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A Summer of Maths (ASoM) 2016

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Reply 60

Kryptonian

This sounds interesting. I'm doing A levels next year but I think it will be fun to learn some of this stuff. :biggrin:

Reply 61

Zacken

Original post by 16Characters....

Hmm. I'm way too nervous about results day to start preparing for uni maths haha.

Agreed!! I'm much the same, bit hypocritical since I made the thread, but it's more for the benefit of the others who are fairly sure they have 1's and S's whilst I'm too nervous about my 2 to do anything properly. :lol:

Reply 62

Zacken

Original post by Kryptonian

This sounds interesting. I'm doing A levels next year but I think it will be fun to learn some of this stuff. :biggrin:

You're welcome to do so, but if you're doing A-Levels next year, then this thread might be a bit out of your reach, since it deals with first year uni work. :smile:

Reply 63

username1162972

Original post by Zacken

You're welcome to do so, but if you're doing A-Levels next year, then this thread might be a bit out of your reach, since it deals with first year uni work. :smile:

Unless he is jim leck.

Posted from TSR Mobile

Reply 64

Insight314

Original post by physicsmaths

Im gna let it sit at my desk and not use it so I know how you feel.

Posted from TSR Mobile

"Consider the equation

z^n = 1

. The coefficient of

z^{n-1}

is the sum of all roots." is that Vieta's formula?

Also, if

w = \exp(\frac{2\pi i}{n})

and

w^n - 1 = (w-1)(1+w+w^2+...+w^{n-1})

where

w \neq 1

, then why is

\frac{w^n - 1}{w -1} = 0

when trying to show that

1 + w + w^2 + ... + w^{n-1} = 0

Reply 65

Insight314

Original post by physicsmaths

Unless he is jim leck.

Posted from TSR Mobile

Or physicsmaths... after his two gap years.

Reply 66

Zacken

Original post by Insight314

"Consider the equation

z^n = 1

. The coefficient of

z^{n-1}

is the sum of all roots." is that Vieta's formula?

Yes (ish). Think about expanding the polynomial

(z-z_1)(z-z_2) \cdots (z-z_n)

. The coefficient of

z^{n-1}

is the (

(-1)^{n-1}

of) the sum of the roots.

Also, if

w = \exp(\frac{2\pi i}{n})

and

w^n - 1 = (w-1)(1+w+w^2+...+w^{n-1})

where

w \neq 1

, then why is

\frac{w^n - 1}{w -1} = 0

when trying to show that

1 + w + w^2 + ... + w^{n-1} = 0

Roots of unity, so

w^n - 1 = 0

but then it follows that the latter bracket must be 0 for

w \neq 1

then geometric series, i.e: sum of the first

n

terms with first term

1

and common ratio

w

(edited 7 years ago)

Reply 67

username1162972

Original post by Insight314

"Consider the equation

z^n = 1

. The coefficient of

z^{n-1}

is the sum of all roots." is that Vieta's formula?

Also, if

w = \exp(\frac{2\pi i}{n})

and

w^n - 1 = (w-1)(1+w+w^2+...+w^{n-1})

where

w \neq 1

, then why is

\frac{w^n - 1}{w -1} = 0

when trying to show that

1 + w + w^2 + ... + w^{n-1} = 0

Geometric sum of the right bracket since one is zero.
Well yeh its application of vietas formula yeh. The wiki page covers it rather well for general terms.

Posted from TSR Mobile

Reply 68

JackHKeynes

Original post by Insight314

"Consider the equation

z^n = 1

. The coefficient of

z^{n-1}

is the sum of all roots." is that Vieta's formula?

Also, if

w = \exp(\frac{2\pi i}{n})

and

w^n - 1 = (w-1)(1+w+w^2+...+w^{n-1})

where

w \neq 1

, then why is

\frac{w^n - 1}{w -1} = 0

when trying to show that

1 + w + w^2 + ... + w^{n-1} = 0

w^n - 1 = 0

by definition, since we is an nth root of unity.

(edited 7 years ago)

Reply 69

Zacken

Original post by JackHKeynes

... by definition, since we is an nth root of unity.

You'll want to use [noparse] and not [noparse][\latex][/noparse].

(it's not by definition, it's because

w

is a

n

th root of unity).

(edited 7 years ago)

Reply 70

username1539513

Original post by Zacken

This looks amazing, such a shame I don't have the brain for advanced maths haha. Can you recommend me any resources over the summer to bridge the gap from GCSE to A-level maths? It's been 5 years since I've done maths GCSE and I acheived a B grade. I'm looking at starting an A-level in maths this september coming if my university back up plan does not work out.

(edited 7 years ago)

Reply 71

16Characters....

Original post by Zacken

I'm sure you'll be fine :-)

I think I might teach myself the rest of M3 and 4. There is no M5 with MEI but they used to go up to M6 so that might be the extention :-)

Reply 72

Zacken

Original post by 16Characters....

I'm sure you'll be fine :-)

I think I might teach myself the rest of M3 and 4. There is no M5 with MEI but they used to go up to M6 so that might be the extention :-)

I've heard some very interesting things about those mechanics modules, stuff like intrinsic co-ordinates and the likes. I think I saw a few resources on the really old A-Level spec with all those things, if I ever track them down again I'll let you know. :-)

Reply 73

Insight314

Right, thanks guys. I don't know why I was this silly not to see the initial point of the argument i.e that it is roots of unity so it must be zero haha.

Btw, check out what I just printed out. :smile: