How to complete this proof involving factorials The Next CEO of Stack OverflowBi-implication...
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How to complete this proof involving factorials
The Next CEO of Stack OverflowBi-implication theorem provingPurely combinatorial proof and simplification of identity involving factorials and summationsProof Involving Rational NumbersChallenge: How to prove this reduction identity for factorials of even numbers?Proving an identity involving factorialsEvaulate a determinant involving factorials.Proof of inequality involving binomial coefficientsProving a combinatorial identity: $sum_{i=0}^m binom{l}{i}binom{m+n-l}{m-i} = binom{m+n}{m}$'Transforming' a function $T(m,n)$ into $a(n)$ in an inductive proofProof by Induction involving Inequality and Factorials as denominators
$begingroup$
Recently I came across the following identity, but if I try proving it with induction, then I get stuck.
$$n! = sum^n_{k=0}(-1)^{n-k}binom{n}{k}(k+1)^n$$
While trying my induction step I get the following:
$$(n+1)! = (n+1)sum^n_{k=0}(-1)^{n-k}binom{n}{k}(k+1)^n$$
$$(n+1)! = sum^n_{k=0}(-1)^{n-k}binom{n+1}{k+1}(k+1)^{n+1}$$
$$vdots$$
$$(n+1)! = sum^{n+1}_{k=0}(-1)^{n+1-k}binom{n+1}{k}(k+1)^{n+1}$$
(Question) How do I complete the proof?
Thanks in advance.
proof-verification proof-writing factorial
asked Feb 11 at 15:25
MaxMax
9711319
$endgroup$
add a comment |
$begingroup$
Recently I came across the following identity, but if I try proving it with induction, then I get stuck.
$$n! = sum^n_{k=0}(-1)^{n-k}binom{n}{k}(k+1)^n$$
While trying my induction step I get the following:
$$(n+1)! = (n+1)sum^n_{k=0}(-1)^{n-k}binom{n}{k}(k+1)^n$$
$$(n+1)! = sum^n_{k=0}(-1)^{n-k}binom{n+1}{k+1}(k+1)^{n+1}$$
$$vdots$$
$$(n+1)! = sum^{n+1}_{k=0}(-1)^{n+1-k}binom{n+1}{k}(k+1)^{n+1}$$
(Question) How do I complete the proof?
Thanks in advance.
proof-verification proof-writing factorial
asked Feb 11 at 15:25
MaxMax
9711319
$endgroup$
add a comment |
$begingroup$
Recently I came across the following identity, but if I try proving it with induction, then I get stuck.
$$n! = sum^n_{k=0}(-1)^{n-k}binom{n}{k}(k+1)^n$$
While trying my induction step I get the following:
$$(n+1)! = (n+1)sum^n_{k=0}(-1)^{n-k}binom{n}{k}(k+1)^n$$
$$(n+1)! = sum^n_{k=0}(-1)^{n-k}binom{n+1}{k+1}(k+1)^{n+1}$$
$$vdots$$
$$(n+1)! = sum^{n+1}_{k=0}(-1)^{n+1-k}binom{n+1}{k}(k+1)^{n+1}$$
(Question) How do I complete the proof?
Thanks in advance.
proof-verification proof-writing factorial
asked Feb 11 at 15:25
MaxMax
9711319
$endgroup$
Recently I came across the following identity, but if I try proving it with induction, then I get stuck.
$$n! = sum^n_{k=0}(-1)^{n-k}binom{n}{k}(k+1)^n$$
While trying my induction step I get the following:
$$(n+1)! = (n+1)sum^n_{k=0}(-1)^{n-k}binom{n}{k}(k+1)^n$$
$$(n+1)! = sum^n_{k=0}(-1)^{n-k}binom{n+1}{k+1}(k+1)^{n+1}$$
$$vdots$$
$$(n+1)! = sum^{n+1}_{k=0}(-1)^{n+1-k}binom{n+1}{k}(k+1)^{n+1}$$
(Question) How do I complete the proof?
