Proof of Logaritmetic Problem: x-1/2x^2 < ln(1+x) < x

[reza]

show that (for all posetive x)
x-1/2 x^2<ln(1+x)<x

i understand it in graf please prove it for my in methemathic

 

[f(x)]

show that (for all posetive x)
x-1/2 x^2<ln(1+x)<x

i understand it in graf please prove it for my in methemathic

does the 1st inequality mean $$x-\frac{x^2}{2} < \ln (1+x)$$ ?
i suppose this will be true for only |x|<1

 

[reza]

yes you right x - (x^2)/2
and in the book (calculus , rechard a.silverman) has written for x>0
Do you need the address of problem?

 

[f(x)]

ok sorry about that...the problem's fine , i mistook the signs.
The identity may be proved by using the expansion series for ln(1+x) but this holds for |x|<1.

 

[VietDao29]

show that (for all posetive x)
x-1/2 x^2<ln(1+x)<x

i understand it in graf please prove it for my in methemathic

Or, you can try to take the derivative and prove that it's either an increasing function, or a decreasing function (Have you covered how to differentiate a function?). It goes like this:

I am going to prove the ln(1 + x) < x for all x > 0 part.
Re-arrange the inequality gives:
ln(1 + x) - x < 0

Let f(x) = ln(1 + x) - x
Now, we have that f(0) = ln(1 + 0) - 0 = 0
$$f'(x) = \frac{1}{1 + x} - 1$$
For x > 0, we have 1 + x > 1, hence 1 / (x + 1) < 1, hence 1 / (x + 1) - 1 < 0, so f'(x) < 0 for all x > 0. f(x) is a decreasing function for x > 0.

So, for x > 0, we have:
f(x) < f(0) <=> f(x) < 0 <=> ln(1 + x) - x < 0 <=> ln(1 + x) < x

 

[reza]

thank you very much but how we can prove the other side

 

[VietDao29]

thank you very much but how we can prove the other side

Have you tried anything? It's pretty much the same.

 

[reza]

f(x)=ln(x+1)-x+x^2/2
f'(x)=1/(x+1)-1+x=x^2/(x+1)>0
f'(x)>0 then it is increasing
So, for x > 0, we have:
f(x) > f(0) <=> f(x) > 0 <=> ln (x+1)-x+x^2/2>0 <=> ln (x+1)>x-x^2/2
is this proof Ok?