How good is your Mathematics?

[Brown, Jon Brow]

What do you get when you differentiate y=3a+5b(squared)

[imull]

y=3+10b

i think. worrying how much i have forgotten since i finished uni lol

[Brown, Jon Brow]

Yep, i hope

[Alfa Fan]

Just done A levels maths, so its pretty good :)

The expression you gave, has 2 variables in it a & b, therefore you would need to inform us what you want to differentiate with respect too.

However, I believe you mean to differentiate the fuction y = 5x2 + 3x

In which case the differential, dy/dx = 10x + 3

[EuroTroll]

What do you get when you differentiate y=3a+5b(squared)

Finally a thread to replace the geography one as the most popular! :up: :p :

I'd venture a guess that it's... oh, never mind. :erm:

[Brown, Jon Brow]

Just done A levels maths, so its pretty good :)

The expression you gave, has 2 variables in it a & b, therefore you would need to inform us what you want to differentiate with respect too.

However, I believe you mean to differentiate the fuction y = 5x2 + 3x

In which case the differential, dy/dx = 10x + 3


Yeah I wanted dy/dx :dozey:

[Alfa Fan]

Now we've solved that one, I'll post something a bit trickier.

y = 5x3 + 4x2 + 7

Give me the turning points of the function

[Brown, Jon Brow]

Is that where Y=0 ?

Diffrential of Y=0

[Alfa Fan]

You need to differentiate the funtion with respect to X. Then do what you suggested ;) The answer don't work out to exact decimals btw.

[Brown, Jon Brow]

Now we've solved that one, I'll post something a bit trickier.

y = 5x3 + 4x2 + 7

Give me the turning points of the function

Is the equation y=5 multiplied by3 etc.....

or y=5cubed etc.....

[Alfa Fan]

cubed, you can't superscribe on the forum sadly.

[Brown, Jon Brow]

y=5xcubed+4xsqd+7

dy/dx= 5xsqd + 8x

at turning point dy/dx = 0
therefore 15x sqrd + 8x = 0

x(15 + 8) = 0

Turning points are (0,0) and (8/15,0)

[Alfa Fan]

points are (0,7) and (-8/15, 7.29)

You missed the effect of the +7 in the first one :)

[Brown, Jon Brow]

Let me check - I might have made a mistake

:s looks like I need to do some revision before I do my Core 2 resit next month.

[Brown, Jon Brow]

(0,7) (8/15, 8.9)

[Alfa Fan]

:s looks like I need to do some revision before I do my Core 2 resit next month.

lol, good luck with that :) Did that in June 2005 and got 70/100

[Alfa Fan]

Here's the working if you wondering where you've gone wrong...

y = 5x3 + 4x2 + 7

dy/dx = 15x2 + 8x

15x2 + 8x = 0

x(15x + 8) = 0

x = 0 15x + 8 = 0

x = 0 x = -8/15

When x=0

y = 5(0)3 + 4(0)2 + 7
y = 7

Therefore the first point is (0,7)

when x=-8/15

y = 5(-8/15)3 + 4(-8/15)2 + 7
y = 7.29.......

Therefore the second point is ~ (-8/15,7.29)

[Brown, Jon Brow]

A box of mass 200kg rests in equilibrium on a plane inclined at 32degrees to the horizontal

calculate the frictional force acting on the box.

[J4MIE]

Squared²
Cubed³

Maths sucks :p :

[harvick#1]

maths sucks indeed :D :up:

although I'm rather good at it for some parts

[Brown, Jon Brow]

Maths is cool!!!! It would be a rubbish world without it.

BTW how do you do squares and cubes

[Erki]

You do differentiating at high school? :eek:
I do it at Uni...oooooops, I mean I'm SUPPOSED to do. :s

MATHS SUCKS!

;)

[AndySpeed]

Maths is cool!!!! It would be a rubbish world without it.

BTW how do you do squares and cubes

There's an oxymoron if ever I saw one!

The world might be slightly less... organised without maths though.

I do cubes in my coffee.

[grassrootsracer]

A box of mass 200kg rests in equilibrium on a plane inclined at 32degrees to the horizontal

calculate the frictional force acting on the box.

I'll post the answer after I return from my physics final in a few hours. If someone hasn't already figured it out.

