Recent content by KMjuniormint5

Looking at it at a different angle take w = 0.5*A - I and find the characteristic equation which is λ2 -λ = 0. We see that λ2 = λ. Plug in w for λ Now for w3 we can do w2*w but that is w*w and that is w. So for w100=w99*w= ... = w*w = w. In the end we see that w = C = [0.5 -0.5; 0.5 -0.5]...

Homework Statement Given the following matrix A = [3 -1; -1 3] Find C = (0.5*A - I)100 Homework Equations Using the knowledge that the Cayley - Hamilton Theorem must satisfy its own characteristic polynomial. The Attempt at a Solution Here the characteristic polynomial is λ2 -...

after going back and doing it the correct way (substitution) I get something of: _{}a_n=\frac{}{}2/\pi*(\frac{}{}1/n*sin^2(x)*sin(nx)+\frac{}{}1/n^2sin(2x)*cos(nx)+\frac{}{}2/n^2\int cos(2x)*cos(nx)dx Using SammyS cos(2x)= 1 - 2sin^2(x) you can see that you get the original function of...

an online integral calculator (numberempire.com) but i forgot the negative sign out front

when I intergrate I get 1/4\pi*(cos(nx)*sin(2x)-2*x*cos(nx)) evaluated over 0 to 2\pi which is still zero

so in this case would you just plug cos(2x) into an to get it to be: an = (1/(∏))*∫(sin^2(x)*(1-2*sin^2(x/2))dx

Homework Statement Find the Fourier series of f(x) = sin^2(x) Homework Equations bn = because f(x) is even ao = (1/(2*∏))*∫(f(x)) (from 0 to 2*∏) an = (1/(∏))*∫(f(x)*cos(x)) (from 0 to 2*∏) The Attempt at a Solution ao = (1/(2*∏))*∫(f(x)) (from 0 to 2*∏) = ao = 1/2 an =...

ok i understand where the confusion comes into play as i was helping someone with physics on here. . . .thanks again for the help and getting me totally frustrated but in the end. . .i understand now where I need to go with problems like this. . .thanks so much kudos to ya

ok so my original thought of 1.45% was within the ballpark of an answer?

ok you that's what started to do but for some odd reason. . .im stupid: Mass(Li)+Mass(Al)=Mass density(Li)*volume(Li)+density(Al)*volume(Al)=(density)*(volume) 0.534*V(Li)+2.7*(2-V(Li))=2.55*2 solving for Volume one gets .13876 mL and to get mass (.534*.13876)=0.0741 grams

[0.534(V(Li))/0.534]+[2.7*(2-V(Li))/2.7]=[5.1/2.55] but in this case doesn't everything cancel?

mass Al = 2.7V(Al) mass Li = .543V(Li) V(Al)=2-V(Li) V(Li)=2-V(Al) putting the 4 equations together if the form of: Volume(Li)+Volume(Al)=2 mL Mass(Li)/Density(Li)+Mass(Al)/Density(Al)=Mass/Density 0.534(2-V(Al))/0.534+2.7*(2-V(Li))/2.7=5.1/2.55 **This is where I think go wrong**

density = mass/volume saying that 1mL of each 2mL total mass of Al = 2.7*1= 2.7 grams mass of Li =0.534*1= 0.534 grams mass of the entire system = (2.55)*2 = 5.1 grams putting that into the equation: Mass(Li)*Volume(Li)+Mass(Al)*Volume(Al)=5.1 solving gets to about 1.45 percent weight of Li?

V(Al)+V(Li)=V(total) V(Al)+V(Li)=5 mL (where 5 is just an arbitrary number) solve for each: V(Al)=5-V(Li) V(Li)=5-V(Al)

where are u setting your zero? the top of the building or the bottom?