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The Proof is Trivial!

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Original post by k9markiii
I liked that middle question can't be bothered with the last one.


For question 362, I made an edit.

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Jkn
11
Original post by james22
That's the one I meant, I'm curious what the solution is because I could not find any way to make simpler. I managed to convert it to several different integrals but they didn't look any nicer.

Well that's pretty much what I did, though it seemed satisfying so I'll type it up anyway:

Original post by Lord of the Flies
Problem 260***

∫−∞∞x−1tan⁡xcos⁡tan⁡x dx\displaystyle \int_{-\infty}^{\infty} x^{-1}\tan x \cos \tan x\,dx

Let f(α)=∫0∞sin⁡(αtan⁡x)x dx\displaystyle f(\alpha)= \int_0^{\infty} \frac{\sin( \alpha \tan x)}{x} \ dx so that ∫−∞∞x−1tan⁡xcos⁡tan⁡x dx=2f′(1)\displaystyle \int_{-\infty}^{\infty} x^{-1} \tan x \cos \tan x \ dx = 2 f'(1)

Taking the Laplace transform, applying Fubini's theorem and then using the standard result (proven by IBP or the exponential representation of sine) for the Laplace transform of sin⁡ax\sin ax, we have

L{f(α)}=∫0∞1x∫0∞e−sαsin⁡(αtan⁡x) dα dx=∫0∞1x(tan⁡xs2+tan⁡2x) dx\displaystyle \mathcal{L}\{f(\alpha)\} = \int_0^{\infty} \frac{1}{x} \int_0^{\infty} e^{-s \alpha} \sin( \alpha \tan x) \ d \alpha \ dx = \int_0^{\infty} \frac{1}{x} \left( \frac{\tan x}{s^2+\tan^2 x} \right) \ dx

Let g(β)=∫0∞1x(tan⁡βxs2+tan⁡2x) dx\displaystyle g(\beta) = \int_0^{\infty} \frac{1}{x} \left( \frac{\tan \beta x}{s^2+\tan^2 x} \right) \ dx so that g(1)=L{f(α)}\displaystyle g(1) = \mathcal{L}\{f(\alpha)\}

⇒g′(β)=∫0∞sec⁡2βxs2+tan⁡2x dx=∫0∞sec⁡2βx(s2−1)+sec⁡2x dx\displaystyle \Rightarrow g'(\beta) = \int_0^{\infty} \frac{\sec^2 \beta x}{s^2+\tan^2 x} \ dx = \int_0^{\infty} \frac{\sec^2 \beta x}{(s^2-1)+\sec^2 x} \ dx

The starter question on an old USAMO paper:

Problem 366*

Find the average of the numbers nsin⁡n∘\displaystyle n \sin n^{\circ} (n=2,4,6,⋯ ,180n=2, 4, 6, \cdots, 180)
(edited 10 years ago)
I have missed this integral.

Solution 260

Let me use the notation of Jkn's post.
∫0π(n+12)sin⁡(αtan⁡x)xdx=12∑k=−nn∫−π2π2sin⁡(αtan⁡x)x+kπdx=12∫−π2π2(∑k=−nnxx2−k2π2)sin⁡(αtan⁡x)dx\begin{aligned} \displaystyle \int_{0}^{\pi(n+\frac{1}{2})} \frac{\sin( \alpha \tan x)}{x}dx &= \frac{1}{2}\sum_{k=-n}^{n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin( \alpha \tan x)}{x+k\pi}dx \\&= \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\sum_{k=-n}^{n} \frac{x}{x^{2}-k^{2}\pi^{2}} \right)\sin( \alpha \tan x)dx \end{aligned}.
Note that ∑k=−nnxx2−k2π2=cot⁡x+O(1n)\displaystyle \sum_{k=-n}^{n} \frac{x}{x^{2}-k^{2}\pi^{2}} = \cot x + O(\frac{1}{n}) .
Hence, we can let n→∞n \to \infty and obtain
f(α)=12∫−π2π2cot⁡xsin⁡(αtan⁡x)dx=tan⁡x↦x12∫−∞∞sin⁡(αx)x(1+x2)dx=12∫−∞∞sin⁡(αx)xdx−12∫−∞∞xsin⁡(αx)1+x2dx=π2−π2e−α\begin{aligned} \displaystyle f(\alpha)& = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cot x \sin(\alpha \tan x)dx \\& \mathop= \limits^{\tan x \mapsto x} \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin( \alpha x)}{x(1+x^{2})} dx \\&= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin(\alpha x)}{x} dx -\frac{1}{2}\int_{-\infty}^{\infty} \frac{x\sin(\alpha x)}{1+x^{2}}dx = \frac{\pi}{2} - \frac{\pi}{2}e^{-\alpha} \end{aligned}.
The last integral is an easy exercise in complex analysis.
(edited 10 years ago)
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Jkn
11
Original post by Mladenov
Unparseable latex formula:

\displaystyle \int_{0}^{\pi(n+\frac{1}{2})} \frac{\sin( \alpha \tan x)}{x}dx &= \frac{1}{2}\sum_{k=-n}^{n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin \alpha \tan x}{x}dx


What's this supposed to be? (its's a typo!)
12∫−∞∞sin⁡(αx)x(1+x2)dx=∫−∞∞sin⁡(αx)xdx−∫−∞∞xsin⁡(αx)1+x2dx=π2−π2e−α\frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin( \alpha x)}{x(1+x^{2})} dx = \int_{-\infty}^{\infty} \frac{\sin(\alpha x)}{x} dx -\int_{-\infty}^{\infty} \frac{x\sin(\alpha x)}{1+x^{2}}dx = \frac{\pi}{2} - \frac{\pi}{2}e^{-\alpha}

You missed out the 12\frac{1}{2}s. Typo?
Original post by Arieisit
Problem 362*

This is actually very similar to an A level question but without the added guidance that they usually provide.

A sports association is planning to construct a running track in the shape of a rectangle surmounted by a semicircle. Let the letter xx represents the length of the rectangular section and rr represents the radius of the semicircle.
Determine the length, xx, that maximises the area enclosed by the track.

EDIT

The perimeter of the track must be 600 metres.

I completely forgot to put that in. Sorry :frown:

600m running track hmm OK different from the 400m standard, nice. Well A=πr2+2rxA=\pi r^2 + 2rx and P=2x+2πrP = 2x+2\pi r where P is 600m. There are a number of ways to do this but it would be neatest with lagrange multipliers I think.

A=πr2+2rxA=\pi r^2 + 2rx

0=2x+2πr−6000 = 2x+2\pi r - 600

A=πr2+2rx+λ(2x+2πr−600)A = \pi r^2 + 2rx + \lambda(2x+2\pi r - 600)

A=πr2+2rx+2λx+2λπr−600λA = \pi r^2 + 2rx + 2\lambda x +2\lambda \pi r - 600\lambda

dAdx=dAdr=dAdλ=0\frac{dA}{dx} = \frac{dA}{dr} = \frac{dA}{d\lambda} = 0

0=2x+2πr−6000 = 2 x +2 \pi r - 600

0=2πr+2x+2λπ0 = 2\pi r+ 2x +2\lambda \pi

0=2r+2λ0 = 2r + 2\lambda

r=−λr = - \lambda

0=2πr+2x−2πr0 = 2\pi r +2x -2\pi r

0=2x0 = 2x

This implies the maximum area is enclosed when x = 0.

This makes sense as a circle contains the most area for a given perimeter of any shape.


Certainly didn't go for a straightforward method but using langrange multipliers is useful when you are adding constraints to a system. It seems an odd answer are you sure you've got the question right?
(edited 10 years ago)
Original post by Jkn
What's this supposed to be? (its's a typo!)

You missed out the 12\frac{1}{2}s. Typo?


Indeed. What I did is ∫nπ−π2nπ+π2sin⁡(αtan⁡x)xdx=∫−π2π2sin⁡(αtan⁡x)x+nπdx\displaystyle \int_{n\pi - \frac{\pi}{2}}^{n\pi+\frac{\pi}{2}} \frac{\sin(\alpha \tan x)}{x}dx = \int_{ -\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin(\alpha \tan x)}{x+n\pi}dx.
(edited 10 years ago)
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Jkn
11
Original post by Mladenov
Indeed. What I did is ∫nπ−π2nπ+π2sin⁡(αtan⁡x)xdx=∫−π2π2sin⁡(αtan⁡x)x+nπdx\displaystyle \int_{n\pi - \frac{\pi}{2}}^{n\pi+\frac{\pi}{2}} \frac{\sin(\alpha \tan x)}{x}dx = \int_{ -\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin(\alpha \tan x)}{x+n\pi}dx.

Wow, that's brilliant! Beautiful solution! :biggrin:

Spoiler



Btw, I'm currently fiddling around trying to see if I can find a closed form for L{log⁡n(x)}\displaystyle \mathcal{L}\{ \log^n (x) \} and have come to a halt at dndxnΓ(x+1)∣x=0\displaystyle \frac{d^n}{dx^n} \Gamma(x+1) \Big|_{x=0} which, despite being able to evaluate ϕ(n)(1)\phi^{(n)}(1) very neatly, I am yet to find a general expression for it (it decays into a mess of polygammas). I've also realised that, if this could be evaluated, we would have a taylor series for the gamma function. Is there a taylor series for the gamma function or am I going into the realm of the unknown?
(edited 10 years ago)
Original post by k9markiii
Certainly didn't go for a straightforward method but using langrange multipliers is useful when you are adding constraints to a system. It seems an odd answer are you sure you've got the question right?


Yes, I am sure this time. My solution is much simpler but I agree, the answer is definitely odd.
Original post by Arieisit
Yes, I am sure this time. My solution is much simpler but I agree, the answer is definitely odd.


I was practising out some fun maths. I know you can do it through simultaneous equations but I was practising some maths I learnt last year to get ready for lectures starting next week.
Original post by Arieisit
Problem 362*

This is actually very similar to an A level question but without the added guidance that they usually provide.

A sports association is planning to construct a running track in the shape of a rectangle surmounted by a semicircle. Let the letter xx represents the length of the rectangular section and rr represents the radius of the semicircle.
Determine the length, xx, that maximises the area enclosed by the track.

EDIT

The perimeter of the track must be 600 metres.

I completely forgot to put that in. Sorry :frown:
You don't need to assign a numerical P here, so long as P is constant.

I suppose you have marked this as * so maybe you just want to show something nice - but surely the answer is trivial when you consider what a circle actually is and what some of its inherent properties are.
Original post by Llewellyn
You don't need to assign a numerical P here, so long as P is constant.

I suppose you have marked this as * so maybe you just want to show something nice - but surely the answer is trivial when you consider what a circle actually is and what some of its inherent properties are.


I think you quoted the wrong person. Lol

EDIT

I reread your post. I was looking for a numerical answer hence P was given.

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(edited 10 years ago)
Original post by Arieisit
Problem 362*

This is actually very similar to an A level question but without the added guidance that they usually provide.

A sports association is planning to construct a running track in the shape of a rectangle surmounted by a semicircle. Let the letter xx represents the length of the rectangular section and rr represents the radius of the semicircle.
Determine the length, xx, that maximises the area enclosed by the track.

EDIT

The perimeter of the track must be 600 metres.


Ignore this: I misread the question and maximised the area for the rectangle surmounted by a semi circle on either side
.

Wrong solution

(edited 10 years ago)
Original post by Khallil
Solution 362

Let A be the area of the track and let P be its perimeter. That means that:

A=πr2+2rxA = \pi r^2 + 2rx

P=600=2x+2πrP = 600 = 2x + 2\pi r

To find the length of x that maximises the area enclosed by the track, it'd be helpful to find the area in terms of x alone.