Thanks in advance.
proof-verification proof-writing factorial
proof-verification proof-writing factorial
asked Feb 11 at 15:25
MaxMax
9711319
asked Feb 11 at 15:25
MaxMax
9711319
edited Mar 18 at 17:33
Max
asked Feb 11 at 15:25
MaxMax
9711319
asked Feb 11 at 15:25
MaxMax
9711319
asked Feb 11 at 15:25
MaxMax
9711319
9711319
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
This is quite tough to show using induction. BUT ...
Using the $[x^k]:f(x)$ to represent the coeficient of $x^k$ in the power series for $f(x)$.
Note that
begin{eqnarray*}
frac{(k+1)^n}{n!} =[x^n]: e^{(k+1)x}.
end{eqnarray*}
Now divide your equation by $n!$ and we have
begin{eqnarray*}
sum_{k-0}^{n} (-1)^{n-k} binom{n}{k}frac{(k+1)^n}{n!} &=&[x^n]: sum_{k-0}^{n} (-1)^{n-k} binom{n}{k} e^{(k+1)x} \&=& [x^n]:e^x(e^x-1)^n =color{red}{1}.
end{eqnarray*}
answered Feb 11 at 15:55
Donald SplutterwitDonald Splutterwit
23k21446
$endgroup$
add a comment |
$begingroup$
Alternatively, and proceeding from a more general result, we have that
the Finite Difference of first order, and the next iterated ones, are defined as
$$
eqalign{
& Delta _{,x} f(x) = f(x + 1) - f(x) cr
& Delta _{,x} ^n f(x) = Delta _{,x} left( {Delta _{,x} ^{n - 1} f(x)} right)
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {left( { - 1} right)^{n - k} binom{n}{k}f(x + k)} cr}
$$
Then, if $f(x)$ is a polynomial of degree $n$
$$
eqalign{
& f(x) = p_{,n} (x) = a_{,n} x^{,n} + a_{,n - 1} x^{,n - 1} + cdots + a_{,0} = cr
& = n!a_{,n} binom{x}{n}
+ left( {n - 1} right)!b_{,n - 1} binom{x}{n-1} + cdots + b_{,0} binom{x}{0} cr}
$$
the $n$th order Difference will leave only the leading coefficient in the Newton expression, i.e.
$$
Delta _{,x} ^n p_{,n} (x) = n!a_{,n} binom{x}{n-n} = n!a_{,n}
$$
This for whichever $p_n(x)$ and in particular for $ x^n$ (then computed at $x=1$).
answered Mar 18 at 18:27
G CabG Cab
20.4k31341
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
This is quite tough to show using induction. BUT ...
Using the $[x^k]:f(x)$ to represent the coeficient of $x^k$ in the power series for $f(x)$.
Note that
begin{eqnarray*}
frac{(k+1)^n}{n!} =[x^n]: e^{(k+1)x}.
end{eqnarray*}
Now divide your equation by $n!$ and we have
begin{eqnarray*}
sum_{k-0}^{n} (-1)^{n-k} binom{n}{k}frac{(k+1)^n}{n!} &=&[x^n]: sum_{k-0}^{n} (-1)^{n-k} binom{n}{k} e^{(k+1)x} \&=& [x^n]:e^x(e^x-1)^n =color{red}{1}.
end{eqnarray*}
answered Feb 11 at 15:55
Donald SplutterwitDonald Splutterwit
23k21446
$endgroup$
add a comment |
$begingroup$
This is quite tough to show using induction. BUT ...
Using the $[x^k]:f(x)$ to represent the coeficient of $x^k$ in the power series for $f(x)$.
