Recent content by iloveannaw

Homework Statement Hi, we're supposed to put the following integral into a power series: \int \frac{arctan(t)}{t} dt with 0 < t < x. Homework Equations n/a The Attempt at a Solution I just want to know whether this step is ok. \int \frac{arctan(t)}{t} dt = \int \frac{1}{t} \int...

thanks, so you think I should start by working from \frac{dv}{dt} = -g -\frac{\gamma}{m}v ? I have already done that, but the question is quite clear it asks for the diff. eq. in terms of \dot{z}(t) and then asks the student to make substitution. And it also asks for type of differential...

Homework Statement A point mass m falls from rest through a height h. The frictional force is given by -\gamma \dot{z} and gravity by -mg. Give the 'equation of motion' (differential equation) for the height z(t).The Attempt at a Solution \ddot{z} = -g - \frac{\gamma}{m} \dot{z} I thought...

yes, that's why I'm so pleased - it just needed a bit of factoring after doing it your way. cheers

thankyou!

Homework Statement the title says it all x \rightarrow 4 for f\left(x\right) = \frac{\sqrt{1+2x}-3}{\sqrt{x}-2} I have multiplied both top and bottom by conjugate, \sqrt{x}+2: f\left(x\right) = \frac{\sqrt{x(1+2x)}+2\sqrt{1+2x} -3\sqrt(x)-6}{x-4}but don't know how to take this further...

Hi, thanks for the reply :smile: Looking more at the given function i can show that it is indeed one-to-one. As to whether it is surjective I'm not really sure what my answer is telling me: f\left(x\right) = \frac{ax+b}{cx+d} = \xi ax+b = cx\xi+d\xi x = \frac{d\xi - b}{a - c\xi}...

ah, i just found out about the bijective requirements for inverse functions – sorry we didn't cover this in class. but I'm a bit confused injection and surjection seem quite the same. Isn't it enough to show that the function is one-to-one??

Homework Statement Given the function f(x) = \frac{ax+b}{cx+d} where f: \mathbb{R} \backslash \left\{ \frac{-d}{c} \right\} \rightarrow \mathbb{R} show that f is either a constant or has an inverse function. I can see why this would be true. If a function takes all real numbers and returns...

thanks for clearing that up :) ok, so the four dimensions of K4 refer only to the space and our matrix A does not necessarily have the same number of dimensions as the resulting vector space ? thanks again

K3 and k4 are vector spaces with bases: K^{3} = \left\langle \vec{e}_{1} , \vec{e}_{2} , \vec{e}_{3} \right\rangle K^{4} = \left\langle \vec{e}^{*}_{1} , \vec{e}^{*}_{2} , \vec{e}^{*}_{3} , \vec{e}^{*}_{4} \right \rangle the elements within a basis are indeed vectors, therefore K3 and k4...

Homework Statement f: K^{3} \rightarrow K^{4} is a linear transformation of vector spaces: K^{3} = \left\langle \vec{e}_{1}, \vec{e}_{2}, \vec{e}_{3} \right\rangle and K^{4} = \left\langle \vec{e}^{*}_{1}, \vec{e}^{*}_{2}, \vec{e}^{*}_{3}, \vec{e}^{*}_{4} \right \rangle as well...

hi guys (gals?) thanks for all the help. the question as is is not solvable. just got an email from the prof saying that the angle between the beam and the horizontal should be 30°, which wasn't stated in the original question.

thanks. I guess you mean that it is static equilibrium. still don't see how this helps in this situation. Taking the x-direction to be parallel to the beam and the y-direction to be perpendicular to this: \Sigma F_{y} = 0 = Fsin \alpha - W_{B}cos \theta - Wcos \theta cos \theta = \frac{Fsin...

Homework Statement Please see attachment. A uniform massive beam attached to a wall at one end and supported by a cable (also attached to wall). On the free end of the beam is a weight. We are supposed to find the tension in the cable. Homework Equations sum of the torques about...