-1
$\begingroup$

If we have any polynomial with degree $n$ so the number of maximas and minimas is equal to $n-1$.

For example $ax^2 + bx + c$ has only one minima and $ax^3+bx^2+cx+d$ has one maxima and one minima but now when we look at the sine curve, it has infinite maximas and minimas so if it has to be represented as a polynomial it would be like $ax^{\infty}+1$(which is still infinity) $+ ................. +$ Constant

So now when we use power rule to differentiate this polynomial we get $\infty \cdot ax^\infty$. So the gradient at every point of sine curve is infinity, but now how should I prove it wrong without geometry's help?

$\endgroup$
  • $\begingroup$ The answer is Taylor polynomials: en.wikipedia.org/wiki/Taylor_series. $\endgroup$ – Floris Claassens Jun 18 at 8:48
  • 2
    $\begingroup$ The actions you want to take are not rigorous enough. For example, what do you mean by representing the $sin$ function as a polynomial? How to differentiate an expression with infinitely many terms (power series)? $\endgroup$ – tonychow0929 Jun 18 at 8:49
  • 1
    $\begingroup$ Not all functions can be represented by polynomials. A large class (but not all) can be represented by power series but they all not simply long polynomials, you need to study limits to understand them well. Also, even though these power series may look like "infinite polynomials", there is no $x^{\infty}$ term. Look up the Taylor series for sin. Differentiate it naively and you will see the Taylor series for cos. Three more times and you will be back where you started. $\endgroup$ – badjohn Jun 18 at 9:04
  • 1
    $\begingroup$ It is not true in general that a polynomial with degree $n$ has $n-1$ minima and maxima. $\endgroup$ – Peter Jun 18 at 9:04
  • $\begingroup$ $\sin(x)$ is a so-called transcendental function. It cannot be represented as a polynomial. The taylor-series does not terminate, hence is not a polynomial. $\endgroup$ – Peter Jun 18 at 9:07
0
$\begingroup$

I'm not sure whether the following are too hard for you.

First, you may want to know that the $\sin$ function is analytic, which means it is locally given by a convergent power series. There are only very few functions which satisfy this property (even though in practice you see a lot). This property is not guaranteed even if a function is differentiable infinitely many times.

Also, we can write $\sin x$ in the following way:

$$\sin x = \sum_{n=0}^\infty (-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$$

(Intuitively, while the power of $x$ is unbounded, the tail terms get smaller and smaller as the coefficients get smaller and smaller.)

In particular, the power series converges uniformly. Let $$f_n = (-1) ^{n}\frac{x^{2n+1}}{(2n+1)!}$$ $\forall n \in \mathbb{N}$. Indeed we have $$f'_n=(-1)^{n}\frac{x^{2n}}{(2n)!}$$. Since $$\cos x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$$

which again is a power series that converges uniformly, we conclude that we can differentiate term-by-term.

$\endgroup$
0
$\begingroup$

It is true that if a polynomial function has $n$ critical points (maxima or minima), then the degree of the polynomial is at least $n+1$.

You can think of $\sin(x)$ as a polynomial of infinite degree. Using Taylor's Theorem, $$ \sin(x) = x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}-\frac{x^{11}}{39916800}+\frac{x^{13}}{6227020800}+\ldots. $$

The denominator of a term with exponent $n$ is $n!$, so the well known summation formula for $\sin(x)$ is $$ \sin(x) = \sum_{m=1}^\infty (-1)^{m-1} \frac{x^{2m-1}}{(2m-1)!} $$ where $2m-1$ is just the odd positive natural numbers.

Now taking the derivative of each term, we get $$ \frac{\partial}{\partial x} \sin(x) = 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\frac{x^8}{40320}-\frac{x^{10}}{3628800}+\frac{x^{ 12}}{479001600}-\ldots $$ which happens to be $\cos(x)$, thus $$ \frac{\partial}{\partial x} \sin(x)=\cos(x). $$

This derivation can be found in several calculus books.

The fact that the denominators grow much faster than the exponents prevents the summation from reaching infinity. This is much like one of Zeno's paradoxes

"Suppose Homer wishes to walk to the end of a path. Before he can get there, he must get halfway there. Before he can get halfway there, he must get a quarter of the way there. Before traveling a quarter, he must travel one-eighth; before an eighth, one-sixteenth; and so on."

The point is that the sum of an infinite number of positive numbers is not necessarily infinite. In the same way, the derivative of an infinite degree polynomial is not necessarily infinite. There is no one term in the expansion of $\sin(x)$ that has infinite degree, and the terms become small enough that their sum is not infinite.

If you think that there are gaps in the reasoning above, you are right. It takes math majors years to get to the point where they can prove each step rigorously.

Hope that helps.

PS: We often down vote questions from people who are just beginning calculus or have not even begun calculus. Is that correct?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.