How to get nth derivative of $e^{x^2/2}$











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I want to calculate the nth derivative of $e^{x^2/2}$. It is as follow:
$$
frac{d}{dx} e^{x^2/2} = x e^{x^2/2} = P_1(x) e^{x^2/2}
$$
$$
frac{d^n}{dx^n} e^{x^2/2} = frac{d}{dx} (P_{n-1}(x) e^{x^2/2}) = (x P_{n-1}(x) + frac{dP_{n-1}}{dx})e^{x^2/2} = P_n(x) e^{x^2/2}
$$
So we get recursive relation of $P_n$:
$$ P_n(x) = xP_{n-1} + frac{dP_{n-1}}{dx}, P_0(x) = 1tag1
$$
My question is how to solve the recursive relation involving function and derivative. I know generating function for recursion like $a_{n+1}=a_{n}+a_{n-1}$. But I am not sure how to solve $(1)$.










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  • 2




    I think we are looking at the Hermite polynomials (coincidence?) with imaginary argument, cf. Eqs. (30) in the link with $x rightarrow i x/sqrt{2}$.
    – hbp
    Oct 19 '15 at 23:03















up vote
7
down vote

favorite
5












I want to calculate the nth derivative of $e^{x^2/2}$. It is as follow:
$$
frac{d}{dx} e^{x^2/2} = x e^{x^2/2} = P_1(x) e^{x^2/2}
$$
$$
frac{d^n}{dx^n} e^{x^2/2} = frac{d}{dx} (P_{n-1}(x) e^{x^2/2}) = (x P_{n-1}(x) + frac{dP_{n-1}}{dx})e^{x^2/2} = P_n(x) e^{x^2/2}
$$
So we get recursive relation of $P_n$:
$$ P_n(x) = xP_{n-1} + frac{dP_{n-1}}{dx}, P_0(x) = 1tag1
$$
My question is how to solve the recursive relation involving function and derivative. I know generating function for recursion like $a_{n+1}=a_{n}+a_{n-1}$. But I am not sure how to solve $(1)$.










share|cite|improve this question




















  • 2




    I think we are looking at the Hermite polynomials (coincidence?) with imaginary argument, cf. Eqs. (30) in the link with $x rightarrow i x/sqrt{2}$.
    – hbp
    Oct 19 '15 at 23:03













up vote
7
down vote

favorite
5









up vote
7
down vote

favorite
5






5





I want to calculate the nth derivative of $e^{x^2/2}$. It is as follow:
$$
frac{d}{dx} e^{x^2/2} = x e^{x^2/2} = P_1(x) e^{x^2/2}
$$
$$
frac{d^n}{dx^n} e^{x^2/2} = frac{d}{dx} (P_{n-1}(x) e^{x^2/2}) = (x P_{n-1}(x) + frac{dP_{n-1}}{dx})e^{x^2/2} = P_n(x) e^{x^2/2}
$$
So we get recursive relation of $P_n$:
$$ P_n(x) = xP_{n-1} + frac{dP_{n-1}}{dx}, P_0(x) = 1tag1
$$
My question is how to solve the recursive relation involving function and derivative. I know generating function for recursion like $a_{n+1}=a_{n}+a_{n-1}$. But I am not sure how to solve $(1)$.










share|cite|improve this question















I want to calculate the nth derivative of $e^{x^2/2}$. It is as follow:
$$
frac{d}{dx} e^{x^2/2} = x e^{x^2/2} = P_1(x) e^{x^2/2}
$$
$$
frac{d^n}{dx^n} e^{x^2/2} = frac{d}{dx} (P_{n-1}(x) e^{x^2/2}) = (x P_{n-1}(x) + frac{dP_{n-1}}{dx})e^{x^2/2} = P_n(x) e^{x^2/2}
$$
So we get recursive relation of $P_n$:
$$ P_n(x) = xP_{n-1} + frac{dP_{n-1}}{dx}, P_0(x) = 1tag1
$$
My question is how to solve the recursive relation involving function and derivative. I know generating function for recursion like $a_{n+1}=a_{n}+a_{n-1}$. But I am not sure how to solve $(1)$.







functions derivatives polynomials recursion






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edited Oct 19 '15 at 22:40

























asked Oct 19 '15 at 22:06









Math Wizard

13.2k11036




13.2k11036








  • 2




    I think we are looking at the Hermite polynomials (coincidence?) with imaginary argument, cf. Eqs. (30) in the link with $x rightarrow i x/sqrt{2}$.
    – hbp
    Oct 19 '15 at 23:03














  • 2




    I think we are looking at the Hermite polynomials (coincidence?) with imaginary argument, cf. Eqs. (30) in the link with $x rightarrow i x/sqrt{2}$.
    – hbp
    Oct 19 '15 at 23:03








2




2




I think we are looking at the Hermite polynomials (coincidence?) with imaginary argument, cf. Eqs. (30) in the link with $x rightarrow i x/sqrt{2}$.
– hbp
Oct 19 '15 at 23:03




I think we are looking at the Hermite polynomials (coincidence?) with imaginary argument, cf. Eqs. (30) in the link with $x rightarrow i x/sqrt{2}$.
– hbp
Oct 19 '15 at 23:03










5 Answers
5






active

oldest

votes

















up vote
6
down vote



accepted










As noted before this is a variant of the Hermite polynomials. Due to subtle differences, we will adapt the standard derivation from Arfken.



Generating function



First
begin{align}
frac{d^n}{dx^n} e^{x^2/2}
&=
lim_{trightarrow 0} frac{d^n}{dx^n} e^{(x+t)^2/2}
tag{1} \
&=
lim_{trightarrow 0} frac{d^n}{dt^n} e^{(x+t)^2/2} \
&=
e^{x^2/2} lim_{trightarrow 0} frac{d^n}{dt^n} e^{xt + t^2/2},
end{align}

This means that the polynomials $P_n(x)$ we are looking for
are just the $n$th coefficients of the Taylor expansion of
$e^{xt+t^2/2}$. In other words,
begin{align}
e^{xt + t^2/2} = sum_{n = 0}^infty frac{P_n(x)}{n!} t^n.
tag{2}
end{align}

The left-hand side is the exponential generating function of $P_n(x)$.



