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A curious connection : What's the function $f(x)$?
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Hello I was wondering what was the function $f$ defines like this :
Let $f(x)$ a continuous and differentiable function such that :
$$f(x)=sum_k=0^inftyfracf'(k)k!(-x)^k$$
In fact can't solve it but it makes a connection between Ramanujan's Master theorem and Frullani's integral via the Fundamental theorem of calculus I explain :
We have :
$$int_0^inftyx^-s-1f(x)dx=Gamma(-s)f'(s)$$
Or :
$$int_0^inftyfracx^-s-1f(x)Gamma(-s)dx=f'(s)$$
Now we use the Fundamental theorem of calculus to get :
$$int_0^sint_0^inftyfracx^-s-1f(x)Gamma(-s)dxds=f(s)-f(0)$$
Now we take the limit to get :
$$lim_stoinftyint_0^sint_0^inftyfracx^-s-1f(x)Gamma(-s)dxds=f(infty)-f(0)$$
Wich is equal to:
$$int_0^inftyfracf(ax)-f(bx)ln(fracab)x$$
So my question is what is the function $f(x)$ , there exists a closed form to this ,is it trivial or not ?
Thanks
Ps:I know it's not very rigorous but I think it's interesting
calculus integration
asked 12 hours ago
FatsWallers
132217
 |Â
show 6 more comments
up vote
16
down vote
favorite
Hello I was wondering what was the function $f$ defines like this :
Let $f(x)$ a continuous and differentiable function such that :
$$f(x)=sum_k=0^inftyfracf'(k)k!(-x)^k$$
In fact can't solve it but it makes a connection between Ramanujan's Master theorem and Frullani's integral via the Fundamental theorem of calculus I explain :
We have :
$$int_0^inftyx^-s-1f(x)dx=Gamma(-s)f'(s)$$
Or :
$$int_0^inftyfracx^-s-1f(x)Gamma(-s)dx=f'(s)$$
Now we use the Fundamental theorem of calculus to get :
$$int_0^sint_0^inftyfracx^-s-1f(x)Gamma(-s)dxds=f(s)-f(0)$$
Now we take the limit to get :
$$lim_stoinftyint_0^sint_0^inftyfracx^-s-1f(x)Gamma(-s)dxds=f(infty)-f(0)$$
Wich is equal to:
$$int_0^inftyfracf(ax)-f(bx)ln(fracab)x$$
So my question is what is the function $f(x)$ , there exists a closed form to this ,is it trivial or not ?
Thanks
Ps:I know it's not very rigorous but I think it's interesting
calculus integration
asked 12 hours ago
FatsWallers
132217
2
Just a quick observation, but $f(x)=0$ satisfies the condition.
– Karn Watcharasupat
12 hours ago
2
@KarnWatcharasupat The set of solutions form a vector space.
– saulspatz
12 hours ago
1
If I'm not wrong (but I'm sure I am) you can differentiate the expression of $f$ to get $f'(x)=-f(x)$ which has as solution $Ce^-x$. Developing few terms you see that $C=0$ so the unique function that satisfies the definition is the constant function $0$ (this why I think I'm misunderstanding something).
– Dog_69
12 hours ago
2
@Dog_69, you might want to verify you differentiation, I don't believe $f'(x) = -f(x)$ unfortunately
– user88595
11 hours ago
2
I am right. I was wrong after all. $f'$ gives the sum $$ -sum_k=0^infty fracf'(k+1)k!(-x)^k$$ which is different from $-f$. Thanks @user88595. It was too easy to be true.
– Dog_69
11 hours ago
 |Â
show 6 more comments
up vote
16
down vote
favorite
up vote
16
down vote
favorite
Hello I was wondering what was the function $f$ defines like this :
Let $f(x)$ a continuous and differentiable function such that :
$$f(x)=sum_k=0^inftyfracf'(k)k!(-x)^k$$
In fact can't solve it but it makes a connection between Ramanujan's Master theorem and Frullani's integral via the Fundamental theorem of calculus I explain :
We have :
$$int_0^inftyx^-s-1f(x)dx=Gamma(-s)f'(s)$$
Or :
$$int_0^inftyfracx^-s-1f(x)Gamma(-s)dx=f'(s)$$
Now we use the Fundamental theorem of calculus to get :
$$int_0^sint_0^inftyfracx^-s-1f(x)Gamma(-s)dxds=f(s)-f(0)$$
Now we take the limit to get :
$$lim_stoinftyint_0^sint_0^inftyfracx^-s-1f(x)Gamma(-s)dxds=f(infty)-f(0)$$
Wich is equal to:
$$int_0^inftyfracf(ax)-f(bx)ln(fracab)x$$
So my question is what is the function $f(x)$ , there exists a closed form to this ,is it trivial or not ?