[janneppi]

A box of mass 200kg rests in equilibrium on a plane inclined at 32degrees to the horizontal

calculate the frictional force acting on the box.
sin(32)*200[kg]*9,81[m/s^2]
=1040[N]
That's not math though, it's physics. ;)

[Brown, Jon Brow]

sin(32)*200[kg]*9,81[m/s^2]
=1040[N]
That's not math though, it's physics. ;)

It's mechanics actually, it's part of physics and maths ;) it was a question in my maths homework. :p :

[Powered by Cosworth]

I'm still amazed by 0.9 Recurring being exactly equal to 1.

[Dave B]

That's numberwang! :D ;)

[Erki]

I'm still amazed by 0.9 Recurring being exactly equal to 1.

1 - 0.(9) = 0.(0) = 0 => 1 == 1

[Erki]

sin(32)*200[kg]*9,81[m/s^2]
=1040[N]
That's not math though, it's physics. ;)

times friction quotient :dozey:

[donKey jote]

how fast and in which direction was the plane flying ? :p :

http://smileys.smileycentral.com/cat/16/16_3_166.gif

[Vivianita]

You do differentiating at high school? :eek:


I learned integrals in high school, and it wasn't one of the best schools :s



A box of mass 200kg rests in equilibrium on a plane inclined at 32degrees to the horizontal

calculate the frictional force acting on the box.

GRRR! I don't want to see more problems like that... I just had my rigid solid mechanics final exam 6 hours ago and I don't wanna know anything about it in some time...

[janneppi]

GRRR! I don't want to see more problems like that... I just had my rigid solid mechanics final exam 6 hours ago and I don't wanna know anything about it in some time...

Try 6 years ago. :D

[Erki]

I learned integrals in high school, and it wasn't one of the best schools :s

Aaah, integrals, one more thing I'm supposed to learn at Uni...

What is it? :confused: :)

[janneppi]

times friction quotient :dozey:
No, i would need the friction quotient if i had calculated the maximum Friction force "gained until the box began moving).
Fmax=μ*cos(32)*200kg*9,81m/s^2

But since the thing rests in equilibrium i don't need to.
I assume that there were no other outside forces than gravity acting on the box, so Fmax=F_move
F_move=sin(32)*200kg*9,81m/s^2
=1040N
We can also calculate the friction quotient in equilibrium,
μ=tan(32)=0.624


That is only if i haven't forgotten everything i should know.

[stevie_gerrard]

Maths :up:

Mechanics :down: Its not even Maths, even if it is in the Maths module, its a waste of time, and tis got no relevance to University Mathematics (So Far anyway)

[Sleeper]

Maths :up:

Mechanics :down: Its not even Maths, even if it is in the Maths module, its a waste of time, and tis got no relevance to University Mathematics (So Far anyway)
It has everythink to do with Mechanical Engineering though.

[stevie_gerrard]

It has everythink to do with Mechanical Engineering though.

True, but you would take Physics as well as Maths if you wanted to do that. as i didnt take Physics, I can moan at how bad mechanics is :p :

[Sleeper]

True, but you would take Physics as well as Maths if you wanted to do that. as i didnt take Physics, I can moan at how bad mechanics is :p :

I reckon you could get away with not having done Physics. When I did Motorsport Tech last year a lot of what I learnt in A-level Physics was near unecessary and mechanics was a lot more valuble.

[Brown, Jon Brow]

I did AS-level Pysics last year as well as Maths, I got a 'B' in Maths but an 'E' in Physics. :s Hopefully I'll never hear of Nutrino's and y=mc(squared) ever again.

[Vivianita]

Maths is cool!!!! It would be a rubbish world without it.

True (Y) and it's cool that let you understand somethings too...

What is a polar bear?
It's a Cartesian bear after a coordinate transform

I couldn't understand it until I take Calculus II at uni :s

or...

There was a big function party, every function was having the time of their lives. In a corner there's e^x all alone. x^2 sees e^x all lonely and approaches to him:
- Hey, e^x, come on, integrate yourself
- What for - says e^x - it makes no difference.

When I took Calculus II that is the course where you learn integrals, my teacher told us that joke every single time he had to integrate e^x for any example.

[donKey jote]

polar bear :laugh:

[Alfa Fan]

I did AS-level Pysics last year as well as Maths, I got a 'B' in Maths but an 'E' in Physics. :s Hopefully I'll never hear of Nutrino's and y=mc(squared) ever again.

Snap..me too.

But I ended up with a C in Physics at the end of A-Level.