Recurrence relations



If we differentiate (2) with respect to $t$,
$$
(x + t) , e^{xt + t^2/2} = sum_{n = 1}^infty frac{P_n(x)}{(n-1)!} t^{n-1}
= sum_{n = 0}^infty frac{P_{n+1}(x)}{n!} t^n.
tag{3}
$$

where the previous n = 0 term is simply equal to zero, allowing for a shift in indices.
Expanding the left-hand side,
begin{align}
(x + t) , e^{xt + t^2/2}
&=
(x + t) sum_{n = 0}^infty frac{ P_n(x) }{n!} t^n
\
&=
sum_{n = 0}^infty frac{ x , P_n(x) }{n!} t^n
+
sum_{n = 0}^infty frac{ P_n(x) }{n!} t^{n+1}
\
&=
sum_{n = 0}^infty frac{ x , P_n(x) }{n!} t^n
+
sum_{n = 1}^infty frac{ n , P_{n-1}(x) }{n!} t^{n}.
tag{4}
end{align}

Comparing the coefficients of $t^n/n!$ in (3) and (4) yields
$$
P_{n+1}(x) = x , P_n(x) + n P_{n-1}(x).
tag{5}
$$



Similarly, by differentiating (2) with respect to $x$, we get
$$
P'_n(x) = n P_{n-1}(x),
tag{6}
$$

which is noted by Barry Cipra.
Combining the two yields the relation by hermes.



Explicit formula



An explicit formula is more readily derived from the generating function instead of the recurrence relations:
begin{align}
e^{xt+t^2/2}
&= e^{xt} , e^{t^2/2}\
&= sum_{s = 0}^infty frac{(xt)^s}{s!} sum_{m = 0}^infty frac{t^{2m}}{2^m , m!} \
&= sum_{n = 0}^infty
left(
sum_{m = 0}^{[n/2]} frac{ n! , x^{n-2m} }{ 2^m , m! , (n-2m)! }
right) frac{t^n}{n!},
end{align}

where $[n/2]$ denotes the largest integer not exceeding $n/2$.
Comparing this to (2), we get
begin{align}
P_n(x) = sum_{m = 0}^{[n/2]} frac{ n! , x^{n-2m} }{ 2^m , m! , (n-2m)! }.
end{align}



Relations to the standard definitions



For reference, $P_n(x)$ is related to the standard Hermite polynomials as
begin{align}
P_n(x)
&= (-i)^n , mathrm{He}_n(ix) \
&= frac{1}{(sqrt{2} , i)^n} , H_nleft(frac{ix}{sqrt{2}}right).
end{align}



Notes



I recently discovered that the problem is related to Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$, and the above solution is essentially the same as this one.






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  • This is a good and clear answer! You deserve some points for it so you shouldn't have made it community wiki:)
    – Winther
    Oct 20 '15 at 0:22










  • @Winther. Thank you, sir. They have the option, and I like options :-)
    – hbp
    Oct 20 '15 at 0:38


















up vote
4
down vote













Try proving (by induction) that



$$P_{n+1}=xP_n+nP_{n-1}$$



(Note, this is equivalent to proving that $P_n'=nP_{n-1}$.)






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    up vote
    3
    down vote













    Here is another method which will work. You have shown by a simple argument that



    $${P_i}(x) = {P'_{i - 1}}(x) + x{P_{i - 1}}(x),,,,,,,,,,,{P_0}(x) = 1,,,,,,,,,,,i = 1,2,...,ntag{1}$$



    Now let's take a look at the derivatives of $f(x)$ directly



    $$eqalign{
    & {f^{(0)}}(x) = {e^{{{{x^2}} over 2}}} cr
    & {f^{(1)}}(x) = x{e^{{{{x^2}} over 2}}} cr
    & {f^{(2)}}(x) = left( {1 + {x^2}} right){e^{{{{x^2}} over 2}}} cr
    & {f^{(3)}}(x) = left( {3x + {x^3}} right){e^{{{{x^2}} over 2}}} cr
    & . cr
    & . cr
    & . cr
    & {f^{(n)}}(x) = {P_n}(x){e^{{{{x^2}} over 2}}} cr} tag{2}$$



    The first few terms suggest the formula ${{P'}_i}(x) = i{P_{i - 1}}(x)$ to be true. We may prove this easily by using $(1)$ and induction. According to $(2)$, it is clear that the formula is true for $i=1$. Now suppose it is true for $i=k$, ${{P'}_k}(x) = k{P_{k - 1}}(x)$, and then we shall prove it is also true for $i=k+1$. For this purpose, consider the following



    $$eqalign{
    & {P_{k + 1}}(x) = {{P'}_k}(x) + x{P_k}(x) = k{P_{k - 1}}(x) + x{P_k}(x) cr
    & {{P'}_{k + 1}}(x) = k{{P'}_{k - 1}}(x) + {left( {x{P_k}(x)} right)^prime } cr
    & ,,,,,,,,,,,,,,,,, = k{{P'}_{k - 1}}(x) + {P_k}(x) + x{{P'}_k}(x) cr
    & ,,,,,,,,,,,,,,,,, = k{{P'}_{k - 1}}(x) + {P_k}(x) + kx{P_{k - 1}}(x) cr
    & ,,,,,,,,,,,,,,,,, = kleft( {{{P'}_{k - 1}}(x) + x{P_{k - 1}}(x)} right) + {P_k}(x) cr
    & ,,,,,,,,,,,,,,,,, = k{P_k}(x) + {P_k}(x) cr
    & ,,,,,,,,,,,,,,,,, = left( {k + 1} right){P_k}(x) cr}tag{3} $$



    Now combining this new result with $(1)$ we can conclude



    $${P_i}(x) = x{P_{i - 1}}(x) + i{P_{i - 2}}(x),,,,,,{P_0}(x) = 1,,,,{P_1}(x)=x,,,,,,i = 2,3,...,ntag{4}$$



    Finally, you can use $(4)$ as a recursive relation to derive $P_n(x)$ by a usual systematic procedure. That's all.






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    • The induction is nice! Thank you.
      – hbp
      Oct 20 '15 at 23:59










    • @hbp: You're welcome. Thank you too as I noticed some typo errors in Eq.(4) by looking to my answer again due to your notification. I just fixed them! :)
      – H. R.
      Dec 23 '15 at 17:02


















    up vote
    2
    down vote













    You could try $P_n(x)=Sigma^n_{i=0} a_{n,i}x^i$



    Then your equation $(1)$ becomes $Sigma^n_{i=0} a_{n,i}x^i=xSigma^{n-1}_{i=0} a_{n-1,i}x^i+Sigma^n_{i=0} ia_{n-1,i}x^{i-1}$



    Comparing coefficients of $x^i$ gives $a_{n,i}=a_{n-1,i-1}+ia_{n-1,i+1}$



    Does that help?






