For reference, we allow scientific calculators for physics exams but not graphing calculators.

@SemiClassical Yes, but getting mad over it is just a waste of time, you can't force people to be polite

@TedShifrin and with right i think.

@Astyx point

@Astyx: I think you're exaggerating my mood on this whole thing.

@Krijn Where i am, calculators are always forbidden in maths exams

@Ted No, I know you're not that mad, worry not :)

I think even allowing scientific calculators is a bit silly, though. If a question requires a specific numerical calculation, it probably isn't a great question.

Testing calculator skills is just silly.

@Semiclassical so true

Students should be able to add basic fractions and do basic algebra, but tests are not the place for ridiculous arithmetic with decimals, etc. I guess in a physics or chemistry class there is no choice. We used slide rules in my day.

Well, one option is to have problems where the arithmetic is simple enough to not require a calculator.

Calculators are also often forbidden in physics exams here @Ted :p

If you ask for a symbolic answer, then the numerical part of that is just sort of pointless.

I also graded leniently re arithmetic. If the problem was set up correctly and mostly correctly executed and there was a small error, I only took off one point out of 15 or 20, typically. For things like reduced echelon form, that often was a headache for me, but I did it.

i don't even know how to calculate $0.1^{0.1}$ by hand

Could someone explain to me why f(x) is x^2 when x is rational, 0 when x is irrational has only one derivative at 0?

@Null I wouldn't count that as 'simple enough' in a physics test.

They ask for 1 digit reliable answers

For a calculus test, by contrast...

Hi ! By any chance is anyone else almost dying of heat ??

Nor do I, @Null. So what? Leave that as your answer.

@user379685: Where is the function continuous?

i.e. express $x^x$ for small $x$

@TedShifrin that's what i'm not sure of

Really?

What does the graph look like?

though, best you can do for $x^x$ is to write that as $e^{-x\ln x}\approx 1-x\ln x$, which still leaves you to compute $\ln x$.

Like y=x^2 and y=0 simultaneously

Why negative, @Semiclassic.

You could see that $2^{10} = 1024\sim 1000$ and go from there

oops

Sorta, @user379685. So is it continuous at $x=1$?

yeah, $e^{x\ln x}\approx 1+x\ln x$

which gives $0.1^{0.1}\approx 1-0.1\ln 10$.

And I don't remember what $\ln 10$ is off the top of my head :)

@TedShifrin uhm no?

2.303

Why, @user379685?

is the dotproduct associative? my thought is yes!

@TedShifrin the function jumps ?

OK, @user379685. So where is the function continuous?

That makes no sense, @Null. Stop and think.

So 0.1^0.1 is roughtly 1 - 0.23 = 0.77.

@TedShifrin i know it should be at 0, but i don't know why

Wait. Prove it is continuous at $0$. Then argue it cannot be continuous anywhere else.

The real answer is more like 0.794, so that's not a great approximation. But it's decent enough.

I have no idea why 4 of my messages were starred today ..

Yeah, I wondered that.

Well, with the 'wrong' one people might be thinking of Trump :P

@TedShifrin mmh, can't we do $\vec{a}\cdot\vec{b}\cdot\vec{c}$, because if we go from left to right, we then have some number, and not a vector anymore?

Why is that ? @Semi

Uh huh @Null.

well...

h

Hi @MikeMiller

G'night, @MikeM.

@TedShifrin you said it makes no sense, I ask why. after $\vec{a}\cdot\vec{b}$ we have some number, instead of a vector.

Right, @Null, I agree.

multiplying a scalar $\lambda$ and a vector $v$ is denoted $\lambda v$, and the dot product of a vector $v$ with a vector $w$ is denoted $v\cdot w$. don't mix up the notations. one could write $(a\cdot b)c$ or $a(b\cdot c)$ for example, they would make sense (and be two different things), but $a\cdot b\cdot c$ does not make sense.

mmh, my question, which i answer myself now, was (ab)c=a(bc).

eh

it doesn't equal

What are these things?

@Semi Wrong

lol

vectors and dotproducts @TedShifrin i just had a misconcetion

Oh.

So unless $c$ is a scalar multiple of $a$, you're out of luck.

so how do i find a orthogonal vector to 2 independent vectors?

What if $b=0$ ?

$(a\times b)\times c$, by contrast, does make sense. But that's precisely because the cross-product of vectors maps two vectors to another vector.

OK, @Astyx wins on that one :D

leaves in pride

(and, moreover, $(a\times b)\times c\neq a\times (b\times c)$ in general)

I'll get you back for that, @Astyx, don't worry.

I'm gonna leave this chat forever just to prevent this

@Astyx that's a long time

fore-eh-vuh

rubs hands gleefully

@Null Where do these vectors live?

That's prettty mean .. @Ted

@TedShifrin 2 independent vectors in the land of the real monsters

Where the real things are.

@Null: In $\Bbb R^3$ or ...

@TedShifrin yep ;)

$R^3$

Then you know how to find a vector orthogonal to two vectors in $\Bbb R^3$.

A way that doesn't exist in higher dimensions.

R^3 is an easy case, yes.

@TedShifrin so in higher dimensions only independent is possible but not orthogonal?

no, @Null: In higher dimensions there are zillions of solutions, but you can't do it the same way.