Thanks
Ps:I know it's not very rigorous but I think it's interesting
calculus integration
asked 12 hours ago
FatsWallers
132217
Hello I was wondering what was the function $f$ defines like this :
Let $f(x)$ a continuous and differentiable function such that :
$$f(x)=sum_k=0^inftyfracf'(k)k!(-x)^k$$
In fact can't solve it but it makes a connection between Ramanujan's Master theorem and Frullani's integral via the Fundamental theorem of calculus I explain :
We have :
$$int_0^inftyx^-s-1f(x)dx=Gamma(-s)f'(s)$$
Or :
$$int_0^inftyfracx^-s-1f(x)Gamma(-s)dx=f'(s)$$
Now we use the Fundamental theorem of calculus to get :
$$int_0^sint_0^inftyfracx^-s-1f(x)Gamma(-s)dxds=f(s)-f(0)$$
Now we take the limit to get :
$$lim_stoinftyint_0^sint_0^inftyfracx^-s-1f(x)Gamma(-s)dxds=f(infty)-f(0)$$
Wich is equal to:
$$int_0^inftyfracf(ax)-f(bx)ln(fracab)x$$
So my question is what is the function $f(x)$ , there exists a closed form to this ,is it trivial or not ?
Thanks
Ps:I know it's not very rigorous but I think it's interesting
calculus integration
asked 12 hours ago
FatsWallers
132217
asked 12 hours ago
FatsWallers
132217
asked 12 hours ago
FatsWallers
132217
asked 12 hours ago
FatsWallers
132217
132217
2
Just a quick observation, but $f(x)=0$ satisfies the condition.
– Karn Watcharasupat
12 hours ago
2
@KarnWatcharasupat The set of solutions form a vector space.
– saulspatz
12 hours ago
1
If I'm not wrong (but I'm sure I am) you can differentiate the expression of $f$ to get $f'(x)=-f(x)$ which has as solution $Ce^-x$. Developing few terms you see that $C=0$ so the unique function that satisfies the definition is the constant function $0$ (this why I think I'm misunderstanding something).
– Dog_69
12 hours ago
2
@Dog_69, you might want to verify you differentiation, I don't believe $f'(x) = -f(x)$ unfortunately
– user88595
11 hours ago
2
I am right. I was wrong after all. $f'$ gives the sum $$ -sum_k=0^infty fracf'(k+1)k!(-x)^k$$ which is different from $-f$. Thanks @user88595. It was too easy to be true.
– Dog_69
11 hours ago
 |Â
show 6 more comments
2
Just a quick observation, but $f(x)=0$ satisfies the condition.
– Karn Watcharasupat
12 hours ago
2
@KarnWatcharasupat The set of solutions form a vector space.
– saulspatz
12 hours ago
1
If I'm not wrong (but I'm sure I am) you can differentiate the expression of $f$ to get $f'(x)=-f(x)$ which has as solution $Ce^-x$. Developing few terms you see that $C=0$ so the unique function that satisfies the definition is the constant function $0$ (this why I think I'm misunderstanding something).
– Dog_69
12 hours ago
2
@Dog_69, you might want to verify you differentiation, I don't believe $f'(x) = -f(x)$ unfortunately
– user88595
11 hours ago
2
I am right. I was wrong after all. $f'$ gives the sum $$ -sum_k=0^infty fracf'(k+1)k!(-x)^k$$ which is different from $-f$. Thanks @user88595. It was too easy to be true.
– Dog_69
11 hours ago
2
2
Just a quick observation, but $f(x)=0$ satisfies the condition.
– Karn Watcharasupat
12 hours ago
Just a quick observation, but $f(x)=0$ satisfies the condition.
– Karn Watcharasupat
12 hours ago
2
2
@KarnWatcharasupat The set of solutions form a vector space.
– saulspatz
12 hours ago
@KarnWatcharasupat The set of solutions form a vector space.
– saulspatz
12 hours ago
1
1
If I'm not wrong (but I'm sure I am) you can differentiate the expression of $f$ to get $f'(x)=-f(x)$ which has as solution $Ce^-x$. Developing few terms you see that $C=0$ so the unique function that satisfies the definition is the constant function $0$ (this why I think I'm misunderstanding something).
