Official TSR Mathematical Society

This is a quote of what was said earlier and failed to parse

Reply 1302

The Muon

2

SimonM

How about the old latex?

What do you mean?

Reply 1303

SimonM

The Muon

What do you mean?

See what I quoted? That was valid latex which doesn't parse where you've written it

Reply 1304

The Muon

2

SimonM

See what I quoted? That was valid latex which doesn't parse where you've written it

Oh I see. Was that just the Latex posted whilst latex as down?

Reply 1305

Let

\alpha

be a root of

x^4 - x^3 +x^2 - x + 1

.

Find the value of

\alpha^{20} - \alpha^{15} + \alpha^{10} - \alpha^{5} + 1

Reply 1306

Daniel Freedman

Let

\alpha

be a root of

x^4 - x^3 +x^2 - x + 1

.

Find the value of

\alpha^{20} - \alpha^{15} + \alpha^{10} - \alpha^{5} + 1

x^4 - x^3 +x^2 - x + 1 = 0 = (x^2 + x^{-2}) - (x + x^{-1}) + 1

2cos(2\theta) - 2cos(\theta) + 1 = 0 \Rightarrow cos(\theta) = \frac{1+\sqrt{5}}{4} \Rightarrow \theta = \frac{\pi}{5}

Then with the use of the useful

e^{i\theta} = cos(\theta) + isin(\theta) \Rightarrow \alpha^{20} - \alpha^{15} + \alpha^{10} - \alpha^{5} + 1 = e^{i\pi 4} - e^{i \pi 3} + e^{i\pi 2} - e^{i\pi} + 1 = 5

Reply 1307

DeanK22

x^4 - x^3 +x^2 - x + 1

= 0 =

(x^2 + x^{-2}) - (x + x^{-1}) + 1

2cos(2\theta) - 2cos(\theta) + 1 = 0 \Rightarrow cos(\theta) = \frac{1+\sqrt{5}}{4} \Rightarrow \theta = \frac{\pi}{5}

Then with the use of the useful

e^{i\theta} = cos(\theta) + isin(\theta) \Rightarrow \alpha^{20} - \alpha^{15} + \alpha^{10} - \alpha^{5} + 1 = e^{i\pi 4} - e^{i \pi 3} + e^{i\pi 2} - e^{i\pi} + 1 = 1

Your answer doesn't match the one I have.

EDIT: You edited.

Reply 1308

Daniel Freedman

I don't completely follow what you've done, and your answer doesn't match the one I have.

... erm. Sorry. I added up incorrectly. The answer should not be one. I think I should be get 5 (will confirm shortly).

Reply 1309

DeanK22

... erm. Sorry. I added up incorrectly. The answer should not be one. I think I should be get 5 (will confirm shortly).

Yes, 5 is right.

Although, I prefer the method I have.

Reply 1310

Daniel Freedman

Yes, 5 is right.

Although, I prefer the method I have.

hmm. I like using this result;

z^n + z^{-n} = 2cos(n\theta)

... Out of curiosity how did you do it?

Reply 1311

DeanK22

hmm. I like using this result;

z^n + z^{-n} = 2cos(n\theta)

... Out of curiosity how did you do it?

Your method is actually pretty nice :smile:

Alternatively,

Unparseable latex formula:

\\ x^5+1 = (x+1)(x^4-x^3+x^2-x+1) \\[br]\\ \implies \alpha^5 + 1 = (\alpha+1)(\alpha^4 - \alpha^3 + \alpha^2 - \alpha + 1) = 0 \\[br]\\ \implies \alpha^5 = -1 \\[br]\\ \implies \alpha^{20} - \alpha^{15} + \alpha^{10} - \alpha^5 + 1 = 1-(-1)+1-(-1)+1=5

Reply 1312

Swayum

Daniel Freedman

Your method is actually pretty nice :smile:

Alternatively,

x^5+1 = (x+1)(x^4-x^3+x^2-x+1)

What made you think to consider x^5 + 1 in the first place?

Reply 1313

SimonM

Swayum

What made you think to consider x^5 + 1 in the first place?

Geometric progressions, equalling zero... looks like roots of unity to me

Reply 1314

Swayum

What made you think to consider x^5 + 1 in the first place?

From the question, knowing something about

\alpha^5

would be very useful.

EDIT: Geometric progressions, as Simon said.

Reply 1315

I'll give credit to the source of this afterwards, as the solution is on there too.

For all real x, evaluate

\displaystyle \sum_{r=0}^{\infty} \frac{\sin^3{(3^r x)}}{3^r}

Reply 1316

Lysdexia

Here's a question I've always wanted an answer to. It's probably very simple for most of you; I'm only a lowly A2 Maths student!

If there are an infinite amount of numbers, and therefore an infinite amount of numbers between 1 and 2, how is it possible to get from 1 to 2? That implies a beginning and an end.

Reply 1317

Lysdexia

Here's a question I've always wanted an answer to. It's probably very simple for most of you; I'm only a lowly A2 Maths student!

If there are an infinite amount of numbers, and therefore an infinite amount of numbers between 1 and 2, how is it possible to get from 1 to 2? That implies a beginning and an end.

What do you mean by "get from 1 to 2"?

Reply 1318

Lysdexia

That 1 is the beginning of a series of an infinite amount of numbers and that 2 somehow causes an abrupt end to this series.

Reply 1319