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Case 1 :$f'(x) = 0$$\implies f(x) = c$, where $c$ is a constant. Substituting in the above equation gives $c=0$ or $c=1$. There are two solutions here and they are $f(x) = 0$ and $f(x) = 1$.

Case 2 :$f'(x) \neq 0$. For our convenience let us denote $f(x)$ with $y$.

The above equation transfers into $y^2 = 2y \dfrac{dy}{dx} +y$. But as we assumed that $y \neq 0$.

It implies that $y = 2 \dfrac{dy}{dx} +1$, which infers that $\dfrac{dx}{2} = \dfrac{dy}{y-1}$. Integrating this both sides we get,
$\dfrac{x}{2} = ln (y-1) -ln(c)$$\implies y-1 = c \times e^{x/2}$$\implies y= c \times e^{x/2} +1$

Now let us find out the value of $c$.

Substitute this in the original equation, then we get

But $e^{x} > 0$ for all $x \in \mathbb{R}$. It implies that $c^2 =0$, i.e. $c=0$. But this implies that $f(x) = 1$, a contradiction, as we assumed that $f'(x) \neq 0$.

So, there are only two solutions i.e. $f(x) =0$ and $f(x) =1$.

Thank you @Pranshu Gaba , For making me realize my mistake.

This means $c = 0$, and therefore $f(x) = 1$ for all values of $x$.

Another solution is $f(x) = 0$ for all values of $x$. These are the only solutions to the given equations.

In the end, we must not forget to substitute our solutions back in the equations to check for extraneous solutions. Also, we must check if the summation on the left hand side converges, otherwise our solution might not be valid.

You've made a very common mistake with such functional equations. When you get to an equation like $f(x) \times ( \ldots ) = 0$, the only conclusion we have is that pointwise$f(x) = 0$ or $(\ldots) = 0$. It is not true that we only have the 2 cases of $f(x) =0$everywhere OR $( \ldots ) = 0$everywhere.

For example, it might be possible that the equation satisfies

$\begin{cases} f(x) = 0 & x \leq 0 \\ ( \ldots ) = 0 & 0 < x < 1 \\ f(x) = 0 & x \geq 1 \\ \end{cases}$

Of course, there is no reason why our domain is split up into intervals.

Note: It is not clear to me why you set your cases as such. That seems extremely arbitrary / unmotivated. Similarly, you will need to consider possibilities of

$\begin{cases} f'(x) = 0 & x \leq 0 \\ f'(x) \neq 0 & 0 < x < 1 \\ f'(x) = 0 & x \geq 1 \\ \end{cases}$

Always factorize when dealing with a functional equation (or any equation). Never just cancel out terms, because that will cause you to miss out solutions.

IE Your third line should have been $f(x) \left[ f'(x) - \frac{f(x) } {2} + \frac{1}{2} \right] = 0$, which makes $f(x) = 0$ for all $x$ as a possible solution.

This means $c = 0$, and therefore $f(x) = 1$ for all values of $x$.

Another solution is $f(x) = 0$ for all values of $x$. These are the only solutions to the given equations.

In the end, we must not forget to substitute our solutions back in the equations to check for extraneous solutions. Also, we must check if the summation on the left hand side converges, otherwise our solution might not be valid.

@Julian Poon
–
I see that you substituted it in the equation $f\left(x\right)^2-2f\left(x\right)f'\left(x\right)=f\left(x\right)$

However, I was referring to the original equation $f(x)+f'(x)+f''(x)+\cdots =(f(x))^2$. Try substituting $f(x) = c\cdot e^{\frac{x}{2}}+1$ in that equation, and you will see that only one value of $c$ satisfies it, that is $c = 0$.

Is this reasoning correct,since derivatives' degrees would be less than that of $f(x)$,and the degree of $[f(x)]^2$ would be two times that of $f(x)$,the only solution could be when $(f(x)=k=\text{constant}$,hence all the derivatives would be $0$ we would have,$k=k^2 \Longrightarrow k=0,1$. This solution is for those functions of $x$ which have only $x$ and its powers(notified by @Surya Prakash ).

To everyone that begin their solutions by differentiating both sides: How do you justify that $\frac{d}{dx} (f(x) + f'(x) + f''(x) + \ldots) = f'(x) + f''(x) + f'''(x) + \ldots$? As far as I know, differentiating term-by-term is only justifiable for finite number of terms.

I'd say that term-by-term differentiation is applicable whenever the number of terms in the series isn't dependent on the function variable. A more formal way to write the original equation would be,

Of course, we assume that $f^{(i)}(x)$ exists for all non-negative integers $i$ to proceed and verify solutions in the end by checking it against the original equation.