– Dog_69
12 hours ago
If I'm not wrong (but I'm sure I am) you can differentiate the expression of $f$ to get $f'(x)=-f(x)$ which has as solution $Ce^-x$. Developing few terms you see that $C=0$ so the unique function that satisfies the definition is the constant function $0$ (this why I think I'm misunderstanding something).
– Dog_69
12 hours ago
2
2
@Dog_69, you might want to verify you differentiation, I don't believe $f'(x) = -f(x)$ unfortunately
– user88595
11 hours ago
@Dog_69, you might want to verify you differentiation, I don't believe $f'(x) = -f(x)$ unfortunately
– user88595
11 hours ago
2
2
I am right. I was wrong after all. $f'$ gives the sum $$ -sum_k=0^infty fracf'(k+1)k!(-x)^k$$ which is different from $-f$. Thanks @user88595. It was too easy to be true.
– Dog_69
11 hours ago
I am right. I was wrong after all. $f'$ gives the sum $$ -sum_k=0^infty fracf'(k+1)k!(-x)^k$$ which is different from $-f$. Thanks @user88595. It was too easy to be true.
– Dog_69
11 hours ago
 |Â
show 6 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
Not a full answer, but a set of equations to exploit.
Let
$$
g(x) = f'(x)=sum_k=1^inftyfrack f'(k)k!(-1)^k x^k-1= sum_k=0^inftyfracg(k+1)k!(-1)^k+1 x^k
$$
But also, via Taylor expansion
$$
g(x) = sum_k=0^inftyfracg^(k)(0)k! x^k
$$
So we obtain the set of equations, for all $k=0 cdots infty$:
$$
g^(k)(0) = g(k+1)(-1)^k+1
$$
The first few are:
$$
g(0) = - g(1)\
g'(0) = g(2)\
g''(0) = - g(3)\
g^(3)(0) = g(4)\
cdots
$$
Can we proceed from here?
answered 11 hours ago
Andreas
6,502834
Let $g(x)=-f'(-x)$ instead to remove the alternating signs, and then we may use the results derived in Is there a function with the property $f(n)=f^(n)(0)$?
– Simply Beautiful Art
9 hours ago
@SimplyBeautifulArt Using your suggestion, one gets $g^(n)(0) = f'(n+1) = - g(-(n+1))$.
– Andreas
8 hours ago
Hm, you're right. Well, using the approach from the link, one can solve what you have here already I believe.
– Simply Beautiful Art
4 hours ago
add a comment |Â
up vote
1
down vote
Considering the operator
$$
calL(f) = sum_k=0^inftyfracf'(k)k!(-x)^k
$$
and choosing
$$
f(x) = e^-W(1) x
$$
with $W(cdot)$ the Lambert function, we have
$$
calL(f) =mbox $-frac1W(1)f$
$$
because
$$
f(x) = 1-x W(1)+frac 12 W(1)^2 x^2-frac 16 W(1)^3 x^3+cdots + fracW(1)^kk!(-x)^k + cdots\
fracf(x)W(1) = -1 + e^-W(1)x-frac 12e^-2W(1)x^2+frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots
$$
with
$$
W(1)^k = e^-k W(1)
$$
I hope it helps.
answered 11 hours ago
Cesareo
5,4592412
2
If $f(x)=e^-ex$, then $e^k neq f'(k) =-ee^-ek$.
– Rócherz
11 hours ago
What you have established, using your proposed $f(x)$, is that $ frac-1W(1)sum_k=0^inftyfracf'(k)k!(-x)^k = 1 - e^-W(1)x+frac 12e^-2W(1)x^2-frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots = f(x) $ where the last relation follows, as you write, from $W(1)^k = e^-k W(1)$. Now, the prefactor $frac-1W(1)$ unfortunately is NOT given in the original task.
– Andreas
7 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Not a full answer, but a set of equations to exploit.
Let
$$
g(x) = f'(x)=sum_k=1^inftyfrack f'(k)k!(-1)^k x^k-1= sum_k=0^inftyfracg(k+1)k!(-1)^k+1 x^k
$$
But also, via Taylor expansion
$$
g(x) = sum_k=0^inftyfracg^(k)(0)k! x^k
$$
So we obtain the set of equations, for all $k=0 cdots infty$:
$$
g^(k)(0) = g(k+1)(-1)^k+1
$$
The first few are:
$$
g(0) = - g(1)\
g'(0) = g(2)\
g''(0) = - g(3)\
g^(3)(0) = g(4)\
cdots
$$
Can we proceed from here?
answered 11 hours ago
Andreas
6,502834
Let $g(x)=-f'(-x)$ instead to remove the alternating signs, and then we may use the results derived in Is there a function with the property $f(n)=f^(n)(0)$?
