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The Proof is Trivial!

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Problem 301**

Prove, given a,c>0;  ac−bd≠0 a,c > 0;\ \ ac-bd\not=0

arctan⁡(ad+bcac−bd)=arctan⁡(ba)+arctan⁡(dc) \arctan\left(\dfrac{ad+bc}{ac-bd}\right) = \arctan\left(\dfrac{b}{a}\right) + \arctan\left(\dfrac{d}{c}\right)
Original post by FireGarden
Problem 301**

Prove, given a,c>0;  ac−bd≠0 a,c > 0;\ \ ac-bd\not=0

arctan⁡(ad+bcac−bd)=arctan⁡(ba)+arctan⁡(dc) \arctan\left(\dfrac{ad+bc}{ac-bd}\right) = \arctan\left(\dfrac{b}{a}\right) + \arctan\left(\dfrac{d}{c}\right)


Is there a nicer way than the rather horrible:

Spoiler

Original post by Smaug123
Is there a nicer way than the rather horrible:

Spoiler



Yes. The formula arises naturally, if you consider the right objects/behaviour. I wrote the question with the nice answer in mind :wink:
Original post by Smaug123
Is there a nicer way than the rather horrible:

Spoiler



That's not that horrible though, it's only 3-4 lines of working.
Original post by james22
That's not that horrible though, it's only 3-4 lines of working.


It's not pretty, though - it's really mechanical :smile:
Reply 2085
Avatar for MW24595
MW24595
Original post by FireGarden
Problem 301**

Prove, given a,c>0;  ac−bd≠0 a,c > 0;\ \ ac-bd\not=0

arctan⁡(ad+bcac−bd)=arctan⁡(ba)+arctan⁡(dc) \arctan\left(\dfrac{ad+bc}{ac-bd}\right) = \arctan\left(\dfrac{b}{a}\right) + \arctan\left(\dfrac{d}{c}\right)


Solution 301

tan⁡(arctan⁡ba+arctan⁡dc)=ba+dc1−bdac=ad−bcac−bd \tan {( \arctan {\frac{b}{a}} + \arctan { \frac{d}{c}})} = \displaystyle \frac{ \frac{b}{a} + \frac {d}{c} } {1- \frac {bd}{ac} } = \frac{ ad -bc}{ac- bd}

Here's a nice one from the Putnam.

Problem 302**

Let F0(x)=lnx. F_0 (x) = ln x. For n≥0 n \ge 0 and x>0, x > 0, let Fn+1(x)=∫0xFn(t) dt F_{n+1}(x) = \displaystyle \int_0^x F_n (t)\ dt .

Evaluate:

lim⁡n→∞n!Fn(1)ln⁡n \lim_{ n \rightarrow \infty} \displaystyle \frac{ n! F_n (1) } { \ln {n} } .
.
(edited 10 years ago)
Original post by MW24595
Solution 301

tan⁡(arctan⁡ba+arctan⁡dc)=ba+dc1−bdac=ad−bcac−bd \tan {( \arctan {\frac{b}{a}} + \arctan { \frac{d}{c}})} = \displaystyle \frac{ \frac{b}{a} + \frac {d}{c} } {1- \frac {bd}{ac} } = \frac{ ad -bc}{ac- bd}

Here's a nice one from the Putnam.

Problem 302**

Let F0(x)=lnx. F_0 (x) = ln x. For n≥0 n \ge 0 and x>0, x > 0, let Fn+1(x)=∫0xFn(t) dt F_{n+1}(x) = \displaystyle \int_0^x F_n (t)\ dt .

Evaluate:

lim⁡n→∞n!Fn(1)ln⁡n \lim_{ n \rightarrow \infty} \displaystyle \frac{ n! F_n (1) } { \ln {n} } .
.

