The Student Room Group

The Proof is Trivial!

Scroll to see replies

Original post by metaltron
I prefer to keep it real :biggrin:.


Perhaps you lack IMAGINAtion :wink:
Reply 1941
Original post by Hasufel
Problem 276

(simple one, really)

if u+v=50u+v=50 which choice of both uu and vv make u×vu \times v as large as posilble?

Alternate solution:

Consider quadratic equation (xu)(xv)=x250x+P=05024P\displaystyle (x-u)(x-v)=x^2-50x+P=0 \Rightarrow 50^2 \ge 4P as the equation has two real roots. P5024\displaystyle \Rightarrow P \le \frac{50^2}{4} and so the maximum value of the product occurs when "b2=4acb^2=4ac" (i.e. when the quadratic has equal roots). This tells us that the maximum value occurs at u=vu=v.

(I have assumed that the roots u and v are real, of course). :tongue:

Another alternate solution:

Using AM-GM, we know that u+v2uvuv5024\displaystyle \frac{u+v}{2} \ge \sqrt{uv} \Rightarrow uv \le \frac{50^2}{4} We know that equality holds when u=vu=v from the property of the AM-GM inequality. Edit: We must also note that u, v are positive and real etc.. which can be deduced both by reasoning and assumption.
(edited 10 years ago)
Original post by bogstandardname
Problem 276 *

Evaluate 02π1esinx+1dx\displaystyle \int_0^{2\pi}\frac{1}{e^{\sin x}+1}\,dx

Spoiler


Solution 276:
I=02π1esin(x)+1 dxI = \displaystyle\int^{2\pi}_0 \dfrac{1}{e^{\sin (x)}+1}\ dx
Let:
x2πx[br]I=02π1esin(x)+1 dx[br]2I=02π1esin(x)+1 dx+02π1esin(x)+1 dx=02π1 dx[br]I=πx \to 2\pi -x[br]\Rightarrow I=\displaystyle\int^{2\pi}_0 \dfrac{1}{e^{-\sin (x)}+1}\ dx[br]\Rightarrow 2I = \displaystyle\int^{2\pi}_0 \dfrac{1}{e^{\sin (x)}+1}\ dx+\displaystyle\int^{2\pi}_0 \dfrac{1}{e^{-\sin (x)}+1}\ dx = \displaystyle\int^{2\pi}_0 1 \ dx[br]\Rightarrow I =\pi
(edited 10 years ago)


As I'm sure you intended, the last part of Problem 271 is giving me a headache :cool:
There are several nice little theorems and properties which will enable the last part to be done in seconds :tongue:
Original post by joostan
I=02π1esin(x)+1 dxI = \displaystyle\int^{2\pi}_0 \dfrac{1}{e^{\sin (x)}+1}\ dx
Let:
x2πx[br]I=02π1esin(x)+1 dx[br]2I=02π1esin(x)+1 dx+02π1esin(x)+1 dx=02π1 dx[br]I=πx \to 2\pi -x[br]\Rightarrow I=\displaystyle\int^{2\pi}_0 \dfrac{1}{e^{-\sin (x)}+1}\ dx[br]\Rightarrow 2I = \displaystyle\int^{2\pi}_0 \dfrac{1}{e^{\sin (x)}+1}\ dx+\displaystyle\int^{2\pi}_0 \dfrac{1}{e^{-\sin (x)}+1}\ dx = \displaystyle\int^{2\pi}_0 1 \ dx[br]\Rightarrow I =\pi


What do you call this method btw?- where you have used a dummy variable and then replace it with the x because it doesnt matter what variable you use.

x2πxx \to 2\pi -x
this bit
Reply 1945
Original post by joostan
As I'm sure you intended, the last part of Problem 271 is giving me a headache :cool:
There are several nice little theorems and properties which will enable the last part to be done in seconds :tongue:

Indeed I did! :colone:

I take it you are referring to digamma representations? Your task now is to find either a proof of one of such representations, the hardest part being to link it with γ\gamma directly from the definition. :colone:
Original post by nahomyemane778
What do you call this method btw?- where you have used a dummy variable and then replace it with the x because it doesnt matter what variable you use.

x2πxx \to 2\pi -x
this bit

I like to think of it as 'limit-flipping'. Everyone seems to think it is hard to spot when to use it but, once you have it in your toolbox, it's just like another other standard technique. :tongue:
Original post by Jkn
Indeed I did! :colone:

I take it you are referring to digamma representations? Your task now is to find either a proof of one of such representations, the hardest part being to link it with γ\gamma directly from the definition. :colone:

Yep. . . The rest of the problem's fairly straightforward, I also found a couple of really nice papers on the Gamma, digamma and polygamma functions :smile: and from several of the properties I can prove it, but I can't prove the properties :colone:

Original post by nahomyemane778
What do you call this method btw?- where you have used a dummy variable and then replace it with the x because it doesnt matter what variable you use.

x2πxx \to 2\pi -x
this bit


I call it introducing a dummy variable
In general:
abf(x) dx=abf(a+bx) dx\displaystyle\int^b_a f(x) \ dx = \displaystyle\int^b_a f(a+b-x) \ dx
On occasion this is a good way to solve integrals
(edited 10 years ago)
Original post by Jkn
Indeed I did! :colone:

I take it you are referring to digamma representations? Your task now is to find either a proof of one of such representations, the hardest part being to link it with γ\gamma directly from the definition. :colone:

I like to think of it as 'limit-flipping'. Everyone seems to think it is hard to spot when to use it but, once you have it in your toolbox, it's just like another other standard technique. :tongue:


Its a really smart trick- is it something that you either have learnt or havent or am i just not so bright- i would never have thought to substitute u as x and replace it as x again with definite integrals
Reply 1948
Original post by joostan
Yep. . . The rest of the problem's fairly straightforward, I also found a couple of really nice papers on the Gamma, digamma and polygamma functions :smile: and from several of the properties I can prove it, but I can't prove the properties :colone:

Don't you just love saying 'papers'.. :colone:
In general:
abf(x) dx=abf(a+bx) dx\displaystyle\int^b_a f(x) \ dx = \displaystyle\int^b_a f(a+b-x) \ dx
On occasion this is a good way to solve integrals :smile:

Don't forget that it can also be very useful with summations and products (as the above formula applies in those cases too).

I find that with 'problems of substance', it tends to be a great way to start. Any problem that is trivialised by such a technique is certainly only of A-level standard (for someone who has seen the technique before). :tongue:
Original post by nahomyemane778
Its a really smart trick- is it something that you either have learnt or havent or am i just not so bright- i would never have thought to substitute u as x and replace it as x again with definite integrals

Well I don't know about other people, but I was never taught it formally. Don't worry if you didn't come up with it yourself. :tongue:
(edited 10 years ago)
Original post by Jkn
Don't you just love saying 'papers'.. :colone:

Don't forget that it can also be very useful with summations and products (as the above formula applies in those cases too).

Interesting, I've never come across a sum that requires me to do that. . .
Original post by Jkn
I find that with 'problems of substance', it tends to be a great way to start. Any problem that is trivialised by such a technique is certainly only of A-level standard (for someone who has seen the technique before). :tongue:

Usually you can just look at a problem and tell whether or not it's useful. But I agree. :smile:
Btw dude for problem 271 are we allowed to use the infinite product definition? I think I can see a solution with that

Original post by Jkn
Ikr! You're going to absolutely destroy at uni next year. Are you applying (or should I saying going) to Cambridge and, if so, which college? :tongue:

Nah, I'm applying to Ox....joking :colone: Of course :wink: I'm not 100% sure on which college yet but I've got a shortlist of: Pembroke, Clare, Kings and Emma.