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      up vote
      2
      down vote













      We have the multiplicative rule for differentiation: $$frac{partial fg}{partial x} = frac{partial f}{partial x} g + f frac{partial g }{partial x}$$



      Also the "chain rule" or rule for function composition: $$frac{partial (g(h))}{partial x} = frac{partial g}{partial h}frac{partial h}{partial x}$$



      So we let $g = exp(h)$, $h = x^2/2$. We see that $frac{partial h}{partial x} = x$ and $frac{partial g}{partial h} = g$



      Now we let $f$ be a polynomial. Differentiating a polynomial and expressing the result is a simple linear thing to do. Each exponent is reduced by one and coefficient multiplied by the old exponent so we can write this as a matrix with the numbers $1,2,3$ in one of the superdiagonals. But multiplication with $x$ is also a matrix operation on the coefficients of a polynomial. So we can rewrite each differentiation as a matrix-vector product and then investigate it's properties in terms of matrix properties and linear algebra.






      share|cite|improve this answer





















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        5 Answers
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        oldest

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        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        6
        down vote



        accepted










        As noted before this is a variant of the Hermite polynomials. Due to subtle differences, we will adapt the standard derivation from Arfken.



        Generating function



        First
        begin{align}
        frac{d^n}{dx^n} e^{x^2/2}
        &=
        lim_{trightarrow 0} frac{d^n}{dx^n} e^{(x+t)^2/2}
        tag{1} \
        &=
        lim_{trightarrow 0} frac{d^n}{dt^n} e^{(x+t)^2/2} \
        &=
        e^{x^2/2} lim_{trightarrow 0} frac{d^n}{dt^n} e^{xt + t^2/2},
        end{align}

        This means that the polynomials $P_n(x)$ we are looking for
        are just the $n$th coefficients of the Taylor expansion of
        $e^{xt+t^2/2}$. In other words,
        begin{align}
        e^{xt + t^2/2} = sum_{n = 0}^infty frac{P_n(x)}{n!} t^n.
        tag{2}
        end{align}

        The left-hand side is the exponential generating function of $P_n(x)$.



        Recurrence relations



        If we differentiate (2) with respect to $t$,
        $$
        (x + t) , e^{xt + t^2/2} = sum_{n = 1}^infty frac{P_n(x)}{(n-1)!} t^{n-1}
        = sum_{n = 0}^infty frac{P_{n+1}(x)}{n!} t^n.
        tag{3}
        $$

        where the previous n = 0 term is simply equal to zero, allowing for a shift in indices.
        Expanding the left-hand side,
        begin{align}
        (x + t) , e^{xt + t^2/2}
        &=
        (x + t) sum_{n = 0}^infty frac{ P_n(x) }{n!} t^n
        \
        &=
        sum_{n = 0}^infty frac{ x , P_n(x) }{n!} t^n
        +
        sum_{n = 0}^infty frac{ P_n(x) }{n!} t^{n+1}
        \
        &=
        sum_{n = 0}^infty frac{ x , P_n(x) }{n!} t^n
        +
        sum_{n = 1}^infty frac{ n , P_{n-1}(x) }{n!} t^{n}.
        tag{4}
        end{align}

        Comparing the coefficients of $t^n/n!$ in (3) and (4) yields
        $$
        P_{n+1}(x) = x , P_n(x) + n P_{n-1}(x).
        tag{5}
        $$



        Similarly, by differentiating (2) with respect to $x$, we get
        $$
        P'_n(x) = n P_{n-1}(x),
        tag{6}
        $$

        which is noted by Barry Cipra.
        Combining the two yields the relation by hermes.



        Explicit formula



        An explicit formula is more readily derived from the generating function instead of the recurrence relations:
        begin{align}
        e^{xt+t^2/2}
        &= e^{xt} , e^{t^2/2}\
        &= sum_{s = 0}^infty frac{(xt)^s}{s!} sum_{m = 0}^infty frac{t^{2m}}{2^m , m!} \
        &= sum_{n = 0}^infty
        left(
        sum_{m = 0}^{[n/2]} frac{ n! , x^{n-2m} }{ 2^m , m! , (n-2m)! }
        right) frac{t^n}{n!},
        end{align}

        where $[n/2]$ denotes the largest integer not exceeding $n/2$.
        Comparing this to (2), we get
        begin{align}
        P_n(x) = sum_{m = 0}^{[n/2]} frac{ n! , x^{n-2m} }{ 2^m , m! , (n-2m)! }.
        end{align}



        Relations to the standard definitions



        For reference, $P_n(x)$ is related to the standard Hermite polynomials as
        begin{align}
        P_n(x)
        &= (-i)^n , mathrm{He}_n(ix) \
        &= frac{1}{(sqrt{2} , i)^n} , H_nleft(frac{ix}{sqrt{2}}right).
        end{align}



        Notes



        I recently discovered that the problem is related to Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$, and the above solution is essentially the same as this one.






        share|cite|improve this answer























        • This is a good and clear answer! You deserve some points for it so you shouldn't have made it community wiki:)
          – Winther
          Oct 20 '15 at 0:22










        • @Winther. Thank you, sir. They have the option, and I like options :-)
          – hbp
          Oct 20 '15 at 0:38















        up vote
        6
        down vote



        accepted










        As noted before this is a variant of the Hermite polynomials. Due to subtle differences, we will adapt the standard derivation from Arfken.



        Generating function



        First
        begin{align}
        frac{d^n}{dx^n} e^{x^2/2}
        &=
        lim_{trightarrow 0} frac{d^n}{dx^n} e^{(x+t)^2/2}
        tag{1} \
        &=
        lim_{trightarrow 0} frac{d^n}{dt^n} e^{(x+t)^2/2} \
        &=
        e^{x^2/2} lim_{trightarrow 0} frac{d^n}{dt^n} e^{xt + t^2/2},
        end{align}

        This means that the polynomials $P_n(x)$ we are looking for
        are just the $n$th coefficients of the Taylor expansion of
        $e^{xt+t^2/2}$. In other words,
        begin{align}
        e^{xt + t^2/2} = sum_{n = 0}^infty frac{P_n(x)}{n!} t^n.
        tag{2}
        end{align}

        The left-hand side is the exponential generating function of $P_n(x)$.