Solution 302:
Consider:
F1(x)=xln⁡(x)−x[br]F2(x)=x22ln⁡(x)−3x24F_1(x)=x\ln(x)-x[br]F_2(x)=\dfrac{x^2}{2}\ln(x)-\dfrac{3x^2}{4}
Suppose then that: Fn(x)F_n(x) is of the form:
xn(ln⁡(x)n!−Cn)x^n \left(\dfrac{\ln(x)}{n!} - C_n \right)
For some constant CnC_n
As it is true for F1(x)F_1(x) and F2(x)F_2(x) Suppose that it is also true for Fk(x)F_k(x)
⇒Fk+1(x)=∫0xtkk!ln⁡(t)−Cktk dt\Rightarrow F_{k+1}(x)=\displaystyle\int^x_0 \dfrac{t^k}{k!}\ln(t)-C_kt^k \ dt
Let:
u=ln⁡(t)⇒u′=1tu=\ln(t) \Rightarrow u'=\dfrac{1}{t}
v′=tkk!⇒v=tk+1(k+1)!v'=\dfrac{t^k}{k!} \Rightarrow v=\dfrac{t^{k+1}}{(k+1)!}
⇒Fk+1=[tk+1ln⁡(t)(k+1)!]0x−∫0xtk+1(k+1)!×1t dt\Rightarrow F_{k+1}=\left[\dfrac{t^{k+1}\ln(t)}{(k+1)!} \right]^x_0 -\displaystyle\int^x_0 \dfrac{t^{k+1}}{(k+1)!}\times \dfrac{1}{t} \ dt−[Cktk+1k+1]0x -\left[C_k\dfrac{t^{k+1}}{k+1} \right]^x_0
Observing that:
lim⁡t→0tk+1ln⁡(t)=0\displaystyle\lim_{t \to 0} t^{k+1}\ln(t)=0
⇒Fk+1(x)=xk+1ln⁡(x)(k+1)!−xk+1(k+1)(k+1)!−Ckxk+1k+1[br]⇒Fk+1(x)=xk+1(ln⁡(x)(k+1)!−Ck+1)\Rightarrow F_{k+1}(x)= \dfrac{x^{k+1}\ln(x)}{(k+1)!}-\dfrac{x^{k+1}}{(k+1)(k+1)!}-C_k \dfrac{x^{k+1}}{k+1}[br]\Rightarrow F_{k+1}(x)=x^{k+1} \left(\dfrac{\ln(x)}{(k+1)!}-C_{k+1} \right)
Where:
Ck+1=Ckk+1+1(k+1)(k+1)!C_{k+1}=\dfrac{C_k}{k+1}+\dfrac{1}{(k+1)(k+1)!}
This is of the same format as was proposed:
∴Fn(x)=xn(ln⁡(x)n!−Cn)∀n∈N\therefore F_{n}(x)=x^n \left(\dfrac{\ln(x)}{n!} - C_n \right) \forall n \in \mathbb{N}
Now using the fact that:
Ck+1=Ckk+1+1(k+1)(k+1)![br]⇒Ck+1=1(k+1)!(k!Ck+1k+1)[br]∴Cn=1n!((n−1)!Cn−1+1n)C_{k+1}=\dfrac{C_k}{k+1}+\dfrac{1}{(k+1)(k+1)!}[br]\Rightarrow C_{k+1}=\dfrac{1}{(k+1)!} \left(k!C_k+\dfrac{1}{k+1} \right)[br]\therefore C_n=\dfrac{1}{n!} \left((n-1)!C_{n-1}+\dfrac{1}{n} \right)
Note that:
(n−1)!Cn−1=((n−2)!Cn−2+1n−1)(n-1)!C_{n-1}= \left((n-2)!C_{n-2}+\dfrac{1}{n-1} \right) and that C0=0C_0=0
∴Cn=1n!(1n+1n−1+...+13+12+1)=\therefore C_n=\dfrac{1}{n!}\left(\dfrac{1}{n}+\dfrac{1}{n-1}+. . . +\dfrac{1}{3}+\dfrac{1}{2}+1 \right)=1n!∑r=1n1r\dfrac{1}{n!}\displaystyle\sum_{r=1}^n \dfrac{1}{r}
So:
Fn(x)=xn(ln⁡(x)n!−1n!∑r=1n1r)F_n(x)=x^n \left(\dfrac{\ln(x)}{n!} -\dfrac{1}{n!}\displaystyle\sum_{r=1}^n \dfrac{1}{r} \right)
Subbing in:
Fn(1)=1n(ln⁡(1)n!−1n!∑r=1n1r=−1n!∑r=1n1r)[br]⇒n!Fn(1)=−∑r=1n1rF_n(1)=1^n \left(\dfrac{\ln(1)}{n!} -\dfrac{1}{n!}\displaystyle\sum_{r=1}^n \dfrac{1}{r}=-\dfrac{1}{n!}\displaystyle\sum_{r=1}^n \dfrac{1}{r} \right)[br]\Rightarrow n!F_n(1)=-\displaystyle\sum_{r=1}^n \dfrac{1}{r}
So:
lim⁡n→∞(n!Fn(1)ln⁡(n))=\displaystyle\lim_{ n \to \infty} \left(\dfrac{ n! F_n (1)} {\ln(n)}\right)=lim⁡n→∞(−1ln⁡(n)∑r=1n1r)\displaystyle\lim_{ n \to \infty} \left( \dfrac{-1}{\ln(n)}\displaystyle\sum_{r=1}^n \dfrac{1}{r} \right)
Note that:
γ=lim⁡n→∞(∑r=1n1r−ln⁡(n))[br]∴γ+lim⁡n→∞ln⁡(n)=lim⁡n→∞(∑r=1n1r)[br]⇒lim⁡n→∞(n!Fn(1)ln⁡(n))=lim⁡n→∞(−γ−ln⁡(n)ln⁡(n))=−1\gamma = \displaystyle\lim_{n \to \infty} \left( \displaystyle\sum_{r=1}^n \dfrac{1}{r}-\ln(n) \right)[br]\therefore \gamma + \displaystyle\lim_{n \to \infty}\ln(n)=\displaystyle\lim_{n \to \infty} \left( \displaystyle\sum_{r=1}^n \dfrac{1}{r}\right)[br]\Rightarrow \displaystyle\lim_{ n \to \infty} \left( \dfrac{ n! F_n (1)} {\ln(n)} \right)=\displaystyle\lim_{ n \to \infty} \left(\dfrac{-\gamma-\ln(n)}{\ln(n)} \right)=-1
(edited 10 years ago)
Reply 2087
Avatar for Jkn
Jkn
11
Original post by MW24595
Problem 302**

