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The Proof is Trivial!

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Problem 205/ **

At the age of three, Zakee begins to learn how to count, and is faced with a problem in his counting book. He believes it is just a trivial problem, after all, he is only three years of age, and so looks at it and tries to solve it. What was the outcome of his attempts? You decide (see below):

Sequence of Real numbers:

x_0, x_1,\ldots, x_n,\ldots

accede to these mathematical conditions

1 = x_{0}\leq x_{1}\leq\cdots\leq x_{n}\leq\cdots

Additionally, let

y_{1}, y_{2},\ldots, y_n,\ldots

be defined as:

y_n =\sum_{k=1}^{n}\frac{1-\frac{x_{k-1}}{x_{k}}}{\sqrt{x_k}}

Prove that:

0\leq y_n < 2

FOR ALL

n

Edit: Question amended due to erratum.

(edited 10 years ago)

Reply 1321

Jkn

Original post by Zakee

Back to the topic:

Problem 205/ **
Prove that:

0\leq y_n < 2

Let

x_i=1 \ \forall i

\Rightarrow y_n = n

let n be greater that 2... ?

Reply 1322

Zakee

Original post by Jkn

Let

x_i=1 \ \forall i

\Rightarrow y_n = n

let n be greater that 2... ?

I was looking at your solution and thought to myself, either Jkn is a prodigious genius with a bajillion IQ points, or I have made a trivial error in my question. Both are correct. :wink:

Y'may want to redo your solution.

(edited 10 years ago)

Reply 1323

Mladenov

Original post by Zakee

Spoiler

You want to make the sequence

x_{n}

strictly increasing for all

n

which are greater than a given number

N

. Otherwise

y_{n}

tends to

\infty

n\to \infty

(edited 10 years ago)

Reply 1324

bananarama2

Original post by Zakee

I was looking at your solution and thought to myself, either Jkn is a prodigious genius with a bajillion IQ points, or I have made a trivial error in my question. Both are correct. :wink:

Y'may want to redo your solution.

I may be being thick again, but doesn't his point still stand? You can still choose n to be greater than 2 if x_i=1....

Reply 1325

Jkn

Original post by Zakee

I was looking at your solution and thought to myself, either Jkn is a prodigious genius with a bajillion IQ points, or I have made a trivial error in my question. Both are correct. :wink:

Y'may want to redo your solution.

Hahhaaha

I wasn't offering a solution, merely pointing out an error :lol:

The error is either in your

y_n

(notice how quickly it can be simplified, perhaps you are missing a rogue square terms) or in the definition of the sequence. Perhaps strictly increasing integers? Something that forces

x_i

to tend to infinity is a necessary (though not sufficient) condition for

y_n

to converge. i.e. we require that each term in the series tends to zero as the underlying parameter approach infinity. :tongue:

Edit: Just checked your latex and say that that was an

x_{k-1}

rather than an

x_k -1

(would be clearer with displaystyle!) Though you would still need a strict increase for some infinite set of

x_i

(as Mladenov has said) :smile:

(edited 10 years ago)

Reply 1326

Zakee

Original post by Jkn

Hahhaaha

I wasn't offering a solution, merely pointing out an error :lol:

The error is either in your

y_n

(notice how quickly it can be simplified, perhaps you are missing a rogue square terms) or in the definition of the sequence. Perhaps strictly increasing integers? Something that forces

x_i

to tend to infinity is a necessary (though not sufficient) condition for

y_n

to converge. i.e. we require that each term in the series tends to zero as the underlying parameter approach infinity. :tongue:

Edit: Just checked your latex and say that that was an

x_{k-1}

rather than an

x_k -1

(would be clearer with displaystyle!) Though you would still need a strict increase for some infinite set of

x_i

(as Mladenov has said) :smile:

Sorry, my Latex abilities are awful, and I have an exam tomorrow. I rushed with writing up the question. My bad, if you can understand what I was trying to say, then I hope you can do it. If not, I will refine the post tomorrow. Sorry for the problem caused.

