Math Problems

Divergent right?

Divergent right?

Yeah and if p>1 then it converges

How about the extra credit

I remember something about a comparison test, but I don't know if that is the original or the offshoot, or even the right one for that matter.

I remember something about a comparison test, but I don't know if that is the original or the offshoot, or even the right one for that matter.

No it is the P-Series test which comes from the Integral test the comparison test is when you compare the series to another series that you already know if it converges or diverges

It's been a while.  Did you do the balloon one?  I'm curious if I was close or not.

It's been a while.  Did you do the balloon one?  I'm curious if I was close or not.

No but is should just be the derivative of the volume.

I did a derivative of radius with respect to volume.

you should end up with (1/4pi)(3)= dr/dt, so what ever the decimal equivalent of 3/4pi is should be your final answer

I could be wrong but I don't think that is right.  How does differentiating volume give you rate of change of radius?

I could be wrong but I don't think that is right.  How does differentiating volume give you rate of change of radius?

you need to differentiate the Volume of a Sphere because you need to relate the radius and the volume. We already know dV/dt so we needed to find dr/dt when r equaled 12

I just realized I did not put the radius on there so it shoul be the decimal equivalent of 3/4pi(12)^2

340cm/min sounds a bit high.

Quote from: Mr. Ireland on January 19, 2009, 07:09:36 PM

Yea, 2+2/2 = 3.

That is not what I said take your scientific calculator a type it just like this 2 + 2 / 2 =
You wrote this: the sum of 2+2 divided by 2

Yea, 2+2/2 = 3.

IT'S THREEEEEE.  I know it's not (2+2)/2.  It's 2+(2/2) WHICH IS EQUAL TO THREE.

I am retarded you are right I dont know how I got messed up there but you are right

340cm/min sounds a bit high.

it should be something like 0.0017cm/min

I am retarded you are right I dont know how I got messed up there but you are right

It's been so long, I wasn't completely sure either.

Nerds.

Also, I'd join in if I wasn't so tired from work right now.

Nerds.

Also, I'd join in if I wasn't so tired from work right now.

I'd join in if I didnt suck at maths.

e^iπ+x=Σ_mⁿ₌₁sin(^mπ/_n)
Solve x for an integer.


Edit: readability

e^iπ+x=Σ_i=1->nsin(iπ/n)
Solve x for an integer.

I could do that, I just dont want to 

Quote from: Mr. Davis on January 20, 2009, 05:05:54 AM

e^iπ+x=Σ_m=1->nsin(mπ/n)
Solve x for an integer.

I could do that, I just dont want to 

Should be able to look at it and know the answer .  Its a celebrity.

Quote from: Robosteve on January 20, 2009, 04:12:01 AM

Nerds.

Also, I'd join in if I wasn't so tired from work right now.

I'd join in if I didnt suck at maths.

Ditto.

God I hate math.

QFT. I just picked up Calculus again.

e^iπ+x=Σ_mⁿ₌₁sin(^mπ/_n)
Solve x for an integer.


Edit: readability

0?

No, but good guess.

e^iπ+x=Σ_mⁿ₌₁sin(^mπ/_n)
Solve x for an integer.


Edit: readability

I'm going to guess 1, but I don't think it's right... it looks like it would have to depend on n, since the RHS looks like it would be larger for larger values of n. Unless it was the sum from 0 to 2n, or the function used was a cosine.

Quote from: Mr. Ireland on January 19, 2009, 07:09:36 PM

Quote from: cbarnett97 on January 19, 2009, 06:50:42 PM

Quote from: Mr. Ireland on January 19, 2009, 06:45:42 PM

Quote from: cbarnett97 on January 19, 2009, 06:41:53 PM

What is two plus two divided by two

3

put it in a scientific calculator exactly as it reads

Yea, 2+2/2 = 3.

That is not what I said take your scientific calculator a type it just like this 2 + 2 / 2 =
You wrote this: the sum of 2+2 divided by 2

The initial post said nothing about scientific calculators.  The answer to the original question is indeed 3 because when a math statement is written out multiplication and division are done first and addition and subtraction second.  So 2+2/2=3 and Mr Ireland was correct.

I'm going to guess 1, but I don't think it's right... it looks like it would have to depend on n, since the RHS looks like it would be larger for larger values of n. Unless it was the sum from 0 to 2n, or the function used was a cosine.

I was leaning towards 1 as well.
I remember that e^iπ = -1
MN is less than N because the opposite of a triangle is less than the hypotenuse. (M < 1)
Σm^n approaches 0 when m<1, as long as n>1.
-1 + x = 0 simplifies to x = 1

e^iπ+x=Σ_mⁿ₌₁sin(^mπ/_n)
Solve x for an integer.

What's the point of the 1 anyways?

Quote from: Robosteve on January 20, 2009, 08:49:22 AM

I'm going to guess 1, but I don't think it's right... it looks like it would have to depend on n, since the RHS looks like it would be larger for larger values of n. Unless it was the sum from 0 to 2n, or the function used was a cosine.

I was leaning towards 1 as well.
I remember that e^iπ = -1
MN is less than N because the opposite of a triangle is less than the hypotenuse. (M < 1)
Σm^n approaches 0 when m<1, as long as n>1.
-1 + x = 0 simplifies to x = 1
....of course I made a few assumptions to get there.

Quote from: Mr. Davis on January 20, 2009, 05:05:54 AM

e^iπ+x=Σ_mⁿ₌₁sin(^mπ/_n)
Solve x for an integer.

What's the point of the 1 anyways?

It looks like series notation maybe so the one should be the lower limit and if n is supposed to be the upper limit then I guess it should be infinity then

Yes, the answer is 1.  The right side of the equation is a Fourier series that sums to 0.  Ignoring if you know what that is, you should be able to drudge through that and figure it out.

That leaves us with eulers identity, which makes x=1.

In retrospect maybe this was a silly little problem.

Quote from: Robosteve on January 20, 2009, 08:49:22 AM

I'm going to guess 1, but I don't think it's right... it looks like it would have to depend on n, since the RHS looks like it would be larger for larger values of n. Unless it was the sum from 0 to 2n, or the function used was a cosine.

I was leaning towards 1 as well.
I remember that e^iπ = -1
MN is less than N because the opposite of a triangle is less than the hypotenuse. (M < 1)
Σm^n approaches 0 when m<1, as long as n>1.
-1 + x = 0 simplifies to x = 1
....of course I made a few assumptions to get there.

Quote from: Mr. Davis on January 20, 2009, 05:05:54 AM

e^iπ+x=Σ_mⁿ₌₁sin(^mπ/_n)
Solve x for an integer.

What's the point of the 1 anyways?

I get used to explicitly setting the lower limit to 1 because I often set it to 0 due to it being used as such in computer science so often.