Prove ln(x) <= x-1 for positive x

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In summary, the natural logarithm is a function that takes a positive number and returns a negative number. If you can prove that the function is always smaller than the number it is applied to, then you have proven that the function is always smaller than the number -1.
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Dank2
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proof ##\ln x##<=x-1 for positive x.
if i show that at the start of the segment, the value in the start of the segment (0, +inf), of inx is smaller than x-1, and the derivative of inx is always smaller than x-1, is that suffice to proof that inx is always smaller than x-1? if not why?
 
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What is inx?
 
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Math_QED said:
What is inx?
Natural logarithm ;)
 
  • #6
Dank2 said:
the value in the start of the segment (0, +inf), of inx is smaller than x-1, AND the derivative of inx is always smaller than x-1,

corrected
 
  • #7
solved. thanks
 
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Dank2 said:
proof ##\ln x##<=x-1 for positive x.
if i show that at the start of the segment, the value in the start of the segment (0, +inf), of inx is smaller than x-1, and the derivative of inx is always smaller than x-1, is that suffice to proof that inx is always smaller than x-1? if not why?
But the derivative of ln(x) is 1/x, which gets huge near 0.
 
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FactChecker said:
But the derivative of ln(x) is 1/x, which gets huge near 0.
that's right, so:
f(x) = x-1 - Inx . f''(x) = 1 - 1/x = 0 ==> x=1. that should be either maximus or minimum
f(1) = 0.
now if we take the second derivative:
f''(x) = 1/x^2
which means the derivative is increasing constantly and f(1) is the minimum point inthe graph, and hence x-1>=Inx
 
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  • #10
It looks like you are ok here.

In my view, the slickest and most intuitive approach to this by far is to use Jensen's Inequality-- as the logarithm is (strictly) negative convex. If you need to, you can derive Jensen's Inequality, and the above relation falls out of it... that is, construct a Taylor Polynomial for the logarithm at x = 1, with the remainder term being O(n^2). I'd suggest using the Lagrange form for the remainder term. Then use the fact that the logarithm has a continuously negative second derivative. Graphically, this means ln(x) will always be below the tangent line for said logarithm evaluated at x = 1, with equality only where x = 1.
 
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FAQ: Prove ln(x) <= x-1 for positive x

1. What does ln(x) mean?

ln(x) is the natural logarithm function, which is the inverse of the exponential function e^x. It represents the power to which e must be raised to equal the given value x.

2. Why is it important to prove ln(x) <= x-1 for positive x?

This inequality is important in mathematics and scientific research because it helps us understand the behavior of exponential functions and their relationship to logarithmic functions. It also has practical applications in fields such as finance and biology.

3. How can we prove ln(x) <= x-1 for positive x?

We can prove this inequality using mathematical techniques such as calculus and algebra. One approach is to consider the function f(x) = x - ln(x) and show that it is always greater than or equal to 1 for positive x. Another approach is to use the Mean Value Theorem for integrals.

4. What are some real-life examples of ln(x) <= x-1 for positive x?

One example is in finance, where this inequality is used to model compound interest. Another example is in biology, where it can be used to model population growth and decay. Additionally, it has applications in physics, chemistry, and other scientific fields.

5. Can this inequality ever be false for positive x?

No, this inequality is always true for positive x. It can be proven using mathematical techniques and has been verified through numerous examples and real-life applications.

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