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Year 12s: what will be the first way you research your future options? >>

The Proof is Trivial!

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Anyone for some analytic number theory? :biggrin:

:biggrin:

Problem 525***

Let

x > 1

be a real number. Prove that that there exists a constant

E

such that

\displaystyle \sum_{n \leq x} \dfrac{1}{n} = \log x + E + O(x^{-1})

Original post by Indeterminate

x

Wot. I'll see if I can wrap my head around that later, looks fun! :biggrin:

:biggrin:

Problem 526*

If

f

is an increasing function then show that

\displaystyle \int_{n-1}^n f(x) \, \mathrm{d}x \leq f(n) \leq \int_{n}^{n+1} f(x) \, \mathrm{d}x

.

Hence approximate

\ln n!

for large

n

.

Problem 527**

Let

M

be a large real number. Explain briefly why there must be exactly one root

w

of the equation

Mx = e^x

with

w > 1

. Why is

\log M

a reasonable approximation to

w

?

By writing

w = \log M + y

find an approximation on

y

and hence improve on

\log M

as an approximation to

w

.

16Characters....

Solution 526

Spoiler

(edited 8 years ago)

Original post by 16Characters....

Solution 526

Very nice, I just realised that I forgot to prove that

\ln x

was strictly increasing in my proof. This was a Pembroke interview question, if you're interested! :smile:

:smile:

16Characters....

Original post by Zacken

Very nice, I just realised that I forgot to prove that

\ln x

was strictly increasing in my proof. This was a Pembroke interview question, if you're interested! :smile:

:smile:

Oh right, nice. I have not been able to find any interview questions for my college so I'll just stick to those Trinity ones for now. There is a similar STEP question to this one in the Advanced Problems in Mathematics booklet (Question 27) which finds estimates to factorials by comparing an exact integral to its Trapezium Rule approximation.

Original post by 16Characters....

Oh right, nice. I have not been able to find any interview questions for my college so I'll just stick to those Trinity ones for now. There is a similar STEP question to this one in the Advanced Problems in Mathematics booklet (Question 27) which finds estimates to factorials by comparing an exact integral to its Trapezium Rule approximation.

St. John's, right? I haven't been able to find any for King's either, so I'm making do with what I find. (although with what I've seen with your mathematical ability, you don't need any prep!)

Yes, I think I recall that vaguely, although I encountered it in a past paper instead of the booklet!

16Characters....

Original post by Zacken

St. John's, right? I haven't been able to find any for King's either, so I'm making do with what I find. (although with what I've seen with your mathematical ability, you don't need any prep!)

Yes, I think I recall that vaguely, although I encountered it in a past paper instead of the booklet!

Yep St John's. And thank you, the same can certainly be said about you. That problem from the Trinity questions, the one you posted here as 527. I don't know where you'd even begin with that one.

Original post by 16Characters....

Yep St John's. And thank you, the same can certainly be said about you. That problem from the Trinity questions, the one you posted here as 527. I don't know where you'd even begin with that one.

I gave it a go on the trinity solutions page, first post on page 2 - but couldn't get further than that.

16Characters....

Original post by Zacken

I gave it a go on the trinity solutions page, first post on page 2 - but couldn't get further than that.

Seen it. I'll leave it to DF to was his way through that one haha.

Solution to Problem 527:

Spoiler

(edited 8 years ago)

Original post by joostan

Solution to Problem 527:

Very, very nice! :smile:

:smile:

I urge you to just copy-paste this into this.

Problem 528:

Which is greater for large

n

,

\displaystyle f(n) = 2^{2^{2^n}}

or

g(n) = 100^{100^n}

, justify your answer.

16Characters....

Problem 529

Evaluate:

\displaystyle \int e^x sech x \ dx

\displaystyle \int e^{-x}sech x \ dx

Solution to 529:
Observe that:

e^xsech(x)=\dfrac{2e^x}{e^x+e^{-x}}=\dfrac{2e^x}{e^{2x}+1}[br]e^{-x}sech(x)=\dfrac{2e^x}{e^x+e^{-x}}=\dfrac{2e^{-2x}}{e^{-2x}+1}

\Rightarrow \displaystyle\int e^xsech(x) \ dx = \displaystyle\int \dfrac{2e^{2x}}{e^{2x}+1} \ dx =\ln(e^{2x}+1)+\mathcal{C}[br]\displaystyle\int e^{-x}sech(x) \ dx = \displaystyle\int \dfrac{2e^{-2x}}{e^{-2x}+1} \ dx =\mathcal{C}-\ln(e^{-2x}+1)[br]

(edited 8 years ago)

Solution to 528

Clearly

log_{2}100 < log_{2}256 = 8

.
From this we can deduce

log_{2}g(n) = 100^{n}log_{2}100 < (8)100^{n}

Taking the logarithm again

log_{2}log_{2}g(n) < log_{2}((8)100^{n}) = 3 + nlog_{2}100 < 3 + 8n

Now taking the logarithm of

f(n)

twice

log_{2}log_{2}f(n) = 2^{n}

2^{n} > 3+ 8n \hspace{3mm} n \geq 6

So for

n \geq 6

f(n) > g(n)

(edited 8 years ago)

Original post by Hauss

Solution to 528
x

I did this, except I used

\log_{10}

instead. You might want to add a quick note about

\ln x

being a strictly increasing (continuous bijective) function so that your assertions ring true.

(edited 8 years ago)

Problem 530: */**
Let

a,b,c,d>0

be arbitrary constants.
Find all possible values of:

S=\dfrac{a}{a+b+d}+\dfrac{b}{a+b+c}+\dfrac{c}{b+c+d}+\dfrac{d}{a+c+d}

(edited 8 years ago)

Original post by joostan

Problem 530: */**
Let

a,b,c,d>0

be arbitrary constants.
Find all possible values of:

S=\dfrac{a}{a+b+d}+\dfrac{b}{a+b+d}+\dfrac{c}{b+c+d}+\dfrac{d}{a+c+d}

Is the denominator of b/etc correct?

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