Last Person To Post Here Wins (Part 24)

AHHH She stole the cup

https://simple.wikipedia.org/wiki/Group_theory
not sure if that's helpful, but a group is basically an algebraic structure. You have a set, and a binary operation which will combine two elements of the set to produce another element of the set. Think of rotational symmetries of the square. If you rotate a square by any multiple of 90 degrees, it's still a square, so they're symmetries. If you combine any two rotations, you still get a rotation. Also, since any rotation has an inverse (which will undo the rotation), since rotation by 0 degrees is an identity element (an element that does nothing), and since composition of these rotations has a nice property called associativity (namely that a*(b*c)=(a*b)*c, which is not true for everything), the rotational symmetries of a square form a group.

https://simple.wikipedia.org/wiki/Group_theory
not sure if that's helpful, but a group is basically an algebraic structure. You have a set, and a binary operation which will combine two elements of the set to produce another element of the set. Think of rotational symmetries of the square. If you rotate a square by any multiple of 90 degrees, it's still a square, so they're symmetries. If you combine any two rotations, you still get a rotation. Also, since any rotation has an inverse (which will undo the rotation), since rotation by 0 degrees is an identity element (an element that does nothing), and since composition of these rotations has a nice property called associativity (namely that a*(b*c)=(a*b)*c, which is not true for everything), the rotational symmetries of a square form a group.

I see. Thanks for explaining everything so simply the finding roots of equations sounds really interesting to me.

That's great to hear. Maths is generally a very hard subject so I think it's probably easy to feel like you regret it at some point. Wow that sounds awesome! What kinds of stuff did you do at the summer school? Good good, it's quite rare for me to find people who love maths

It's a reference to mathematics. Oooh what's group theory about?

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I didnt come on here expecting to see discussion about my favourite area, so I will recommend this book, free in PDF form :https://marinazahara22.files.wordpress.com/2013/10/i-n-herstein-topics-in-algebra-2nd-edition-1975-wiley-international-editions-john-wiley-and-sons-wie-1975.pdf

I didnt come on here expecting to see discussion about my favourite area, so I will recommend this book, free in PDF form :https://marinazahara22.files.wordpress.com/2013/10/i-n-herstein-topics-in-algebra-2nd-edition-1975-wiley-international-editions-john-wiley-and-sons-wie-1975.pdf

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what is the question your trying to solve?

A quick question to you guys:

If I have used the gradient function to find a single stationary point on a cubic curve, how would I go about proving that it is a point of inflection?
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A point of inflection is where the second derivative changes sign (not just where it equals 0). Find the 2nd derivative and substitute in a value of x or whatever the variable is either side of the stationary point. If its a point of inflection, there will be a sign change from one value to the other.

Sorry, I don't quite understand this. What is the 2nd derivative?

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If you differentiate the equation (y) once, you get $\dfrac{ \mathrm{d}y}{ \mathrm{d}x}$ which is the first derivative. If you differentiate this again you get an expression for $\dfrac{ \mathrm{d}^{2}y}{ \mathrm{d}x^{2}}$ which is the second derivative. (Likewise $\dfrac{ \mathrm{d}^{n}y}{ \mathrm{d}x^{n}}$ is known as the nth derivative )

So from post #6353, the second derivative would be 6x - 30?

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yup

what did you get to be your stationary point? You should find that the second derivative is also zero in this case (not always true, just to be clear), and that if you increase x from that point, the second derivative will be positive, and if you decrease it, it will be negative, so it changes sign at that point.

(5, 131) which doesn't look right to me

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where did the 131 come from? You should find that there is only one stationary point at x=5.

The 131 was meant to be the y value when x=5

Thanks, I think I understand how to do it now
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