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Nothing personal. The computers have a hard time with them for some reason. You know how computers are ... Temperamental.

What do you have against unicorns?

@@ End of Day Wrap - Monday 2/6/14   Sydney, Australia   Time 10:45 pm 

Hi Everyone,

Our great answers today were provided by CPhill, Alan, GoldenLeaf, admin, Stu, reinout-g, TaviSage, Kitty,3, Itoen, Feitan and Bertie.  Thank you all.

Feitan and TaviSage are both new.  We welcome you to web2.0calc forum and we hope you like it here.

Oh, Admin has introduced an avatar feature.  A number of members have told me that they love it. Thank you Andre.  Your efforts, and the improvements that they bring, are much appreciated.  

I have checked all posts up to but not including the current pages. (there is at least one that i want to return to tonight) 

Here are a couple of the more memorable threads.  

Great probability question - and comedy for the day!  

The chain rule is very basic and I do not think that it is expained to students very well.

Do you know how you can cancel in fractions

eg

speed*time = distance

because

$$\frac{6km}{hour}\times\frac{7hour}{1}=42km$$  Because the hours cancel out!

With calulus the d in front of the x and the y etc just stands for difference.  You can cancel out the units just like I did in the above example.

So for instance

$$\frac{dy}{du}\times \frac{du}{dx}\times \frac{dx}{dt}=\frac{dy}{dt}$$     The du's and dx's cancel out.

That is the chain rule!

--That's very impressive!

I couldn't play it untill I was 6 --at least not without mistakes.

The teacher would whack me with the fiddlestick when I made a mistake. She was mean!!

Her name was Melody, and she had perfect pitch. She knew how to use the bow too!

20/100*17180

3436$

Yes Stu, if two lines are perpendicular then the product of their gradients will be -1

$$m_1m_2=-1\\\\
therefore\\\\
m_2=\dfrac{-1}{m_1}$$

The normal to a curve is the name given to a line that is perpendicular to the tangent of the curve at that point.

That is;  The normal is perpendicular to the tangent.

440/x+348=356

x*440/x+348*x=356*x

440+348*x=356x-348x

440/8=8x/8

x=55

WOW I could play Sibelius's Violin Concerto when I was 4.  I must have been a slow learner!

Hey Alan,

you're right, but they are not initial values.

They are constants which can be calculated for given initial values for x and y

g(7) = 7² - 5 = 49 - 5 = 44

f(44) = 44 + 6 = 50

Replace the t by 2x

f(2x) = 3*(2x)² - 2 

f(2x) = 3*4x² - 2

f(2x) = 12x² - 2

Perhaps I've misunderstood, but if c1 and c2 are initial values (n=0) for x and y then (for n = 1):

These don't look the same to me!

Subtracting lambda from each of the diagonal elements and expanding the determinant along the top row gets you

$$(3-\lambda)\{(2-\lambda)(3-\lambda)-1\}-(3-\lambda)=0.$$

(3 - lambda) comes out naturally as a common factor.

If you choose to expand the whole lot though you get -19lambda rather than -20lambda.

Thank you Alan, that was really helpfull :)

Now that I come to think of it,

for the eigenvalues it is always true that

$$\lambda_1 + \lambda_2 + \lambda_3 = trace (A)$$

and

$$\lambda_1*\lambda_2*\lambda_3 = det(A)$$

In this case

$$det(A) = 12\\
trace(A) = 8\\$$

I get  $$-\lambda^3+8\lambda^2-19\lambda+12=0$$

Multiply through by -1 (not essential - I just find it easier that way!)

$$\lambda^3-8\lambda^2+19\lambda-12=0$$

If we assume that, because you are doing it by hand, the solutions are integers, then consider

$$(\lambda-a)(\lambda-b)(\lambda-c)$$

The three constants must multiply together such that $$abc=12$$

If they are each integers then we have either 1, 2 and 6 or 1, 3 and 4 (ok a couple of them might have negative signs attached!).

The sum of the three must also be 8, and since 1+2+6 is not 8, and 1+3+4 is, there is a good chance that the three eigenvalues are 1, 3 and 4.

Plugging each of these into the LHS of the first equation above does indeed give zero.

edit: Aaah I found it 

Hey Feitan,

Nice answer!

I haven't seen you before on the forum are you new here?

I was just wondering about the part of the signifant numbers...

Isn't it true that since 2g has only 1 significant number the final answer should also have only 1 significant number?

Reinout 

Should be -19 rather than -20, further, (lambda-3) comes out naturally as a common factor when you expand the determinant.

I learned that when I was 2 years old 

That is great!   I was 2 1/2, but I learnd the quadratic formula when I was 3.

A positive number is always greater than a negative number so it must be true.

What is the : sign doing in there?  I do not understand the question!

Thanks Kitty

YES the question asker had it right in the first place!

That is not nice Kalecia!  If you can't say something nice don't say anything at all!