You might have to be careful with the initial conditions required to apply the theorem. It is not immediately apparent why $\lim \sum f_i = F \Rightarrow \sum f'_i = F'$.

What you have to justify is the interchange of limits:

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## Comments

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TopNewest$f(x) + f'(x) + f''(x) + \ldots = (f(x))^2$

Differentiating it and adding $f(x)$ on both sides we get,

$f(x) + f'(x) + f''(x) + \ldots = 2f(x) \times f'(x) +f(x)$

$\implies (f(x))^2 = 2f(x) \times f'(x) + f(x)$

Case 1 :$f'(x) = 0$ $\implies f(x) = c$, where $c$ is a constant. Substituting in the above equation gives $c=0$ or $c=1$. There are two solutions here and they are $f(x) = 0$ and $f(x) = 1$.Case 2 :$f'(x) \neq 0$. For our convenience let us denote $f(x)$ with $y$.The above equation transfers into $y^2 = 2y \dfrac{dy}{dx} +y$. But as we assumed that $y \neq 0$.

It implies that $y = 2 \dfrac{dy}{dx} +1$, which infers that $\dfrac{dx}{2} = \dfrac{dy}{y-1}$. Integrating this both sides we get, $\dfrac{x}{2} = ln (y-1) -ln(c)$ $\implies y-1 = c \times e^{x/2}$ $\implies y= c \times e^{x/2} +1$

Now let us find out the value of $c$.

Substitute this in the original equation, then we get

$(1+c e^{x/2}) + \dfrac{c}{2}e^{x/2} + \dfrac{c}{4}e^{x/2} + \ldots = (1+c e^{x/2})^2 = 1+2ce^{x/2} + c^2 e^{x}$

$c e^{x/2} (1+ \dfrac{1}{2} + \dfrac{1}{4} + \ldots ) = 2ce^{x/2} + c^2 e^x \\ 2c e^{x/2} = 2ce^{x/2} + c^2 e^x \\ c^2 e^x = 0$

But $e^{x} > 0$ for all $x \in \mathbb{R}$. It implies that $c^2 =0$, i.e. $c=0$. But this implies that $f(x) = 1$, a contradiction, as we assumed that $f'(x) \neq 0$.

So, there are only two solutions i.e. $f(x) =0$ and $f(x) =1$.

Thank you @Pranshu Gaba , For making me realize my mistake.

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Hello Surya,

I think you have made an error in integration in Case 2. The power of $e$ is $+ \frac{x}{2}$ and not $- \frac{x}{2}$.

Also, I think you are missing one crucial step: When we substitute $f( x) = c \cdot e^{\frac{x}{2} } + 1$ in the given equation, we get

$1 + 2 c \cdot e^{\frac{x}{2}} = 1 + 2c \cdot e^{\frac{x}{2} } + c^{2} \cdot e^{x}$

This means $c = 0$, and therefore $f(x) = 1$ for all values of $x$.

Another solution is $f(x) = 0$ for all values of $x$. These are the only solutions to the given equations.

In the end, we must not forget to substitute our solutions back in the equations to check for extraneous solutions. Also, we must check if the summation on the left hand side converges, otherwise our solution might not be valid.

Log in to reply

You've made a very common mistake with such functional equations. When you get to an equation like $f(x) \times ( \ldots ) = 0$, the only conclusion we have is that

pointwise$f(x) = 0$ or $(\ldots) = 0$. It is not true that we only have the 2 cases of $f(x) =0$everywhereOR $( \ldots ) = 0$everywhere.For example, it might be possible that the equation satisfies

$\begin{cases} f(x) = 0 & x \leq 0 \\ ( \ldots ) = 0 & 0 < x < 1 \\ f(x) = 0 & x \geq 1 \\ \end{cases}$

Of course, there is no reason why our domain is split up into intervals.

Note: It is not clear to me why you set your cases as such. That seems extremely arbitrary / unmotivated. Similarly, you will need to consider possibilities of

$\begin{cases} f'(x) = 0 & x \leq 0 \\ f'(x) \neq 0 & 0 < x < 1 \\ f'(x) = 0 & x \geq 1 \\ \end{cases}$

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It is easy to see that

$f\left(x\right)^2-2f\left(x\right)f'\left(x\right)=f\left(x\right)$

Since $2f\left(x\right)f'\left(x\right)=\frac{d}{dx}f\left(x\right)^2=f'\left(x\right)+f''\left(x\right)+f'''\left(x\right)...$

Rearranging gives:

$f'\left(x\right)=\frac{f\left(x\right)}{2}-\frac{1}{2}$

Solving this differential equation yields $f\left(x\right)=c\cdot e^{\frac{x}{2}}+1$, where $c$ is a constant that comes with integration.

