
answered Mar 18 at 7:21
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
answered Mar 18 at 7:21
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
answered Mar 18 at 7:21
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
answered Mar 18 at 7:21
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
70.8k53170
add a comment |
add a comment |
$begingroup$
Calculations-free argument:
$$fhbox evenimplies f'hbox oddimplies
f''hbox evenimpliescdotsimplies
f^(1001)hbox oddimplies f^(1001)(0) = 0.$$
answered Mar 18 at 9:06
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
1
$begingroup$
(+1) I was just about to post a very similar answer.
$endgroup$
– robjohn♦
Mar 18 at 9:09
add a comment |
$begingroup$
Calculations-free argument:
$$fhbox evenimplies f'hbox oddimplies
f''hbox evenimpliescdotsimplies
f^(1001)hbox oddimplies f^(1001)(0) = 0.$$
answered Mar 18 at 9:06
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
1
$begingroup$
(+1) I was just about to post a very similar answer.
$endgroup$
– robjohn♦
Mar 18 at 9:09
add a comment |
$begingroup$
Calculations-free argument:
$$fhbox evenimplies f'hbox oddimplies
f''hbox evenimpliescdotsimplies
f^(1001)hbox oddimplies f^(1001)(0) = 0.$$
answered Mar 18 at 9:06
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
$endgroup$
Calculations-free argument:
$$fhbox evenimplies f'hbox oddimplies
f''hbox evenimpliescdotsimplies
f^(1001)hbox oddimplies f^(1001)(0) = 0.$$
answered Mar 18 at 9:06
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
answered Mar 18 at 9:06
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
answered Mar 18 at 9:06
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
answered Mar 18 at 9:06
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.8k42971
34.8k42971
1
$begingroup$
(+1) I was just about to post a very similar answer.
$endgroup$
– robjohn♦
Mar 18 at 9:09
add a comment |
1
$begingroup$
(+1) I was just about to post a very similar answer.
$endgroup$
– robjohn♦
Mar 18 at 9:09
1
1
$begingroup$
(+1) I was just about to post a very similar answer.
$endgroup$
– robjohn♦
Mar 18 at 9:09
$begingroup$
(+1) I was just about to post a very similar answer.
$endgroup$
– robjohn♦
Mar 18 at 9:09
add a comment |
$begingroup$
Alternatively:
$$beginalignf(x)&=frac12+3x^2=frac12sqrt2left(frac1sqrt2-sqrt3xi+frac1sqrt2+sqrt3xiright);\
f'(x)&=frac12sqrt2left(frac(-1)(-sqrt3i)(sqrt2-sqrt3xi)^2+frac(-1)(sqrt3i)(sqrt2+sqrt3xi)^2right); f'(0)=0;\
f''(x)&=frac12sqrt2left(frac(-1)(-2)(-sqrt3i)^2(sqrt2-sqrt3xi)^3+frac(-1)(-2)(sqrt3i)^2(sqrt2+sqrt3xi)^3right); f''(0)=-frac32;\
vdots\
f^(2n+1)(x)&=frac12sqrt2left(frac(-1)^2n+1(2n+1)!(-sqrt3i)^2n+1(sqrt2-sqrt3xi)^2n+2+frac(-1)^2n+1(2n+1)!(sqrt3i)^2n+1(sqrt2+sqrt3xi)^2n+2right); \
f^(2n+1)(0)&=0.endalign$$
answered Mar 18 at 11:33
farruhotafarruhota
21.6k2842
$endgroup$
add a comment |
$begingroup$
Alternatively:
$$beginalignf(x)&=frac12+3x^2=frac12sqrt2left(frac1sqrt2-sqrt3xi+frac1sqrt2+sqrt3xiright);\
f'(x)&=frac12sqrt2left(frac(-1)(-sqrt3i)(sqrt2-sqrt3xi)^2+frac(-1)(sqrt3i)(sqrt2+sqrt3xi)^2right); f'(0)=0;\
f''(x)&=frac12sqrt2left(frac(-1)(-2)(-sqrt3i)^2(sqrt2-sqrt3xi)^3+frac(-1)(-2)(sqrt3i)^2(sqrt2+sqrt3xi)^3right); f''(0)=-frac32;\
vdots\
f^(2n+1)(x)&=frac12sqrt2left(frac(-1)^2n+1(2n+1)!(-sqrt3i)^2n+1(sqrt2-sqrt3xi)^2n+2+frac(-1)^2n+1(2n+1)!(sqrt3i)^2n+1(sqrt2+sqrt3xi)^2n+2right); \
f^(2n+1)(0)&=0.endalign$$
answered Mar 18 at 11:33
farruhotafarruhota
21.6k2842
$endgroup$
add a comment |
$begingroup$
Alternatively:
$$beginalignf(x)&=frac12+3x^2=frac12sqrt2left(frac1sqrt2-sqrt3xi+frac1sqrt2+sqrt3xiright);\
f'(x)&=frac12sqrt2left(frac(-1)(-sqrt3i)(sqrt2-sqrt3xi)^2+frac(-1)(sqrt3i)(sqrt2+sqrt3xi)^2right); f'(0)=0;\
f''(x)&=frac12sqrt2left(frac(-1)(-2)(-sqrt3i)^2(sqrt2-sqrt3xi)^3+frac(-1)(-2)(sqrt3i)^2(sqrt2+sqrt3xi)^3right); f''(0)=-frac32;\
vdots\
f^(2n+1)(x)&=frac12sqrt2left(frac(-1)^2n+1(2n+1)!(-sqrt3i)^2n+1(sqrt2-sqrt3xi)^2n+2+frac(-1)^2n+1(2n+1)!(sqrt3i)^2n+1(sqrt2+sqrt3xi)^2n+2right); \
f^(2n+1)(0)&=0.endalign$$
answered Mar 18 at 11:33
farruhotafarruhota
21.6k2842
$endgroup$
Alternatively:
$$beginalignf(x)&=frac12+3x^2=frac12sqrt2left(frac1sqrt2-sqrt3xi+frac1sqrt2+sqrt3xiright);\
f'(x)&=frac12sqrt2left(frac(-1)(-sqrt3i)(sqrt2-sqrt3xi)^2+frac(-1)(sqrt3i)(sqrt2+sqrt3xi)^2right); f'(0)=0;\
f''(x)&=frac12sqrt2left(frac(-1)(-2)(-sqrt3i)^2(sqrt2-sqrt3xi)^3+frac(-1)(-2)(sqrt3i)^2(sqrt2+sqrt3xi)^3right); f''(0)=-frac32;\
vdots\
f^(2n+1)(x)&=frac12sqrt2left(frac(-1)^2n+1(2n+1)!(-sqrt3i)^2n+1(sqrt2-sqrt3xi)^2n+2+frac(-1)^2n+1(2n+1)!(sqrt3i)^2n+1(sqrt2+sqrt3xi)^2n+2right); \
f^(2n+1)(0)&=0.endalign$$
answered Mar 18 at 11:33
farruhotafarruhota
21.6k2842
answered Mar 18 at 11:33
farruhotafarruhota
21.6k2842
answered Mar 18 at 11:33
farruhotafarruhota
21.6k2842
answered Mar 18 at 11:33
farruhotafarruhota
21.6k2842
21.6k2842
add a comment |
add a comment |
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