Prove that a function is indefinitely differentiableFinding the general term of two related recurrence relationsPiecewise interpolation with derivatives that is also twice differentiableShow that there is a series in R^infinity has some term greater than or equal to 1/n but that also is arbitrary close to the zero sequence.Uniform convergence to a differentiable functionFinding the general formula for $a_n+1=2^n a_n +4$, where $a_1=1$.When does the limit of derivatives coincide with the derivative of the limit function?Solving a recurrence with an alternating pattern e.g. do $mathrm X$, then do $mathrm Y$; repeatProve that if $f$ is derivable in an interval $I$ and $f'(a)=0$ and $f''(a) ne 0$, then $a$ must be a local minimum or maximum of the function.How to prove that a function has directional derivative?Formula for an m derivative of a function from $mathbbR^ntomathbbR$
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Prove that a function is indefinitely differentiable
Finding the general term of two related recurrence relationsPiecewise interpolation with derivatives that is also twice differentiableShow that there is a series in R^infinity has some term greater than or equal to 1/n but that also is arbitrary close to the zero sequence.Uniform convergence to a differentiable functionFinding the general formula for $a_n+1=2^n a_n +4$, where $a_1=1$.When does the limit of derivatives coincide with the derivative of the limit function?Solving a recurrence with an alternating pattern e.g. do $mathrm X$, then do $mathrm Y$; repeatProve that if $f$ is derivable in an interval $I$ and $f'(a)=0$ and $f''(a) ne 0$, then $a$ must be a local minimum or maximum of the function.How to prove that a function has directional derivative?Formula for an m derivative of a function from $mathbbR^ntomathbbR$
$begingroup$
Show that if
$$f: mathbbR to mathbbR$$ is differentiable in second order which satisfies the equation $$f'' =f'+f$$ then $f$ is indefinitely differentiable.
I was thinking to write the $n$the derivative as
$$f^(n)=a_nf+b_nf', a_n=b_n-1, b_n=a_n-1+b_n-1 $$
I calculated a few derivatives so I think this must be the form of the sequence, but I don't know how to solve it and I cannot see any other pattern to write the general term.
calculus sequences-and-series functions derivatives recurrence-relations
edited 8 hours ago
Chase Ryan Taylor
4,53621531
asked 9 hours ago
AndreiAndrei
134
$endgroup$
add a comment |
$begingroup$
Show that if
$$f: mathbbR to mathbbR$$ is differentiable in second order which satisfies the equation $$f'' =f'+f$$ then $f$ is indefinitely differentiable.
I was thinking to write the $n$the derivative as
$$f^(n)=a_nf+b_nf', a_n=b_n-1, b_n=a_n-1+b_n-1 $$
I calculated a few derivatives so I think this must be the form of the sequence, but I don't know how to solve it and I cannot see any other pattern to write the general term.
calculus sequences-and-series functions derivatives recurrence-relations
edited 8 hours ago
Chase Ryan Taylor
4,53621531
asked 9 hours ago
AndreiAndrei
134
$endgroup$
add a comment |
$begingroup$
Show that if
$$f: mathbbR to mathbbR$$ is differentiable in second order which satisfies the equation $$f'' =f'+f$$ then $f$ is indefinitely differentiable.
I was thinking to write the $n$the derivative as
$$f^(n)=a_nf+b_nf', a_n=b_n-1, b_n=a_n-1+b_n-1 $$
I calculated a few derivatives so I think this must be the form of the sequence, but I don't know how to solve it and I cannot see any other pattern to write the general term.
calculus sequences-and-series functions derivatives recurrence-relations
edited 8 hours ago
Chase Ryan Taylor
4,53621531
asked 9 hours ago
AndreiAndrei
134
$endgroup$
Show that if
$$f: mathbbR to mathbbR$$ is differentiable in second order which satisfies the equation $$f'' =f'+f$$ then $f$ is indefinitely differentiable.