Note that
begin{eqnarray*}
frac{(k+1)^n}{n!} =[x^n]: e^{(k+1)x}.
end{eqnarray*}
Now divide your equation by $n!$ and we have
begin{eqnarray*}
sum_{k-0}^{n} (-1)^{n-k} binom{n}{k}frac{(k+1)^n}{n!} &=&[x^n]: sum_{k-0}^{n} (-1)^{n-k} binom{n}{k} e^{(k+1)x} \&=& [x^n]:e^x(e^x-1)^n =color{red}{1}.
end{eqnarray*}
answered Feb 11 at 15:55
Donald SplutterwitDonald Splutterwit
23k21446
$endgroup$
add a comment |
$begingroup$
This is quite tough to show using induction. BUT ...
Using the $[x^k]:f(x)$ to represent the coeficient of $x^k$ in the power series for $f(x)$.
Note that
begin{eqnarray*}
frac{(k+1)^n}{n!} =[x^n]: e^{(k+1)x}.
end{eqnarray*}
Now divide your equation by $n!$ and we have
begin{eqnarray*}
sum_{k-0}^{n} (-1)^{n-k} binom{n}{k}frac{(k+1)^n}{n!} &=&[x^n]: sum_{k-0}^{n} (-1)^{n-k} binom{n}{k} e^{(k+1)x} \&=& [x^n]:e^x(e^x-1)^n =color{red}{1}.
end{eqnarray*}
answered Feb 11 at 15:55
Donald SplutterwitDonald Splutterwit
23k21446
$endgroup$
This is quite tough to show using induction. BUT ...
Using the $[x^k]:f(x)$ to represent the coeficient of $x^k$ in the power series for $f(x)$.
Note that
begin{eqnarray*}
frac{(k+1)^n}{n!} =[x^n]: e^{(k+1)x}.
end{eqnarray*}
Now divide your equation by $n!$ and we have
begin{eqnarray*}
sum_{k-0}^{n} (-1)^{n-k} binom{n}{k}frac{(k+1)^n}{n!} &=&[x^n]: sum_{k-0}^{n} (-1)^{n-k} binom{n}{k} e^{(k+1)x} \&=& [x^n]:e^x(e^x-1)^n =color{red}{1}.
end{eqnarray*}
answered Feb 11 at 15:55
Donald SplutterwitDonald Splutterwit
23k21446
edited Mar 16 at 18:23
answered Feb 11 at 15:55
Donald SplutterwitDonald Splutterwit
23k21446
answered Feb 11 at 15:55
Donald SplutterwitDonald Splutterwit
23k21446
answered Feb 11 at 15:55
Donald SplutterwitDonald Splutterwit
23k21446
23k21446
add a comment |
add a comment |
$begingroup$
Alternatively, and proceeding from a more general result, we have that
the Finite Difference of first order, and the next iterated ones, are defined as
$$
eqalign{
& Delta _{,x} f(x) = f(x + 1) - f(x) cr
& Delta _{,x} ^n f(x) = Delta _{,x} left( {Delta _{,x} ^{n - 1} f(x)} right)
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {left( { - 1} right)^{n - k} binom{n}{k}f(x + k)} cr}
$$
Then, if $f(x)$ is a polynomial of degree $n$
$$
eqalign{
& f(x) = p_{,n} (x) = a_{,n} x^{,n} + a_{,n - 1} x^{,n - 1} + cdots + a_{,0} = cr
& = n!a_{,n} binom{x}{n}
+ left( {n - 1} right)!b_{,n - 1} binom{x}{n-1} + cdots + b_{,0} binom{x}{0} cr}
$$
the $n$th order Difference will leave only the leading coefficient in the Newton expression, i.e.
$$
Delta _{,x} ^n p_{,n} (x) = n!a_{,n} binom{x}{n-n} = n!a_{,n}
$$
This for whichever $p_n(x)$ and in particular for $ x^n$ (then computed at $x=1$).
