Difficult Maths/Physics Problems Help Thread

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I hope this question makes sense. I adapted it from something different.

Using integration, construct bounds for the sum

\displaystyle \sum^{n}_{i=1} \dfrac{1}{\sqrt{n}}

If you've never seen something like this before, then this will help:

Spoiler

I might have got the wrong end of the stick but where you looking for something like this?

Spoiler

I'm not sure if it holds because integration isn't just over the integers.

Also the question isn't very specific.
For example the lower bound could be the integral of 1 between 1 and n.
And the upper bound be the integral of x between 1 and n. So the bounds are n-1 and (n^2-1)/2.

Reply 361

0x2a

Original post by Elcor

Odd functions came up in the Imperial exam, which I wouldn't have known about if it wasn't for this thread, so cheers!

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What were the questions about if I may ask? And I hope you did well :smile:

Reply 362

Gawain

Original post by Gawain

I might have got the wrong end of the stick but where you looking for something like this?

Spoiler

The upper bound was meant to be the integral of 1/(x^0.5) from 1 to
n which gives 2(n^0.5) -2.
And the lower bound
From 1 to n too.
So 2(n+1)^0.5 -2(2^0.5)

Reply 363

Elcor

Original post by 0x2a

What were the questions about if I may ask? And I hope you did well :smile:

It was only C3 and C4 maths really, and thanks! I don't really like the course so I'm not too fussed.

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Reply 364

Gawain

Original post by 0x2a

Spoiler

Another something like this is:

Prove that

\displaystyle \sum^{\infty}_{n =1} \dfrac{1}{n}

diverges by comparing it to an appropriate integral.

Is it sufficient to show that integrating 1/x from 1 to infinity is infinite?

Reply 365

0x2a

Original post by Gawain

Is it sufficient to show that integrating 1/x from 1 to infinity is infinite?

If you can show that it is sufficient then sure.

This is not a cheeky reply or anything. My solution to the question is the same as yours

Reply 366

ThatPerson

Original post by 0x2a

Spoiler

Another something like this is:

Prove that

\displaystyle \sum^{\infty}_{n =1} \dfrac{1}{n}

diverges by comparing it to an appropriate integral.

Spoiler

Is this right?

Unparseable latex formula:

\displaystyle \sum^{\infty}_{n =1} \dfrac{1}{n} = \int^{\infty}_1 \dfrac{1}{x} \ dx = ln(x)\right^{\infty}_1

Which is unbounded as

\displaystyle \lim_{x \to \infty} \ln(x)

diverges.

Reply 367

0x2a

Original post by ThatPerson

Spoiler

Is this right?

Unparseable latex formula:

\displaystyle \sum^{\infty}_{n =1} \dfrac{1}{n} = \int^{\infty}_1 \dfrac{1}{x} \ dx = ln(x)\right^{\infty}_1

Which is unbounded as

\displaystyle \lim_{x \to \infty} \ln(x)

diverges.

Spoiler

Reply 368

Gawain

1) In a group of N people, some of them shake hands. Prove that there are at least two of that group that have shaken the same number of hands.

2) A ruler is stood up parallel to a wall. As it slides/falls down the wall and lies flat on the ground, what path does its mid point follow?

3) Find the equation of a parabola that has the all of following lines as tangents:
y=x+1
y=2x
y=-x

(edited 9 years ago)

Reply 369

ThatPerson

Original post by Gawain

Spoiler

Original post by 0x2a

Spoiler

Ah, I see. Thanks.

Reply 370

Gawain

Original post by ThatPerson

Spoiler

Ah, I see. Thanks.

It's not

Spoiler

.

If you draw a bigger diagram it should be easier to spot.
Then proving it is an exercise in geometry.

Reply 371

Krollo

Original post by Gawain

1.

Assume that everyone shakes a different number of hands. The maximum hands one can shake is N-1, if you shake hands with everyone. If everyone shakes a different number of hands, then one of the N people must shake hands with noone else, but someone else shakes hands with everyone. Contradiction - so at least two people shake the same number of hands.

My explanation is dire, but I think my point still stands.

2. Guessing here, but is it the arc of a circle?

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(edited 9 years ago)

Reply 372

Elcor