[janneppi]

Can some one exlain vector analysis(rotors, hobbits and whatnotts) and Fourier math to me in 10 minutes so that even i understand. :)

[donKey jote]

probably not but I remember using them and thinking they were well cool tools :D

what do you want to know?
http://smileys.smileycentral.com/cat/16/16_3_166.gif

[janneppi]

Enough to pass two courses ;)

[donKey jote]

in 10 minutes? :eek:
what's for dessert ?

[janneppi]

The vector thingy is a long time friend of mine, i can't seem to be able to pass it. Fourier seems to be just as hard for me.

[Vivianita]

Fourier is really easy... if it is what I'm thinking it is... We are talking about Fourier series, right?

[donKey jote]

or transforms?

[Vivianita]

or transforms?

But it is Laplace transforms

[Alfa Fan]

Back on topic ;)

Give me the estimated area under the curve

y = 4x3 + 5x2 - 2x + 1

Between x=6 and x=2

(a) Using Integration

(b) Using Trapezium Rule (h=1)

(c) Then state the error in (b) as a percentage

[Vivianita]

Back on topic ;)

Give me the estimated area under the curve

y = 4x3 + 5x2 - 2x + 1

Between x=6 and x=2

(a) Using Integration

(b) Using Trapezium Rule (h=1)

(c) Then state the error in (b) as a percentage


(a) -1597.7 aprox

(b) Don't remember that trapezium thing

[Erki]

Bye bye :eek: :s

[Alfa Fan]

Almost correct, might just be rounding errors...

INTEGRATION METHOD (Exact Answer)

y = 4x3 + 5x2 - 2x + 1

Integrate (can't show an integral on this, so a / will signify an integral)

/(6,2) x4 + 5/3x3 - x2 + x

[1296+360-36+6]-[16+13r1r3-4+2]

1626 - 27r1r3 = 1598r2r3

TRAPEZIUM METHOD (Approx Answer)

Area ~= 1/2 x h x [y0 + 2(y1+y2+...+yn-1) + yn]
1/2 x 1 x [49 + 2(148 + 329 + 626) + 1033]
1/2 x [49 + 2186 + 1033]
~= 1634

To work out the error, its simple

1598r2r3 / 1634 *100 = 2.16%

:)

[Brown, Jon Brow]

An easy question to start this thread again.......

8³ x 4² = ?#?

Give your answer in exact form ;)

[donKey jote]

8Kb
http://smileys.smileycentral.com/cat/16/16_3_166.gif

[Brown, Jon Brow]

write 4ln13+ln64 as a single log

[donKey jote]

ln(13^4*64)

[jens]

Oh dear. In primary school I was pretty good at maths, but after that I have gradually lost all my knowledge and now I even can't read any of your given tasks as it puts my brain into a total confusion!

[akv89]

How come everyone like Calculus so much?

x^5 +5x^4 + 10x^3 + 10x^2 - 5x + 1 = 10

Solve for x. (Anyone who noticed the minus sign before the 5x, it is supposed to be negative)

[tinchote]

How come everyone like Calculus so much?

x^5 +5x^4 + 10x^3 + 10x^2 - 5x + 1 = 10

Solve for x. (Anyone who noticed the minus sign before the 5x, it is supposed to be negative)


Are you sure you the signs go that way?

If the answer is no, then the five roots are -1, -1+10^(1/4), -1-10^(1/4), -1+i 10^(1/4), -1-i 10^(1/4). But I wouldn't have known how to find them without the computer.

[McLeagle]

cubed, you can't superscribe on the forum sadly.

Computerese for "exponent" is the caret symbol: ^

"Five ex cubed" would be 5x^3 or, better, 5*X^3

iirc

Cheers

[millencolin]

i failed maths when i was at high school... so this will be first, and last, post in this thread


in fact id rather get kicked in the testicles by a soccer player 17 times in a row than to re-do my final maths exam

[McLeagle]

Integrate (can't show an integral on this, so a / will signify an integral)