A=1π(9000−x2)A = \dfrac{1}{\pi} \left( 9000 - x^2 \right)

dAdx=−2xπ\dfrac{dA}{dx} = \dfrac{-2x}{\pi}

The maximum value of the area occurs when the first derivative is equal to 0 and the second derivative is less than 0.

dAdx=0⇒x=0\dfrac{dA}{dx} = 0 \Rightarrow x=0

d2Adx2=−2π<0\dfrac{d^2A}{dx^2} = \dfrac{-2}{\pi} < 0

∴\therefore The area is maximised when the length xx is equal to 0 and the track is circular.


I have a slight qualm - your equations are for a circle when it is quite specifically a semi-circle that is mentioned.
Original post by Arieisit
Problem 362*
Original post by DJMayes
I have a slight qualm - your equations are for a circle when it is quite specifically a semi-circle that is mentioned.


Damn it! I read the question but my brain told me that the rectangle is surmounted by a semi circle on either side, like the race tracks in the Olympics :redface:

I hope this is right:

Solution 362

Let A be the area of the track and let P be its perimeter. That means that:

A=12πr2+2rxA = \dfrac{1}{2} \pi r^2 + 2rx

P=600=2r+2x+πrP = 600 = 2r + 2x + \pi r

To find the length of x that maximises the area enclosed by the track, it'd be helpful to find the area in terms of x alone.

A=1(2+π)2(180000π+2400x−x2(8+2π))A = \dfrac{1}{(2+\pi)^2} \left( 180000\pi + 2400x - x^2 \left(8+2\pi \right) \right)

dAdx=2400−2x(8+2π)(2+π)2\dfrac{dA}{dx} = \dfrac{2400- 2x(8+2\pi)}{(2+\pi)^2}

The maximum value of the area occurs when the first derivative is equal to 0 and the second derivative is less than 0.

dAdx=0 ⇒ 2400=2x(8+2π) ⇒ 1200=x(8+2π) ⇒ x=6004+π\dfrac{dA}{dx} = 0 \ \Rightarrow \ 2400=2x(8+2\pi) \ \Rightarrow \ 1200=x(8+2\pi) \ \Rightarrow \ x = \dfrac{600}{4+\pi}

d2Adx2=−2(8+2π)(2+π)2<0\dfrac{d^2A}{dx^2} = \dfrac{-2(8+2\pi)}{(2+\pi)^2} < 0

∴\therefore The area is maximised when the length xx is equal to 6004+π\dfrac{600}{4+\pi}
(edited 10 years ago)
Since nobody attempted Problem 364 I'll provide a hint or two.

Problem 364***

Determine the extreme values of

f(x,y,z,u,v,w)f(x,y,z,u,v,w) = 11+x+u\frac{1}{1+x+u} ++ 11+y+v\frac{1}{1+y+v} ++ 11+z+w\frac{1}{1+z+w}


where xyzxyz = a3a^3, uvwuvw = b3b^3, and x,y,z,u,v,w>0.x,y,z,u,v,w>0.

Spoiler




Problem 367**

If InI_n = ∫0π4tannx dx,n≥2\displaystyle\int^\frac{\pi}{4}_0 tan^nx\ dx, n \geq 2 find I4I_4

Problem 368**

Given that ImI_m = ∫(cosmx)(sin3x)dx\displaystyle\int (cos^mx) (sin3x) dx and JmJ_m = ∫(cosmx)(sin2x)dx,\displaystyle\int (cos^mx) (sin2x) dx, prove that (m+3)Im=mJm−1−(cosmx)(cos3x).(m+3) I_m = mJ_{m-1} - (cos^mx) (cos3x).
(edited 10 years ago)
Original post by Arieisit


Problem 368**

Given that ImI_m = ∫(cosmx)(sin3x)dx\displaystyle\int (cos^mx) (sin3x) dx and JmJ_m = ∫(cosmx)(sin2x)dx,\displaystyle\int (cos^mx) (sin2x) dx, prove that (m+3)Im=mJm−1−(cosmx)(cos3x).(m+3) I_m = mJ_{m-1} - (cos^mx) (cos3x).


