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The Proof is Trivial!

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Surely using more knives decreases the force applied by just one knife making it harder to cut. Then again LOTF can't do mechanics.?

HAHAHAHAHAHHAHA! :lol:

WHY IS THIS SO FUNNY :lol:

Original post by bananarama2

BTW I hope you realise I've resorted to pictures of Emma more exam motivation :tongue:

*Approving nod* :wink:

Original post by miketree

So how would you recommend getting a more problem-solving focused mathematical view? If that is a right way to put it.

Lots of STEP I papers! Loooooooooooads! **** loads! If you can start occasionally getting grade 2s by the time you go to uni (if not now) which corresponds to roughly 50% (I think?), then you will likely be of a standard comparable to the upper quarter of the Russell Group university students at your time of entry! :smile:

Reply 1461

ukdragon37

Original post by miketree

I self-studied my A-level maths and further maths.

Then I'm sure you can try self-studying beyond that too.

Original post by miketree

But surely practicing for exams such as AEA, STEP, etc. are just learning for an exam again, just exams that are at a higher level?

They key difference is that the techniques they require have no set curriculum. Maybe that's not the case as much for the AEA, but STEP for example is explicitly stated not to have a curriculum even though it may be guided by what is taught in A-level. To make the difference more obvious - in A-level you might have a question on integration by parts for the last ten years except two or three, and you know if it comes up again as long as you know how to do integration by parts then you can do the question. In STEP a question combines different techniques, and it's entirely unsurprising if the question that comes up is totally unique and the combination or application of techniques it requires has never come up before. It's up to you to figure out what best to apply during the exam, and that is a skill valuable to a maths degree you can learn to improve. In some questions in STEP it even teaches you a novel technique through a simple example at the start of the question, and then you are tested on whether you can apply it in more difficult contexts for the rest of the question. (This is the same format by the way as many maths interview questions for Oxbridge.)

(edited 10 years ago)

Reply 1462

ukdragon37

Original post by bananarama2

I don't follow...sorry :tongue:

If you go to Cornell surely you are going to one of those many places? Wow...I wish I could say that.

I guess I'd get to see the rest of the US though more easily, which was one of the reasons why I decided to go in the first place. :moon:

Reply 1463

Mladenov

Original post by MW24595

Spoiler

I am in Bulgaria, Dostoevsky and Tolstoy are in our literature programme. :biggrin:

Then, methinks, you have been following wrong way since the first time you refuted her sublime ideas. :tongue:

Solution 213

Let us, instead, consider

\displaystyle f(z)=\frac{\pi \sec \pi z}{z^{5}}

over some nice rectangular contour. Call it

\displaystyle \prod_{R}

Clearly,

\displaystyle \int_{\prod_{R}} f(z) \to 0

R \to \infty

.
Hence,

\displaystyle -\sum_{n= -\infty}^{\infty} Res(f(z),z_{n}) = Res(f(z),0)

.
A bit of arithmetic shows that

\displaystyle Res(f(z),0)=\frac{5}{4!}\pi^{5}

; it is transparent that

\displaystyle Res(f(z),z_{n}) = 2^{7}\frac{(-1)^{n+1}}{(2n+1)^{5}}

Therefore,

\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{5}} = \frac{5}{4!2^{6}}\pi^{5}

.

Another way to solve this problem:

\displaystyle \int_{C_{n}} \frac{dz}{(2z+1)^{5}\sin \pi z} = 2i\pi \left(Res(-\frac{1}{2}) + \sum_{j=-n}^{n} Res(f(z),z_{j}) \right)

, where

C_{n}

is an arbitrary sequence of Jordan curves, which contain the circle

K(O,R)

from some point, and which do not intersect poles of

f(z)

Reply 1464

bananarama2

Original post by ukdragon37

I guess I'd get to see the rest of the US though more easily, which was one of the reasons why I decided to go in the first place. :moon:

I have to say I've never been that attracted to travelling the US. I think there are other places I'd visit first, but then again you want an education as well :tongue:

Reply 1465

miketree

Original post by ukdragon37

...

Thanks for the input.

