Find a function such that $f^{(i)}(x)=i,i=1,2,cdots,n.$
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0
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Give $n$-th derivatives, can we find a function ?
For example, if $f^{(i)}(x)=i,i=1,2,cdots,n.$ , can we construct a function such that it satisfies the condition.
It just occurs in my mind. I didn't know if it has a solution.
calculus functions
asked Nov 23 at 2:56
LOIS
3638
add a comment |
up vote
0
down vote
favorite
Give $n$-th derivatives, can we find a function ?
For example, if $f^{(i)}(x)=i,i=1,2,cdots,n.$ , can we construct a function such that it satisfies the condition.
It just occurs in my mind. I didn't know if it has a solution.
calculus functions
asked Nov 23 at 2:56
LOIS
3638
1
If you need $f(z)$ with $f^{(i)}(x)=i$ for $iinBbb{Z}_{geq 0}$ (just at $z=x$ with fixed $x$), the Taylor series gives you $$f(z)=sum_{i=1}^{infty}frac{i(z-x)^i}{i!}=(z-x)e^{z-x}.$$
– metamorphy
Nov 23 at 3:07
As you can see, no one understands what do you really mean.
– metamorphy
Nov 23 at 3:13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Give $n$-th derivatives, can we find a function ?
For example, if $f^{(i)}(x)=i,i=1,2,cdots,n.$ , can we construct a function such that it satisfies the condition.
It just occurs in my mind. I didn't know if it has a solution.
calculus functions
asked Nov 23 at 2:56
LOIS
3638
Give $n$-th derivatives, can we find a function ?
For example, if $f^{(i)}(x)=i,i=1,2,cdots,n.$ , can we construct a function such that it satisfies the condition.
It just occurs in my mind. I didn't know if it has a solution.
calculus functions
calculus functions
asked Nov 23 at 2:56
LOIS
3638
asked Nov 23 at 2:56
LOIS
3638
asked Nov 23 at 2:56
LOIS
3638
asked Nov 23 at 2:56
LOIS
3638
asked Nov 23 at 2:56
LOIS
3638
3638
1
If you need $f(z)$ with $f^{(i)}(x)=i$ for $iinBbb{Z}_{geq 0}$ (just at $z=x$ with fixed $x$), the Taylor series gives you $$f(z)=sum_{i=1}^{infty}frac{i(z-x)^i}{i!}=(z-x)e^{z-x}.$$
– metamorphy
Nov 23 at 3:07
As you can see, no one understands what do you really mean.
– metamorphy
Nov 23 at 3:13
add a comment |
1
If you need $f(z)$ with $f^{(i)}(x)=i$ for $iinBbb{Z}_{geq 0}$ (just at $z=x$ with fixed $x$), the Taylor series gives you $$f(z)=sum_{i=1}^{infty}frac{i(z-x)^i}{i!}=(z-x)e^{z-x}.$$
– metamorphy
Nov 23 at 3:07
As you can see, no one understands what do you really mean.
– metamorphy
Nov 23 at 3:13
1
1
If you need $f(z)$ with $f^{(i)}(x)=i$ for $iinBbb{Z}_{geq 0}$ (just at $z=x$ with fixed $x$), the Taylor series gives you $$f(z)=sum_{i=1}^{infty}frac{i(z-x)^i}{i!}=(z-x)e^{z-x}.$$
– metamorphy
Nov 23 at 3:07
If you need $f(z)$ with $f^{(i)}(x)=i$ for $iinBbb{Z}_{geq 0}$ (just at $z=x$ with fixed $x$), the Taylor series gives you $$f(z)=sum_{i=1}^{infty}frac{i(z-x)^i}{i!}=(z-x)e^{z-x}.$$
– metamorphy
Nov 23 at 3:07
As you can see, no one understands what do you really mean.
– metamorphy
Nov 23 at 3:13
As you can see, no one understands what do you really mean.
– metamorphy
Nov 23 at 3:13
add a comment |
4 Answers
4
active
oldest
votes
up vote
1
down vote
It sounds like you're asking for a function $f$, such that
$$f'=1$$
$$f''=2$$
$$f'''=3$$
$$...$$
But if $f'$ is a constant, then $f''=0$ (and so are all higher derivatives), so no such function can exist.
I suppose we can't rule out functions with domains/codomains other than the Real numbers without a little more work, but we'd still need $f$ to be differentiable.
Am I answering the right question?
answered Nov 23 at 3:07
ShapeOfMatter
313
add a comment |
up vote
1
down vote
Assume that $f$ is such that it satisfies this condition. In particular, this implies that $f^{(1)}(x)=1$ for all $x$ and hence $f^{(2)}(x)=0$ for all $x$, a contradiction. So, such an $f$ does not exist.
answered Nov 23 at 3:08
Jimmy R.