I was thinking to write the $n$the derivative as
$$f^(n)=a_nf+b_nf', a_n=b_n-1, b_n=a_n-1+b_n-1 $$
I calculated a few derivatives so I think this must be the form of the sequence, but I don't know how to solve it and I cannot see any other pattern to write the general term.
calculus sequences-and-series functions derivatives recurrence-relations
calculus sequences-and-series functions derivatives recurrence-relations
edited 8 hours ago
Chase Ryan Taylor
4,53621531
asked 9 hours ago
AndreiAndrei
134
edited 8 hours ago
Chase Ryan Taylor
4,53621531
asked 9 hours ago
AndreiAndrei
134
edited 8 hours ago
Chase Ryan Taylor
4,53621531
edited 8 hours ago
Chase Ryan Taylor
4,53621531
edited 8 hours ago
Chase Ryan Taylor
4,53621531
4,53621531
asked 9 hours ago
AndreiAndrei
134
asked 9 hours ago
AndreiAndrei
134
asked 9 hours ago
AndreiAndrei
134
134
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since $f$ is twice differentiable, $f''=f'+f$ is the sum of two differentiable functions, so $f'''$ exists and is equal to $f'''=f''+f'$. By induction you can now show that it's infinitely differentiable, as $f^(n)=f^(n-1)+f^(n-2)$, implying that $f^(n+1)$ exists.
answered 8 hours ago
Alex R.Alex R.
25.8k12455
$endgroup$
$begingroup$
Can we simply neglect the coefficients of f and f' ?
$endgroup$
– Andrei
8 hours ago
$begingroup$
@Andrei What coefficients are you talking about?
$endgroup$
– DonAntonio
8 hours ago
3
$begingroup$
@Andrei I see what you are trying to do, but you are overthinking the problem. $$dfracddxf'' = dfracddx(f'+f)$$ Then $$f''' = f''+f'$$ You don't need to then convert $f''$ into $f'+f$. You can immediately apply induction. This is the base case.
$endgroup$
– InterstellarProbe
8 hours ago
add a comment |
$begingroup$
You wanted to know the general form for $f^(n)$ in terms of $f$ and $f'$. The solution Alex R. provided is the correct solution. But, if you just want to see what form the $n$-th derivative takes (in terms of $f$ and $f'$), you have:
$$f^(n) = a_nf' + b_nf$$
Take the derivative derivatives:
$$f^(n+1) = a_nf''+b_nf' = a_n(f'+f)+b_nf' = (a_n+b_n)f' + a_nf$$
From this, we get:
$$b_n+1 = a_n$$
$$a_n+1 = a_n+b_n$$
So, we have:
$$a_n+2 = a_n+1+b_n+1 = a_n+1+a_n$$
This is the Fibonacci recurrence relation. From this, we obtain:
$$f^(n) = F_n+1f'+F_nf, nge 0$$
where $$F_0=1, F_1=0$$ and $$F_n+2 = F_n+1+F_n, nge 0$$ is the Fibonacci sequence.
answered 8 hours ago
InterstellarProbeInterstellarProbe
4,311931
$endgroup$
$begingroup$
a very interesting approach to acquire some extra information
$endgroup$
– Chase Ryan Taylor
8 hours ago
add a comment |
$begingroup$
Another way (different way) to look at this is to think of it as an ODE:
$$ fracd^2 fdx^2 - fracdfdx - f = 0 $$
This is a pretty simple ODE which has the solution of:
$$ f(x) = a_1 e^fracsqrt5 + 1 2x + a_2e^-fracsqrt5 + 1 2x $$
Where $a_1$ and $a_2$ are some constant. Since we know that exponential functions are infinite differentiable, this imply that $f(x)$ is infinite differentiable since it is just linear combination of exponential functions.
answered 6 hours ago
user209663user209663
40129
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $f$ is twice differentiable, $f''=f'+f$ is the sum of two differentiable functions, so $f'''$ exists and is equal to $f'''=f''+f'$. By induction you can now show that it's infinitely differentiable, as $f^(n)=f^(n-1)+f^(n-2)$, implying that $f^(n+1)$ exists.
answered 8 hours ago
Alex R.Alex R.
25.8k12455
$endgroup$
$begingroup$
Can we simply neglect the coefficients of f and f' ?
$endgroup$
– Andrei
8 hours ago
$begingroup$
@Andrei What coefficients are you talking about?