Let F0(x)=lnx. F_0 (x) = ln x. For n≥0 n \ge 0 and x>0, x > 0, let Fn+1(x)=∫0xFn(t) dt F_{n+1}(x) = \displaystyle \int_0^x F_n (t)\ dt .

Evaluate:

lim⁡n→∞n!Fn(1)ln⁡n \lim_{ n \rightarrow \infty} \displaystyle \frac{ n! F_n (1) } { \ln {n} } .

If only they had Putnam in the UK..

Solution 302 (2)

Calculating the first few terms we conjecture that Fn(x)=1n(xFn−1(x)−xnn!)\displaystyle F_n(x)=\frac{1}{n} \left(xF_{n-1}(x)-\frac{x^n}{n!} \right).

This is true for n=1 as F1(x)=xF0(x)−x\displaystyle F_1 (x)=xF_0 (x)-x. Integrating both sides, using integration by parts, formalizing by noting that lim⁡n→0nln⁡(n)=0\displaystyle \lim_{n \to 0} n \ln(n) = 0 and factorising Fn+1(x)F_{n+1}(x) on the left hand side shows that, under this hypothesis, we have that:

Fn+1(x)=1n(∫0xtFn−1(t) dt−xn+1(n+1)!)=1n(x∫0xFn−1(t) dt−∫0x∫0yFn−1(t) dt dy−xn+1(n+1)!)\displaystyle \begin{aligned} F_{n+1} (x) = \frac{1}{n} \left( \int_0^x t F_{n-1}(t) \ dt - \frac{x^{n+1}}{(n+1)!} \right) = \frac{1}{n} \left(x \int_0^x F_{n-1} (t) \ dt - \int_0^x \int_0^y F_{n-1} (t) \ dt \ dy - \frac{x^{n+1}}{(n+1)!} \right) \end{aligned}
=1n(xFn(x)−Fn+1(x)−xn+1(n+1)!)=1n+1(xFn(x)−xn+1(n+1)!)\displaystyle \begin{aligned} = \frac{1}{n} \left( x F_n (x) - F_{n+1} (x) - \frac{x^{n+1}}{(n+1)!} \right) = \frac{1}{n+1} \left( x F_n (x) - \frac{x^{n+1}}{(n+1)!} \right) \end{aligned}
which completes the induction.

This then gives nFn(1)=Fn−1(x)−1n! ⇒n!Fn(1)=(n−1)!Fn−1(1)−1n=⋯=F0(1)−(1+12+⋯+1n)=−Hn\displaystyle \begin{aligned} n F_n (1) = F_{n-1} (x) - \frac{1}{n!}\ \Rightarrow n! F_n(1) = (n-1)! F_{n-1} (1) - \frac{1}{n} = \cdots =F_0 (1) -\left(1+\frac{1}{2}+ \cdots + \frac{1}{n} \right)= -H_n \end{aligned}

where HnH_n denotes the nth harmonic number.

Using the asymptotic relation lim⁡n→∞Hn=γ+ln⁡(n)\displaystyle \lim_{n \to \infty} H_n = \gamma + \ln (n), which follows directly from the definition of the Euler Mascheroni constant, we get:

lim⁡n→∞n!Fn(1)ln⁡(n)=−lim⁡n→∞γ+ln⁡(n)ln⁡(n)=−1 □\displaystyle \lim_{n \to \infty} \frac{n! F_n (1)}{\ln(n)} = - \lim_{n \to \infty} \frac{\gamma + \ln(n)}{\ln(n)} = -1 \ \square as the logarithmic function is obviously divergent under this limit.