Im=∫cos⁡mx⏟usin⁡3x⏟v′ dx\displaystyle I_{m}=\int \underbrace{ \cos^m x }_{\mathrm{u}} \underbrace{ \sin 3x }_{\mathrm{v'}} \,dx

Im=−13cos⁡3xcos⁡mx−m3∫(sin⁡xcos⁡3x)cos⁡m−1x dx\displaystyle I_{m}=- \frac{1}{3} \cos 3x \cos^m x -\frac{m}{3} \int (\sin x \cos 3x) \cos^{m-1} x \, dx

But note sin⁡(3x−x)=sin⁡3xcos⁡x−cos⁡3xsin⁡x\sin (3x-x) = \sin 3x \cos x - \cos 3x \sin x

So cos⁡3xsin⁡x=sin⁡3xcos⁡x−sin⁡2x\cos 3x \sin x = \sin 3x \cos x - \sin 2x

So 3Im=−cos⁡3xcos⁡mx−m(Im−Jm−1)\displaystyle 3I_{m}=-\cos 3x \cos^m x -m \left( I_m - J_{m-1} \right)

So (m+3)Im=mJm−1−(cosmx)(cos3x)(m+3) I_m = mJ_{m-1} - (cos^mx) (cos3x)
Was my response to problem 362 correct? :smile:

Original post by Arieisit
Problem 367**

If InI_n = ∫0π4tannx dx,n≥2\displaystyle\int^\frac{\pi}{4}_0 tan^nx\ dx, n \geq 2 find I4I_4


Solution 367

In=∫0π4tannx dx=∫0π4tan2x tann−2x dx=∫0π4(sec2x−1)tann−2x dx=∫0π4sec2x tann−2x dx−In−2\begin{aligned} I_n & = \displaystyle\int^\frac{\pi}{4}_0 tan^nx\ dx \\ & = \displaystyle\int^\frac{\pi}{4}_0 tan^2x \ tan^{n-2}x\ dx \\ & = \displaystyle\int^\frac{\pi}{4}_0 \left( sec^2x - 1 \right) tan^{n-2}x\ dx \\ & = \displaystyle\int^\frac{\pi}{4}_0 sec^2x \ tan^{n-2}x\ dx - I_{n-2} \end{aligned}

u=tanx ⇒ dudx=sec2xu=tan x \ \Rightarrow \ \dfrac{du}{dx} = sec^2x

x=0⇒u=0, x=π4⇒u=1x=0 \Rightarrow u=0, \ x=\dfrac{\pi}{4} \Rightarrow u=1

∴In=∫01un−2 du−In−2=[un−1n−1]01−In−2=1n−1−In−2\begin{aligned} \therefore I_n & = \displaystyle\int^{1}_0 u^{n-2}\ du - I_{n-2} \\ & = \Big[ \dfrac{u^{n-1}}{n-1} \Big]_{0}^1 - I_{n-2} = \dfrac{1}{n-1} - I_{n-2} \end{aligned}

I4=13−I2=13−(1−I0)=I0−23=π4−23\begin{aligned} I_4 & = \frac{1}{3} - I_2 = \frac{1}{3} - \left( 1 - I_0 \right) = I_0 - \frac{2}{3} = \frac{\pi}{4} - \frac{2}{3} \end{aligned}
(edited 10 years ago)
Original post by Khallil
Was my response to problem 362 correct? :smile:

I thought this thread wasn't about getting the right answer :tongue:

I got a similar answer if that helps :rolleyes:

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Original post by Arieisit
Original post by Khallil
Was my response to problem 362 correct? :smile:


I thought this thread wasn't about getting the right answer :tongue:


Indeed it isn't :tongue:

Original post by Arieisit
I got a similar answer if that helps :rolleyes:

How similar? :pierre:
Original post by Khallil
How similar? :pierre:


Lets just say that yours looks more elegant. :colondollar:



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