Thread: symmetrical even functions

Prove that all even functions are a linear subspace on the reals. I think you can do this by proving closure of addition and multiplication. Since the sum and scalar product of even functions is still an even function then this is a linear space.
it's interesting if you take the integral of an even function it would just be double the integral of the positive side of the function since it is symmetrical about 0-there must be other properties one could deduce

prove that all rational functions f/g where degree of f<=degree of g are a linear subspace on the reals:

What exactly is your question?

prove that all rational functions f/g where degree of f<=degree of g are a linear subspace on the reals
I tried synthetic division and I think the result will have a degree of (g-f)^-1, but if g=0 which is in the domain then this function is undefined and not in the set so it isn't a subspace?
maybe Lang Algebra is a good text to read?

Originally Posted by mathnerd15

if g=0 which is in the domain then this function is undefined and not in the set so it isn't a subspace?

You need to prove that the sum of two rational functions (as well as a product by a number) is in the subspace. When you are proving this, you can assume that the functions you are given are defined at least somewhere, i.e., their denominators are not zero. You need to prove that the denominator of the sum is not identical to zero and that the restriction on the degrees is fulfilled.

Let $f,g,f',g'$ be such polynomials that $\deg f\le\deg g$ and $\deg f'\le\deg g'$. Express $\frac{f}{g}+\frac{f'}{g'}$ as one fraction. What can you say about the degrees of the numerator and the denominator?

the 1.5 Apostol problem says that f=0 is possible. by the way why do you reference the derivatives of f, g?
by the product rule if both f and g are functions of x, the maximum value of the numerator will be 2(deg g)-1 and also the denominator will have 2(deg g)-1 giving a degree of 0? but the problem doesn't say that f and g are both functions of x, maybe you have g(x), f(y)...
while in the case that f=0 then the rational function will equal 0 and have a degree of 0...

Originally Posted by mathnerd15

by the way why do you reference the derivatives of f, g?

I don't.

Originally Posted by mathnerd15

by the product rule if both f and g are functions of x, the maximum value of the numerator will be 2(deg g)-1

Numerator of what?

Originally Posted by mathnerd15

but the problem doesn't say that f and g are both functions of x, maybe you have g(x), f(y)...

The problem deals with rational functions on reals, probably one-argument functions.

oh the numerator of the addition of f/g and f'/g' wil be fg'+f'g. do you mean that f' is the derivative of f or deg of f
anyway thanks a lot

In post #4, I wrote:

Originally Posted by emakarov

Let $f,g,f',g'$ be such polynomials that $\deg f\le\deg g$ and $\deg f'\le\deg g'$.

Usually when we say, "Let $x$, $y$ be sets (or matrices, or numbers, or other mathematical objects) such that $P(x,y)$", it means that we consider a situation where two independent objects $x$ and $y$ are given subject to the restriction $P(x,y)$. Such language is not used when $y$ is completely determined by $x$. In the case where $y$ is a function of $x$, e.g., $y=f(x)$, we say, "Let $x$ be a set (or some other object) such that $P(x,f(x))$ holds". The derivative of a polynomial is completely determined by that polynomial, so saying "Let $f$, $f'$ be polynomials" intending for $f'$ to determine the derivative of $f$ is like saying, "Consider two polynomials. In fact, the second polynomial is not arbitrary; it is the derivative of the first one". This is a bad style. Therefore, I was considering four polynomials $f,g,f',g'$ subject to restrictions that $\deg f\le\deg g$ and $\deg f'\le\deg g'$. However, to avoid confusion, let's rename them to $f_1$, $g_1$, $f_2$ and $g_2$.

Originally Posted by mathnerd15

oh the numerator of the addition of f/g and f'/g' wil be fg'+f'g.