        Recurrence relations



        If we differentiate (2) with respect to $t$,
        $$
        (x + t) , e^{xt + t^2/2} = sum_{n = 1}^infty frac{P_n(x)}{(n-1)!} t^{n-1}
        = sum_{n = 0}^infty frac{P_{n+1}(x)}{n!} t^n.
        tag{3}
        $$

        where the previous n = 0 term is simply equal to zero, allowing for a shift in indices.
        Expanding the left-hand side,
        begin{align}
        (x + t) , e^{xt + t^2/2}
        &=
        (x + t) sum_{n = 0}^infty frac{ P_n(x) }{n!} t^n
        \
        &=
        sum_{n = 0}^infty frac{ x , P_n(x) }{n!} t^n
        +
        sum_{n = 0}^infty frac{ P_n(x) }{n!} t^{n+1}
        \
        &=
        sum_{n = 0}^infty frac{ x , P_n(x) }{n!} t^n
        +
        sum_{n = 1}^infty frac{ n , P_{n-1}(x) }{n!} t^{n}.
        tag{4}
        end{align}

        Comparing the coefficients of $t^n/n!$ in (3) and (4) yields
        $$
        P_{n+1}(x) = x , P_n(x) + n P_{n-1}(x).
        tag{5}
        $$



        Similarly, by differentiating (2) with respect to $x$, we get
        $$
        P'_n(x) = n P_{n-1}(x),
        tag{6}
        $$

        which is noted by Barry Cipra.
        Combining the two yields the relation by hermes.



        Explicit formula



        An explicit formula is more readily derived from the generating function instead of the recurrence relations:
        begin{align}
        e^{xt+t^2/2}
        &= e^{xt} , e^{t^2/2}\
        &= sum_{s = 0}^infty frac{(xt)^s}{s!} sum_{m = 0}^infty frac{t^{2m}}{2^m , m!} \
        &= sum_{n = 0}^infty
        left(
        sum_{m = 0}^{[n/2]} frac{ n! , x^{n-2m} }{ 2^m , m! , (n-2m)! }
        right) frac{t^n}{n!},
        end{align}

        where $[n/2]$ denotes the largest integer not exceeding $n/2$.
        Comparing this to (2), we get
        begin{align}
        P_n(x) = sum_{m = 0}^{[n/2]} frac{ n! , x^{n-2m} }{ 2^m , m! , (n-2m)! }.
        end{align}



        Relations to the standard definitions



        For reference, $P_n(x)$ is related to the standard Hermite polynomials as
        begin{align}
        P_n(x)
        &= (-i)^n , mathrm{He}_n(ix) \
        &= frac{1}{(sqrt{2} , i)^n} , H_nleft(frac{ix}{sqrt{2}}right).
        end{align}



        Notes



        I recently discovered that the problem is related to Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$, and the above solution is essentially the same as this one.






        share|cite|improve this answer























        • This is a good and clear answer! You deserve some points for it so you shouldn't have made it community wiki:)
          – Winther
          Oct 20 '15 at 0:22










        • @Winther. Thank you, sir. They have the option, and I like options :-)
          – hbp
          Oct 20 '15 at 0:38













        up vote
        6
        down vote



        accepted







        up vote
        6
        down vote



        accepted






        As noted before this is a variant of the Hermite polynomials. Due to subtle differences, we will adapt the standard derivation from Arfken.



        Generating function



        First
        begin{align}
        frac{d^n}{dx^n} e^{x^2/2}
        &=
        lim_{trightarrow 0} frac{d^n}{dx^n} e^{(x+t)^2/2}
        tag{1} \
        &=
        lim_{trightarrow 0} frac{d^n}{dt^n} e^{(x+t)^2/2} \
        &=
        e^{x^2/2} lim_{trightarrow 0} frac{d^n}{dt^n} e^{xt + t^2/2},
        end{align}

        This means that the polynomials $P_n(x)$ we are looking for
        are just the $n$th coefficients of the Taylor expansion of
        $e^{xt+t^2/2}$. In other words,
        begin{align}
        e^{xt + t^2/2} = sum_{n = 0}^infty frac{P_n(x)}{n!} t^n.
        tag{2}
        end{align}

        The left-hand side is the exponential generating function of $P_n(x)$.



        Recurrence relations



        If we differentiate (2) with respect to $t$,
        $$
        (x + t) , e^{xt + t^2/2} = sum_{n = 1}^infty frac{P_n(x)}{(n-1)!} t^{n-1}
        = sum_{n = 0}^infty frac{P_{n+1}(x)}{n!} t^n.
        tag{3}
        $$

        where the previous n = 0 term is simply equal to zero, allowing for a shift in indices.
        Expanding the left-hand side,
        begin{align}
        (x + t) , e^{xt + t^2/2}
        &=
        (x + t) sum_{n = 0}^infty frac{ P_n(x) }{n!} t^n
        \
        &=
        sum_{n = 0}^infty frac{ x , P_n(x) }{n!} t^n
        +
        sum_{n = 0}^infty frac{ P_n(x) }{n!} t^{n+1}
        \
        &=
        sum_{n = 0}^infty frac{ x , P_n(x) }{n!} t^n
        +
        sum_{n = 1}^infty frac{ n , P_{n-1}(x) }{n!} t^{n}.
        tag{4}
        end{align}

        Comparing the coefficients of $t^n/n!$ in (3) and (4) yields
        $$
        P_{n+1}(x) = x , P_n(x) + n P_{n-1}(x).
        tag{5}
        $$



        Similarly, by differentiating (2) with respect to $x$, we get
        $$
        P'_n(x) = n P_{n-1}(x),
        tag{6}
        $$

        which is noted by Barry Cipra.
        Combining the two yields the relation by hermes.



        Explicit formula



        An explicit formula is more readily derived from the generating function instead of the recurrence relations:
        begin{align}
        e^{xt+t^2/2}
        &= e^{xt} , e^{t^2/2}\
        &= sum_{s = 0}^infty frac{(xt)^s}{s!} sum_{m = 0}^infty frac{t^{2m}}{2^m , m!} \
        &= sum_{n = 0}^infty
        left(
        sum_{m = 0}^{[n/2]} frac{ n! , x^{n-2m} }{ 2^m , m! , (n-2m)! }
        right) frac{t^n}{n!},
        end{align}

        where $[n/2]$ denotes the largest integer not exceeding $n/2$.
        Comparing this to (2), we get
        begin{align}
        P_n(x) = sum_{m = 0}^{[n/2]} frac{ n! , x^{n-2m} }{ 2^m , m! , (n-2m)! }.
        end{align}



        Relations to the standard definitions



        For reference, $P_n(x)$ is related to the standard Hermite polynomials as
        begin{align}
        P_n(x)
        &= (-i)^n , mathrm{He}_n(ix) \
        &= frac{1}{(sqrt{2} , i)^n} , H_nleft(frac{ix}{sqrt{2}}right).
        end{align}



        Notes



        I recently discovered that the problem is related to Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$, and the above solution is essentially the same as this one.






        share|cite|improve this answer














        As noted before this is a variant of the Hermite polynomials. Due to subtle differences, we will adapt the standard derivation from Arfken.



