The Student Room Group

A Summer of Maths

Scroll to see replies

Original post by Blutooth
Didn't want to use L'Hopital?


I tend to use it when I am lazy or desperate; it's a powerful tool though. :tongue:
Question:

One can easily observe that 20+50=30+402^0 + 5^0 = 3^0 + 4^0 and 21+51=31+412^1 + 5^1 = 3^1 + 4^1.

Are there any other real numbers xx such that 2x+5x=3x+4x2^x + 5^x = 3^x + 4^x?


Required: (undergraduate)

Spoiler

Question:

The unit cube

C={(x,y,z)  0x,y,z1}\mathbf{C} = \{ (x,y,z)\ |\ 0 \leq x,y,z \leq 1\}

is sliced along the planes x=y,y=zx = y, y = z and z=xz = x.

How many pieces are there?



Any interesting questions? :tongue: Is everyone getting bored already? :smile:
Reply 103
Can anyone here solve this? I suspect most of you will be able to, but having not done A2 maths, Cartesian vectors are a bit vague to me. Also, if you fancy explaining slowly how you did each step as if you were explaining to an amateur, that might help me get my head round it :tongue:

This story takes place on the infamous Planar Slope which, as it's name suggests, is a flat sloping plane. Conveniently for us, the grid reference system in use was a simple (x,y,z) Cartesian system.

Dot Product was out searching for an object and found herself at position (17, -3, 4).

The object she was after was located on a footpath. The footpath is perfectly straight in the direction (4,5,-6) and passes through point (16,19,-4).

Dot knew that the object is located on the paths at a distance of ab.c (to 1 decimal place) from her and at location (d,e,f).

Interestingly this was in fact at the closest point on the path to her, although Dot had zero knowledge of this.

Find (d,e,f) and ab.c

(Note: I don't have the answer to this but I'm hoping there'll be a general consensus as to what the right answer is)
Original post by jack.hadamard
Question:

The unit cube

C={(x,y,z)  0x,y,z1}\mathbf{C} = \{ (x,y,z)\ |\ 0 \leq x,y,z \leq 1\}

is sliced along the planes x=y,y=zx = y, y = z and z=xz = x.

How many pieces are there?



Any interesting questions? :tongue: Is everyone getting bored already? :smile:


6 surely since their only intersection is the line x=y=zx=y=z? I've had too much to drink to be able to actually picture it in my mind! :tongue:
Some fun: Euler bricks and the search for a perfect cuboid.
Reply 106
you have the following weights: 6,9,5.
and an old fashioned weighing balance.

is it possible to weigh out a 1? How?

What combination minimises the total number (not weight) of weights used?

Spoiler


same question, but
suppose you only have the following weights: 51, 12, 67
(edited 11 years ago)
I'll throw in a purer question to balance out with the other problems on this page (some of you may recognise this):

Question

s1,s2,s3  ...s_1,s_2,s_3\;... is a strictly increasing sequence of positive integers such that ss1,ss2,ss3  ...s_{s_1},s_{s_2},s_{s_3}\;... and ss1+1,ss2+1,ss3+1  ...s_{s_1+1},s_{s_2+1},s_{s_3+1}\;... are both in arithmetic progression.

Prove that s1,s2,s3  ...s_1,s_2,s_3\;... is also in arithmetic progression.
Original post by jj193
you have the following weights: 6,9,5.
and an old fashioned weighing balance.

is it possible to weigh out a 1? How?

What combination minimises the total number (not weight) of weights used?

Spoiler



you don't need much effort for this one - just put the 6 on one pan and the 5 on the other.
Original post by Lord of the Flies
...

Prove that s1,s2,s3  ...s_1,s_2,s_3\;... is also in arithmetic progression.



Hmm.. any hints on this?

Spoiler

Original post by jack.hadamard
Hmm.. any hints on this?

Spoiler



Spoiler



I'm not very good with sequences, but that's what makes them fun...
(edited 11 years ago)
Prove that any positive number xx is equal to the infinite sum:

i=0(logx)ii! \displaystyle \sum _{i=0}^\infty \dfrac{(\log x)^i}{i!}

Hint

Original post by electriic_ink
Prove that any positive number xx is equal to the infinite sum:

i=0(logx)ii! \displaystyle \sum _{i=0}^\infty \dfrac{(\log x)^i}{i!}

Hint



Spoiler

Two squares on an 8*8 chessboard are called touching if they have at least a common vertex. Determine if it is possible for a king to begin in some square and visit all squares exactly once in such a way that all moves except the first are made into squares touching an even number of squares already visited.

Hint

Original post by jack.hadamard
Hmm.. any hints on this?

Spoiler



Spoiler

Original post by electriic_ink
Prove that any positive number xx is equal to the infinite sum:

i=0(logx)ii! \displaystyle \sum _{i=0}^\infty \dfrac{(\log x)^i}{i!}

Hint



Helpful hint! :tongue:
Original post by Lord of the Flies

Spoiler



Spoiler

More sequences?

Question:

Let bb be a real number. For every integer m0m \geq 0, define the sequence {am(i)}\{a_m(i)\} by the conditions

i)  am(0)=b/2m\ a_m(0) = b/2^m

ii)  am(i+1)=(am(i))2+2am(i)\ a_m(i + 1) = (a_m(i))^2 + 2a_m(i)


where i=0,1,2,...i = 0, 1, 2, ... is a non-negative integer.


Find limnan(n)\displaystyle \lim_{n \to \infty} a_n(n).


Hint:

Spoiler

Original post by jack.hadamard

Spoiler


Spoiler


Original post by jack.hadamard

Spoiler


Spoiler


Original post by jack.hadamard

Spoiler


Spoiler

(edited 11 years ago)
Original post by jack.hadamard
More sequences?

Question:

Let bb be a real number. For every integer m0m \geq 0, define the sequence {am(i)}\{a_m(i)\} by the conditions

i)  am(0)=b/2m\ a_m(0) = b/2^m

ii)  am(i+1)=(am(i))2+2am(i)\ a_m(i + 1) = (a_m(i))^2 + 2a_m(i)


where i=0,1,2,...i = 0, 1, 2, ... is a non-negative integer.


Find limnan(n)\displaystyle \lim_{n \to \infty} a_n(n).


Hint:

Spoiler



Spoiler


I feel like there may be a mistake in my evaluation of the limit - if so, give me a hint!
(edited 11 years ago)

Quick Reply

Latest