Edit: Dammit, too slow!
(edited 10 years ago)
Reply 2088
Avatar for Jkn
Jkn
11
Original post by MW24595
I just came across this. There's an off reference in the section titled "Integration Problems." Read the last paragraph.

http://en.wikipedia.org/wiki/Gamma_function

Oh cheers, must've missed that! (That page is one of my 'most-visited' :lol:)
Original post by MW24595
Solution 300

Nice solution broseph! Here is your prize! :biggrin:
Reply 2089
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MW24595
Original post by Jkn
Oh cheers, must've missed that! (That page is one of my 'most-visited' :lol:)

Nice solution broseph! Here is your prize! :biggrin:


Roflol. I'm surprised you did.
Interesting that the reference was almost bang-on though.

Lol, thanks. Ha :tongue:

Found anything else of interest lately?
Reply 2090
Avatar for MW24595
MW24595
Original post by FireGarden
Problem 301**

Prove, given a,c>0;  ac−bd≠0 a,c > 0;\ \ ac-bd\not=0

arctan⁡(ad+bcac−bd)=arctan⁡(ba)+arctan⁡(dc) \arctan\left(\dfrac{ad+bc}{ac-bd}\right) = \arctan\left(\dfrac{b}{a}\right) + \arctan\left(\dfrac{d}{c}\right)


I just realized.

Alternate Solution 301

Let z1=a+ib, z_1 = a+ ib, and z2=c+id z_2 = c + id .

Then, z1=a+bi=eiarctan⁡ba z_1 = a+ bi = \displaystyle e^{i \arctan{ \frac{b}{a}}} , and, z2=eiarctan⁡dc z_2 = \displaystyle e^{i \arctan {\frac{d}{c}}} .

[br][br]⇒z1z2=(a+ib)(c+id)=(ac−bd)+i(ad+bc)=eiarctan⁡ad+bcac−bd[br][br][br][br]\Rightarrow z_1 z_2 = (a+ib)(c+id) = (ac- bd) + i(ad + bc) = e^ { i \arctan{\frac{ad+bc}{ac-bd}}}[br][br].

But, z1z2=eiarctan⁡ba+iarctan⁡dc=eiarctan⁡ad+bcac−bd z_1 z_2 = \displaystyle e^{i \arctan{ \frac{b}{a}} + i \arctan{ \frac{d}{c}}} = e^ { i \arctan{\frac{ad+bc}{ac-bd}}} .

Taking logarithms, we have:

arctan⁡ad+bcac−bd=arctan⁡ba+arctan⁡dc\displaystyle \arctan{\frac{ad+bc}{ac-bd}} = \arctan{ \frac{b}{a}} + \arctan{ \frac{d}{c}} .
(edited 10 years ago)
Just posted this in one of the regular threads.. thought you guys might like it :biggrin:

Problem 303**

Give a justification for the claim 1∞>2 1^{\infty} > 2
Original post by MW24595
I just realized.

Alternate Solution 301

Let z1=a+ib, z_1 = a+ ib, and z2=c+id z_2 = c + id .

Then, z1=a+bi=eiarctan⁡ba z_1 = a+ bi = \displaystyle e^{i \arctan{ \frac{b}{a}}} , and, z2=eiarctan⁡dc z_2 = \displaystyle e^{i \arctan {\frac{d}{c}}} .

[br][br]⇒z1z2=(a+ib)(c+id)=(ac−bd)+i(ad+bc)=eiarctan⁡ad+bcac−bd[br][br][br][br]\Rightarrow z_1 z_2 = (a+ib)(c+id) = (ac- bd) + i(ad + bc) = e^ { i \arctan{\frac{ad+bc}{ac-bd}}}[br][br].

But, z1z2=eiarctan⁡ba+iarctan⁡dc=eiarctan⁡ad+bcac−bd z_1 z_2 = \displaystyle e^{i \arctan{ \frac{b}{a}} + i \arctan{ \frac{d}{c}}} = e^ { i \arctan{\frac{ad+bc}{ac-bd}}} .

Taking logarithms, we have:

arctan⁡ad+bcac−bd=arctan⁡ba+arctan⁡dc\displaystyle \arctan{\frac{ad+bc}{ac-bd}} = \arctan{ \frac{b}{a}} + \arctan{ \frac{d}{c}} .