33k42157
add a comment |
up vote
0
down vote
If $f^{(i)}(x)=i$, then $f^{(i-1)}(x)=ix+k=i-1$, so this could work for specific values of $x$, but it will not hold in general
answered Nov 23 at 3:04
Seth
43612
add a comment |
up vote
0
down vote
If I understand correctly you are asking if we can always construct a function such that its first derivative is equal to the constant function $x mapsto 1$ , it second derivative equal to $ x mapsto 2 $, etc.
Well this is impossible because if for a given $n$ we have $f^{(n)} = x mapsto n $ then the derivative of this function would be identically zero. And then $f^{(n-1)}$ would be a polynomial and not a constant function equal to $n-1$
Otherwise if you were wondering if we can always find a function such that $f^{(n)} = x mapsto n $ for a unique n, then the answer is yes and is any polynomial of degree $n$ such that the highest monomial is $frac{x^n}{(n-1)!}$ for instance.
answered Nov 23 at 3:11
Q.Reindeerson
1011
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
up vote
1
down vote
It sounds like you're asking for a function $f$, such that
$$f'=1$$
$$f''=2$$
$$f'''=3$$
$$...$$
But if $f'$ is a constant, then $f''=0$ (and so are all higher derivatives), so no such function can exist.
I suppose we can't rule out functions with domains/codomains other than the Real numbers without a little more work, but we'd still need $f$ to be differentiable.
Am I answering the right question?
answered Nov 23 at 3:07
ShapeOfMatter
313
add a comment |
up vote
1
down vote
It sounds like you're asking for a function $f$, such that
$$f'=1$$
$$f''=2$$
$$f'''=3$$
$$...$$
But if $f'$ is a constant, then $f''=0$ (and so are all higher derivatives), so no such function can exist.
I suppose we can't rule out functions with domains/codomains other than the Real numbers without a little more work, but we'd still need $f$ to be differentiable.
Am I answering the right question?
answered Nov 23 at 3:07
ShapeOfMatter
313
add a comment |
up vote
1
down vote
up vote
1
down vote
It sounds like you're asking for a function $f$, such that
$$f'=1$$
$$f''=2$$
$$f'''=3$$
$$...$$
But if $f'$ is a constant, then $f''=0$ (and so are all higher derivatives), so no such function can exist.
I suppose we can't rule out functions with domains/codomains other than the Real numbers without a little more work, but we'd still need $f$ to be differentiable.
Am I answering the right question?
answered Nov 23 at 3:07
ShapeOfMatter
313
It sounds like you're asking for a function $f$, such that
$$f'=1$$
$$f''=2$$
$$f'''=3$$
$$...$$
But if $f'$ is a constant, then $f''=0$ (and so are all higher derivatives), so no such function can exist.
I suppose we can't rule out functions with domains/codomains other than the Real numbers without a little more work, but we'd still need $f$ to be differentiable.
Am I answering the right question?
answered Nov 23 at 3:07
ShapeOfMatter
313
answered Nov 23 at 3:07
ShapeOfMatter
313
answered Nov 23 at 3:07
ShapeOfMatter
313
answered Nov 23 at 3:07
ShapeOfMatter
313
313
add a comment |
add a comment |
up vote
1
down vote
Assume that $f$ is such that it satisfies this condition. In particular, this implies that $f^{(1)}(x)=1$ for all $x$ and hence $f^{(2)}(x)=0$ for all $x$, a contradiction. So, such an $f$ does not exist.
answered Nov 23 at 3:08
Jimmy R.
33k42157
add a comment |
up vote
1
down vote
Assume that $f$ is such that it satisfies this condition. In particular, this implies that $f^{(1)}(x)=1$ for all $x$ and hence $f^{(2)}(x)=0$ for all $x$, a contradiction. So, such an $f$ does not exist.
answered Nov 23 at 3:08
Jimmy R.
33k42157
add a comment |
up vote
1
down vote
up vote
1
down vote
Assume that $f$ is such that it satisfies this condition. In particular, this implies that $f^{(1)}(x)=1$ for all $x$ and hence $f^{(2)}(x)=0$ for all $x$, a contradiction. So, such an $f$ does not exist.
answered Nov 23 at 3:08
Jimmy R.
33k42157
Assume that $f$ is such that it satisfies this condition. In particular, this implies that $f^{(1)}(x)=1$ for all $x$ and hence $f^{(2)}(x)=0$ for all $x$, a contradiction. So, such an $f$ does not exist.
answered Nov 23 at 3:08
Jimmy R.
33k42157
answered Nov 23 at 3:08
Jimmy R.
33k42157
answered Nov 23 at 3:08
Jimmy R.
33k42157
answered Nov 23 at 3:08
Jimmy R.
