Logaritmetic problem

1. May 8, 2007

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

2. May 8, 2007

f(x)

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

Last edited: May 8, 2007
3. May 8, 2007

reza

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

4. May 9, 2007

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.

5. May 9, 2007

VietDao29

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

6. May 12, 2007

reza

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

7. May 12, 2007

VietDao29

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

8. May 14, 2007

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?

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