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# A Summer of Maths Tweet

Maths and statistics discussion, revision, exam and homework help.

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1. Re: A Summer of Maths
(Original post by electriic_ink)
Let z=y'.

Also quite easy: How many (real) solutions does have? Knowledge required: GCSE & below.
Thanks

(and none surely?)
2. Re: A Summer of Maths
(Original post by Brit_Miller)
Thanks

(and none surely?)
Yeah
3. Re: A Summer of Maths
(Original post by 4ever_drifting)
Love these kind of problems Just a note though - could you not observe that you need the product of the roots of the quadratic equation, which is given by c/a in 0=ax^2+bx+c, so you know the answer is 26 without having to find the roots then multiply them together?
You could in this case because the quadratic simultaneous equations you get when substituting for X and when substituting for Y are the same. I preferred not to take that shortcut on the off-chance that substituting for Y generated a different quadratic than substituting for X.

I think you could probably solve it geometrically as well, the method I posted was just the one I prefer to use.
4. Re: A Summer of Maths
(Original post by DJMayes)
At the Nottingham University Open Day there was a "Maths Trail" with several interesting questions on it. The questions ranged from requiring a working knowledge of arithmetic progressions and combinations to lowest common multiples and counting squares; and more emphasis was put on thinking about them than slogging through endless manipulation. I thought I'd share one with you. The question is of the kind that could be set in C1, but is an interesting one:

A rectangle is inscribed inside a circle of radius 6 units such that each of the vertices of the rectangle lie on the circumference of the circle. Given that the perimeter of the rectangle is 28 units, what is the area?

Required Knowledge:
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- Pythagoras' Theorem

Hints:
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What information are you given? You have both the perimeter of the rectangle, and the radius of the circle it is inscribed inside. How can you relate these in terms of the length and width of the rectangle?
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Consider the diagonal of the rectangle

Full Solution:
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Let X and Y represent the length and width of the rectangle. Using this and the perimeter we can automatically deduce an equation:

2X + 2Y = 28

Which simplifies to X + Y = 14

As we are told that the radius of the circle is 6, we also know that its diameter must be 12. This diameter forms the diagonal of the rectangle. From this, another equation can be deduced using Pythagoras' Theorem:

X^2 + Y^2 = 144

We now have a set of simultaneous equations to solve. Re-arranging the first to leave Y in terms of X and substituting in the second equation leaves you with this:

X^2 + (14 - X)^2 = 144

This expands to:

X^2 + X^2 - 28X + 196 = 144

Which then simplifies to:

2X^2 - 28X + 52 = 0

or:

X^2 - 14X + 26 = 0

You can then complete the square or use the quadratic formula to get the result:

X = 7 + rt(23) or X = 7 - rt(23)

Using this you can then confirm that Y has the same values (i.e. if X = 7 + rt(23) then Y = 7 + rt(23) and vice-versa)

Now you know both side lengths, you simply multiply them together to get the area. The final answer is 26.
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This could be made a lot neater. Starting as you did, x + y = 14, and x^2 + y^2 = 144. Now see that (x+y)^2 = x^2 + y^2 + 2xy and so area = xy = 0.5(196 - 144) = 26
5. Re: A Summer of Maths
(Original post by hassi94)
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This could be made a lot neater. Starting as you did, x + y = 14, and x^2 + y^2 = 144. Now see that (x+y)^2 = x^2 + y^2 + 2xy and so area = xy = 0.5(196 - 144) = 26
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That's a very clever trick for doing it actually! Unfortunately I have an insistence on solving all simultaneous equations (Even linear ones) through substitution instead of subtraction or manipulation which prevents me from spotting things like this
6. Re: A Summer of Maths
This one is a bit tricky, but I think it has something to teach; use hints if you get stuck.

{*} Question:

Let be a function with the property that there exists such that

for all .

Prove that is periodic.

{**} Required: (A-level)

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A function is periodic, with period , if for every in its domain one has where is an integer.

{***} Hints:

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(1)
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(2)
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How does look like?

7. Re: A Summer of Maths
(Original post by hassi94)
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This could be made a lot neater. Starting as you did, x + y = 14, and x^2 + y^2 = 144. Now see that (x+y)^2 = x^2 + y^2 + 2xy and so area = xy = 0.5(196 - 144) = 26
wasnt this in the online lecture thing?
8. Re: A Summer of Maths
(Original post by james22)
What is the derivative of y=x^x?

What is the derivatie and inverse of y=x^x^x^x^... (an infinite string of x's)

Here x^x^x=x^(x^x) not (x^x)^x

Also for what values of x does y exist?
Some thoughts under cruel assumptions.

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Without any justification, I claim that .