        Generating function



        First
        begin{align}
        frac{d^n}{dx^n} e^{x^2/2}
        &=
        lim_{trightarrow 0} frac{d^n}{dx^n} e^{(x+t)^2/2}
        tag{1} \
        &=
        lim_{trightarrow 0} frac{d^n}{dt^n} e^{(x+t)^2/2} \
        &=
        e^{x^2/2} lim_{trightarrow 0} frac{d^n}{dt^n} e^{xt + t^2/2},
        end{align}

        This means that the polynomials $P_n(x)$ we are looking for
        are just the $n$th coefficients of the Taylor expansion of
        $e^{xt+t^2/2}$. In other words,
        begin{align}
        e^{xt + t^2/2} = sum_{n = 0}^infty frac{P_n(x)}{n!} t^n.
        tag{2}
        end{align}

        The left-hand side is the exponential generating function of $P_n(x)$.



        Recurrence relations



        If we differentiate (2) with respect to $t$,
        $$
        (x + t) , e^{xt + t^2/2} = sum_{n = 1}^infty frac{P_n(x)}{(n-1)!} t^{n-1}
        = sum_{n = 0}^infty frac{P_{n+1}(x)}{n!} t^n.
        tag{3}
        $$

        where the previous n = 0 term is simply equal to zero, allowing for a shift in indices.
        Expanding the left-hand side,
        begin{align}
        (x + t) , e^{xt + t^2/2}
        &=
        (x + t) sum_{n = 0}^infty frac{ P_n(x) }{n!} t^n
        \
        &=
        sum_{n = 0}^infty frac{ x , P_n(x) }{n!} t^n
        +
        sum_{n = 0}^infty frac{ P_n(x) }{n!} t^{n+1}
        \
        &=
        sum_{n = 0}^infty frac{ x , P_n(x) }{n!} t^n
        +
        sum_{n = 1}^infty frac{ n , P_{n-1}(x) }{n!} t^{n}.
        tag{4}
        end{align}

        Comparing the coefficients of $t^n/n!$ in (3) and (4) yields
        $$
        P_{n+1}(x) = x , P_n(x) + n P_{n-1}(x).
        tag{5}
        $$



        Similarly, by differentiating (2) with respect to $x$, we get
        $$
        P'_n(x) = n P_{n-1}(x),
        tag{6}
        $$

        which is noted by Barry Cipra.
        Combining the two yields the relation by hermes.



        Explicit formula



        An explicit formula is more readily derived from the generating function instead of the recurrence relations:
        begin{align}
        e^{xt+t^2/2}
        &= e^{xt} , e^{t^2/2}\
        &= sum_{s = 0}^infty frac{(xt)^s}{s!} sum_{m = 0}^infty frac{t^{2m}}{2^m , m!} \
        &= sum_{n = 0}^infty
        left(
        sum_{m = 0}^{[n/2]} frac{ n! , x^{n-2m} }{ 2^m , m! , (n-2m)! }
        right) frac{t^n}{n!},
        end{align}

        where $[n/2]$ denotes the largest integer not exceeding $n/2$.
        Comparing this to (2), we get
        begin{align}
        P_n(x) = sum_{m = 0}^{[n/2]} frac{ n! , x^{n-2m} }{ 2^m , m! , (n-2m)! }.
        end{align}



        Relations to the standard definitions



        For reference, $P_n(x)$ is related to the standard Hermite polynomials as
        begin{align}
        P_n(x)
        &= (-i)^n , mathrm{He}_n(ix) \
        &= frac{1}{(sqrt{2} , i)^n} , H_nleft(frac{ix}{sqrt{2}}right).
        end{align}



        Notes



        I recently discovered that the problem is related to Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$, and the above solution is essentially the same as this one.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 at 1:21


























        community wiki





        7 revs, 2 users 96%
        hbp













        • This is a good and clear answer! You deserve some points for it so you shouldn't have made it community wiki:)
          – Winther
          Oct 20 '15 at 0:22










        • @Winther. Thank you, sir. They have the option, and I like options :-)
          – hbp
          Oct 20 '15 at 0:38


















        • This is a good and clear answer! You deserve some points for it so you shouldn't have made it community wiki:)
          – Winther
          Oct 20 '15 at 0:22










        • @Winther. Thank you, sir. They have the option, and I like options :-)
          – hbp
          Oct 20 '15 at 0:38
















        This is a good and clear answer! You deserve some points for it so you shouldn't have made it community wiki:)
        – Winther
        Oct 20 '15 at 0:22




        This is a good and clear answer! You deserve some points for it so you shouldn't have made it community wiki:)
        – Winther
        Oct 20 '15 at 0:22












        @Winther. Thank you, sir. They have the option, and I like options :-)
        – hbp
        Oct 20 '15 at 0:38




        @Winther. Thank you, sir. They have the option, and I like options :-)
        – hbp
        Oct 20 '15 at 0:38










        up vote
        4
        down vote













        Try proving (by induction) that



        $$P_{n+1}=xP_n+nP_{n-1}$$



        (Note, this is equivalent to proving that $P_n'=nP_{n-1}$.)






        share|cite|improve this answer

























          up vote
          4
          down vote













          Try proving (by induction) that



          $$P_{n+1}=xP_n+nP_{n-1}$$



          (Note, this is equivalent to proving that $P_n'=nP_{n-1}$.)






          share|cite|improve this answer























            up vote
            4
            down vote










            up vote
            4
            down vote









            Try proving (by induction) that



            $$P_{n+1}=xP_n+nP_{n-1}$$



            (Note, this is equivalent to proving that $P_n'=nP_{n-1}$.)






            share|cite|improve this answer












            Try proving (by induction) that



            $$P_{n+1}=xP_n+nP_{n-1}$$



            (Note, this is equivalent to proving that $P_n'=nP_{n-1}$.)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 19 '15 at 22:27