integral = ampersand#8747; >>>replace 'ampersand' with &

if I just type it proper, you won't see the code: ∫

Let me try this:
[code=php:vygogobr]<span class="syntaxdefault"></span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#8747;*</span><span class="syntaxdefault"></span>[/code:vygogobr]
~~~~~~~~~~~~
Some others:
[code=php:vygogobr]<span class="syntaxdefault"></span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#178;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#179;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#177;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#402;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#916;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#931;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#8747;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#8756;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#8800;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#8804;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#8805;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#9824;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#9827;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#9829;
</span><span class="syntaxkeyword">&</span><span class="syntaxcomment">#9830;*</span><span class="syntaxdefault"></span>[/code:vygogobr]
² - superscript 2 ---edit - sonnofa... code gone; it's 178 between &# and ; (no spaces)
³ - superscript 3 ------ 179
± ---------- 177
ƒ ------402
Δ
Σ
∫
∴
≠
≤
≥
♠
♣
♥
♦

Very strange - codes are back after I edit?????? Also, this should work with code tags, but doesn't; that's why the PHP tags!!

[tony_yeboah]

5x + 10 = 35
x + 10a = 35
xa + 20 = 35

what is x and a?

[Brown, Jon Brow]

x = 5

a = 3

[tinchote]

x = 5

a = 3

and there is no need for the third equation.

[akv89]

Are you sure you the signs go that way?

If the answer is no, then the five roots are -1, -1+10^(1/4), -1-10^(1/4), -1+i 10^(1/4), -1-i 10^(1/4). But I wouldn't have known how to find them without the computer.

Yes I am sure. But since people have proceeded on i will post the solution anyway.
x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1 is equal to (x+1)^5. Those who know Pascal's triangle would be able to figure this out. My question switched the signs on the 5x. So x^5 + 5x^4 + 10x^3 + 10x^2 - 5x + 1 = 10.
So in order to make it look like the expansion of (x+1)^5 I add 10x to both sides. Thus I get:
x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1 = 10x + 10
or (x+1)^5 = 10(x+1)
(x+1)^5 - 10(x+1) = 0
(x+1)((x+1)^4 - 10)= 0
So x = -1 or (x+1)^4 = 10
x+1 = +/-fourth root of 10 or 10^(1/4) (I was only looking for real solutions).
x = +/- 10^(1/4) - 1 or -1 which are the real solutions that tinchote posted.

[tinchote]

Yes I am sure. But since people have proceeded on i will post the solution anyway.
x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1 is equal to (x+1)^5. Those who know Pascal's triangle would be able to figure this out. My question switched the signs on the 5x. So x^5 + 5x^4 + 10x^3 + 10x^2 - 5x + 1 = 10.
So in order to make it look like the expansion of (x+1)^5 I add 10x to both sides. Thus I get:
x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1 = 10x + 10
or (x+1)^5 = 10(x+1)
(x+1)^5 - 10(x+1) = 0
(x+1)((x+1)^4 - 10)= 0
So x = -1 or (x+1)^4 = 10
x+1 = +/-fourth root of 10 or 10^(1/4) (I was only looking for real solutions).
x = +/- 10^(1/4) - 1 or -1 which are the real solutions that tinchote posted.

Nice :)

[veeten]

♠
♣
♥
♦


either it's Poker night, or it's "Lucky Charms". :D

... they're 'magically delicious', you know... ;) :p

[akv89]

I've got another question: Completely expand the following expression
(x-a)(x-b)(x-c)...(x-z) :D

[tinchote]

I've got another question: Completely expand the following expression
(x-a)(x-b)(x-c)...(x-z) :D


The shortest way I know is:



( x - a ) ( x - b ) ( x - c ) . . . ( x - z )


;) :D

[libra65]

Happy Pi Day to all of you Mathletes out there!! This morning on the news, the anchorman reminded us that it is Pi Day--3.14. I wouldn't have caught the reference at 6AM if he didn't have a picture of the Pi symbol in his hand when he said it. BTW- the morning news crew in Philly is a little screwy.

[akv89]

The shortest way I know is:



( x - a ) ( x - b ) ( x - c ) . . . ( x - z )


;) :D

Fortunately there is a shorter way :D
In the expression (x-a)(x-b)(x-c)...(x-z),
the 24th term is (x-x) which is equal to 0.
Therefore, since 0 multiplied by anything is 0, the expansion of the entire expression is equal to 0.

[donKey jote]

:up: :laugh: :laugh:

[Brown, Jon Brow]

I don't know what any of you are talking about :confused:

Is this Phd level maths?

[akv89]

I don't know what any of you are talking about :confused:

Is this Phd level maths?

