Let's discuss Mathematics

I read your first post more closely
There are two problems:
You have forgotten the multiplication rule (though you use it correctly there) and
You derivated tanx wrong
Again, I'll show you later though

 

Okay, this is true. I think I remember the multiplication rule now somehow but the first part of the calculation is to derivate x which is 1 and then it is 1 times the tan part. So basicly in the first operation we didn't have to derivate the tan part but the counting goes right if I put the D of the inner function there even though we didn't derivate the outer function. Is that how it is supposed to work?

Also I cannot see how I derivated the tan2x wrong.

And I don't know what you mean by later but I have to add that this is for practicing to a math test which is tomorrow so it would help a lot more if you could check it out today. Helping is voluntary of course but I still wanted to tell this fact

 

No

You're not supposed to derivate the inner function if the outer isn't.

Derivative of tanx is 1/(cosx)^2

I'll show you the whole thing in I hope within 2 hours. Maybe it'll be 3 but I'll try to avoid that

 

Thanks for the help.

I don't know why the counting goes right if I do the wrong thing (this is just a part of one big counting though). Maybe we'll find out in a few hours

However, I am quite sure that tan(x) can be also derivated to 1+tan(x)^2 . I have used it successfully in other calculations...

 

It's not that you do the wrong thing, it's that you doubt yourself

But a mistake on my part, 1/(cosx)^2 ans 1+(tanx)^2 are identities, ,so they're both right.

But I on my way home now, so I'm a little over an hour (I promise) I'll show you why the first one is the more obvious solution (basically I'll write it on a piece of paper and scan it )

 

I shure hope this is understandable

Spoiler :

 

Thanks a lot. Now I will try the calculation again using those tips. I got the main points but your writing is quite... special .

I understand now that it was all good exept that I added the ' of inner function when it was not needed.

I will post you back here when I have tried that. Thank you really much, you are really helpful. I can't believe how someone can help a weirdo like me, lol

 

Hey, thanks a lot for your time and effort

The calculation worked when I removed the D of the inner function (the one that was not needed). The answer happened to be the same when I did it the wrong way or the right way so that is why I was so confused.

Cheers!

 

Now I shall ask for help

an integral
the integrand is:
(x-x0) * e^(-mw/h * (x-x0)^2) * (1+sqrt(2mw/h)*(x-x0) + 2mw/h * (x-x0)^2)
x is the variable, everything else is constants
the limits are from minus to plus infinite

how does one solve this (by hand)?

 

(disclaimer: non-helpful post)

So with the integrand being (x-x0)^2*[(e^stuff)*1+sqrt(other stuff)+2mw/h]?

I hate west calculus, not going to help here (can't), but after the stuff e is raised in are turned to log, and assuming x0 is one of x's variations, can't you do the antiderivative stuff using the fundamental theorem of west calculus?

I haven't yet felt like approaching that again (disliked it in the end of highschool as well)

 

here is a more clear representation


this doesn't look like the fundamental theorem can be used easily

 

1. Get rid of the parentheses inside the integral.
2. You have then an integral of sum of three things. You can integrate the things separately.
3. For the first one the multiplier of e^stuff is the derivative of [stuff], almost, so you know how to integrate it.
4. For the others this doesn't work. Use partial integration instead to get the power of multiplier what you want. You may have to do that many times.
5. First though, substitute x for x-x_0. It makes calculations easier and doesn't require any extra hassle, since you're integrating from -inf to inf.

 

yeah introduce a new variable t = x-x0 and for the sake of simplicity, put C = mw/h since it's repetitive. If you have issues, I can write it on paper and scan it, but it won't be until tonight

 

Does anyone know if the notation L^2(G) is?
Its coming up with regards to Class functions and I'm not sure if its just a different notation for Class(G) or if it means something else

 

I'm not sure what you mean by class, but if G is a set, that means the set of measurable functions whose square has finite integral.

For example (suppose G=[0,1]):

f is in L^2 ([0,1])
if and only if
f : [0,1] -> R is measurable and \int f^2 < infinity.

 

A football player scores when performing a penalty kick with the probability of 80% (0.8). How many times he has to try until he will get a goal with probability of 99% (0.99) or more?

Answer:

Spoiler :

At least 3 times

I don't get it... I tried to use log but that went wrong. It just doesn't work to power 0.8 (like 0.8^2) because the answer will be less than 0.8....

Doing revision for a math test. Let's see if someone is that fast, lol

 

80% probability of scoring a goal with one kick:
20% probability of missing

P = 1 - 0.2^1 = 0.8

Three kicks

P = 1 - 0.2^3 = 0.992
99.2%

 

Thanks, I did not read this in time and kinda failed the test but at least I know how to do it now ^_^

 

Here's two puzzles:

1. You want to cut a piece of Brie cheese in two equal pieces. The piece is a segment of a circle with the angle \alpha. Because of the preferences of the cheese eaters the other piece should contain the tip and the other the arc. Where do you cut?

2. What is the average distance of two points in a square?

 

I worked it out in Excel but got no useful insight at all. You can put that on my tombstone