Let $a$ and $b$ be two vectors. Then what is $a\cdot (b - {(a\cdot b)\over(a\cdot a)} a)$ ?

@Astyx: But he wants a vector orthogonal to both $a$ and $b$.

Applying this twice will give him what he want I think

Or rather might give him what he wants

Oh, way too hard.

Surely

You wouldn't want to do it with $a$ and $b$, though. you'd want to do it with $a$ and $v$, then $b$ and $v'$.

actually i only "have" to give an independant one. But i want to give the orthogonal one jsut for the funs since independent is easy

Do you have the coordinates of your vectors ? Or is it more abstract than that ?

Hint: What's the orthogonal vector to the unit vectors $e_x,e_y$, and how could you produce this vector?

$a$ and $b$ span a plan, just pick one from the orthogonal line

(2,1,1),(3,2,2) @Astyx

(51,5,7) will suffice as independent, PROBABLY lol^^

I doubt so

(it wouldnt be orthogonal, but independent)

What nice vectors live in the plane generated by these two vectors ? @Null

(-1,0,0)

actually

(1,0,0) is even nicer, but okay

yep

since (1,0,0) is in there, there's another nice vector in there as well

And what other one ? (not quite as nice, but still)

And what can you deduce form the fact that (1,0,0) lives in that plane ?

(0,1,1)?

Yep

ding.

so (0,1,0) would be orthogonal i think

which is orthogonal to (1,0,0), amusingly

No

yeah, no.

"amusingly" :D

need the dot product to be zero.

Now let (x,y,z) be a vector orthogonal to both your vectors

You can think of this geometrically, though. Try drawing the vectors (1,0,0) and (0,1,1)

so, a=(1,0,0), b=(0,1,1). and i need vector c, thats orthogonal to both of them?

Yes

(And ponder why this is true, and sufficient)

To get this in a brute force way, it's best to start with figuring out what vectors are orthogonal to (1,0,0)

(0,a,b) ?

Right.

Or write $c = (x,y,z)$ and solve $\begin{cases} c\cdot(1,0,0) = 0\\c\cdot(0,1,1)=0\end{cases}$

Anything that doesn't have an x-component.

but what's (0,a,b).(0,1,1) ?

(0,-a,a)

yeah.

and really, any a!=0 will work for that.

Hey, how can I express $ \int \dfrac {1} {1+x^4} $ as a power series ? , I know $ \dfrac {1} {1+x^4} $ can be represented as $ ar^n $, but what do I do with the integral ?

thanks @Semiclassical @Astyx

Maks, you can integrate the series termwise

represent the integrand as a geometric series and then integrate term-by-term

My pleasure @Null

hey semic :)

mmh, so n dimensional coordinate systems are really bad to represent stuff?

(where n>3)

@Semiclassical $ \dfrac {1} {1+x^4} $ should be $ -x^4 $ or $ -x^n $ ??

uh

What?

My bad, I was mixings concepts

I can't represent $ \dfrac {1} {1+x^4}$ on any other (helpful) way

what do you know about geometric series?

you were on the right track, I suspect, but you said it in a confusing way

I know they are of the form $ ar^n $ , and they converge to the number $ \dfrac {a} {1-r} $ when $ | r | < 1 $

right. So what's r here?

So I was thinking that with $ \dfrac {1} {1+x^4} $, $ a = 1, r = x^4 $

Then $ \dfrac {1} {1+x^4} = (x^4)^n $ ??

Not $r=x^4$. you were on a better track earlier.

$ r \neq x^4 $ ?

Right. $1-(x^4)\neq 1+x^4$, after all.

Ohh right!

$ r = -(x^4) $

?

right.

So what does that tell you about $\frac{1}{1+x^4}$?

$ \dfrac {1} {1+x^4} = -(x^4)^n $ ?

Not literally, no.

@Maks You are integrating over a segment of ]-1,1[ right ?

After all, when $x=0$ that'd mean $1=0$.

@Astyx its an indefinite integral

Right

The meaning of a geometric series is not $\frac{a}{1-r}=ar^n$.

You might be saying something else for simplicity, but I want the actual expression.

@Semiclassical That's what I was having trouble with

Then look up the definition of geometric series.

I know that $ \lim_{x \to \infty} \sum ar^n = \dfrac {a} {1-r} $

No.

That make sno sense whatsoever

That is most definitely not true.

to only state a liniarly independent vector to (2,1,1),(3,2,2) one could say: (a,b,c) with $b\not=c$?

now ?

Nop

if you mean $\sum_{n=1}^x$, yes.

Let me think what I'm saying

Yes

@Null yes

That's what I'm trying to say

@Semi You mean $\infty$ right ?

not $x$

He had $\lim_{x\to\infty}$ in there as well.

when x tends to $ \infty $

So .. lets see I'm not wrong here

Ah sure, my bad !

In other words, it's not that $\frac{a}{1-r}=ar^n$. It's that $\frac{a}{1-r}$ is the sum of $a r^n$ over all integer $n\geq 0$

$ \dfrac {1} {1+x^4} = \sum_{n=0}^{\infty} -(x^4)^n $ ?

Typo on my part earlier: Should've been from $n=0$ , not $n=1$.

What if n = 1 ?