$endgroup$
– DonAntonio
8 hours ago
3
$begingroup$
@Andrei I see what you are trying to do, but you are overthinking the problem. $$dfracddxf'' = dfracddx(f'+f)$$ Then $$f''' = f''+f'$$ You don't need to then convert $f''$ into $f'+f$. You can immediately apply induction. This is the base case.
$endgroup$
– InterstellarProbe
8 hours ago
add a comment |
$begingroup$
Since $f$ is twice differentiable, $f''=f'+f$ is the sum of two differentiable functions, so $f'''$ exists and is equal to $f'''=f''+f'$. By induction you can now show that it's infinitely differentiable, as $f^(n)=f^(n-1)+f^(n-2)$, implying that $f^(n+1)$ exists.
answered 8 hours ago
Alex R.Alex R.
25.8k12455
$endgroup$
$begingroup$
Can we simply neglect the coefficients of f and f' ?
$endgroup$
– Andrei
8 hours ago
$begingroup$
@Andrei What coefficients are you talking about?
$endgroup$
– DonAntonio
8 hours ago
3
$begingroup$
@Andrei I see what you are trying to do, but you are overthinking the problem. $$dfracddxf'' = dfracddx(f'+f)$$ Then $$f''' = f''+f'$$ You don't need to then convert $f''$ into $f'+f$. You can immediately apply induction. This is the base case.
$endgroup$
– InterstellarProbe
8 hours ago
add a comment |
$begingroup$
Since $f$ is twice differentiable, $f''=f'+f$ is the sum of two differentiable functions, so $f'''$ exists and is equal to $f'''=f''+f'$. By induction you can now show that it's infinitely differentiable, as $f^(n)=f^(n-1)+f^(n-2)$, implying that $f^(n+1)$ exists.
answered 8 hours ago
Alex R.Alex R.
25.8k12455
$endgroup$
Since $f$ is twice differentiable, $f''=f'+f$ is the sum of two differentiable functions, so $f'''$ exists and is equal to $f'''=f''+f'$. By induction you can now show that it's infinitely differentiable, as $f^(n)=f^(n-1)+f^(n-2)$, implying that $f^(n+1)$ exists.
answered 8 hours ago
Alex R.Alex R.
25.8k12455
answered 8 hours ago
Alex R.Alex R.
25.8k12455
answered 8 hours ago
Alex R.Alex R.
25.8k12455
answered 8 hours ago
Alex R.Alex R.
25.8k12455
25.8k12455
$begingroup$
Can we simply neglect the coefficients of f and f' ?
$endgroup$
– Andrei
8 hours ago
$begingroup$
@Andrei What coefficients are you talking about?
$endgroup$
– DonAntonio
8 hours ago
3
$begingroup$
@Andrei I see what you are trying to do, but you are overthinking the problem. $$dfracddxf'' = dfracddx(f'+f)$$ Then $$f''' = f''+f'$$ You don't need to then convert $f''$ into $f'+f$. You can immediately apply induction. This is the base case.
$endgroup$
– InterstellarProbe
8 hours ago
add a comment |
$begingroup$
Can we simply neglect the coefficients of f and f' ?
$endgroup$
– Andrei
8 hours ago
$begingroup$
@Andrei What coefficients are you talking about?
$endgroup$
– DonAntonio
8 hours ago
3
$begingroup$
@Andrei I see what you are trying to do, but you are overthinking the problem. $$dfracddxf'' = dfracddx(f'+f)$$ Then $$f''' = f''+f'$$ You don't need to then convert $f''$ into $f'+f$. You can immediately apply induction. This is the base case.
$endgroup$
– InterstellarProbe
8 hours ago
$begingroup$
Can we simply neglect the coefficients of f and f' ?
$endgroup$
– Andrei
8 hours ago
$begingroup$
Can we simply neglect the coefficients of f and f' ?
$endgroup$
– Andrei
8 hours ago
$begingroup$
@Andrei What coefficients are you talking about?
$endgroup$
– DonAntonio
8 hours ago
$begingroup$
@Andrei What coefficients are you talking about?
$endgroup$
– DonAntonio
8 hours ago
3
3
$begingroup$
@Andrei I see what you are trying to do, but you are overthinking the problem. $$dfracddxf'' = dfracddx(f'+f)$$ Then $$f''' = f''+f'$$ You don't need to then convert $f''$ into $f'+f$. You can immediately apply induction. This is the base case.
