Derivatives to Infinity

Find all infinitely differentiable functions f:RRf:\mathbb{R}\to\mathbb{R} that satisfy f(x)+f(x)+f(x)+=(f(x))2f(x)+f'(x)+f''(x)+\cdots =(f(x))^2

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Note by Daniel Liu
3 years, 11 months ago

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

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

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

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

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

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

The above equation transfers into y2=2ydydx+yy^2 = 2y \dfrac{dy}{dx} +y. But as we assumed that y0y \neq 0.

It implies that y=2dydx+1y = 2 \dfrac{dy}{dx} +1, which infers that dx2=dyy1\dfrac{dx}{2} = \dfrac{dy}{y-1}. Integrating this both sides we get, x2=ln(y1)ln(c)\dfrac{x}{2} = ln (y-1) -ln(c)     y1=c×ex/2\implies y-1 = c \times e^{x/2}     y=c×ex/2+1\implies y= c \times e^{x/2} +1

Now let us find out the value of cc.

Substitute this in the original equation, then we get

(1+cex/2)+c2ex/2+c4ex/2+=(1+cex/2)2=1+2cex/2+c2ex (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}

cex/2(1+12+14+)=2cex/2+c2ex2cex/2=2cex/2+c2exc2ex=0c 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 ex>0e^{x} > 0 for all xRx \in \mathbb{R}. It implies that c2=0c^2 =0, i.e. c=0c=0. But this implies that f(x)=1f(x) = 1, a contradiction, as we assumed that f(x)0f'(x) \neq 0.

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


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

Surya Prakash - 3 years, 11 months ago

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Hello Surya,

I think you have made an error in integration in Case 2. The power of ee is +x2 + \frac{x}{2} and not x2 - \frac{x}{2} .

Also, I think you are missing one crucial step: When we substitute f(x)=cex2+1 f( x) = c \cdot e^{\frac{x}{2} } + 1 in the given equation, we get

1+2cex2=1+2cex2+c2ex 1 + 2 c \cdot e^{\frac{x}{2}} = 1 + 2c \cdot e^{\frac{x}{2} } + c^{2} \cdot e^{x}

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

Another solution is f(x)=0 f(x) = 0 for all values of xx. 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.

Pranshu Gaba - 3 years, 11 months ago

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

For example, it might be possible that the equation satisfies

{f(x)=0x0()=00<x<1f(x)=0x1 \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

{f(x)=0x0f(x)00<x<1f(x)=0x1 \begin{cases} f'(x) = 0 & x \leq 0 \\ f'(x) \neq 0 & 0 < x < 1 \\ f'(x) = 0 & x \geq 1 \\ \end{cases}

Calvin Lin Staff - 3 years, 11 months ago

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It is easy to see that

f(x)22f(x)f(x)=f(x)f\left(x\right)^2-2f\left(x\right)f'\left(x\right)=f\left(x\right)

Since 2f(x)f(x)=ddxf(x)2=f(x)+f(x)+f(x)...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(x)=f(x)212f'\left(x\right)=\frac{f\left(x\right)}{2}-\frac{1}{2}

Solving this differential equation yields f(x)=cex2+1f\left(x\right)=c\cdot e^{\frac{x}{2}}+1, where cc is a constant that comes with integration.

Julian Poon - 3 years, 11 months ago

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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)[f(x)f(x)2+12]=0 f(x) \left[ f'(x) - \frac{f(x) } {2} + \frac{1}{2} \right] = 0 , which makes f(x)=0f(x) = 0 for all xx as a possible solution.

Calvin Lin Staff - 3 years, 11 months ago

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Hi Julian,

I think you are missing one crucial step: When we substitute f(x)=cex2+1 f( x) = c \cdot e^{\frac{x}{2} } + 1 in the given equation, we get

1+2cex2=1+2cex2+c2ex 1 + 2 c \cdot e^{\frac{x}{2}} = 1 + 2c \cdot e^{\frac{x}{2} } + c^{2} \cdot e^{x}

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

Another solution is f(x)=0 f(x) = 0 for all values of xx. 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.

Pranshu Gaba - 3 years, 11 months ago

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

(c2ex+1+2cex2)(c2ex+cex2)=cex2+1\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

Julian Poon - 3 years, 11 months ago

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@Julian Poon I see that you substituted it in the equation f(x)22f(x)f(x)=f(x)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)+=(f(x))2f(x)+f'(x)+f''(x)+\cdots =(f(x))^2. Try substituting f(x)=cex2+1 f(x) = c\cdot e^{\frac{x}{2}}+1 in that equation, and you will see that only one value of cc satisfies it, that is c=0c = 0 .

Pranshu Gaba - 3 years, 11 months ago

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@Pranshu Gaba Oh yes, yeah I just realised. Thanks

Julian Poon - 3 years, 11 months ago

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Is this reasoning correct,since derivatives' degrees would be less than that of f(x)f(x),and the degree of [f(x)]2[f(x)]^2 would be two times that of f(x)f(x),the only solution could be when (f(x)=k=constant(f(x)=k=\text{constant},hence all the derivatives would be 00 we would have,k=k2k=0,1k=k^2 \Longrightarrow k=0,1. This solution is for those functions of xx which have only xx and its powers(notified by @Surya Prakash ).

Adarsh Kumar - 3 years, 11 months ago

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@Daniel Liu @Calvin Lin

Adarsh Kumar - 3 years, 11 months ago

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(ddxf(x))2=d2dx2f(x)2\left(\frac{d}{dx}f(x)\right)^{2} = \frac{d^{2}}{dx^{2}} f(x)^{2} is not true for most functions.

Julian Poon - 3 years, 11 months ago

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No it is not. ×2= \infty \times 2 = \infty and =2 \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.

Calvin Lin Staff - 3 years, 11 months ago

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Ohk sir.Thanx for correcting me!

Adarsh Kumar - 3 years, 11 months ago

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To everyone that begin their solutions by differentiating both sides: How do you justify that ddx(f(x)+f(x)+f(x)+)=f(x)+f(x)+f(x)+\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.

Ivan Koswara - 3 years, 11 months ago

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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,

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

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

Note that since nn isn't dependent on the function variable xx, you can differentiate both sides to get,

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

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

2f(x)f(x)+f(x)=limni=0nf(i)(x)=(f(x))22f(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)f^{(i)}(x) exists for all non-negative integers ii to proceed and verify solutions in the end by checking it against the original equation.

Prasun Biswas - 3 years, 11 months ago

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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 limfi=Ffi=F \lim \sum f_i = F \Rightarrow \sum f'_i = F' .

What you have to justify is the interchange of limits:

limddxfi=limddxfi=ddxlimfi \lim \sum \frac{d}{dx} f_i = \lim \frac{d}{dx} \sum f_i = \frac{d}{dx} \lim \sum f_i

Calvin Lin Staff - 3 years, 11 months ago

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0,1

Rui-Xian Siew - 3 years, 11 months ago

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Are those the only solutions? Why?

Calvin Lin Staff - 3 years, 11 months ago

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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.

Rui-Xian Siew - 3 years, 11 months ago

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