Spivak's Calculus: sum of derivatives of a polynomia

In summary, the conversation discusses a proof involving a polynomial function of even degree that is always greater than or equal to zero. The solution involves using induction and the first coefficient of the polynomial. The conversation also mentions a book that the speaker enjoys.
  • #1
AlwaysCurious
33
0

Homework Statement


Let f be a polynomial function of degree n such that f(x) ≥ 0 for all x (note that n must be even). Prove that f + f' +f'' + f''' + ... + f^(n) ≥ 0.


Homework Equations



I believe that is all - the derivative of some term in the polynomial ax^n is anx^(n-1).

The Attempt at a Solution



I attempted induction on n (stepping up by two, not one, each time), actually writing out the final summed polynomial in terms of its coefficients, etc. but have made no real progress.

Thank you for hints/solutions in advance!
(if you have a suggestive hint, I actually would appreciate that MORE than a solution, but if you want to give a solution I'd still be happy).
 
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  • #2
Never mind, I found the solution a while ago. For anyone curious, here it is:

Note that the first coefficient of f must be positive, otherwise f(x) would be negative for large enough x.

Let Q(x) = (f + f' + f'' +... +f^(n))(x). Note that the first coefficient of Q is the first coefficient of f, so Q is positive for large enough (in absolute value) x, so Q(x) is positive for all x such that x<A for some number A, and Q(x) is positive for all x such that x>B for some number B. Let x2>B .

If Q(x) is positive for all x, then we're done. But assume false , i.e there is some interval that is a subset of (A,B) such that Q is negative on said interval, in particular say Q(n) < 0 for some n such that n<B.

Then by the continuity of Q there is a z such that n<z<x2 such that Q(z) = 0, and z is the closest root to n, i.e Q is negative on (n,z). Then again by the continuity and differentiability of Q, there is a y[itex]\ni[/itex](n,z) such that Q'(y) = [itex]\displaystyle\frac{Q(z) - Q(n) = -Q(n)}{z-n}[/itex], so Q'(y) is positive. Because y[itex]\ni[/itex](n,z), Q(y) is negative. So Q(y) - Q'(y) is negative.

But notice that Q(y) -Q'(y) = (f + f' + f'' +...)(y) - (f' + f'' + f''' +...)(y) = f(x). So the existence of an n such that Q(n) < 0 implies that f is negative at some y which is false.

I really like this book!
 

Related to Spivak's Calculus: sum of derivatives of a polynomia

1. What is Spivak's Calculus?

Spivak's Calculus, also known as "Calculus on Manifolds," is a rigorous mathematical approach to calculus that focuses on the study of multivariable functions and their derivatives. It was developed by Michael Spivak and is commonly used in advanced mathematics and physics courses.

2. What is the sum of derivatives of a polynomial in Spivak's Calculus?

In Spivak's Calculus, the sum of derivatives of a polynomial is known as the "total derivative." It is the derivative of a function with respect to all of its variables, rather than just one variable. This concept is useful for understanding how a function changes in response to changes in multiple variables.

3. How is Spivak's Calculus different from traditional calculus?

Spivak's Calculus is more abstract and rigorous than traditional calculus, which is typically based on limits and derivatives. In Spivak's Calculus, the focus is on the manipulation and analysis of functions and their derivatives, rather than just finding their values at a specific point.

4. What are some real-world applications of Spivak's Calculus?

Spivak's Calculus is used in a variety of fields, including physics, engineering, economics, and computer science. It can be used to model and understand complex systems and phenomena, such as fluid dynamics, chemical reactions, and economic markets.

5. Is Spivak's Calculus difficult to learn?

Spivak's Calculus can be challenging for those who are new to the subject, as it requires a solid understanding of mathematical concepts such as functions, derivatives, and multivariable calculus. However, with dedication and practice, it can be mastered, and the rewards of understanding this powerful tool for mathematical analysis are well worth the effort.

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