$endgroup$
– InterstellarProbe
8 hours ago
$begingroup$
@Andrei I see what you are trying to do, but you are overthinking the problem. $$dfracddxf'' = dfracddx(f'+f)$$ Then $$f''' = f''+f'$$ You don't need to then convert $f''$ into $f'+f$. You can immediately apply induction. This is the base case.
$endgroup$
– InterstellarProbe
8 hours ago
add a comment |
$begingroup$
You wanted to know the general form for $f^(n)$ in terms of $f$ and $f'$. The solution Alex R. provided is the correct solution. But, if you just want to see what form the $n$-th derivative takes (in terms of $f$ and $f'$), you have:
$$f^(n) = a_nf' + b_nf$$
Take the derivative derivatives:
$$f^(n+1) = a_nf''+b_nf' = a_n(f'+f)+b_nf' = (a_n+b_n)f' + a_nf$$
From this, we get:
$$b_n+1 = a_n$$
$$a_n+1 = a_n+b_n$$
So, we have:
$$a_n+2 = a_n+1+b_n+1 = a_n+1+a_n$$
This is the Fibonacci recurrence relation. From this, we obtain:
$$f^(n) = F_n+1f'+F_nf, nge 0$$
where $$F_0=1, F_1=0$$ and $$F_n+2 = F_n+1+F_n, nge 0$$ is the Fibonacci sequence.
answered 8 hours ago
InterstellarProbeInterstellarProbe
4,311931
$endgroup$
$begingroup$
a very interesting approach to acquire some extra information
$endgroup$
– Chase Ryan Taylor
8 hours ago
add a comment |
$begingroup$
You wanted to know the general form for $f^(n)$ in terms of $f$ and $f'$. The solution Alex R. provided is the correct solution. But, if you just want to see what form the $n$-th derivative takes (in terms of $f$ and $f'$), you have:
$$f^(n) = a_nf' + b_nf$$
Take the derivative derivatives:
$$f^(n+1) = a_nf''+b_nf' = a_n(f'+f)+b_nf' = (a_n+b_n)f' + a_nf$$
From this, we get:
$$b_n+1 = a_n$$
$$a_n+1 = a_n+b_n$$
So, we have:
$$a_n+2 = a_n+1+b_n+1 = a_n+1+a_n$$
This is the Fibonacci recurrence relation. From this, we obtain:
$$f^(n) = F_n+1f'+F_nf, nge 0$$
where $$F_0=1, F_1=0$$ and $$F_n+2 = F_n+1+F_n, nge 0$$ is the Fibonacci sequence.
answered 8 hours ago
InterstellarProbeInterstellarProbe
4,311931
$endgroup$
$begingroup$
a very interesting approach to acquire some extra information
$endgroup$
– Chase Ryan Taylor
8 hours ago
add a comment |
$begingroup$
You wanted to know the general form for $f^(n)$ in terms of $f$ and $f'$. The solution Alex R. provided is the correct solution. But, if you just want to see what form the $n$-th derivative takes (in terms of $f$ and $f'$), you have:
$$f^(n) = a_nf' + b_nf$$
Take the derivative derivatives:
$$f^(n+1) = a_nf''+b_nf' = a_n(f'+f)+b_nf' = (a_n+b_n)f' + a_nf$$
From this, we get:
$$b_n+1 = a_n$$
$$a_n+1 = a_n+b_n$$
So, we have:
$$a_n+2 = a_n+1+b_n+1 = a_n+1+a_n$$
This is the Fibonacci recurrence relation. From this, we obtain:
$$f^(n) = F_n+1f'+F_nf, nge 0$$
where $$F_0=1, F_1=0$$ and $$F_n+2 = F_n+1+F_n, nge 0$$ is the Fibonacci sequence.