Is there anything I could really do before I jump straight into STEP etc? I had a look at those questions last year and all the questions seem pretty alien.

Reply 1466

james22

Problem 218 *
Using the operations addition and multiplication as often as required, and using multiplcation only 3 times (no other operations allowed) write (a+bi)(c+di) in the form e+fi

Reply 1467

miketree

Original post by james22

Problem 218 *
Using the operations addition and multiplication as often as required, and using multiplcation only 3 times (no other operations allowed) write (a+bi)(c+di) in the form e+fi

Solution 218
(don't know if this is right or not)

Spoiler

Reply 1468

Jkn

Original post by james22

Problem 218 *
Using the operations addition and multiplication as often as required, and using multiplcation only 3 times (no other operations allowed) write (a+bi)(c+di) in the form e+fi

I'm confused... what do you want us to do? :tongue:

I don't believe it makes sense to assign a "number of times" to a process of multiplication. Here, only one multiplication is performed if you chose to see it that way.

(a+ib)(c+id)=(ac-bd)+i(ad+bc)

It just occurred to me that I don't think I have ever even considered this! (Though I had been ignorantly using it for the last two years...)

Problem 219*

Prove that, when evaluating

\displaystyle \int_a^b u(x) v'(x) \ dx

using integration by parts (the 'parts' being

u(x)

and

v'(x)

), we may disregard an arbitrary constant when integrating

v'(x)

.

So that, for example, if

v'(x)=3x^2+1

, writing

v(x)=x^3+x

will consistently yield the correct answer.

(edited 10 years ago)

Reply 1469

ukdragon37

Original post by Jkn

I'm confused... what do you want us to do? :tongue:

I don't believe it makes sense to assign a "number of times" to a process of multiplication. Here, only one multiplication is performed if you chose to see it that way.

Solution 218 ...?

(a+ib)(c+id)=(ac-bd)+i(ad+bc)

Original post by miketree

Solution 218
(don't know if this is right or not)

Spoiler

He means given a, b, c and d you need to compute e and f from them with only three multiplications. Both of you have clearly used four.

Original post by miketree

Thanks for the input.

Is there anything I could really do before I jump straight into STEP etc? I had a look at those questions last year and all the questions seem pretty alien.

You can try the AEA, but most people who start STEP actually does just jump straight in to its style of questions. The books "Advanced Problems in Core Mathematics" and "Advanced Problems in Mathematics" published by one of the examiners (Siklos) gives examples and solutions in additional to the thought processes involved. Other than that some STEP questions are easier than others, and I'm sure people here can point you to them.

Original post by james22

Problem 218 *
Using the operations addition and multiplication as often as required, and using multiplcation only 3 times (no other operations allowed) write (a+bi)(c+di) in the form e+fi

Solution 218

Unfortunately you picked a time when a CompSci is present :tongue:

Spoiler

(edited 10 years ago)

Reply 1470

Mladenov

Problem 221**

Let

x,y,z

be sides of a triangle. Show that

\displaystyle \frac{(x+z-y)^{4}}{x(x+y-z)}+\frac{(x+y-z)^{4}}{y(y+z-x)} + \frac{(y+z-x)^{4}}{z(z+x-y)} \ge xy+yz+zx

.

Problem 222**

Find all

f: \mathbb{R} \to \mathbb{R}

which satisfy

f(x^{2})+f(xy)= f(x)f(y)+yf(x)+xf(x+y)

.

C'mon problem 219 is trivial,

C(u(b)-u(a))

cancels as it is equal to

\displaystyle \int_{a}^{b} Cu'(x)dx

(edited 10 years ago)

Reply 1471

bananarama2

Original post by Jkn

It just occurred to me that I don't think I have ever even considered this! (Though I had been ignorantly using it for the last two years...)

Don't you find it quite pleasing when a thought strikes you, not one that is really complex, just one that makes you think: "I hadn't considered that."

Reply 1472

miketree

Original post by ukdragon37

I looked at STEP I 2012 qu 1 and it seems so basic but I have no idea what to do? any hints?