Yes. So,
\[
\frac{f_1}{g_1}+\frac{f_2}{g_2}=\frac{f_1g_2+f_2g_ 1}{g_1g_2}.
\]
Let us define
\begin{align*}
m_1&=\deg f_1\\
m_2&=\deg f_2\\
n_1&=\deg g_1\\
n_2&=\deg g_2
\end{align*}
Can we express $\deg(f_1g_2+f_2g_1)$ through $m_i$, $n_i$ in all cases? If not, what is an upper bound on $\deg(f_1g_2+f_2g_1)$? Can we express $\deg(g_1g_2)$ through $n_1,n_2$? How do these two numbers compare with each other?

thanks so much! I'm sorry it seems like an easy problem
the upperbound of the numerator would have degree n1+n2 and in one case the resulting fraction has degree num<degree denominator
for this case the roots are also symmetrically placed on a circle in the complex plane as you know.
1. in the case that m1=n1, m2=n2 then deg num=deg denom
2. in the case that m1<n1, m2<n2, then (m1+n2) or (m2+n1)<(n1+n2) so this seems to prove closure under addition with the resultant polynomial in the set..
3. for the case of scalar multiplication, it doesn't affect the deg f, deg g so all af/g are also in the set and this is satisfied
by the way do you mean it's bad style if f and f' are dependent on one another or share the same domain?

so for 1.5.5 this isn't a linear subspace since h(0)=f(0)+g(0)=2+h(1), which is not closed under addition whereas
1.5.3 f(0)+g(0)=h(0)=f(1)+g(1)=h(1) and
ah(0)=af(0)+ag(0)=a(f(1)+g(1))=a(h(1)), and the only requirement is that the set of functions satisfy this requirement for values 0, 1. so by closure under add/mul this is a linear subspace since all of the 10 axioms can be reduced to these two

Originally Posted by mathnerd15

the upperbound of the numerator would have degree n1+n2

"Of the degree of the numerator". Yes, $n_1+n_2$ is an upper bound. Since $\deg(g_1g_2)=n_1+n_2$, it follows that $\deg(f_1g_2+f_2g_1)\le\deg(g_1g_2)$, so $f_1/g_1+f_2/g_2$ again has the property that the degree of the numerator does not exceed the degree of the denominator.

It is possible to write a more accurate estimate for the degree of the numerator, but this is not necessary for this problem. We have $\deg(f_1g_2)=\deg(f_1)\deg(g_2)=m_1+n_2$ and similarly $\deg(f_2g_1)=m_2+n_1$. Therefore,
\[
\deg(f_1g_2+f_2g_1)\le\max(m_1+n_2,m_2+n_1).
\]

Originally Posted by mathnerd15

by the way do you mean it's bad style if f and f' are dependent on one another or share the same domain?

I did not say anything about sharing the same domain. And if $f'=\frac{df}{dx}$, there is nothing bad about this, of course. I said it is bad style to write, "Let $f,f'$ be some polynomials" when in fact $f'$ is not some polynomial, but a very specific one, namely, the derivative of $f$. In such case, one should write, "Let $f$ be a polynomial..."; then one can refer to both $f$ and $f'$.

Originally Posted by mathnerd15

so for 1.5.5 this isn't a linear subspace since h(0)=f(0)+g(0)=2+h(1), which is not closed under addition whereas
1.5.3 f(0)+g(0)=h(0)=f(1)+g(1)=h(1) and
ah(0)=af(0)+ag(0)=a(f(1)+g(1))=a(h(1)), and the only requirement is that the set of functions satisfy this requirement for values 0, 1. so by closure under add/mul this is a linear subspace since all of the 10 axioms can be reduced to these two

Am I supposed to know what 1.5.3, 1.5.5 and $h(x)$ are?

thanks so much!!! I guess I just should focus on analyzing the problem and it would be easier. maybe isn't the numerator going to be just the max of (m2+n1), (m1+n2) since you add polynomials?
anyways one person posted on a forum that it's better to learn analysis instead of study calculus again with Apostol

Originally Posted by mathnerd15

isn't the numerator going to be just the max of (m2+n1), (m1+n2) since you add polynomials?

If $m_1+n_2=m_2+n_1$ and the leading terms of $f_1g_2$ and $f_2g_1$ cancel out, then $\deg(f_1g_2+f_2g_1)<\max(m_1+n_2,m_2+n_1)$. For example, the degree of the numerator of
\[
\frac{x+1}{x+2}-\frac{x+3}{x+4}
\]
is 0 and not 2.