            Barry Cipra

            58.8k653123




            58.8k653123






















                up vote
                3
                down vote













                Here is another method which will work. You have shown by a simple argument that



                $${P_i}(x) = {P'_{i - 1}}(x) + x{P_{i - 1}}(x),,,,,,,,,,,{P_0}(x) = 1,,,,,,,,,,,i = 1,2,...,ntag{1}$$



                Now let's take a look at the derivatives of $f(x)$ directly



                $$eqalign{
                & {f^{(0)}}(x) = {e^{{{{x^2}} over 2}}} cr
                & {f^{(1)}}(x) = x{e^{{{{x^2}} over 2}}} cr
                & {f^{(2)}}(x) = left( {1 + {x^2}} right){e^{{{{x^2}} over 2}}} cr
                & {f^{(3)}}(x) = left( {3x + {x^3}} right){e^{{{{x^2}} over 2}}} cr
                & . cr
                & . cr
                & . cr
                & {f^{(n)}}(x) = {P_n}(x){e^{{{{x^2}} over 2}}} cr} tag{2}$$



                The first few terms suggest the formula ${{P'}_i}(x) = i{P_{i - 1}}(x)$ to be true. We may prove this easily by using $(1)$ and induction. According to $(2)$, it is clear that the formula is true for $i=1$. Now suppose it is true for $i=k$, ${{P'}_k}(x) = k{P_{k - 1}}(x)$, and then we shall prove it is also true for $i=k+1$. For this purpose, consider the following



                $$eqalign{
                & {P_{k + 1}}(x) = {{P'}_k}(x) + x{P_k}(x) = k{P_{k - 1}}(x) + x{P_k}(x) cr
                & {{P'}_{k + 1}}(x) = k{{P'}_{k - 1}}(x) + {left( {x{P_k}(x)} right)^prime } cr
                & ,,,,,,,,,,,,,,,,, = k{{P'}_{k - 1}}(x) + {P_k}(x) + x{{P'}_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = k{{P'}_{k - 1}}(x) + {P_k}(x) + kx{P_{k - 1}}(x) cr
                & ,,,,,,,,,,,,,,,,, = kleft( {{{P'}_{k - 1}}(x) + x{P_{k - 1}}(x)} right) + {P_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = k{P_k}(x) + {P_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = left( {k + 1} right){P_k}(x) cr}tag{3} $$



                Now combining this new result with $(1)$ we can conclude



                $${P_i}(x) = x{P_{i - 1}}(x) + i{P_{i - 2}}(x),,,,,,{P_0}(x) = 1,,,,{P_1}(x)=x,,,,,,i = 2,3,...,ntag{4}$$



                Finally, you can use $(4)$ as a recursive relation to derive $P_n(x)$ by a usual systematic procedure. That's all.






                share|cite|improve this answer























                • The induction is nice! Thank you.
                  – hbp
                  Oct 20 '15 at 23:59










                • @hbp: You're welcome. Thank you too as I noticed some typo errors in Eq.(4) by looking to my answer again due to your notification. I just fixed them! :)
                  – H. R.
                  Dec 23 '15 at 17:02















                up vote
                3
                down vote













                Here is another method which will work. You have shown by a simple argument that



                $${P_i}(x) = {P'_{i - 1}}(x) + x{P_{i - 1}}(x),,,,,,,,,,,{P_0}(x) = 1,,,,,,,,,,,i = 1,2,...,ntag{1}$$



                Now let's take a look at the derivatives of $f(x)$ directly



                $$eqalign{
                & {f^{(0)}}(x) = {e^{{{{x^2}} over 2}}} cr
                & {f^{(1)}}(x) = x{e^{{{{x^2}} over 2}}} cr
                & {f^{(2)}}(x) = left( {1 + {x^2}} right){e^{{{{x^2}} over 2}}} cr
                & {f^{(3)}}(x) = left( {3x + {x^3}} right){e^{{{{x^2}} over 2}}} cr
                & . cr
                & . cr
                & . cr
                & {f^{(n)}}(x) = {P_n}(x){e^{{{{x^2}} over 2}}} cr} tag{2}$$



                The first few terms suggest the formula ${{P'}_i}(x) = i{P_{i - 1}}(x)$ to be true. We may prove this easily by using $(1)$ and induction. According to $(2)$, it is clear that the formula is true for $i=1$. Now suppose it is true for $i=k$, ${{P'}_k}(x) = k{P_{k - 1}}(x)$, and then we shall prove it is also true for $i=k+1$. For this purpose, consider the following



                $$eqalign{
                & {P_{k + 1}}(x) = {{P'}_k}(x) + x{P_k}(x) = k{P_{k - 1}}(x) + x{P_k}(x) cr
                & {{P'}_{k + 1}}(x) = k{{P'}_{k - 1}}(x) + {left( {x{P_k}(x)} right)^prime } cr
                & ,,,,,,,,,,,,,,,,, = k{{P'}_{k - 1}}(x) + {P_k}(x) + x{{P'}_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = k{{P'}_{k - 1}}(x) + {P_k}(x) + kx{P_{k - 1}}(x) cr
                & ,,,,,,,,,,,,,,,,, = kleft( {{{P'}_{k - 1}}(x) + x{P_{k - 1}}(x)} right) + {P_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = k{P_k}(x) + {P_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = left( {k + 1} right){P_k}(x) cr}tag{3} $$



                Now combining this new result with $(1)$ we can conclude



                $${P_i}(x) = x{P_{i - 1}}(x) + i{P_{i - 2}}(x),,,,,,{P_0}(x) = 1,,,,{P_1}(x)=x,,,,,,i = 2,3,...,ntag{4}$$



                Finally, you can use $(4)$ as a recursive relation to derive $P_n(x)$ by a usual systematic procedure. That's all.






                share|cite|improve this answer























                • The induction is nice! Thank you.
                  – hbp
                  Oct 20 '15 at 23:59










                • @hbp: You're welcome. Thank you too as I noticed some typo errors in Eq.(4) by looking to my answer again due to your notification. I just fixed them! :)
                  – H. R.
                  Dec 23 '15 at 17:02













                up vote
                3
                down vote










                up vote
                3
                down vote









                Here is another method which will work. You have shown by a simple argument that



                $${P_i}(x) = {P'_{i - 1}}(x) + x{P_{i - 1}}(x),,,,,,,,,,,{P_0}(x) = 1,,,,,,,,,,,i = 1,2,...,ntag{1}$$



