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Integrate 1/x^2

Within the boundaries of 1 and -1.

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Reply 1
Avatar for Square
Square
13
1/x^2=x^(-2)
Why is the answer negative?

Spoiler

Reply 3
Avatar for Vivisteiner
Vivisteiner
OP
@Square: Yes, now integrate it within the boundaries of 1 and -1 to give me a number.


@TheTallOne: Dammit, its a trick question! A kinda lame one, I know.
Reply 4
Avatar for fL3X
fL3X
>_>
I thought it can't be done when the domain of integration includes 0?
Reply 6
Avatar for SimonM
SimonM
18
Vivisteiner
^Yes, now integrate it within the boundaries of 1 and -1 to give me a number.


It doesn't converge. Happy?
Reply 7
Avatar for Chaoslord
Chaoslord
16
ans = math error :yep:
Reply 8
Avatar for SimonM
SimonM
18
Chaoslord
ans = math error :yep:


Not quite sure, but isn't it over 9000?
Reply 9
Avatar for Vivisteiner
Vivisteiner
OP
^lol, CORRECT!

Ok fine, I'll give you another;

calculate e^i*pi
Vivisteiner
^lol, CORRECT!

Ok fine, I'll give you another;

calculate e^i*pi


That's in a simple enough form.
Reply 11
Avatar for SimonM
SimonM
18
Vivisteiner
^lol, CORRECT!

Ok fine, I'll give you another;

calculate e^i*pi


Why exactly?

eiπ=−1e^{i \pi} = -1
Reply 12
Avatar for Agrippa
Agrippa
8
Vivisteiner
^lol, CORRECT!

Ok fine, I'll give you another;

calculate e^i*pi


Presumably you mean ei*pi, rather than ei*pi.

The answer is -1. Back to you - prove it :wink:
Reply 13
Avatar for DeanK2
DeanK2
Agrippa
Presumably you mean ei*pi, rather than ei*pi.

The answer is -1. Back to you - prove it :wink:


I'd be surprised if someone at interview was asked to expand e^(ix) then note the taylor series for the trig functions (although it is quite a pleasing result).
Reply 14
Avatar for Vivisteiner
Vivisteiner
OP
Lol, it was a trick. e^(i*pi) = -1 but e^i*pi is already in its simplest form.


And I don't know how to ruddy well prove it. xD
Reply 15
Avatar for Swayum
Swayum
18
A hack way of doing it is to differentiate

f(x) = e^(ix)/(cosx + isinx) and hence work out what f(x) is.
Reply 16
Avatar for SimonM
SimonM
18
Vivisteiner
Lol, it was a trick. e^(i*pi) = -1 but e^i*pi is already in its simplest form.


http://xkcd.com/169/

All there is to say on things like that
Reply 17
Avatar for DeanK2
DeanK2
Vivisteiner
Lol, it was a trick. e^(i*pi) = -1 but e^i*pi is already in its simplest form.


And I don't know how to ruddy well prove it. xD


A strange maths troll indeed. An interview with your kind of questions actually would worry me.
Reply 18
Avatar for Vivisteiner
Vivisteiner
OP
Lmao, I'm just kinda bored so I asked inane questions and sounded stupid even though I'm obviously incredibly smart and can even differentiate x^x^x^x^x^x in under ten seconds with respect to x.
Reply 19
Avatar for SimonM
SimonM
18
Vivisteiner
Lmao, I'm just kinda bored so I asked inane questions and sounded stupid even though I'm obviously incredibly smart and can even differentiate x^x^x^x^x^x in under ten seconds with respect to x.


Sodding LaTeX

x^x^x^x^x^x (x^(-1+x^x^x^x)+x^x^x^x^x Log[x] (x^(-1+x^x^x)+x^x^x^x Log[x] (x^(-1+x^x)+x^x^x Log[x] (x^(-1+x)+x^x Log[x] (1+Log[x])))))

Done

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