1997 question 6 (1 Viewer)

[Bellow]

1997 4u maths
6a) The serues 1-x^2+x^4-.....+x^4n has 2n+1 terms
i) Explain why

1-x^2+x^4-.....+x^4n = [1+x^(4n+2)]/[1+x^2]

ii) Hence show that

1/[1+x^2] <= 1-x^2+x^4-......+x^4n <= 1/[1+x^2] + x^(4n+2)

iii) Hence show that, if 0<=y<=1, then

inv.tan y <= y - (y^3)/3 + (y^5)/5 - ..... + [y^(4n+1)]/[4n+1] <= inv.tan y + 1/[4n+3]

iv) Deduce that 0 < (1 - 1/3 - 1/5 - ..... + 1/1001) - pi/4 < 10^-3

need help on this thx very much =)

 

[Necros87]

well the first 1 is a geometric progression (yes that thing that we did in 2u)
dont ask me how to explain it, slept through that topic, just thaught this might help

 

[KFunk]

Another question to attach to this one: Whenever they have these questions where you're summing a series they always make the assumption that |x| < 1 (where x is the common multiplied difference), why is that?

 

[Bellow]

Another question to attach to this one: Whenever they have these questions where you're summing a series they always make the assumption that |x| < 1 (where x is the common multiplied difference), why is that?

i have no idea sorry, but i really need help on inequalities.....very weak at it.....i dunno how to approach some of them.

 

[Necros87]

*turns to solutions*
1-x2+x4-... +x4n=1-x2+(x2)2-...+(x2)2n
this is the sum of a GP with a=1
r=-x2 and N=2n+1

Sn = a(1-r^n)
1-r

=1*[1-(1-x2)^(2n+1)]
1-(-x2)

=1+(x2)^(2n+1)
1+x

(since (-x2)^(2n+1)+-(x2)^2n+1) because 2n+1 is odd)

=1+x^(4n+2)
1+x2

 

[Bellow]

*turns to solutions*
1-x2+x4-... +x4n=1-x2+(x2)2-...+(x2)2n
this is the sum of a GP with a=1
r=-x2 and N=2n+1

Sn = a(1-r^n)
1-r

=1*[1-(1-x2)^(2n+1)]
1-(-x2)

=1+(x2)^(2n+1)
1+x

(since (-x2)^(2n+1)+-(x2)^2n+1) because 2n+1 is odd)

=1+x^(4n+2)
1+x2

yeh i got that part sorry didnt mention that to u but thx anyway.....u kno how to do parts iii) and iv)?.......even part ii), i can do it but im not sure if im correct.

 

[KFunk]

For part two:

x² + 1 &ge; 1 ... and ... 0 &le; x⁴ⁿ⁺² (even power)

This gives you:

0 &le; x⁴ⁿ⁺²/(1 + x²) &le; x⁴ⁿ⁺²

1/(1+x²) &le; (x⁴ⁿ⁺²+1)/(1 + x²) &le; x⁴ⁿ⁺² + 1/(1+x²) ..... by adding 1/(1+x²) to each

and of course the middle part is equal to the series as proved in i)

 

[KFunk]

for iii) integrate each part of the inequality w.r.t x between 0 and y

1/[1+x²] &le; 1-x²+x⁴-...+x⁴ⁿ &le; 1/[1+x²] + x⁴ⁿ⁺²

&int; 1/[1+x²] dx &le; &int; 1-x²+x⁴-...+x⁴ⁿ dx &le; &int; 1/[1+x²] + x⁴ⁿ⁺² dx (between 0 and y for all)

tan^-1y &le; y - y³/3 + y⁵/5 -...+ y⁴ⁿ⁺¹/(4n+1) &le; tan^-1y + y⁴ⁿ⁺³/(4n+3) after subbing in the limits of integration

tan^-1y &le; y - y³/3 + y⁵/5 -...+ y⁴ⁿ⁺¹/(4n+1) &le; tan^-1y + 1/(4n+3)

The last step is because y⁴ⁿ⁺³/(4n+3) &le; 1/(4n+3) since 0 &le; y &le; 1 as defined.

 

[Bellow]

For part two:

x² + 1 &ge; 1 ... and ... 0 &le; x⁴ⁿ⁺² (even power)

This gives you:

0 &le; x⁴ⁿ⁺²/(1 + x²) &le; x⁴ⁿ⁺²

1/(1+x²) &le; (x⁴ⁿ⁺²+1)/(1 + x²) &le; x⁴ⁿ⁺² + 1/(1+x²) ..... by adding 1/(1+x²) to each

and of course the middle part is equal to the series as proved in i)

cool thx, any ideas on iii) and iv)

 

[KFunk]

Give iv) ago by letting y = 1 and choosing an appropriate value of n (i.e. so that 4n+1 = 1001)


EDIT: Also, for part iii) the reason you can integrate between 0 and y (where 0 &le; y &le; 1) is because in summing the geometric series in the form (x⁴ⁿ⁺²+1)/(1 + x²) you're making an assumption that |x| < 1 so the inequality holds across that interval 0 &le; x < 1 i.e. you can integrate over it as well.

 

[Bellow]

Give iv) ago by letting y = 0 and choosing an appropriate value of n (i.e. so that 4n+1 = 1001)


EDIT: Also, for part iii) the reason you can integrate between 0 and y (where 0 &le; y &le; 1) is because in summing the geometric series in the form (x⁴ⁿ⁺²+1)/(1 + x²) you're making an assumption that |x| < 1 so the inequality holds across that interval 0 &le; x < 1 i.e. you can integrate over it as well.

ohhh ok.....but then if i let y=0,

0 < 0 - 0/3 +1/5 - ...... + 0/4n+1 < 1/4n+3

did u mean let y=1, since theres a - pi/4 in the middle inequality?

 

[KFunk]

did u mean let y=1, since theres a - pi/4 in the middle inequality?

Haha, my bad , yeh I meant y=1. well spotted.

 

[Bellow]

did u approximate n?....
if 4n+1=1001
n=250 (for the mid inequality)

but 4(250)+3 is not equal to 10^-3 (as in the right inequality)

......unless i did something wrong....

 

[robbo_145]

4(250) + 3 =1003 > 1000
that should help