                Now let's take a look at the derivatives of $f(x)$ directly



                $$eqalign{
                & {f^{(0)}}(x) = {e^{{{{x^2}} over 2}}} cr
                & {f^{(1)}}(x) = x{e^{{{{x^2}} over 2}}} cr
                & {f^{(2)}}(x) = left( {1 + {x^2}} right){e^{{{{x^2}} over 2}}} cr
                & {f^{(3)}}(x) = left( {3x + {x^3}} right){e^{{{{x^2}} over 2}}} cr
                & . cr
                & . cr
                & . cr
                & {f^{(n)}}(x) = {P_n}(x){e^{{{{x^2}} over 2}}} cr} tag{2}$$



                The first few terms suggest the formula ${{P'}_i}(x) = i{P_{i - 1}}(x)$ to be true. We may prove this easily by using $(1)$ and induction. According to $(2)$, it is clear that the formula is true for $i=1$. Now suppose it is true for $i=k$, ${{P'}_k}(x) = k{P_{k - 1}}(x)$, and then we shall prove it is also true for $i=k+1$. For this purpose, consider the following



                $$eqalign{
                & {P_{k + 1}}(x) = {{P'}_k}(x) + x{P_k}(x) = k{P_{k - 1}}(x) + x{P_k}(x) cr
                & {{P'}_{k + 1}}(x) = k{{P'}_{k - 1}}(x) + {left( {x{P_k}(x)} right)^prime } cr
                & ,,,,,,,,,,,,,,,,, = k{{P'}_{k - 1}}(x) + {P_k}(x) + x{{P'}_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = k{{P'}_{k - 1}}(x) + {P_k}(x) + kx{P_{k - 1}}(x) cr
                & ,,,,,,,,,,,,,,,,, = kleft( {{{P'}_{k - 1}}(x) + x{P_{k - 1}}(x)} right) + {P_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = k{P_k}(x) + {P_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = left( {k + 1} right){P_k}(x) cr}tag{3} $$



                Now combining this new result with $(1)$ we can conclude



                $${P_i}(x) = x{P_{i - 1}}(x) + i{P_{i - 2}}(x),,,,,,{P_0}(x) = 1,,,,{P_1}(x)=x,,,,,,i = 2,3,...,ntag{4}$$



                Finally, you can use $(4)$ as a recursive relation to derive $P_n(x)$ by a usual systematic procedure. That's all.






                share|cite|improve this answer














                Here is another method which will work. You have shown by a simple argument that



                $${P_i}(x) = {P'_{i - 1}}(x) + x{P_{i - 1}}(x),,,,,,,,,,,{P_0}(x) = 1,,,,,,,,,,,i = 1,2,...,ntag{1}$$



                Now let's take a look at the derivatives of $f(x)$ directly



                $$eqalign{
                & {f^{(0)}}(x) = {e^{{{{x^2}} over 2}}} cr
                & {f^{(1)}}(x) = x{e^{{{{x^2}} over 2}}} cr
                & {f^{(2)}}(x) = left( {1 + {x^2}} right){e^{{{{x^2}} over 2}}} cr
                & {f^{(3)}}(x) = left( {3x + {x^3}} right){e^{{{{x^2}} over 2}}} cr
                & . cr
                & . cr
                & . cr
                & {f^{(n)}}(x) = {P_n}(x){e^{{{{x^2}} over 2}}} cr} tag{2}$$



                The first few terms suggest the formula ${{P'}_i}(x) = i{P_{i - 1}}(x)$ to be true. We may prove this easily by using $(1)$ and induction. According to $(2)$, it is clear that the formula is true for $i=1$. Now suppose it is true for $i=k$, ${{P'}_k}(x) = k{P_{k - 1}}(x)$, and then we shall prove it is also true for $i=k+1$. For this purpose, consider the following



                $$eqalign{
                & {P_{k + 1}}(x) = {{P'}_k}(x) + x{P_k}(x) = k{P_{k - 1}}(x) + x{P_k}(x) cr
                & {{P'}_{k + 1}}(x) = k{{P'}_{k - 1}}(x) + {left( {x{P_k}(x)} right)^prime } cr
                & ,,,,,,,,,,,,,,,,, = k{{P'}_{k - 1}}(x) + {P_k}(x) + x{{P'}_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = k{{P'}_{k - 1}}(x) + {P_k}(x) + kx{P_{k - 1}}(x) cr
                & ,,,,,,,,,,,,,,,,, = kleft( {{{P'}_{k - 1}}(x) + x{P_{k - 1}}(x)} right) + {P_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = k{P_k}(x) + {P_k}(x) cr
                & ,,,,,,,,,,,,,,,,, = left( {k + 1} right){P_k}(x) cr}tag{3} $$



                Now combining this new result with $(1)$ we can conclude



                $${P_i}(x) = x{P_{i - 1}}(x) + i{P_{i - 2}}(x),,,,,,{P_0}(x) = 1,,,,{P_1}(x)=x,,,,,,i = 2,3,...,ntag{4}$$



                Finally, you can use $(4)$ as a recursive relation to derive $P_n(x)$ by a usual systematic procedure. That's all.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Oct 21 '15 at 10:58

























                answered Oct 19 '15 at 22:48









                H. R.

                9,40593262




                9,40593262












                • The induction is nice! Thank you.
                  – hbp
                  Oct 20 '15 at 23:59










                • @hbp: You're welcome. Thank you too as I noticed some typo errors in Eq.(4) by looking to my answer again due to your notification. I just fixed them! :)
                  – H. R.
                  Dec 23 '15 at 17:02


















                • The induction is nice! Thank you.
                  – hbp
                  Oct 20 '15 at 23:59










                • @hbp: You're welcome. Thank you too as I noticed some typo errors in Eq.(4) by looking to my answer again due to your notification. I just fixed them! :)
                  – H. R.
                  Dec 23 '15 at 17:02
















                The induction is nice! Thank you.
                – hbp
                Oct 20 '15 at 23:59




                The induction is nice! Thank you.
                – hbp
                Oct 20 '15 at 23:59












                @hbp: You're welcome. Thank you too as I noticed some typo errors in Eq.(4) by looking to my answer again due to your notification. I just fixed them! :)
                – H. R.
                Dec 23 '15 at 17:02