Thanks man! :biggrin: I've got into the habit of extending everything now, it's so satisfying! I've fallen in love with Euler's reflection formula btw! :biggrin: I was messing about with the problem as I was trying to work through Mladenov's contour integration method, and then I though I would ass that to the mix. I then thought that I probably looked like a bit of a dick as I was bringing nothing new to the table and then realised that, whist repeated differentiation tends to be tedious, the Taylor's series was an awesome shortcut! :biggrin: We don't use Taylor's (or Laurent) series enough on this thread, perhaps you could dig up some fun questions?

I've still yet to get properly acquainted with Euler's reflection formula :emo: Maybe because I haven't done that many integrals involving the Beta/ Gamma functions yet :lol: Haha, I see! I'll try and have a look around for some Taylor series problems then :wink: Feel like a bit of a dick not having posted that many problems in a while :colondollar:

Cheers but I felt like a bit of an idiot right after I finished as I realised that it was entirely unnecessary given the fact that all I had to do was use a trigonometric substitution to trivially evaluate the appropriate value of the Beta function (though I left it there as the digamma method of evaluating that integral was interesting in itself :colondollar:). Incidentally, it was that precise integral that was the entirety of STEP II Q2 2013 (which makes it really satisfying that I found it of my own accord) :smile:

I'm looking forward to being taught how to evaluate the Gaussian next year :lol: I'll stroll up like "yeah I found like 10 ways to do this ever the summer :pierre:"

Ha, it's still a nice method and it doesn't involve all that polar coordinate jazz :wink: Yeah, I saw that on the STEP paper! Was nice seeing the Beta function pop up and it made for a sweet question :biggrin:

Lol, at the rate you guys are going on the ASOM thread, I assume that's going to happen on a lot of classes :tongue:
I hope we are not being too narrow by being obsessed with integration and special functions..

Oh, btw, if you liked some of those posts you might like the one I wrote about the E-M constant (etc..) a few pages back on ASOM :smile:

I sometimes feel like I'm neglecting other areas of maths by spending too much time on integrals :sadnod: Then I remember how awesome they are :colone: :sogood: Nah, seriously there are other areas of maths I wanna look and improve at over this summer (mainly to try and get to BMO2 next year) :biggrin:

I'll have a look around but I haven't been really keeping up to date with that thread :P
Reply 1951
Original post by joostan
Usually you can just look at a problem and tell whether or not it's useful. But I agree. :smile:

I disagree. There are also ones where you need to map xx1x \mapsto x^{-1} and there was even one problem where I had to map x1x+1x \mapsto \frac{1}{x+1} to take it from [0,1] to [0,\infty] which I thought was really hard to spot (though perhaps that's a slightly different thing as I am generalising to all problems that are susceptible to 'limits tricks'). I also remember another that invalid integrating cosxx2+x+1\frac{\cos x}{x^2+x+1} or something like that over some limits that was easy to spot but didn't, at first, seem all that useful. :tongue:
Original post by Felix Felicis
Btw dude for problem 271 are we allowed to use the infinite product definition? I think I can see a solution with that

Hmm, I suppose so (derive it from the general theorem using residues if you are feeling brave?). You are still king to have to do something with that ghastly limit definition though! :biggrin:
Nah, I'm applying to Ox....joking :colone:

LOL :lol:
Of course :wink: I'm not 100% sure on which college yet but I've got a shortlist of: Pembroke, Clare, Kings and Emma.

Ah, that's am extremely good shortlist! What was your reason for discarding Trinity then? :tongue:
I've still yet to get properly acquainted with Euler's reflection formula :emo: Maybe because I haven't done that many integrals involving the Beta/ Gamma functions yet :lol: Haha, I see! I'll try and have a look around for some Taylor series problems then :wink: Feel like a bit of a dick not having posted that many problems in a while :colondollar:

I swear you've done loads! :lol: You best do, you should feel like a dick! I'm pretty sure I have posted at least a fifth of the problems on this thread.. :lol:
Ha, it's still a nice method and it doesn't involve all that polar coordinate jazz :wink: Yeah, I saw that on the STEP paper! Was nice seeing the Beta function pop up and it made for a sweet question :biggrin:

Very true. The proof of the Gamma-Beta link though.. :| (beautiful theorem no doubt!)

