Let the nth derivative of f(x) equal a non-zero constant c.
Knowing n and c, what if anything can be said about f?
Hint: apply the Fundamental Theorem of Calculus.
Hardmode: Only knowing c, what can be said about f?
n+1th derivative = 0
>>7746024
X is an nth order polynomial
>>7746050
f(x)*
>>7746050
Is that all?
>>7746024
F is an nth degree polynomial.
F is asymptotic to (cx^n)/(n)! as x->infinity.
F has n roots over the reals.
F is continuous, differentiable, analytic, etc.
F lives in the vector space of kth degree polynomials with k greater than or equal to n.
>>7746086
Over the complex numbers I mean
>>7746086
>the vector space of kth degree polynomials with k greater than or equal to n.
Not a vector space. You probably meant "k LESS than or equal to n".
a_1 is c/(n!)
f(x) is not a periodic function
>>7746036
The naturalized, as to say most invasive, rhythm of f is that of a condition of x=1/[{even/odd}] and that the construct of the .0 in 0.0{even/odd} resulting from the 1 is that of a polynomial set that determines the attributes of a Fibonacci or spiral like endowment through the course of the nth derivative though the focus of c making the factor of f(x) that of a irrational within each numerator of an irrational, as persay polynomial set
>>7746097
No it'd live in the vector space of polynomials of degree greater than or equal to n, over the complex numbers.
>>7746234
>the vector space of polynomials of degree greater than or equal to n
This is not a vector space. The sum of two polynomials can have degree less than n if the leading terms cancel out.
>>7746256
And then the coordinate with respect to the element in the natural basis (1, x, x^2,...,x^k) of the leading term (x^k in this case) would be zero. This is perfectly fine.
Just as a vector in R^3 can be represented as ai+bj+ck. One of these coefficients could be 0 and then the vector would read something like ai+ck.
This is still a vector in R^3 where i,j, and k are the standard basis in R^3.
Get what I'm sayin friend?
There is also an obvious isomorphism between the vector space of kth degree polynomials (over the reals in this case) and R^3, check it yourself.
>>7746067
Yes, you lose information as you take derivatives. You always try to make up for this when integrating with a +C in single variable, and a function in multivariable.
>>7746316
no it's not perfectly fine...
if P1 = x^n
P2 = 1-x^n
P1 and P2 are in your "space", but P1+P2 is 1, which is not a polynomial of degree greater than or equal to n.
>>7746368
Ah I see where the misunderstanding is coming from, if you look back to my original post I said
>F lives in the vector space of kth degree polynomials with k greater than or equal to n
I should have said
>F lives in a vector space of kth degree polynomials with k greater than or equal to n
I was meaning to say, that f exists as a vector in a vector space of polynomials and since it has degree n, the vector space has to be generated by a basis (1,x,x^2,...,x^n,...,x^k) where k is greater than n.
I didn't mean that the vector space only contains polynomials with degree greater than or equal to n.
My bad.
>>7746388
And obviously any basis will do but the one I listed is just the natural choice.
>>7746067
Cant you also say that the coefficient of X is c*n! ?
>>7746097
>not aware that polynomials of degree k form a vector space
Brush up on your intro lin alg anon. This is first year stuff.
