A freshman calculus question, ask, ask!!

f(x)-(1 2)f(x 2)=x 2 Let x=x 2, and then multiply 1 2 on both sides of the equation

1 2) f(x 2) - (1 4) f(x 4) = x 2 8 Continue repeating the above process.

1/4)f(x/4)-(1/8)f(x/8)=x^2/64

1/2^(n-1)]f[1/2^(n-1)]-1/2^n)f(x/2^n)=x^2/(2^3n)

All of the above add up.

f(x)-f(x/2^n)/2^n=x^2[1+1/8+1/64+..1/(2^3n)]

x^2[(1(1-(1/8)^(n+1))]/[1-1/8]

8 7) x 2 n->infinite(1 8) (n+1)=0

When n approaches infinity, an n can always be taken such that when n > n, x 2 n belongs to a neighborhood of x=0.

So f(x 2 n) is bounded and (1 2 n) is infinitesimal, so lim n-> infinity f(x 2 n) 2 n=0 (bounded function multiplied by infinitesimal limit is 0).

So f(x)-0=(8 7)x 2

f(x)=(8/7)*x^2

As a hint, f(x)-f(x 2) 2=x 2f(x 2)-f(x 4) 2=x 2 4->1 2f(x 2)-f(x 4) 4=x 2 8

f(x/4)-f(x/8)/2=x^2/16->f(x/4)/4-f(x/8)/8=x^2/64

And so on, and then add all the equations together.

Solution: The differential equation is y"=(1+y'2y, which becomes 2y"/(1+y'²)=1/y

(y'²)'=2y'y"The equation is reconverted to 2y'y"/(1+y'²)=y'y, the integral on both sides has ln(1+y'²)=ln|y|+ln|a|(a is any non-zero constant) has 1+y'²=ay，y'=±√(ay-1)，dy/√(ay-1)=±dx，2 [√ay-1)]/a=±x+c/a

c is an arbitrary constant).

The general solution of the equation is y=(ax+c) 4a+1 a, and the general solution of the equation can be reduced to .

y=ax²/4+cx/2+c²/4a+1/a

lim(x tends to 0) f(x) divided by (absolute value x) = a exists.

f(0)=0;

lim(x tends to 0)f'(x) = lim [f(x)-f(0)]/(x-0) = lim f(x)/x ;

When f(x) is derivable at x=0, we get: lim(x tends to 0 negative) f'(x) = lim(x tends to 0 positive) f'(x)

a = lim -f(0-)/|x| = lim f(0+)/|x| = a ∴a=0

This proves the sufficiency, and the above process is reversed to necessity.

(1) .Let f(x) = f(x) -xf(1 2) =f(1 2) -1 2 =1 2>0f(1 ) = f(1) -1 =-1<0 So: f(1 2) *f(1) <0

From the mediator theorem, in (1 2,1), there must be f( )= 0 both: f( )= ;

2).Let f(x) = f(x) -x

f(1 2) =f(1 2) -1 2 =1 2>0f(1 ) = f(1) -1 =-1<0 So: f(1 2) *f(1) <0

According to the mediator value theorem, in (1 2,1), there must be f( ) = 0 and f(0) = 0

Using Roll's theorem on [0, ] for f(x), there is (0, ) such that f'( = 0

Both: f'（η）= 1

3).Let g(x) = exp(- x)*f(x) again: g(0) = 0, g( ) = 0

According to Rawl's theorem, for any real number , there must be x0 (0, ) such that :g'(x0) =exp(-λx0)*[f'(x0) -1 - f(x0)-x0]] =0

exp(-x0)>0

Both: f'(x0)- f(x0)-x0]=1exp represents the natural logarithm. exp(-x) is the -x power of exp.

A hint to you, the mean theorem.