33k42157
33k42157
add a comment |
add a comment |
up vote
0
down vote
If $f^{(i)}(x)=i$, then $f^{(i-1)}(x)=ix+k=i-1$, so this could work for specific values of $x$, but it will not hold in general
answered Nov 23 at 3:04
Seth
43612
add a comment |
up vote
0
down vote
If $f^{(i)}(x)=i$, then $f^{(i-1)}(x)=ix+k=i-1$, so this could work for specific values of $x$, but it will not hold in general
answered Nov 23 at 3:04
Seth
43612
add a comment |
up vote
0
down vote
up vote
0
down vote
If $f^{(i)}(x)=i$, then $f^{(i-1)}(x)=ix+k=i-1$, so this could work for specific values of $x$, but it will not hold in general
answered Nov 23 at 3:04
Seth
43612
If $f^{(i)}(x)=i$, then $f^{(i-1)}(x)=ix+k=i-1$, so this could work for specific values of $x$, but it will not hold in general
answered Nov 23 at 3:04
Seth
43612
answered Nov 23 at 3:04
Seth
43612
answered Nov 23 at 3:04
Seth
43612
answered Nov 23 at 3:04
Seth
43612
43612
add a comment |
add a comment |
up vote
0
down vote
If I understand correctly you are asking if we can always construct a function such that its first derivative is equal to the constant function $x mapsto 1$ , it second derivative equal to $ x mapsto 2 $, etc.
Well this is impossible because if for a given $n$ we have $f^{(n)} = x mapsto n $ then the derivative of this function would be identically zero. And then $f^{(n-1)}$ would be a polynomial and not a constant function equal to $n-1$
Otherwise if you were wondering if we can always find a function such that $f^{(n)} = x mapsto n $ for a unique n, then the answer is yes and is any polynomial of degree $n$ such that the highest monomial is $frac{x^n}{(n-1)!}$ for instance.
answered Nov 23 at 3:11
Q.Reindeerson
1011
add a comment |
up vote
0
down vote
If I understand correctly you are asking if we can always construct a function such that its first derivative is equal to the constant function $x mapsto 1$ , it second derivative equal to $ x mapsto 2 $, etc.
Well this is impossible because if for a given $n$ we have $f^{(n)} = x mapsto n $ then the derivative of this function would be identically zero. And then $f^{(n-1)}$ would be a polynomial and not a constant function equal to $n-1$
Otherwise if you were wondering if we can always find a function such that $f^{(n)} = x mapsto n $ for a unique n, then the answer is yes and is any polynomial of degree $n$ such that the highest monomial is $frac{x^n}{(n-1)!}$ for instance.
answered Nov 23 at 3:11
Q.Reindeerson
1011
add a comment |
up vote
0
down vote
up vote
0
down vote
If I understand correctly you are asking if we can always construct a function such that its first derivative is equal to the constant function $x mapsto 1$ , it second derivative equal to $ x mapsto 2 $, etc.
Well this is impossible because if for a given $n$ we have $f^{(n)} = x mapsto n $ then the derivative of this function would be identically zero. And then $f^{(n-1)}$ would be a polynomial and not a constant function equal to $n-1$
Otherwise if you were wondering if we can always find a function such that $f^{(n)} = x mapsto n $ for a unique n, then the answer is yes and is any polynomial of degree $n$ such that the highest monomial is $frac{x^n}{(n-1)!}$ for instance.
answered Nov 23 at 3:11
Q.Reindeerson
1011
If I understand correctly you are asking if we can always construct a function such that its first derivative is equal to the constant function $x mapsto 1$ , it second derivative equal to $ x mapsto 2 $, etc.
Well this is impossible because if for a given $n$ we have $f^{(n)} = x mapsto n $ then the derivative of this function would be identically zero. And then $f^{(n-1)}$ would be a polynomial and not a constant function equal to $n-1$
Otherwise if you were wondering if we can always find a function such that $f^{(n)} = x mapsto n $ for a unique n, then the answer is yes and is any polynomial of degree $n$ such that the highest monomial is $frac{x^n}{(n-1)!}$ for instance.
answered Nov 23 at 3:11
Q.Reindeerson
1011
answered Nov 23 at 3:11
Q.Reindeerson
1011
answered Nov 23 at 3:11
Q.Reindeerson
1011
answered Nov 23 at 3:11
Q.Reindeerson
1011
1011
add a comment |
add a comment |
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1
If you need $f(z)$ with $f^{(i)}(x)=i$ for $iinBbb{Z}_{geq 0}$ (just at $z=x$ with fixed $x$), the Taylor series gives you $$f(z)=sum_{i=1}^{infty}frac{i(z-x)^i}{i!}=(z-x)e^{z-x}.$$
– metamorphy
Nov 23 at 3:07
As you can see, no one understands what do you really mean.
– metamorphy
Nov 23 at 3:13