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Miklós Schweitzer
1957 Miklós Schweitzer
6
6
Part of
1957 Miklós Schweitzer
Problems
(1)
Miklós Schweitzer 1957- Problem 6
Source:
10/16/2015
6. Let
f
(
x
)
f(x)
f
(
x
)
be an arbitrary function, differentiable infinitely many times. Then the
n
n
n
th derivative of
f
(
e
x
)
f(e^{x})
f
(
e
x
)
has the form
d
n
d
x
n
f
(
e
x
)
=
∑
k
=
0
n
a
k
n
e
k
x
f
(
k
)
(
e
x
)
\frac{d^{n}}{dx^{n}}f(e^{x})= \sum_{k=0}^{n} a_{kn}e^{kx}f^{(k)}(e^{x})
d
x
n
d
n
f
(
e
x
)
=
∑
k
=
0
n
a
kn
e
k
x
f
(
k
)
(
e
x
)
(
n
=
0
,
1
,
2
,
…
n=0,1,2,\dots
n
=
0
,
1
,
2
,
…
).From the coefficients
a
k
n
a_{kn}
a
kn
compose the sequence of polynomials
P
n
(
x
)
=
∑
k
=
0
n
a
k
n
x
k
P_{n}(x)= \sum_{k=0}^{n} a_{kn}x^{k}
P
n
(
x
)
=
∑
k
=
0
n
a
kn
x
k
(
n
=
0
,
1
,
2
,
…
n=0,1,2,\dots
n
=
0
,
1
,
2
,
…
)and find a closed form for the function
F
(
t
,
x
)
=
∑
n
=
0
∞
P
n
(
x
)
n
!
t
n
.
F(t,x) = \sum_{n=0}^{\infty} \frac{P_{n}(x)}{n!}t^{n}.
F
(
t
,
x
)
=
∑
n
=
0
∞
n
!
P
n
(
x
)
t
n
.
(S. 22)
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