Certainly! I wrote a little note to the examiner about the wide applicability of what we had proven! :smile:
Lol, at the rate you guys are going on the ASOM thread, I assume that's going to happen on a lot of classes :tongue:

I hope so! :colone: Not tempted to join?
I sometimes feel like I'm neglecting other areas of maths by spending too much time on integrals :sadnod: Then I remember how awesome they are :colone: :sogood: Nah, seriously there are other areas of maths I wanna look and improve at over this summer (mainly to try and get to BMO2 next year) :biggrin:

I'll have a look around but I haven't been really keeping up to date with that thread :P

They literally are! It's one of the few examples I know of of 'pure' problem solving (by which I mean problems that are easily stated in a non-ambiguous standard way, etc..). Perhaps skills with things like that would be useful if went in to fractional calculus or something like that? :tongue:

Good luck, if you don't get through, pay to go in! You're going to love it when you find it :colone:
(edited 10 years ago)
Original post by Jkn
I disagree. There are also ones where you need to map xx1x \mapsto x^{-1} and there was even one problem where I had to map x1x+1x \mapsto \frac{1}{x+1} to take it from [0,1] to [0,\infty] which I thought was really hard to spot (though perhaps that's a slightly different thing as I am generalising to all problems that are susceptible to 'limits tricks'). I also remember another that invalid integrating cosxx2+x+1\frac{\cos x}{x^2+x+1} or something like that over some limits that was easy to spot but didn't, at first, seem all that useful. :tongue:


There's no doubt that changing limits is a good and useful thing to do, that's not what I was saying. Only that the introduction of a dummy variable is, as far as I'm aware limited. :smile:
Original post by Hasufel
Problem 276

(simple one, really)

if u+v=50u+v=50 which choice of both uu and vv make u×vu \times v as large as posilble?

We provide a proof by gratuitous use of Lagrange multipliers.

Spoiler

Reply 1954
Original post by Felix Felicis
x

Impressive stuff man, well done. Unofruntely for you, you are not quite finished.

Pedantry: for the final limit you are correct but you seem unsure. To justify it, simplify appreciate that you can subtract the (n+1)th harmonic number without changing the sum.

The question asks that you link together 4 things. You have linked together 3. Whilst I accept that you may want to define the Gamma function differently, I can imagine that the way you would have evaluated the first integral (which I'm assuming seemed too trivial for you to need to type) would be from using the integral definition of the Gamma function.

You logically can chose whichever definition you want so long as, in any piece of mathematics, you adhere you only one definition. You, hence, must show that the product definition of the Gamma function is equivalent to the integral definition, or find a way to link the first integral using your new definition.

Have fun :wink:
Original post by Jkn
Impressive stuff man, well done. Unofruntely for you, you are not quite finished.

Pedantry: for the final limit you are correct but you seem unsure. To justify it, simplify appreciate that you can subtract the (n+1)th harmonic number without changing the sum.

The question asks that you link together 4 things. You have linked together 3. Whilst I accept that you may want to define the Gamma function differently, I can imagine that the way you would have evaluated the first integral (which I'm assuming seemed too trivial for you to need to type) would be from using the integral definition of the Gamma function.

You logically can chose whichever definition you want so long as, in any piece of mathematics, you adhere you only one definition. You, hence, must show that the product definition of the Gamma function is equivalent to the integral definition, or find a way to link the first integral using your new definition.

Have fun :wink:

Crap :facepalm: I'd been spending so much time trying to show Γ(1)=γ\Gamma ' (1) = - \gamma I completely forgot about the original problem and there was an integral in it. Oh dear :colondollar: Hang on...
Reply 1956
Original post by Felix Felicis
Crap :facepalm: I'd been spending so much time trying to show Γ(1)=γ\Gamma ' (1) = - \gamma I completely forgot about the original problem and there was an integral in it. Oh dear :colondollar: Hang on...