Reply 1327

Jkn

Original post by Zakee

No worries :smile:

Tbf, I should stop coming on here! Too many exams! :lol:

Luckily now finished my English and Spanish but now I've got Decision and Physics and **** to worry about :frown:

Not to mention STEP! Drownnninnnnggggg! :frown:

Reply 1328

Zakee

Original post by Jkn

No worries :smile:

Tbf, I should stop coming on here! Too many exams! :lol:

Luckily now finished my English and Spanish but now I've got Decision and Physics and **** to worry about :frown:

Not to mention STEP! Drownnninnnnggggg! :frown:

I've got Physics tomorrow and half the syllabus to learn. :colonhash:

. I'm off man, take care. x_x

Reply 1329

Jkn

Original post by Zakee

I've got Physics tomorrow and half the syllabus to learn. :colonhash:

. I'm off man, take care. x_x

****, I've got my physics next week and haven't finished the syllabus either :colondollar:

The answers to most questions can be figured out form the context though :lol:

On the past paper I tried though there were all of these questions asking tricky mathematical things where they, in fact, want you to count squares on a graph, fml I always end up whipping out the calculus and taking longer than intended! :frown:

Good luck!

Reply 1330

Zakee

Original post by Jkn

****, I've got my physics next week and haven't finished the syllabus either :colondollar:

The answers to most questions can be figured out form the context though :lol:

Good luck!

Precisely man. One of them was to do with Hooke's law, and I worked out the area of the graph by approximation through integrals. Instead they wanted counting squares. Damn Physics exams. :wink:

Reply 1331

metaltron

Original post by Zakee

I've got Physics tomorrow and half the syllabus to learn. :colonhash:

. I'm off man, take care. x_x

I'll join that club as well :frown:

Reply 1332

Felix Felicis

Original post by metaltron

I'll join that club as well :frown:

Same o_O

Reply 1333

bananarama2

I'm not impressed guys :l

Reply 1334

Zakee

Original post by bananarama2

I'm not impressed guys :l

I actually never ended up revising in the end. Damn. :frown:

Reply 1335

metaltron

Original post by bananarama2

I'm not impressed guys :l

Yeah, physics is fun as well though! Just a bit late to start revising...

Reply 1336

Mladenov

Well, I see what kind of problems people prefer.

Problem 206*

Evaluate

\displaystyle \int_{0}^{\pi} \frac{\ln(1+a\cos x)}{\cos x} \, dx

, where

|a| <1

.

Problem 207**

Let

\alpha_{i}

i \in \{1,\cdots,k\}

be numbers such that for any two sequences

(a_{n})_{n \ge1}

and

(b_{n})_{n\ge 1}

, which satisfy the relation

\displaystyle b_{n}=a_{n}+\sum_{i=1}^{k} \alpha_{i}a_{n-i}

, for all

n>k

, from the convergence of

(b_{n})_{n\ge 1}

follows the convergence of

(a_{n})_{n \ge1}

. Prove that all the roots of the polynomial

\displaystyle x^{k}+ \sum_{i=1}^{k} \alpha_{i}x^{k-i}

have absolute values which are less than

1

.

Problem 208***

Let

f : [a,b] \to \mathbb{R}

C^{2}

, and suppose that

\displaystyle 0<h<\frac{b-a}{2}

. Then,

\displaystyle \int_{a}^{b} |f'(x)|^{p}dx \le 2^{p}h^{p-1}(b-a)\int_{a}^{b}|f''(x)|^{p}dx+ \frac{2^{2p}}{h^{p}}\int_{a}^{b}|f(x)|^{p}dx

(p>1)

Reply 1337

bananarama2

Solution 206

\displaystyle I(a)= \int_0^{\pi} \frac{\ln (1+a \cos x )}{\cos x}dx

\displaystyle \frac{dI}{da} = \int_0^{\pi} \frac{1}{1+a \cos x} dx

Weierstrass sub http://en.wikipedia.org/wiki/Weierstrass_substitution

\displaystyle = \int_0^{\infty} \frac{1}{1+a \frac{1-t^2}{1+t^2}} \frac{2 dt}{1+t^2}

\displaystyle = \int_0^{\infty} \frac{1}{(1+t^2)+a (1-t^2) } 2dt

\displaystyle = \int_0^{\infty} \frac{1}{ (1-a)t^2 +(1+a) } 2dt

\displaystyle = \left[ -\frac{2 \arctan\sqrt{\frac{1-a}{a+1}}t}{\sqrt{(1+a)(1-a)}} \right]_0^\infty

\displaystyle = \frac{\pi}{\sqrt{(1-a)(1+a)}}

\displaystyle = \frac{\pi}{\sqrt{1-a^2} }

\displaystyle I = \pi \arcsin (a) +c

a=0 I=0 C=0

\displaystyle I = \pi \arcsin (a)

I'm unconvinced.