                @hbp: You're welcome. Thank you too as I noticed some typo errors in Eq.(4) by looking to my answer again due to your notification. I just fixed them! :)
                – H. R.
                Dec 23 '15 at 17:02










                up vote
                2
                down vote













                You could try $P_n(x)=Sigma^n_{i=0} a_{n,i}x^i$



                Then your equation $(1)$ becomes $Sigma^n_{i=0} a_{n,i}x^i=xSigma^{n-1}_{i=0} a_{n-1,i}x^i+Sigma^n_{i=0} ia_{n-1,i}x^{i-1}$



                Comparing coefficients of $x^i$ gives $a_{n,i}=a_{n-1,i-1}+ia_{n-1,i+1}$



                Does that help?






                share|cite|improve this answer



























                  up vote
                  2
                  down vote













                  You could try $P_n(x)=Sigma^n_{i=0} a_{n,i}x^i$



                  Then your equation $(1)$ becomes $Sigma^n_{i=0} a_{n,i}x^i=xSigma^{n-1}_{i=0} a_{n-1,i}x^i+Sigma^n_{i=0} ia_{n-1,i}x^{i-1}$



                  Comparing coefficients of $x^i$ gives $a_{n,i}=a_{n-1,i-1}+ia_{n-1,i+1}$



                  Does that help?






                  share|cite|improve this answer

























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    You could try $P_n(x)=Sigma^n_{i=0} a_{n,i}x^i$



                    Then your equation $(1)$ becomes $Sigma^n_{i=0} a_{n,i}x^i=xSigma^{n-1}_{i=0} a_{n-1,i}x^i+Sigma^n_{i=0} ia_{n-1,i}x^{i-1}$



                    Comparing coefficients of $x^i$ gives $a_{n,i}=a_{n-1,i-1}+ia_{n-1,i+1}$



                    Does that help?






                    share|cite|improve this answer














                    You could try $P_n(x)=Sigma^n_{i=0} a_{n,i}x^i$



                    Then your equation $(1)$ becomes $Sigma^n_{i=0} a_{n,i}x^i=xSigma^{n-1}_{i=0} a_{n-1,i}x^i+Sigma^n_{i=0} ia_{n-1,i}x^{i-1}$



                    Comparing coefficients of $x^i$ gives $a_{n,i}=a_{n-1,i-1}+ia_{n-1,i+1}$



                    Does that help?







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                    edited Oct 19 '15 at 22:23

























                    answered Oct 19 '15 at 22:14









                    tomi

                    6,18611132




                    6,18611132






















                        up vote
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                        down vote













                        We have the multiplicative rule for differentiation: $$frac{partial fg}{partial x} = frac{partial f}{partial x} g + f frac{partial g }{partial x}$$



                        Also the "chain rule" or rule for function composition: $$frac{partial (g(h))}{partial x} = frac{partial g}{partial h}frac{partial h}{partial x}$$



                        So we let $g = exp(h)$, $h = x^2/2$. We see that $frac{partial h}{partial x} = x$ and $frac{partial g}{partial h} = g$



                        Now we let $f$ be a polynomial. Differentiating a polynomial and expressing the result is a simple linear thing to do. Each exponent is reduced by one and coefficient multiplied by the old exponent so we can write this as a matrix with the numbers $1,2,3$ in one of the superdiagonals. But multiplication with $x$ is also a matrix operation on the coefficients of a polynomial. So we can rewrite each differentiation as a matrix-vector product and then investigate it's properties in terms of matrix properties and linear algebra.






                        share|cite|improve this answer

























                          up vote
                          2
                          down vote













                          We have the multiplicative rule for differentiation: $$frac{partial fg}{partial x} = frac{partial f}{partial x} g + f frac{partial g }{partial x}$$



                          Also the "chain rule" or rule for function composition: $$frac{partial (g(h))}{partial x} = frac{partial g}{partial h}frac{partial h}{partial x}$$



                          So we let $g = exp(h)$, $h = x^2/2$. We see that $frac{partial h}{partial x} = x$ and $frac{partial g}{partial h} = g$



                          Now we let $f$ be a polynomial. Differentiating a polynomial and expressing the result is a simple linear thing to do. Each exponent is reduced by one and coefficient multiplied by the old exponent so we can write this as a matrix with the numbers $1,2,3$ in one of the superdiagonals. But multiplication with $x$ is also a matrix operation on the coefficients of a polynomial. So we can rewrite each differentiation as a matrix-vector product and then investigate it's properties in terms of matrix properties and linear algebra.






                          share|cite|improve this answer























                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            We have the multiplicative rule for differentiation: $$frac{partial fg}{partial x} = frac{partial f}{partial x} g + f frac{partial g }{partial x}$$



                            Also the "chain rule" or rule for function composition: $$frac{partial (g(h))}{partial x} = frac{partial g}{partial h}frac{partial h}{partial x}$$



                            So we let $g = exp(h)$, $h = x^2/2$. We see that $frac{partial h}{partial x} = x$ and $frac{partial g}{partial h} = g$



                            Now we let $f$ be a polynomial. Differentiating a polynomial and expressing the result is a simple linear thing to do. Each exponent is reduced by one and coefficient multiplied by the old exponent so we can write this as a matrix with the numbers $1,2,3$ in one of the superdiagonals. But multiplication with $x$ is also a matrix operation on the coefficients of a polynomial. So we can rewrite each differentiation as a matrix-vector product and then investigate it's properties in terms of matrix properties and linear algebra.






                            share|cite|improve this answer












                            We have the multiplicative rule for differentiation: $$frac{partial fg}{partial x} = frac{partial f}{partial x} g + f frac{partial g }{partial x}$$



                            Also the "chain rule" or rule for function composition: $$frac{partial (g(h))}{partial x} = frac{partial g}{partial h}frac{partial h}{partial x}$$



                            So we let $g = exp(h)$, $h = x^2/2$. We see that $frac{partial h}{partial x} = x$ and $frac{partial g}{partial h} = g$



                            Now we let $f$ be a polynomial. Differentiating a polynomial and expressing the result is a simple linear thing to do. Each exponent is reduced by one and coefficient multiplied by the old exponent so we can write this as a matrix with the numbers $1,2,3$ in one of the superdiagonals. But multiplication with $x$ is also a matrix operation on the coefficients of a polynomial. So we can rewrite each differentiation as a matrix-vector product and then investigate it's properties in terms of matrix properties and linear algebra.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Oct 21 '15 at 13:04









                            mathreadler

                            14.7k72160




                            14.7k72160






























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