This is the solution I had in mind. Well done sir!
Original post by FireGarden
Just posted this in one of the regular threads.. thought you guys might like it :biggrin:

Problem 303**

Give a justification for the claim 1∞>2 1^{\infty} > 2


This question doesn't really make sense, do you mean we need to find a function, f(x), such that as x->infinity f(x)->1 but f(x)^x->c>2?
Original post by james22
This question doesn't really make sense, do you mean we need to find a function, f(x), such that as x->infinity f(x)->1 but f(x)^x->c>2?


You could do it that way if you like. I thought the question would make sense, since I'm asking about an indeterminate form, and thus it can take many values - the question is to find some way to approach the IF so that its value is larger than two, as you correctly interpreted. Though your approach is not the only one!
(edited 10 years ago)
Original post by FireGarden
Just posted this in one of the regular threads.. thought you guys might like it :biggrin:

Problem 303**

Give a justification for the claim 1∞>2 1^{\infty} > 2


Solution 303:
Note that:
lim⁡x→∞(tan⁡(π4+1x))x=1∞\displaystyle\lim_{x \to \infty} \left(\tan \left(\dfrac{\pi}{4} +\dfrac{1}{x} \right) \right)^x=1^{\infty}
Now let:
y=lim⁡x→∞(tan⁡(π4+1x))x[br]⇒ln⁡(y)=lim⁡x→∞xln⁡(tan⁡(π4+1x))=lim⁡x→∞ln⁡(tan⁡(π4+1x))1x[br]⇒ln⁡(y)=lim⁡t→0ln⁡(tan⁡(π4+t))ty=\displaystyle\lim_{x \to \infty} \left(\tan \left(\dfrac{\pi}{4} +\dfrac{1}{x} \right) \right)^x[br]\Rightarrow \ln(y)=\displaystyle\lim_{x \to \infty} x\ln \left(\tan \left(\dfrac{\pi}{4} +\dfrac{1}{x} \right) \right)=\displaystyle\lim_{x \to \infty}\dfrac{\ln \left(\tan \left(\dfrac{\pi}{4} +\dfrac{1}{x} \right) \right)}{\frac{1}{x}}[br]\Rightarrow \ln(y)=\displaystyle\lim_{t \to 0} \dfrac{\ln \left(\tan \left(\dfrac{\pi}{4} +t \right) \right)}{t}
Applying L'Hopital's rule:
ln⁡(y)=lim⁡t→0(csc⁡(t+π4)×sec⁡(t+π4)1)\ln(y)=\displaystyle\lim_{t \to 0} \left(\dfrac{\csc \left(t+\frac{\pi}{4} \right) \times \sec \left(t+\frac{\pi}{4} \right)}{1}\right)
⇒ln⁡(y)=lim⁡t→0(2sec⁡(2t)1)=2[br]⇒y=e2\Rightarrow \ln(y)=\displaystyle\lim_{t \to 0} \left(\dfrac{2\sec(2t)}{1}\right)=2[br]\Rightarrow y=e^2
Now doing some bad maths:
lim⁡x→∞(tan⁡(π4+1x))x=1∞=e2>2\displaystyle\lim_{x \to \infty} \left(\tan \left(\dfrac{\pi}{4} +\dfrac{1}{x} \right) \right)^x=1^{\infty}=e^2>2
(edited 10 years ago)
...Well that sure does it, but what a meal!

Spoiler

Original post by FireGarden
...Well that sure does it, but what a meal!

Spoiler



Yeah, I came across the limit problem above earlier, so pretty much just copied and pasted it :tongue:
I considered;
(xx+1)x\left(\dfrac{x}{x+1} \right)^x but it didn't work so I gave up. :facepalm:
Reply 2098
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Jkn
11
Original post by Jkn
Problem 161**/***

Let k be an integer greater than 1. Suppose a0>0a_0>0 and an+1=an+1ank\displaystyle a_{n+1}=a_{n}+\frac{1}{\sqrt[k]{a_{n}}} for n>0n>0

Evaluate lim⁡n→∞ank+1nk\displaystyle\lim_{n \to {\infty}} \frac{a^{k+1}_n}{n^k}

I hadn't actually tried this problem when I posted it (as it was a 'problem 6' from the Putnam exam and I was hunting around for things that were going to challenge people), but I've given it a go and just managed to solve it! Really enjoyable so I hope people take a crack at it! :biggrin:
Reply 2099
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Jkn
11
Original post by FireGarden
Just posted this in one of the regular threads.. thought you guys might like it :biggrin:

Problem 303**

Give a justification for the claim 1∞>2 1^{\infty} > 2

Original post by joostan
Solution 303

Original post by FireGarden
...Well that sure does it, but what a meal!

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