(edited 10 years ago)

Reply 1338

Mladenov

Original post by bananarama2

Spoiler

Well done.

Problem 209*

Evaluate

\displaystyle \int_{0}^{\infty} \frac{A_{1}\cos a_{1}x + \cdots + A_{k}\cos a_{k}x}{x}dx

when

a_{i}>0

for all

i \in \{1,2, \cdots,k \}

and

A_{1}+\cdots+A_{k}=0

.

Problem 210**

Find

\displaystyle \int_{0}^{1} \frac{(1-x^{\alpha})(1-x^{\beta})}{(1-x)\ln x} \, dx

, for

\alpha, \beta >-1

and

\alpha+\beta >-1

Reply 1339

MW24595

Original post by Mladenov

Problem 210**

Find

\displaystyle \int_{0}^{1} \frac{(1-x^{\alpha})(1-x^{\beta})}{(1-x)\ln x} \, dx

, for

\alpha, \beta >-1

and

\alpha+\beta >-1

Solution 210

Firstly, note that:

\frac{1-x^{\alpha}}{1-x} = \sum_{i=0}^{\alpha -1} x^i

When

\alpha = 0

, then we just use the infinite expansion of

(1+x)^{-1}

which is allowed since x only varies between 0 to 1 in the integral.

Unparseable latex formula:

[br][br]\Rightarrow I = \displaystyle\int_0^1 \frac{(1-x^{\alpha})(1-x^{\beta})}{(1-x) \ln x}\ dx = \displaystyle\int_0^1 \frac{(1-x^{\alpha})}{\ln x} \displaystyle\sum_{i=0}^{\beta -1} x^i \ dx [br][br]\beginaligned I = \displaystyle\sum_{i=0}^{\beta -1} \displaystyle\int_0^1 \frac{(1-x^{\alpha})}{\ln x} x^i \ dx[br][br]\beginaligned I = \displaystyle\sum_{i=0}^{\beta -1} \displaystyle\int_0^1 \frac{(x^i-x^{\alpha +i})}{\ln x} \ dx[br]

Now, just consider the integral, for an arbitrary value of i, and let,

x = e^t

\displaystyle\int_0^1 \frac{(x^i-x^{\alpha +i})}{\ln x} \ dx = \displaystyle \int_{-\infty}^0 \frac {e^{(i+1)t}- e^{(\alpha + i+1)t}}{t} \ dt[br][br]

Here, notice that,

\displaystyle \int_{\alpha+i+1}^{i+1} e^{ty}\ dy = \frac{e^{(i+1)t}- e^{(\alpha + i+1)t}}{t}[br][br]

So,

Unparseable latex formula:

\displaystyle \int_{-\infty}^0 \frac {e^{(i+1)t}- e^{(\alpha + i+1)t}}{t} \ dt = \displaystyle\int_{-\infty}^0 \displaystyle\int_{\alpha+i+1}^{i+1} e^{ty}\ dy\ dt = \displaystyle\int_{\alpha+i+1}^{i+1} \displaystyle\int_{-\infty}^0 e^{ty}\ dt\ dy[br][br]\Rightarrow \displaystyle\int_{\alpha+i+1}^{i+1} \displaystyle\int_{-\infty}^0 e^{ty}\ dt\ dy = \displaystyle\int_{\alpha+i+1}^{i+1} \frac{1}{y} \ dy = \ln {(\frac{i+1}{\alpha+i+1})}[br][br]\Rightarow \displaystyle\int_0^1 \frac{(x^i-x^{\alpha +i})}{\ln x} \ dx = \ln {(\frac{i+1}{\alpha+i+1})}[br][br]

Now, substituting this into the original integral,

Unparseable latex formula:

[br][br]\beginaligned I = \displaystyle\sum_{i=0}^{\beta -1} \displaystyle\int_0^1 \frac{(x^i-x^{\alpha +i})}{\ln x} \ dx = \displaystyle\sum_{i=0}^{\beta -1} \ln {(\frac{i+1}{\alpha+i+1})}[br][br][br]\beginaligned I = \displaystyle\sum_{i=1}^{\beta} \ln {(\frac{i}{\alpha+i})}[br][br][br]\beginaligned I = \displaystyle \ln {\frac{\beta !}{(\alpha+1)...(\alpha+ \beta)}}[br][br][br]\beginaligned I = - \ln \binom{\alpha+\beta}{\alpha}[br]

(edited 10 years ago)