Using the Lagrange Remainder Theorem (advanced calc/real analysis)

[kidairbag]

Hi everyone. This is my first post here and I was wondering if any of you could help me.

The question is to prove that
$$1 + \frac{x}{3} - \frac{x^2}{9} < (1 + x)^\frac{1}{3} < 1 + \frac{x}{3}$$ if x>0.

The question is in a section on the lagrange remainder theorem. The fact that the first expression is less than the third is self evident but I'm assuming to show that the middle expression falls between the two involves using the lagrange remainder theorem on that expression. I don't really know how exactly I would go about squeezing it between the two expressions, however, so I'm looking for some pointers on that.

Thanks!

Edit: Nevermind. I was able to figure it out by using the first and second degree taylor polynomials of the second expression and determining sign of the remainder. Thanks though!

 

[mighty2000]

Hi there,

Great question! To prove that the middle expression falls between the first and third expressions, we can use the Lagrange remainder theorem to show that the error term in the middle expression is smaller than the error terms in the first and third expressions.

Let's start by writing out the Taylor series expansions for each expression:

1 + \frac{x}{3} - \frac{x^2}{9} = 1 + \frac{x}{3} - \frac{x^2}{9} + \frac{x^3}{27} + ...

(1 + x)^\frac{1}{3} = 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{1}{27}x^3 + ...

1 + \frac{x}{3} = 1 + \frac{x}{3} + 0 + 0 + ...

We can see that the first and third expressions are the first and second degree Taylor polynomials for the middle expression. Now, let's consider the error term for each expression:

1 + \frac{x}{3} - \frac{x^2}{9} + \frac{x^3}{27} + ... = 1 + \frac{x}{3} - \frac{x^2}{9} + R_2(x)

(1 + x)^\frac{1}{3} = 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{1}{27}x^3 + R_3(x)

1 + \frac{x}{3} = 1 + \frac{x}{3} + R_1(x)

Where R_n(x) represents the remainder term for the nth degree Taylor polynomial. We can see that for the middle expression, the remainder term is R_3(x), which is the smallest among the three expressions. This means that the middle expression is closer to the true value than the first and third expressions.

Therefore, we can conclude that 1 + \frac{x}{3} - \frac{x^2}{9} < (1 + x)^\frac{1}{3} < 1 + \frac{x}{3} for x>0.

I hope this helps! Let me know if you have any further questions or if you need any clarification. Good luck with your studies!