The questions I've posted come from any recollection i have of previous math competitions that my high school takes part in.
Here's another:
x^2 + y^2 = 132
x + y = 12
Determine the value of x*y

[tinchote]

Fortunately there is a shorter way :D
In the expression (x-a)(x-b)(x-c)...(x-z),
the 24th term is (x-x) which is equal to 0.
Therefore, since 0 multiplied by anything is 0, the expansion of the entire expression is equal to 0.

Awsome :up: :D

[DonnieDarco]

You people are all insane.................ly clever ;) :D

[tinchote]

The questions I've posted come from any recollection i have of previous math competitions that my high school takes part in.
Here's another:
x^2 + y^2 = 132
x + y = 12
Determine the value of x*y


If you square the second one, you get (x+y)^2=144. This you write as x^2+y^2+2xy=144, and subtracting the first equation you get 2xy=10, so xy=5.

[akv89]

If you square the second one, you get (x+y)^2=144. This you write as x^2+y^2+2xy=144, and subtracting the first equation you get 2xy=10, so xy=5.

Your method is correct. But there was an arithmetic error when you subtracted x^2 + y^2 (which equals 132) from (x+y)^2 (which equals 144). So 2xy = 12 instead of 10 and x*y = 6.
It's always the simple things that bug smart people :p :

[tinchote]

It's always the simple things that bug smart people :p :

Indeed! And all the worse when you are a professional mathematician :D

[donKey jote]

hehe tincho's fingers were too big for his calculator :p :

do keep 'em coming akv :up: :up:

a professional donkey :D

http://smileys.smileycentral.com/cat/16/16_3_166.gif

[akv89]

hehe tincho's fingers were too big for his calculator :p :

do keep 'em coming akv :up: :up:

a professional donkey :D

http://smileys.smileycentral.com/cat/16/16_3_166.gif

ok, I'm trying to think of questions that test people's logical skills instead of their ability to remember difficult formulas.

Evaluate the product of:
log3(base2)*log4(base3)*log5(base4)*log6(base5)*lo g7(base 6)

This was one of the simpler problem that I found in a book called the Art of Problem Solving, a really good math tutorial for those who have a strong interest in math.

[akv89]

Anyone?

Here's a hint: use the change of base rule for logs

[McLeagle]

Anyone?

Here's a hint: use the change of base rule for logs


****SPOILER****






Using your hint, it becomes:

log3/log2 · log4/log3 · log5/log4 · log6/log5 · log7/log6 ≅ 2.807354922

... if I've done my arithmetic right

[donKey jote]

log7(base2)

... or viceversa :p :

[akv89]

log7(base2)

... or viceversa :p :

log7(base2) is the correct answer. McLeagle probably has the correct answer, but I guess I shoul have mentioned that no calculators are allowed. :p :
The key was the expression that McLeagle had after using the change of base rule. Notice that all of the logs except log7 in the numerator and log2 in the denominator cancel out. Hence the answer becomes log7/log2. The simplified form is log7(base2)

[Alfa Fan]

This is really hard and I would be impressed if someone solves it without looking stuff up....

dy/dt = 1 + t^2 + y^2 +t^2*y^2

Integrate to give y= f(x) in explicit form.

[tinchote]

This is really hard and I would be impressed if someone solves it without looking stuff up....

dy/dt = 1 + t^2 + y^2 +t^2*y^2

Integrate to give y= f(x) in explicit form.


If it is "f(x)" it is simpler, so I will assume you want "f(t)".

In the RHS, you first factor y^2 and then 1+t^2, to get

y' = (1+t^2)(1+y^2)

Then you have 1/(1+y^2) y' = 1+t^2. Integrating, you get

arctan(y) = t + t^3/3. So, in the end,

y(t) = tan ( t + t^3/3).

[Alfa Fan]

You've missed out a Pi term in the answer.

[tinchote]

You've missed out a Pi term in the answer.

No I haven't. I missed the integration constant, if you want, so the most general answer would be

y(t) = tan(t + t^3/3 + c),

but c could be any number.

[veeten]

McLeagle probably has the correct answer, but I guess I shoul have mentioned that no calculators are allowed. :p :

Boooooo!!! Unfair! :p :

[SOD]

integrate:

dx/(exp[x]-1) with respect to x :D

[tinchote]

integrate:

dx/(exp[x]-1) with respect to x :D


It's not that difficult, but I'll let someone else do the honours :)

[Quattroporte]

WHAT DO YOU MEAN IT'S NOT THAT DIFFICULT????????? MY HEAD NEARLY EXPLODED JUST READING IT!!! :eek:

[tinchote]

WHAT DO YOU MEAN IT'S NOT THAT DIFFICULT????????? MY HEAD NEARLY EXPLODED JUST READING IT!!! :eek:

Well, I guess it depends on the point of view :)

When you see an integrand like that, you try and look for a substitution; since the exponential is its own derivative, it behaves well with respect to substitutions. So one way would be to blindly try the substitution u=exp[x]-1. Since that's not the way I think when I see that integral, here's the way that first came to my mind:

multiply above and below by exp[x], then you have to integrate exp[x] dx / (exp[2x] - exp[x]). Then do the substitution u=exp[x], and the integrand becomes du/(u^2-u). This is the canonical "partial fractions" example: it can be written as (1/(u-1) - 1/u ) du. integrating we get ln[u-1]-ln[u], and remembering that u=exp[x], we get

int dx/exp[x] = ln[exp[x] - 1] - x

(well, and "plus constant" if you want ;) )

The substitution I mentioned at the beginning leads to almost the same computations: you won't get rid of the partial fractions.

[SOD]

well done Tinchote :up:

[donKey jote]

hey tin now how about mapping out the E8 Lie-Group :p :

http://smileys.smileycentral.com/cat/16/16_3_166.gif

[akv89]

Well, I guess it depends on the point of view :)

When you see an integrand like that, you try and look for a substitution; since the exponential is its own derivative, it behaves well with respect to substitutions. So one way would be to blindly try the substitution u=exp[x]-1. Since that's not the way I think when I see that integral, here's the way that first came to my mind:

multiply above and below by exp[x], then you have to integrate exp[x] dx / (exp[2x] - exp[x]). Then do the substitution u=exp[x], and the integrand becomes du/(u^2-u). This is the canonical "partial fractions" example: it can be written as (1/(u-1) - 1/u ) du. integrating we get ln[u-1]-ln[u], and remembering that u=exp[x], we get

int dx/exp[x] = ln[exp[x] - 1] - x

(well, and "plus constant" if you want ;) )

The substitution I mentioned at the beginning leads to almost the same computations: you won't get rid of the partial fractions.

That was a lot easier than expected. For some reason I thought that the sqaure brackets represented the greatest integer function and was mystified as to how I can integrate something with a greatest integer function.

[tinchote]

That was a lot easier than expected. For some reason I thought that the sqaure brackets represented the greatest integer function and was mystified as to how I can integrate something with a greatest integer function.

But if the square brackets are the greatest integer function, it becomes a lot easier: because then the function is constant on each interval of the form (k,k+1] for k integer (so it looks "like a ladder"), and when you integrate you get a linear function on each of those intervals. You would also have to state the problem in a clearer way, because for example the function will fail to be defined on the interval (-1,0].

[akv89]

But if the square brackets are the greatest integer function, it becomes a lot easier: because then the function is constant on each interval of the form (k,k+1] for k integer (so it looks "like a ladder"), and when you integrate you get a linear function on each of those intervals. You would also have to state the problem in a clearer way, because for example the function will fail to be defined on the interval (-1,0].

brainfart :p :

[tinchote]

hey tin now how about mapping out the E8 Lie-Group :p :

http://smileys.smileycentral.com/cat/16/16_3_166.gif

Hey! I don't have time for those simple computational things ;) :p :

[SOD]

taken from the almanac, but I think their solution is incorrect.

"A motor car is three times as old as its tires were when it was as old as the tires are now. When its tires are as old as the car is now, the car will be a year older than the tires are now. What are the present ages of car and tires?"

[akv89]

The motor car is three and a half years old and the tyres are one year old.
Solution:
Let m be the current age of the car and t the current age of the tires. The car would be as old as the tires currently are m-t years ago. The age of the tires at that time would be t-(m-t) or 2t-m. Therefore, your first statement : "A motor car is three times as old as its tires were when it was as old as the tires are now" translates to:
m = 3(2t-m)
m = 6t - 3m
m = 3/2t
The tires will be as old as the car m-t years in the future. At that point, the age of the car would therefore be m + (m-t), or 2m - t.
Hence, your second statement:"When its tires are as old as the car is now, the car will be a year older than the tires are now" translates to:
2m-t - t = 1
2m - 2t = 1
Substituting for m from the first equation, i get
2(3/2t) - 2t = 1
3t - 2t = 1
t = 1
m = 3/2
:)

[tinchote]

My only comment would be the 3/2 years is a year and a half ;) :D

[akv89]

My only comment would be the 3/2 years is a year and a half ;) :D

:arrows:
Don't blame me. I did that problem after depleting my brain of oxygen by running intervals.

[tinchote]

:arrows:
Don't blame me. I did that problem after depleting my brain of oxygen by running intervals.


:D

[SOD]

ah shoot. I misread the question :o

[SOD]

If one side of the bottom layer of a triangular pyramid of bowling balls has 12 balls, how many are there in the whole pyramid? [