So you have the function f(x)=\frac{x}{3x+1} and you know that a function is differentiable at a if its derivative exists at a. You also know that
\left.\frac{df}{dx}\right|_{a}\equiv \lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}=\frac{\frac{a+h}{3(a+h)+1}-\frac{a}{3a+1}}{h}
If you simplify...
You can think of it like an infinitesimal form of the Euclidean distance formula. For a function f(t)=\langle x_1(t),x_2(t),x_3(t),\ldots\rangle
\sum_a^b \sqrt { \Delta x_1^2 + \Delta x_2^2+\Delta x_3^2+\ldots } \longrightarrow s=\int_{a}^{b} \sqrt { dx_1^2 + dx_2^2+dx_3^2+\ldots} =...
I figured it out now using a Do[] command, but my point was that if you want to multiply many matrices and not write out the long stretch of A1.A2.A3.A4....AN, then you cannot use the product command on Mathematica because that will just multiply the matrices element-wise. I wanted to know if...
I am trying to compute the following,
\prod_{j=0}^{N-1}\left[\hat{I}+\hat{M(j)}\left(\frac{T}{N}\right)\right]
where \hat{I}, \hat{M(j)} are matrices. My problem is that Mathematica interprets this product as element-wise with respect to the matrices, but I of course want it to use matrix...
If you go by the fact that the order of any alternating group is n!/2 then you would have that the order of A_2 is 2!/2=1 and therefore it's just the trivial group consisting of the identity element. Anything in the form of (wx) would be an odd permutation and therefore not in A_2
Yes since the element of length is (assuming two dimensions)
ds=\sqrt{dx^2+dy^2}
If you integrate you get the length of the curve s from [a,b]
s=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx
If you mean numerically finding the value of a root you can use Newton's Method; say you want to find the square root of two numerically, then whatever that number is can be treated as a variable x as such...
x=\sqrt{2} \Rightarrow x^2=2
and by Newtons method we use the recursive formula...
If it has a 45 degree angle and a right angle then both the adjacent side and the opposite side will be equal so the opposite length (height) will be 50 as well. You could still do it mathematically though...
\tan\theta=\frac{\text{opp}}{\text{adj}} \Rightarrow \tan45=1=\frac{\text{opp}}{50}...
There are some parts of linear algebra where you need calculus--for example taking the norm of a vector in L2 you need to know integration which you learn for the most part in calculus 2--but for the most part I think you can learn linear algebra without a previous knowledge of calculus.