Trying to prove that for x real, ln(1+x)<=x

  • Thread starter penguin007
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In summary, the conversation is about proving the inequality ln(1+x)<=x for real numbers. The solution involves using the concavity of the function x->ln(1+x) and finding the Taylor series expansion for e^x. However, the person is trying to find a solution using only the concavity of the function.
  • #1
penguin007
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Homework Statement



Hi everyone,
I'm trying to proove that for x real, ln(1+x)<=x

Homework Equations





The Attempt at a Solution



I know that I can solve this problem by introducing f:=x->ln(1+x)-x and studying f, but I'd like to use x->ln(1+x)'s concavity so I tried to find lambda, x and y in :
ln(1+(lambda)*x+(1-lambda)*y)>=(lambda)*ln(1+x)+(1-lambda)*ln(1+y)...


Any help is welcome...
 
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  • #2


You are trying too hard.

Do you know the taylor series expansion for e^x? That is all you need.
 
  • #3


I find your solution quite interesting too (I had not thought about it) but I'd like to find a solution using the concavity of this function x:=->ln(1+x) (though I admit it may be too hard for that question).
 

FAQ: Trying to prove that for x real, ln(1+x)<=x

What does ln(1+x) mean?

ln(1+x) is the natural logarithm function, also known as the logarithm base e. It is the inverse of the exponential function e^x, and it represents the power to which the base e must be raised to obtain the value of (1+x).

How is ln(1+x) related to x?

The relationship between ln(1+x) and x is that ln(1+x) is always less than or equal to x for any real value of x. This means that the function ln(1+x) is always "less steep" than the function x, and it can be visualized as a curve that approaches the x-axis but never touches it.

Why is it important to prove that ln(1+x)<=x for x real?

This inequality is important because it is a fundamental property of the natural logarithm function. It allows us to simplify complex mathematical expressions and make calculations easier. It also has many applications in fields such as calculus, economics, and statistics.

How can we prove that ln(1+x)<=x for x real?

One way to prove this inequality is by using the Maclaurin series expansion for ln(1+x). This involves expressing ln(1+x) as an infinite sum of terms, and then showing that each term is always less than or equal to x. Another approach is to use the derivative of ln(1+x) and show that it is always less than or equal to 1 for x real.

Is ln(1+x)<=x always true for x real?

Yes, this inequality holds true for all real values of x. It can be proven mathematically using different techniques as mentioned in the previous question. Additionally, it can also be verified by plugging in different values of x into the original inequality and observing that it holds true for all cases.

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