answered 8 hours ago
InterstellarProbeInterstellarProbe
4,311931
$endgroup$
You wanted to know the general form for $f^(n)$ in terms of $f$ and $f'$. The solution Alex R. provided is the correct solution. But, if you just want to see what form the $n$-th derivative takes (in terms of $f$ and $f'$), you have:
$$f^(n) = a_nf' + b_nf$$
Take the derivative derivatives:
$$f^(n+1) = a_nf''+b_nf' = a_n(f'+f)+b_nf' = (a_n+b_n)f' + a_nf$$
From this, we get:
$$b_n+1 = a_n$$
$$a_n+1 = a_n+b_n$$
So, we have:
$$a_n+2 = a_n+1+b_n+1 = a_n+1+a_n$$
This is the Fibonacci recurrence relation. From this, we obtain:
$$f^(n) = F_n+1f'+F_nf, nge 0$$
where $$F_0=1, F_1=0$$ and $$F_n+2 = F_n+1+F_n, nge 0$$ is the Fibonacci sequence.
answered 8 hours ago
InterstellarProbeInterstellarProbe
4,311931
edited 8 hours ago
answered 8 hours ago
InterstellarProbeInterstellarProbe
4,311931
answered 8 hours ago
InterstellarProbeInterstellarProbe
4,311931
answered 8 hours ago
InterstellarProbeInterstellarProbe
4,311931
4,311931
$begingroup$
a very interesting approach to acquire some extra information
$endgroup$
– Chase Ryan Taylor
8 hours ago
add a comment |
$begingroup$
a very interesting approach to acquire some extra information
$endgroup$
– Chase Ryan Taylor
8 hours ago
$begingroup$
a very interesting approach to acquire some extra information
$endgroup$
– Chase Ryan Taylor
8 hours ago
$begingroup$
a very interesting approach to acquire some extra information
$endgroup$
– Chase Ryan Taylor
8 hours ago
add a comment |
$begingroup$
Another way (different way) to look at this is to think of it as an ODE:
$$ fracd^2 fdx^2 - fracdfdx - f = 0 $$
This is a pretty simple ODE which has the solution of:
$$ f(x) = a_1 e^fracsqrt5 + 1 2x + a_2e^-fracsqrt5 + 1 2x $$
Where $a_1$ and $a_2$ are some constant. Since we know that exponential functions are infinite differentiable, this imply that $f(x)$ is infinite differentiable since it is just linear combination of exponential functions.
answered 6 hours ago
user209663user209663
40129
$endgroup$
add a comment |
$begingroup$
Another way (different way) to look at this is to think of it as an ODE:
$$ fracd^2 fdx^2 - fracdfdx - f = 0 $$
This is a pretty simple ODE which has the solution of:
$$ f(x) = a_1 e^fracsqrt5 + 1 2x + a_2e^-fracsqrt5 + 1 2x $$
Where $a_1$ and $a_2$ are some constant. Since we know that exponential functions are infinite differentiable, this imply that $f(x)$ is infinite differentiable since it is just linear combination of exponential functions.
answered 6 hours ago
user209663user209663
40129
$endgroup$
add a comment |
$begingroup$
Another way (different way) to look at this is to think of it as an ODE:
$$ fracd^2 fdx^2 - fracdfdx - f = 0 $$
This is a pretty simple ODE which has the solution of:
$$ f(x) = a_1 e^fracsqrt5 + 1 2x + a_2e^-fracsqrt5 + 1 2x $$
Where $a_1$ and $a_2$ are some constant. Since we know that exponential functions are infinite differentiable, this imply that $f(x)$ is infinite differentiable since it is just linear combination of exponential functions.
answered 6 hours ago
user209663user209663
40129
$endgroup$
Another way (different way) to look at this is to think of it as an ODE:
$$ fracd^2 fdx^2 - fracdfdx - f = 0 $$
This is a pretty simple ODE which has the solution of:
$$ f(x) = a_1 e^fracsqrt5 + 1 2x + a_2e^-fracsqrt5 + 1 2x $$
Where $a_1$ and $a_2$ are some constant. Since we know that exponential functions are infinite differentiable, this imply that $f(x)$ is infinite differentiable since it is just linear combination of exponential functions.
answered 6 hours ago
user209663user209663
40129
answered 6 hours ago
user209663user209663
40129
answered 6 hours ago
user209663user209663
40129
answered 6 hours ago
user209663user209663
40129
40129
add a comment |
add a comment |
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