– Simply Beautiful Art
9 hours ago
@SimplyBeautifulArt Using your suggestion, one gets $g^(n)(0) = f'(n+1) = - g(-(n+1))$.
– Andreas
8 hours ago
Hm, you're right. Well, using the approach from the link, one can solve what you have here already I believe.
– Simply Beautiful Art
4 hours ago
add a comment |Â
up vote
4
down vote
Not a full answer, but a set of equations to exploit.
Let
$$
g(x) = f'(x)=sum_k=1^inftyfrack f'(k)k!(-1)^k x^k-1= sum_k=0^inftyfracg(k+1)k!(-1)^k+1 x^k
$$
But also, via Taylor expansion
$$
g(x) = sum_k=0^inftyfracg^(k)(0)k! x^k
$$
So we obtain the set of equations, for all $k=0 cdots infty$:
$$
g^(k)(0) = g(k+1)(-1)^k+1
$$
The first few are:
$$
g(0) = - g(1)\
g'(0) = g(2)\
g''(0) = - g(3)\
g^(3)(0) = g(4)\
cdots
$$
Can we proceed from here?
answered 11 hours ago
Andreas
6,502834
Let $g(x)=-f'(-x)$ instead to remove the alternating signs, and then we may use the results derived in Is there a function with the property $f(n)=f^(n)(0)$?
– Simply Beautiful Art
9 hours ago
@SimplyBeautifulArt Using your suggestion, one gets $g^(n)(0) = f'(n+1) = - g(-(n+1))$.
– Andreas
8 hours ago
Hm, you're right. Well, using the approach from the link, one can solve what you have here already I believe.
– Simply Beautiful Art
4 hours ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Not a full answer, but a set of equations to exploit.
Let
$$
g(x) = f'(x)=sum_k=1^inftyfrack f'(k)k!(-1)^k x^k-1= sum_k=0^inftyfracg(k+1)k!(-1)^k+1 x^k
$$
But also, via Taylor expansion
$$
g(x) = sum_k=0^inftyfracg^(k)(0)k! x^k
$$
So we obtain the set of equations, for all $k=0 cdots infty$:
$$
g^(k)(0) = g(k+1)(-1)^k+1
$$
The first few are:
$$
g(0) = - g(1)\
g'(0) = g(2)\
g''(0) = - g(3)\
g^(3)(0) = g(4)\
cdots
$$
Can we proceed from here?
answered 11 hours ago
Andreas
6,502834
Not a full answer, but a set of equations to exploit.
Let
$$
g(x) = f'(x)=sum_k=1^inftyfrack f'(k)k!(-1)^k x^k-1= sum_k=0^inftyfracg(k+1)k!(-1)^k+1 x^k
$$
But also, via Taylor expansion
$$
g(x) = sum_k=0^inftyfracg^(k)(0)k! x^k
$$
So we obtain the set of equations, for all $k=0 cdots infty$:
$$
g^(k)(0) = g(k+1)(-1)^k+1
$$
The first few are:
$$
g(0) = - g(1)\
g'(0) = g(2)\
g''(0) = - g(3)\
g^(3)(0) = g(4)\
cdots
$$
Can we proceed from here?
answered 11 hours ago
Andreas
6,502834
answered 11 hours ago
Andreas
6,502834
answered 11 hours ago
Andreas
6,502834
answered 11 hours ago
Andreas
6,502834
6,502834
Let $g(x)=-f'(-x)$ instead to remove the alternating signs, and then we may use the results derived in Is there a function with the property $f(n)=f^(n)(0)$?
– Simply Beautiful Art
9 hours ago
@SimplyBeautifulArt Using your suggestion, one gets $g^(n)(0) = f'(n+1) = - g(-(n+1))$.
– Andreas
8 hours ago
Hm, you're right. Well, using the approach from the link, one can solve what you have here already I believe.
– Simply Beautiful Art
4 hours ago
add a comment |Â
Let $g(x)=-f'(-x)$ instead to remove the alternating signs, and then we may use the results derived in Is there a function with the property $f(n)=f^(n)(0)$?
– Simply Beautiful Art
9 hours ago
@SimplyBeautifulArt Using your suggestion, one gets $g^(n)(0) = f'(n+1) = - g(-(n+1))$.