Maybe the inverse will be

Now, note that , and hence,

which gives as a final result

9. Re: A Summer of Maths
I wonder how people on here find Analysis in general.

Do you like it (assuming you have seen some), or do you think it is difficult to understand and apply?
10. Re: A Summer of Maths
This one is a bit tricky, but I think it has something to teach; use hints if you get stuck.

{*} Question:

Let be a function with the property that there exists such that

for all .

Prove that is periodic.

{**} Required: (A-level)

Spoiler:
Show

A function is periodic, with period , if for every in its domain one has where is an integer.

{***} Hints:

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(1)
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(2)
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How does look like?

Didn't seem that hard.
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Let f(x)=F
subbing x for x+w we obtain f(x+2w)=(2F-5)/(F-2)
subbing x for x+w again we obtain f(x+3w)=(-3F+5)/(-F+1)
subbing x for x+w again we obtain f(x+4w)=F=f(x)
so clearly f(x)=f(x+4nw) for every n. Period is 4w.
Thus we are done.
Last edited by Blutooth; 03-07-2012 at 02:20.
11. Re: A Summer of Maths
Some thoughts under cruel assumptions.

Spoiler:
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Without any justification, I claim that .

Maybe the inverse will be

Now, note that , and hence,

which gives as a final result

You have the right inverse and are right in the simplification of the original equation. I cannot remember the derivative but your answer looks about right. Any luck with the values of x for which this converges?
12. Re: A Summer of Maths
I wonder how people on here find Analysis in general.

Do you like it (assuming you have seen some), or do you think it is difficult to understand and apply?
I find analysis a bit boring to be honest. Perhaps because it is not my strong point by any means, I don't know. I prefer "even purer maths"... They seem to have more depth and involve more creativity.
13. Re: A Summer of Maths
(Original post by Blutooth)
...
Can you use a spoiler, please. Thanks.

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Can take the values or ?

(Original post by james22)
Any luck with the values of x for which this converges?
I don't have a good reason to believe what I derived works, so I have only rough ideas.

(Original post by Lord of the Flies)
I find analysis a bit boring to be honest. I prefer "even purer maths"... They seem to have more depth and involve more creativity.
I have studied a bit of it, and I find it interesting; some of Cauchy's stuff is amazing and I would say it is creative.
However, I do get stuck from time to time -- I thought I understood uniform continuity two months ago, but now I have to read it again.
14. Re: A Summer of Maths
Can you use a spoiler, please. Thanks.

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Can take the values or ?

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No, by considering the denominator of f(x+2w), f(x+3w) etc. orplugging f(x+w)=3 and working out the value of f(x).
Last edited by Blutooth; 03-07-2012 at 02:11.
15. Re: A Summer of Maths
{*} Question:

The polynomial is irreducible over .

i) By completing the square, show that is not irreducible over the set of real numbers.

Hence, derive the Sophie Germain algebraic identity

by starting from the left-hand side.

ii) Evaluate

{**} Required:

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A polynomial is said to be irreducible over a set if it cannot be factored into polynomials with coefficients from the given set.

As an example,

is irreducible over the set of rational numbers denoted by .

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16. Re: A Summer of Maths
(Original post by james22)
You have the right inverse and are right in the simplification of the original equation. I cannot remember the derivative but your answer looks about right. Any luck with the values of x for which this converges?
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Domain and convergence

clearly both and are undefined.

and strictly increases strictly increases so is injective for

and strictly increases for and strictly decreases for .

Also, . Therefore is undefined for since it would return exactly two values.

Additionally, since the maximum of occurs only for has exactly one solution so is injective for and is undefined for all

Therefore convergence occurs when or

If that is what you meant by convergence?
Last edited by Lord of the Flies; 03-07-2012 at 09:04.
17. Re: A Summer of Maths
Here's an easy one!

Question

Evaluate

Required

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L'Hôpital's rule
Last edited by Lord of the Flies; 03-07-2012 at 09:42.
18. Re: A Summer of Maths
(Original post by Lord of the Flies)
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Domain and convergence

clearly both and are undefined.

and strictly increases strictly increases so is injective for

and strictly increases for and strictly decreases for .

Also, . Therefore is undefined for since it would return exactly two values.

Additionally, since the maximum of occurs only for has exactly one solution so is injective for and is undefined for all

Therefore convergence occurs when or

If that is what you meant by convergence?
That is what I mean by convergence.

You are almost right, but it converges for all 0<x<e^(1/e)
19. Re: A Summer of Maths
(Original post by james22)
That is what I mean by convergence.

You are almost right, but it converges for all 0<x<e^(1/e)
Really? I don't see how... When returns two values, no?
20. Re: A Summer of Maths
(Original post by Lord of the Flies)
Here's an easy one!

Question

Evaluate

Required

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L'Hôpital's rule
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Is it undefined? I make lim x->0 f'(x)/g'(x) 0/0, so obviously g'(x)=0