4u Mathematics Marathon v2.0 (2 Viewers)

[LoneShadow]

If you're trying to figure out the answer to this question by drawing lines back and forward you are making it hard for yourself. The reason i gave this question is that you can do it in one line and about 5 seconds if you use your head a bit...

damn. 2 years of uni didn't teach me to use my head. Never had heard of infinite sums.


btw ... learn to see exaggerations when u come across them.

 

[haque]

I have an interesting question:

Prove that Sigma (from k=0 to n) coskx={cos(nx/2)sin((x(n+1))/2)}/sin(x/2)

and sigma (from k=0 to n) sinkx={sin(nx/2)sin((x(n+1))/2)}/sin(x/2)

note: there are lots of ways to show those two identities.but the main ways are by using mathematical induction or by using the property of complex numbers (Euler's equation e^(ix)=cosx+isinx)). but mathematical induction needs more work so i suggest to use euler's equation

Here's the answer for the first part, can't be bothered for the second part-i mean the proofs for both q iare almost the same so someone else can do it-eulers' notation is just that exp(ix)=cosx+isinx we use this later.

cosx=(exp(ix) +exp(-ix))/2 and so if we sum all the terms as given we get
sum of whatever given=1 +1/2(exp(ix)+......................+exp(inx)) +1/2(exp(-ix)+........................exp(-inx)) these are 2 separate GPs so simplifying this we get
sum=1/2(exp((n+1)ix) -1)/(exp(ix)-1) +1/2(exp(-(n+1)ix -1)/(exp(-ix)-1)

=1/(4sinx/2(-sinx/2+icosx/2))*(cos(n+1)x+isin(n+1)x-cosnx+isinnx-1+cosx+isinx) note this is the numerator
=1/(4sinx/2(-sinx/2+icosx/2))*(2sin(2n+1)x/2(icosx/2-sinx/2) +((cosx-1)+
isinx)/(2(cosx-1 +isinx))
=(sin(2n+1)x/2 +sinx/2)/(2sinx/2)
=sin(n+1)x/2cosnx/2/sin(x/2) QED-I've skipped alot of steps but the q was more algebraic in nature than anything-once u recognise the fact it's a gp ect it comes out nicely with sums to products being used later.

 

[vafa]

ok, You do not need to be bothered to do next part. you can do both parts at once. but as i said before there are many ways to prove these identities and that was your way. for more calrification download my solution.

If any one interested in the mathematical induction proof, then please let me know so I can give you the solution by mathematical induction as well.

http://s6.quicksharing.com/v/3687989/Euler.pdf.html

 

[haque]

I don't have that much time to type up a similar proof etc not with tsp plus all the tutorials and assignemnets i haven't done-besides once an 07er sees the hints they'll be able to do it themselves-the biggest obstacle for them would be euler's notation

 

[jb_nc]

Can you give the proof by induction too please?

 

[vafa]

For the proof by mathematical induction download the solution from the below URL:

http://s23.quicksharing.com/v/4639522/1Induction.pdf.html

 

[HappyFeet]

Sounds like a pretty neat activity. Try this one:

a^2+b^2+c^2 >= 4 \sqrt{3}S

where a,b,c are the 3 sides of a triangle and S is the area.

 

[andykillz]

• There are 5 houses in 5 different colors
• In each house lives a person with a different nationality
• These 5 owners drink a certain type of beverage, smoke a certain brand of cigar, and keep a certain pet
• No owners have the same pet, smoke the same brand of cigar or drink the same drink.

Here's the question: Who owns the fish?

1. The Brit lives in a red house
2. The Swede keeps dogs as pets
3. The Dane drinks tea
4. The green house is on the left of the white house
5. The green house owner drinks coffee
6. The person who smokes Pall Mall rears birds
7. The owner of the yellow house smokes Dunhill
8. The man living in the house right in the middle drinks milk
9. The Norwegian lives in the first house
10. The man who smokes Blend lives next door to the one who keeps cats.
11. The man who keeps horses lives next door to the man who smokes Dunhill
12. The owner who smokes Blue Master drinks beer
13. The German smokes Prince
14. The Norwegian lives next to the blue house
15. The man who smokes Blend has a neighbor who drinks water

 

[ssglain]

• There are 5 houses in 5 different colors
• In each house lives a person with a different nationality
• These 5 owners drink a certain type of beverage, smoke a certain brand of cigar, and keep a certain pet
• No owners have the same pet, smoke the same brand of cigar or drink the same drink.

Here's the question: Who owns the fish?

1. The Brit lives in a red house
2. The Swede keeps dogs as pets
3. The Dane drinks tea
4. The green house is on the left of the white house
5. The green house owner drinks coffee
6. The person who smokes Pall Mall rears birds
7. The owner of the yellow house smokes Dunhill
8. The man living in the house right in the middle drinks milk
9. The Norwegian lives in the first house
10. The man who smokes Blend lives next door to the one who keeps cats.
11. The man who keeps horses lives next door to the man who smokes Dunhill
12. The owner who smokes Blue Master drinks beer
13. The German smokes Prince
14. The Norwegian lives next to the blue house
15. The man who smokes Blend has a neighbor who drinks water

The German owns the fish, although I'm not entirely confident. This is the order of things as I worked them out:
1st house: yellow, Norwegian, cats, water, Dunhill
2nd house: blue, Dane, horses, tea, Blend
3rd house: red, Brit, birds, milk, Pall Mall
4th house: green, German, fish, coffee, Prince
5th house: white, Swede, dogs, beer, Blue Master

 

[ravdawg]

Yup, that what i got

 

[jdcb4]

it takes the cars 1500/(5+10) seconds to meet. in that time, the plane hasnt stopped moving. therefore, the plane has travelled 20 * 1500/(5+10) meters.

20*1500/15 = 20*100 = 2000m = 2 km.

Cars will lmeet after slow car has travelled 1500 and fast 3000, i.e. 300 seconds... plane travels at 20m/s total distance tavelled is 20*300 = 6000m =6km
No?

 

[AMorris]

Happy Feet's question is quite nice
use the cosine rule and the area of the triangle with sine thingamy.

We get:

RTP: a^2 + b^2 + a^2 + b^2 - 2abcosC >= 2*rt(3)*absinC

RTP: a^2 + b^2 >= abcosC + rt(3)*absinC

RTP: a^2 + b^2 >= ab(cosC + rt(3)*sinC)

but we know by AM-GM a^2 + b^2 >= 2ab

thus STP: (sufficient to prove)

2ab >= ab(cosC + rt(3)*sinC)

2 >= cosC + rt3*sinC

now by our auxillary angle stuff cosC +rt3*sinC = 2sin(C + @) where I dont care what @ is (but I think its 30)

and since sinX <= 1
-> 2sin(C + @) <= 2 and we are done