– Andreas
8 hours ago
Hm, you're right. Well, using the approach from the link, one can solve what you have here already I believe.
– Simply Beautiful Art
4 hours ago
Let $g(x)=-f'(-x)$ instead to remove the alternating signs, and then we may use the results derived in Is there a function with the property $f(n)=f^(n)(0)$?
– Simply Beautiful Art
9 hours ago
Let $g(x)=-f'(-x)$ instead to remove the alternating signs, and then we may use the results derived in Is there a function with the property $f(n)=f^(n)(0)$?
– Simply Beautiful Art
9 hours ago
@SimplyBeautifulArt Using your suggestion, one gets $g^(n)(0) = f'(n+1) = - g(-(n+1))$.
– Andreas
8 hours ago
@SimplyBeautifulArt Using your suggestion, one gets $g^(n)(0) = f'(n+1) = - g(-(n+1))$.
– Andreas
8 hours ago
Hm, you're right. Well, using the approach from the link, one can solve what you have here already I believe.
– Simply Beautiful Art
4 hours ago
Hm, you're right. Well, using the approach from the link, one can solve what you have here already I believe.
– Simply Beautiful Art
4 hours ago
add a comment |Â
up vote
1
down vote
Considering the operator
$$
calL(f) = sum_k=0^inftyfracf'(k)k!(-x)^k
$$
and choosing
$$
f(x) = e^-W(1) x
$$
with $W(cdot)$ the Lambert function, we have
$$
calL(f) =mbox $-frac1W(1)f$
$$
because
$$
f(x) = 1-x W(1)+frac 12 W(1)^2 x^2-frac 16 W(1)^3 x^3+cdots + fracW(1)^kk!(-x)^k + cdots\
fracf(x)W(1) = -1 + e^-W(1)x-frac 12e^-2W(1)x^2+frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots
$$
with
$$
W(1)^k = e^-k W(1)
$$
I hope it helps.
answered 11 hours ago
Cesareo
5,4592412
2
If $f(x)=e^-ex$, then $e^k neq f'(k) =-ee^-ek$.
– Rócherz
11 hours ago
What you have established, using your proposed $f(x)$, is that $ frac-1W(1)sum_k=0^inftyfracf'(k)k!(-x)^k = 1 - e^-W(1)x+frac 12e^-2W(1)x^2-frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots = f(x) $ where the last relation follows, as you write, from $W(1)^k = e^-k W(1)$. Now, the prefactor $frac-1W(1)$ unfortunately is NOT given in the original task.
– Andreas
7 hours ago
add a comment |Â
up vote
1
down vote
Considering the operator
$$
calL(f) = sum_k=0^inftyfracf'(k)k!(-x)^k
$$
and choosing
$$
f(x) = e^-W(1) x
$$
with $W(cdot)$ the Lambert function, we have
$$
calL(f) =mbox $-frac1W(1)f$
$$
because
$$
f(x) = 1-x W(1)+frac 12 W(1)^2 x^2-frac 16 W(1)^3 x^3+cdots + fracW(1)^kk!(-x)^k + cdots\
fracf(x)W(1) = -1 + e^-W(1)x-frac 12e^-2W(1)x^2+frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots
$$
with
$$
W(1)^k = e^-k W(1)
$$
I hope it helps.
answered 11 hours ago
Cesareo
5,4592412
2
If $f(x)=e^-ex$, then $e^k neq f'(k) =-ee^-ek$.
– Rócherz
11 hours ago
What you have established, using your proposed $f(x)$, is that $ frac-1W(1)sum_k=0^inftyfracf'(k)k!(-x)^k = 1 - e^-W(1)x+frac 12e^-2W(1)x^2-frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots = f(x) $ where the last relation follows, as you write, from $W(1)^k = e^-k W(1)$. Now, the prefactor $frac-1W(1)$ unfortunately is NOT given in the original task.