Mwuhahaha :colone:

Omg, read this!

I starts off: "The Gamma function of Euler often is thought as the only function which interpolates the factorial numbers n! = 1,2,6,24,... This is far from being true".

Ohhhhh ****.
PROBLEM 277

Derive (pardon the pun) a series representation, excluding the constant of integration, for:

xn(In(x))mdx\displaystyle \int x^{n}(In(x))^{m}dx

(Hint: one way is to incorporate derivatives into your answer)
(edited 10 years ago)
Original post by Hasufel
PROBLEM 277

Derive (pardon the pun) a series representation, excluding the constant of integration, for:

xn(In(x))mdx\displaystyle \int x^{n}(In(x))^{m}dx

(Hint: one way is to incorporate derivatives into your answer)


Is the supposed to be a natural log?
Reply 1959
Problem 278***

Evaluate limxΓ(x+1)xxex2πx\displaystyle \lim_{x \to \infty} \frac{\Gamma(x+1)}{x^x e^{-x} \sqrt{2 \pi x}} for xRx \in \mathbb{R}.

Comment also on what happens when xCx \in \mathbb{C} and x|x| \to \infty.

Problem 279*/**

Evalute limx0+(lnx)3(arctan(ln(x+x2))+π2)+(lnx)2\displaystyle \lim_{x \to 0^{+}} ( \ln x)^3 \left( \arctan ( \ln (x+x^2))+\frac{\pi}{2} \right) + (\ln x)^2

Problem 280*

Prove that sinh1x=n=0(1)n(2n)!4n(n!)2(2n+1)x2n+1\displaystyle \sinh^{-1} x = \sum_{n=0}^{\infty} \frac{(-1)^n (2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1} for x1|x| \le 1.

Original post by peter12345
x

Courtesy of "peter12345":

Problem 281
*

Find an infinite series representation for cosπx\displaystyle \cos \frac{\pi}{x} and sinπx\displaystyle \sin \frac{\pi}{x} involving non-trancendental numbers (ignoring, of course, the possible transcendence of x). A transcendental number is a number that cannot be expressed as the root of a polynomial equation with integer coefficients.

Problem 282**

Prove or disprove the statement that 011+x212x213x4(sinxx)2 dx\displaystyle \int_0^1 \frac{\sqrt{1+x^2}-\frac{1}{2} x^2-1-3x^4}{(\sin x - x)^2} \ dx is convergent. If the integral is convergent, what value does it converge to?

Problem 283***

Prove that n11n=γ\displaystyle \sum_{n \ge 1}^{\Re} \frac{1}{n} = \gamma, where \Re denotes a 'Ramanujan Summation' (a technique used by Srinivasa Ramanujan to assign meaningful values to divergent series).

Hence, or otherwise, comment humorously upon the reaction of English professors that he sent letters to that included such series (without specification that they were 'Ramanujan Summations', as they are now known).

Problem 284*/**/***

Evaluate n=qn(1+qn+1)(1+qn+2)\displaystyle \sum_{n=-\infty}^{\infty} \frac{q^n}{(1+q^{n+1})(1+q^{n+2})}

Problem 285**/***

Find, with proof, d12dx12xn\displaystyle \frac{d^{\frac{1}{2}}}{d x^{\frac{1}{2}}} x^n, where we define the half derivative to be the unique operator HH such that H2f(x)=Df(x)\displaystyle H^2 f(x)=D f(x) where DD denotes the differential operator.

Verify also, by applying this operation twice, that your relation is consistent with ddxxn\displaystyle \frac{d}{dx} x^n.

Problem 286***

Evaluate n=19Γ(n10)\displaystyle \prod_{n=1}^{9} \Gamma \left( \frac{n}{10} \right)

Can the result be easily generalised in any interesting way? If so, provide proof of your assertions.
(edited 10 years ago)

Quick Reply

Latest