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Always factorize when dealing with a functional equation (or any equation). Never just cancel out terms, because that will cause you to miss out solutions.

IE Your third line should have been $f(x) \left[ f'(x) - \frac{f(x) } {2} + \frac{1}{2} \right] = 0$, which makes $f(x) = 0$ for all $x$ as a possible solution.

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Hi Julian,

I think you are missing one crucial step: When we substitute $f( x) = c \cdot e^{\frac{x}{2} } + 1$ in the given equation, we get

$1 + 2 c \cdot e^{\frac{x}{2}} = 1 + 2c \cdot e^{\frac{x}{2} } + c^{2} \cdot e^{x}$

This means $c = 0$, and therefore $f(x) = 1$ for all values of $x$.

Another solution is $f(x) = 0$ for all values of $x$. These are the only solutions to the given equations.

In the end, we must not forget to substitute our solutions back in the equations to check for extraneous solutions. Also, we must check if the summation on the left hand side converges, otherwise our solution might not be valid.

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I do not get you, I did substitute it back into the equation and got this:

$\left(c^2e^x+1+2ce^{\frac{x}{2}}\right)-\left(c^2e^x+ce^{\frac{x}{2}}\right)=ce^{\frac{x}{2}}+1$

And hence, I was sure of my answer. Except, I did not consider the f(x)=constant case

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$f\left(x\right)^2-2f\left(x\right)f'\left(x\right)=f\left(x\right)$

I see that you substituted it in the equationHowever, I was referring to the original equation $f(x)+f'(x)+f''(x)+\cdots =(f(x))^2$. Try substituting $f(x) = c\cdot e^{\frac{x}{2}}+1$ in that equation, and you will see that only one value of $c$ satisfies it, that is $c = 0$.

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Is this reasoning correct,since derivatives' degrees would be less than that of $f(x)$,and the degree of $[f(x)]^2$ would be two times that of $f(x)$,the only solution could be when $(f(x)=k=\text{constant}$,hence all the derivatives would be $0$ we would have,$k=k^2 \Longrightarrow k=0,1$. This solution is for those functions of $x$ which have only $x$ and its powers(notified by @Surya Prakash ).

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@Daniel Liu @Calvin Lin

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$\left(\frac{d}{dx}f(x)\right)^{2} = \frac{d^{2}}{dx^{2}} f(x)^{2}$ is not true for most functions.

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No it is not. $\infty \times 2 = \infty$ and $\infty = \infty ^ 2$.

In other words, you have (unnecessarily) restricted yourself to looking at finite degree polynomials, which is a very small subset of all functions.

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Ohk sir.Thanx for correcting me!

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To everyone that begin their solutions by differentiating both sides: How do you justify that $\frac{d}{dx} (f(x) + f'(x) + f''(x) + \ldots) = f'(x) + f''(x) + f'''(x) + \ldots$? As far as I know, differentiating term-by-term is only justifiable for finite number of terms.

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I'd say that term-by-term differentiation is applicable whenever the number of terms in the series isn't dependent on the function variable. A more formal way to write the original equation would be,

$\lim_{n\to\infty}\sum_{i=0}^n f^{(i)}(x)=(f(x))^2$

where $f^{(i)}(x)$ denotes the $i^{\textrm{th}}$ derivative of $f(x)$ and $f^{(0)}(x)=f(x)$.

Note that since $n$ isn't dependent on the function variable $x$, you can differentiate both sides to get,

$\lim_{n\to\infty}\sum_{i=1}^{n+1} f^{(n)}(x)=2f(x)f^\prime(x)$

Since $n\to\infty$, we also have $n+1\to\infty$ and you can rewrite it as,

$2f(x)f^\prime(x)+f(x)=\lim_{n\to\infty}\sum_{i=0}^n f^{(i)}(x)=(f(x))^2$

Of course, we assume that $f^{(i)}(x)$ exists for all non-negative integers $i$ to proceed and verify solutions in the end by checking it against the original equation.

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You might have to be careful with the initial conditions required to apply the theorem. It is not immediately apparent why $\lim \sum f_i = F \Rightarrow \sum f'_i = F'$.

What you have to justify is the interchange of limits:

$\lim \sum \frac{d}{dx} f_i = \lim \frac{d}{dx} \sum f_i = \frac{d}{dx} \lim \sum f_i$

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0,1

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Are those the only solutions? Why?

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Haha I think the others had explained it in great detail. At first I just feel that f(X) should be constants, then I got the constants.

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