– Andreas
7 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Considering the operator
$$
calL(f) = sum_k=0^inftyfracf'(k)k!(-x)^k
$$
and choosing
$$
f(x) = e^-W(1) x
$$
with $W(cdot)$ the Lambert function, we have
$$
calL(f) =mbox $-frac1W(1)f$
$$
because
$$
f(x) = 1-x W(1)+frac 12 W(1)^2 x^2-frac 16 W(1)^3 x^3+cdots + fracW(1)^kk!(-x)^k + cdots\
fracf(x)W(1) = -1 + e^-W(1)x-frac 12e^-2W(1)x^2+frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots
$$
with
$$
W(1)^k = e^-k W(1)
$$
I hope it helps.
answered 11 hours ago
Cesareo
5,4592412
Considering the operator
$$
calL(f) = sum_k=0^inftyfracf'(k)k!(-x)^k
$$
and choosing
$$
f(x) = e^-W(1) x
$$
with $W(cdot)$ the Lambert function, we have
$$
calL(f) =mbox $-frac1W(1)f$
$$
because
$$
f(x) = 1-x W(1)+frac 12 W(1)^2 x^2-frac 16 W(1)^3 x^3+cdots + fracW(1)^kk!(-x)^k + cdots\
fracf(x)W(1) = -1 + e^-W(1)x-frac 12e^-2W(1)x^2+frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots
$$
with
$$
W(1)^k = e^-k W(1)
$$
I hope it helps.
answered 11 hours ago
Cesareo
5,4592412
edited 7 hours ago
answered 11 hours ago
Cesareo
5,4592412
answered 11 hours ago
Cesareo
5,4592412
answered 11 hours ago
Cesareo
5,4592412
5,4592412
2
If $f(x)=e^-ex$, then $e^k neq f'(k) =-ee^-ek$.
– Rócherz
11 hours ago
What you have established, using your proposed $f(x)$, is that $ frac-1W(1)sum_k=0^inftyfracf'(k)k!(-x)^k = 1 - e^-W(1)x+frac 12e^-2W(1)x^2-frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots = f(x) $ where the last relation follows, as you write, from $W(1)^k = e^-k W(1)$. Now, the prefactor $frac-1W(1)$ unfortunately is NOT given in the original task.
– Andreas
7 hours ago
add a comment |Â
2
If $f(x)=e^-ex$, then $e^k neq f'(k) =-ee^-ek$.
– Rócherz
11 hours ago
What you have established, using your proposed $f(x)$, is that $ frac-1W(1)sum_k=0^inftyfracf'(k)k!(-x)^k = 1 - e^-W(1)x+frac 12e^-2W(1)x^2-frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots = f(x) $ where the last relation follows, as you write, from $W(1)^k = e^-k W(1)$. Now, the prefactor $frac-1W(1)$ unfortunately is NOT given in the original task.
– Andreas
7 hours ago
2
2
If $f(x)=e^-ex$, then $e^k neq f'(k) =-ee^-ek$.
– Rócherz
11 hours ago
If $f(x)=e^-ex$, then $e^k neq f'(k) =-ee^-ek$.
– Rócherz
11 hours ago
What you have established, using your proposed $f(x)$, is that $ frac-1W(1)sum_k=0^inftyfracf'(k)k!(-x)^k = 1 - e^-W(1)x+frac 12e^-2W(1)x^2-frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots = f(x) $ where the last relation follows, as you write, from $W(1)^k = e^-k W(1)$. Now, the prefactor $frac-1W(1)$ unfortunately is NOT given in the original task.
– Andreas
7 hours ago
What you have established, using your proposed $f(x)$, is that $ frac-1W(1)sum_k=0^inftyfracf'(k)k!(-x)^k = 1 - e^-W(1)x+frac 12e^-2W(1)x^2-frac 16 e^-3W(1)x^3+cdots + frace^-kW(1)(-x)^kk!+cdots = f(x) $ where the last relation follows, as you write, from $W(1)^k = e^-k W(1)$. Now, the prefactor $frac-1W(1)$ unfortunately is NOT given in the original task.
– Andreas
7 hours ago
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2
Just a quick observation, but $f(x)=0$ satisfies the condition.
– Karn Watcharasupat
12 hours ago
2
@KarnWatcharasupat The set of solutions form a vector space.
– saulspatz
12 hours ago
1
If I'm not wrong (but I'm sure I am) you can differentiate the expression of $f$ to get $f'(x)=-f(x)$ which has as solution $Ce^-x$. Developing few terms you see that $C=0$ so the unique function that satisfies the definition is the constant function $0$ (this why I think I'm misunderstanding something).
– Dog_69
12 hours ago
2
@Dog_69, you might want to verify you differentiation, I don't believe $f'(x) = -f(x)$ unfortunately
– user88595
11 hours ago
2
I am right. I was wrong after all. $f'$ gives the sum $$ -sum_k=0^infty fracf'(k+1)k!(-x)^k$$ which is different from $-f$. Thanks @user88595. It was too easy to be true.
– Dog_69
11 hours ago