answered Mar 18 at 18:27
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
Alternatively, and proceeding from a more general result, we have that
the Finite Difference of first order, and the next iterated ones, are defined as
$$
eqalign{
& Delta _{,x} f(x) = f(x + 1) - f(x) cr
& Delta _{,x} ^n f(x) = Delta _{,x} left( {Delta _{,x} ^{n - 1} f(x)} right)
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {left( { - 1} right)^{n - k} binom{n}{k}f(x + k)} cr}
$$
Then, if $f(x)$ is a polynomial of degree $n$
$$
eqalign{
& f(x) = p_{,n} (x) = a_{,n} x^{,n} + a_{,n - 1} x^{,n - 1} + cdots + a_{,0} = cr
& = n!a_{,n} binom{x}{n}
+ left( {n - 1} right)!b_{,n - 1} binom{x}{n-1} + cdots + b_{,0} binom{x}{0} cr}
$$
the $n$th order Difference will leave only the leading coefficient in the Newton expression, i.e.
$$
Delta _{,x} ^n p_{,n} (x) = n!a_{,n} binom{x}{n-n} = n!a_{,n}
$$
This for whichever $p_n(x)$ and in particular for $ x^n$ (then computed at $x=1$).
answered Mar 18 at 18:27
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
Alternatively, and proceeding from a more general result, we have that
the Finite Difference of first order, and the next iterated ones, are defined as
$$
eqalign{
& Delta _{,x} f(x) = f(x + 1) - f(x) cr
& Delta _{,x} ^n f(x) = Delta _{,x} left( {Delta _{,x} ^{n - 1} f(x)} right)
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {left( { - 1} right)^{n - k} binom{n}{k}f(x + k)} cr}
$$
Then, if $f(x)$ is a polynomial of degree $n$
$$
eqalign{
& f(x) = p_{,n} (x) = a_{,n} x^{,n} + a_{,n - 1} x^{,n - 1} + cdots + a_{,0} = cr
& = n!a_{,n} binom{x}{n}
+ left( {n - 1} right)!b_{,n - 1} binom{x}{n-1} + cdots + b_{,0} binom{x}{0} cr}
$$
the $n$th order Difference will leave only the leading coefficient in the Newton expression, i.e.
$$
Delta _{,x} ^n p_{,n} (x) = n!a_{,n} binom{x}{n-n} = n!a_{,n}
$$
This for whichever $p_n(x)$ and in particular for $ x^n$ (then computed at $x=1$).
answered Mar 18 at 18:27
G CabG Cab
20.4k31341
$endgroup$
Alternatively, and proceeding from a more general result, we have that
the Finite Difference of first order, and the next iterated ones, are defined as
$$
eqalign{
& Delta _{,x} f(x) = f(x + 1) - f(x) cr
& Delta _{,x} ^n f(x) = Delta _{,x} left( {Delta _{,x} ^{n - 1} f(x)} right)
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {left( { - 1} right)^{n - k} binom{n}{k}f(x + k)} cr}
$$
Then, if $f(x)$ is a polynomial of degree $n$
$$
eqalign{
& f(x) = p_{,n} (x) = a_{,n} x^{,n} + a_{,n - 1} x^{,n - 1} + cdots + a_{,0} = cr
& = n!a_{,n} binom{x}{n}
+ left( {n - 1} right)!b_{,n - 1} binom{x}{n-1} + cdots + b_{,0} binom{x}{0} cr}
$$
the $n$th order Difference will leave only the leading coefficient in the Newton expression, i.e.
$$
Delta _{,x} ^n p_{,n} (x) = n!a_{,n} binom{x}{n-n} = n!a_{,n}
$$
This for whichever $p_n(x)$ and in particular for $ x^n$ (then computed at $x=1$).
answered Mar 18 at 18:27
G CabG Cab
20.4k31341
answered Mar 18 at 18:27
G CabG Cab
20.4k31341
answered Mar 18 at 18:27
G CabG Cab
20.4k31341
answered Mar 18 at 18:27
G CabG Cab
20.4k31341
20.4k31341
add a comment |
add a comment |
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