Find a function such that $f^{(i)}(x)=i,i=1,2,cdots,n.$











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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







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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















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







share|cite|improve this question


















  • 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













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







share|cite|improve this question













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






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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














  • 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










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?






share|cite|improve this answer




























    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.






    share|cite|improve this answer




























      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






      share|cite|improve this answer




























        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.






        share|cite|improve this answer





















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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?






          share|cite|improve this answer

























            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?






            share|cite|improve this answer























              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?






              share|cite|improve this answer












              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?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 23 at 3:07









              ShapeOfMatter

              313




              313






















                  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.






                  share|cite|improve this answer

























                    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.






                    share|cite|improve this answer























                      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.






                      share|cite|improve this answer












                      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.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 23 at 3:08









                      Jimmy R.

                      33k42157




                      33k42157






















                          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






                          share|cite|improve this answer

























                            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






                            share|cite|improve this answer























                              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






                              share|cite|improve this answer












                              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







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 23 at 3:04









                              Seth

                              43612




                              43612






















                                  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.






                                  share|cite|improve this answer

























                                    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.






                                    share|cite|improve this answer























                                      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.






                                      share|cite|improve this answer












                                      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.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 23 at 3:11









                                      Q.Reindeerson

                                      1011




                                      1011






























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