Reply 1473

ukdragon37

Original post by miketree

I looked at STEP I 2012 qu 1 and it seems so basic but I have no idea what to do? any hints?

You can try asking in the STEP prep thread, where STEP questions will always be welcome. Sorry I'm conserving brainpower right now for my dissertation and I'm only on TSR for breaks when I'm stuck on something. :tongue:

Also beware that's a recent exam question so people might prefer to converse with you under a spoiler because others may want to keep that question unseen for a mock.

(edited 10 years ago)

Reply 1474

Slumpy

Original post by Mladenov

Problem 221**

Find all

f: \mathbb{R} \to \mathbb{R}

which satisfy

f(x^{2})+f(xy)= f(x)f(y)+yf(x)+xf(x+y)

x=0 gives y+f(y)=2, so f(x)=2-x. Quick check shows this satisfies the original equation.
Feels like I must've missed something!
Ah, of course, I've entirely omitted when f(0)=0. Give me a couple of minutes...

Edit; f(0)=0, set y=0, f(x^2)=xf(x), so f(x)=+/-x

(edited 10 years ago)

Reply 1475

Jkn

Original post by bananarama2

Don't you find it quite pleasing when a thought strikes you, not one that is really complex, just one that makes you think: "I hadn't considered that."

Yes, exactly! Whilst the proof is trivial ( :colone:

), it is an example of the kinds of things that improve your understanding of the problem-solving algorithms you are taught in school :tongue:

Original post by ukdragon37

He means given a, b, c and d you need to compute e and f from them with only three multiplications. Both of you have clearly used four.

But how is 'number of multiplications' a rigorous concept? Clearly a more rigorous definition is needed here as multiplications are essentially built up from addition so, no matter which way you do it, it can always only be considered 'one' multiplication. Also, we use certain techniques to perform multiplication such as, if we were multiplying a large number by 12, splitting 12 into 10+2. We can also multiply an arbitrary number of times by 'one'... ?

Reply 1476

Jkn

Original post by Mladenov

Problem 220**

Let

x,y,z

be sides of a triangle. Show that

\displaystyle \frac{(y+z-x)^{4}}{x(x+y-z)}+\frac{(z+x-y)^{4}}{y(y+z-x)} + \frac{(y+z-x)^{4}}{x(z+x-y)} \ge xy+yz+zx

Is this correct? :tongue:

(It lacks the symmetry the inequalities you set tend to have :colone:

)

Reply 1477

ukdragon37

Original post by Jkn

What he didn't mention although assumed was the number of additions and subtractions is required to be bounded by a constant. ab as a summed b times is not bounded.

Original post by Jkn

Also, we use certain techniques to perform multiplication such as, if we were multiplying a large number by 12, splitting 12 into 10+2. We can also multiply an arbitrary number of times by 'one'... ?

It would have been better phrased as "Imagine you have a calculator such that every non-number button is broken except decimal point, equals, addition, subtraction and multiplication keys, but you can only press multiply three times before it breaks also." :tongue:

Incidentally there is a point to these restrictions - they are representative of the conditions in a processor where this recipe could be a shortcut.

(edited 10 years ago)

Reply 1478

bananarama2

Original post by ukdragon37

Incidentally there is a point to these restrictions - they are representative of the conditions in a processor where this recipe could be a shortcut.

I wondered whether this is what you were hinting at when you said about a compsci being around.

Reply 1479

Jkn

Original post by ukdragon37

What he didn't mention although assumed was the number of additions and subtractions is required to be bounded by a constant. ab as a summed b times is not bounded.

I don't know why that would be assumed, it changes everything and is crucial to the question! That's like asking "how likely is it that I will go inside the white house today?" and forgetting to say that we are assuming I'm Barack Obama :lol:

Incidentally there is a point to these restrictions - they are representative of the conditions in a processor where this recipe could be a shortcut.

Well now it's nice and well-defined, I am happy! :wink:

Hmm, that's quite cool :tongue:

Is optimising mathematical processes in this way the kinds of things you do in compsci, or is it programming and stuff like that? I literally have no idea what computer scientists even do... :ninja: