Analysis and Calculus

what is it? i know hyperbolic geometry is a type of non-euclidean geometry; does it have anythign to do with that? if so, is there a trig for sphereical geometry?

Does anyone know a good book for lay people on tensor calculus? I have tried reading a grad book on it and it was focused primarily on the algebra of tensors...rather disappointing really. Or, on the other hand, does anyone know tensor analysis? I have tried googling it several times.

questions is find the interval of increasing and decreasing H(x) = (x^2-1)^3 H'(x) = 6x(x^2 - 1)^2 is my crital numbers x = 0, x= sq root 1, x= sq root of -1 when i am plugging them in numbers i plug them in to the deritive example -2 it = a negitive number decreasing ,-.5 it is decreasing, .5 increasing , 2 increasing them for concave up or down i do the same thing and get H"(x) = 12x(x^2-1) are these concepts correct? thanks joe

Hi, Wondering if someone could point me in the right direction for this question. It is an assignment question, so I don't want the answer, just some help and point in the right direction. Heres what I have so far: Find indefinite integral of: h(u) = sin^2 ( 1/6 u ) Now I'm presuming I need to use the double angle formula cos(2x) = 1 - 2sin^2 x to which I have: sin^2 x = 1/2 (1 - cos (2x)) sin^2 (1/6 u ) = 1/2 (1 - cos (2 (1/6 u)) Therefore I get the answer integral sin^2(1/6u) = integral (1/2 - 1/2 cos (2 (1/6 u ))) =1/2x - 3/2 sin (1/3 u ) + c Is that anywhere close?

I haven't taken calculus yet, but this sounds like a calculus problem (maybe precalculus or algebra 2) I want to know any techniques to solving this problem. Suppose there are variables A, B, C, D, E, F, G, and H, where A, B, and C are less than 25. .3A + .8B = D .2A + .1 B + .2C +.4F= E (A+B+C)/3=F G=D/F H=E/F Find values A, B, and C that maximize G + H.

So, can anyone solve for y in terms of x? yx=xy I sure can't solve it, and my TI-89 couldn't solve it either.

hello: here is my problem: if a function f(x)is continuous in [0,1],and its scope is also [0,1],then there must be one point e to make f(e)=e if i use the theorem a continuous function can have any return value between upper and lower bound,then i can have e belong to [0,1],and there will be a vulue a to make f(a)(a also belong to the same definition scope[0,1])to have a value e,but how can i promise the there can must be a chance in which a=e any comment is appreciated.

Hello Can any body please give me a good site giveing me the basics of learning and understanging caculas.

hey could someone please give me a bit of help with this big O notation bussiness, it makes not much sense to me, (mainly in terms of calucating limit sof indeterminate form, and how the big O's cancel etc.) anyways any help or pointers to a good website that explains the stuff would be great! Thanks Guys and Gals Sarah

I'm trying to find some general things about orbits in gravitational systems in multiple dimensions. I've already found that it has to all be in one plane (easy enough), and now I'm assuming the following: [math]F = \frac{g*m_1*m_2}{d^{#dim-1}}[/math] [math]F_{x} = F*\frac{x}{d}[/math] [math]F_{y} = F*\frac{y}{d}[/math] So: [math]\frac{d^{2}y}{y*dt^2}=\frac{d^{2}x}{x*dt^2}[/math] [math]Let |x| = e^{f(t)}[/math] [math]|y|=e^{g(t)}[/math] [math]\frac{dx}{dt}=f'(t)e^{f(t)}[/math] [math]\frac{d^{2}x}{dt^{2}}=f''(t)e^{f(t)}+f'(t)^{2}e^{f(t)}[/math] [math]\frac{dy}{dt}=g'(t)e^{g(t)}[/math] [math]\frac{d^{2}y}{dt^{2}}=g''(t)e^{g(t)}+g'(t)^{2}e^{g(t)}[/math] So: [mat…

i don't see why L'Hop's rule doesnt work by just straight forward application in this limit? the solution i have subs in a value so that the limit has m->0 ? Thanks guys, sorry if this seems like another inane question

Ok, the instructions are to find the area of the region enclosed by the curves... x=y^3-4y^2 +3y and x=y^2-y the coordinates aren't given, but I have worked the y coordinated out to be 0, 1, and 3. This problem throws me off because of its in x in terms of y, instead of vice versa. Can anyone help me out as to how to approach graphing it. Thanks

alright so I have a textbook on quantum mechanics that I'm going through and it comes to the schrodinger equation and tells me to use a wavefunction equal to [math]f(x)exp(-iEt/hbar)[/math] I insert that into the schrodinger equation wich is i* hbar *(partial derivative of the wavefunction with respect to t) = -hbar^2 /2m *(the second partial derivative of the wave function with respect to x) it said that this simplified to P.S. sorry about latex Ill play with it and see if I can figure out how to work it d^2 f(x)/dx^2 + (k^2)f(x)=0 this simplification shouldn't work this way because the wave function is of the form X(x)Y(y) right?

Can someone please show me the exact steps to finding the F(x) of: y = {sec x tan x dx

I'm not sure what different methods there are out there to compute the following series: [math] \sum_{n=0}^{n=\infty} \frac{(-1)^n}{(2n+1)^3} [/math]

In another thread, there was a stage where the following derivative was to be taken: [math] \frac{d}{dn} log (n) = \frac{1}{n} [/math] The problem was that n was restricted to the natural number system, not the real number system. But, if there is an error in the differential calculus, or with the limit concept in general, then perhaps there was no error. In this thread, I simply want to investigate whether or not you can take the 'derivative' above. Let f(x) denote an arbitrary function of the variable x. The difference operator is defined as follows: [math] \Delta f(x) \equiv f(x+h) - f(x) [/math] h is called the step size. In the case…

can someone check my solution to the limit as n goes to infinity of n!^(1/n) L=n!^(1/n) Ln(L)=1/nSUM(Ln(n-i),i,0,n-1) use lhoptais rule Ln(L)=Sum(1/(n-i),i,0,n-1) Ln(L)=Sum(1/(1-i/n)*1/n,i,0,n-1) definiton of intergral LnL=int(1/(1-x),0,1) LnL=-Ln(1-x) from 0-1 LnL=Infinity-0 L=infinity

Ive got the question S x^2cos(x) dx But by using S u(x)v'(x)= u(x)v(x)-S u'(x)v(x) dx So using v(x)= Sin(x) and u(x)= x^2 Im left with x^2Sin(x)- S 2xsin(x) dx with the end part being a product also can someone help me?

Someone tell me if my math is correct? [math] n^n[/math] acts like [math]n!e^n[/math], as follows: [math] \frac{(n+1)^{n+1}}{n^n} = (n+1)\frac{(n+1)^n}{n^n}[/math] [math]=(n+1)\frac{n^n+n*n^{n-1}+\frac{n(n-1)}{2}*n^{n-2}+...}{n^n}[/math] [math]=(n+1)\frac{n^n*(1+1+\frac{1}{2}+...)+n^{n-1}*(-\frac{1}{2}-\frac{3}{6}-\frac{6}{24}-\frac{10}{120}-...)}{n^n}[/math] The last terms can be dropped out because they contain a [math]\frac{1}{n}[/math] term. [math]=(n+1)\frac{n^n*e}{n^n}=e*(n+1)[/math] so after a while, the two act similarly. In this case, similarly means that if you take the ln of both and divide them, the quotient will approach one, as shown below: n [ma…

An extra-credit question at the end of my calc final today: What is the opposite of tree? This doesn't require a math based answer. The best I could do was log(tree) Just wondering what you guys can come up with?

Im having some difficulty in grasping how to do the cover up rule with 3 factors as the denominator, Like (x-2)/(x+1)(x+2)(x+4) For the x+1 factor would i put -2 in for the (x+4), -4 in for the (x-2) factor and that would solve for the numerator, how do i do it, can someone give me a good explantaion on how to do it Cheers

Differentiating a distance or displacement time graph will give the velocity at the point considered which i am sure everybody is aware, differentiating again will give the acceleration of the said object. These facts are fine and dandy but what if i was to start with a whole bunch of data for both distance and time co-ordinates eg: (1,2.6) , (1.26, 3) , (2.57, 3.32) and so on. How do i go about turning a series of co-ordinates into an equation so that i can differentiate and find the velocity at any point? (not considering straight lines lol). Its probably more practical to take the tangent at the point considered but i know this can be inaccurate.

I'm afraid i am not seeing something help would be apreciated. you toss a stone upwards off a 63m cliff at 8 meters per second how long til it hits the ground? here is what i have [math]a=-9.8\frac{m}{s}[/math] then the velocity using antiderivatives is [math]v=-9.8t=8[/math] and distance [math]s=-4.9t^2+8t+63[/math] then if [math]s=0[/math] [math]t=4.49sec[/math] my question is that if you throw a stone upwards off a cliff then wouldn't the C value in the s formula be higher than that? what am i missing?

hey peoples , if anyone is willing could the please give me few hints for this question.... i think i have done part a) i get for part a) ~5.271 units but i have no idea how to "use" the extra information given in part b) ... anyways aghh been tring for hours and hours, lol , you know the feeling Sarah

Hey can i ask if my answers to this question are correct...? a) yes, because f'(x) = 0 at x = 0 b) yes, because there exists a tangent line at x = 0 (because of part a) and the concavity is of a different sign on both sides of x = 0 (i.e. f''(x) is -ve one left and +ve on right) c) no it is not true, because f''(x) = 2 for all real x in the domain d) no, as can been seen in this question Sarah