4U Revising Game (2 Viewers)

[tommykins]

Actual question: show that for x>0, x- x^3/3 < inverse tan x < x- x^3/3 + x^5/3.
Let f(x) = arctan(x) - x + (x^3)/3
f'(x) = 1/(1+x^2) - 1 + x^2
f'(x) = (1+(x^4-1))/(1+x^2)
f'(x) = (x^4)/(1+x^2)
Therefore, f'(x) > 0 for x > 0.
f(0) = arctan(0) - 0 + 0
f(0) = 0
Since f'(x) > 0 for x > 0,
f(x) > 0 for x > 0.
Hence,
arctan(x) - x + (x^3)/3 > 0
arctan(x) > x - (x^3)/3 --(1)
Let g(x) = x - (x^3)/3 + (x^5)/3 - arctan(x)
g'(x) = 1 - x^2 + (5/3)x^4 - 1/(1+x^2)
3g'(x) = 3 - 3x^2 + 5x^4 - 3/(1+x^2)
3g'(x) = ((3 - 3x^2 + 5x^4)(1+x^2)-3)/(1+x^2)
g'(x) = (3 - 3x^2 + 5x^4 + 3x^2 - 3x^4 + 5x^6 - 3)/3(1+x^2)
g'(x) = (5x^6+2x^4)/3(1+x^2)
Therefore, g'(x) > 0 for x > 0.
g(0) = 0 - 0 + 0 - arctan(0)
g(0) = 0
Since g'(x) > 0 for x > 0,
g(x) > 0 for x > 0.
Hence,
x - (x^3)/3 + (x^5)/3 - arctan(x) > 0
x - (x^3)/3 + (x^5)/3 > arctan(x) --(2)
By comparing (1) and (2),
x- x^3/3 < arctan(x) < x- x^3/3 + x^5/3 for x > 0

interesting approach.

 

[duy.le]

again...a question?

 

[3.14159potato26]

New question:
The equation x^3 + px - 1 = 0 has three real, non-zero roots a,b,c.
Find the value of:
a^2 + b^2 + c^2
a^4 + b^4 + c^4
and hence show that p must be strictly negative.

 

[tommykins]

Prove by induction -

1 + 1/2^2 + 1/3^2 + ...+1/n^2 <= 2-1/n

 

[u-borat]

and a followup question to tommy's:

prove that:


1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/n^2 <1.99

 

[duy.le]

New question:
The equation x^3 + px - 1 = 0 has three real, non-zero roots a,b,c.
Find the value of:
a^2 + b^2 + c^2
a^4 + b^4 + c^4
and hence show that p must be strictly negative.

hummmm... i was able to prove p was negative with only a^2+b^2+c^2 [=-2p]
and did anyone else get a^4+b^4+c^4 = 2p^2

 

[u-borat]

^^yea p is still negative, negative squared is positive.

 

[duy.le]

dude u-borat how do u do those types of questions... like of course the rhs is easy to get but the lhs is rather weird. ive seen it before but dont have a general approach to these questions. any tips?

 

[tacogym27101990]

i a^4+b^4+c^4 to be -2p^2

i did
x^3=1-px
so x^4=x-px^2
thn subbed in a,b,c
and then went from there

 

[3.14159potato26]

Prove by induction -
1 + 1/2^2 + 1/3^2 + ...+1/n^2 <= 2-1/n

When n = 1,
LHS = 1
RHS = 2 - 1/1 = 1
Since LHS <= RHS, it is true for n = 1.
Assume that it is true for n = k,
i.e. 1 + 1/2^2 + 1/3^2 + ...+1/k^2 <= 2-1/k
Consider
1 + 1/2^2 + 1/3^2 + ...+1/k^2 + 1/(k+1)^2
<= 2 - 1/k + 1/(k+1)^2
= 2 - 1/k + 1/(k+1)(k+1)
<= 2 - 1/k + 1/k(k+1) ---(because k+1 > k)
= 2 - (k+1)/k(k+1) + 1/k(k+1)
= 2 - k/k(k+1)
= 2 - 1/(k+1)
1 + 1/2^2 + 1/3^2 + ...+1/k^2 + 1/(k+1)^2 <= 2 - 1/(k+1)
Therefore, it is true for n = k + 1.
Since it is true for n = 1 and n = k+1 whenever it is true for n = k, therefore it is true for n = 2,3 and so on.

 

[3.14159potato26]

prove that:
1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/n^2 <1.99

Um....for n = 1, you get 1.45 < 1, which is false. So does 1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/n^2 <1.99 say that it is true for any arbitary n, or are you talking about the limiting sum of 1 + 1/2^2 + 1/3^2 + ...+1/n^2?

 

[u-borat]

dude u-borat how do u do those types of questions... like of course the rhs is easy to get but the lhs is rather weird. ive seen it before but dont have a general approach to these questions. any tips?

man i dunno, its in the 1987 hsc paper but i don't have the answers and have no idea how to do it.

 

[u-borat]

na sorry man, instead of the n, its a 99

1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2 <1.99


fuck sorry for typing all these questions up wrong. :/

edit; the 2nd half of that inequality is easy. someone do the first bit .

 

[tacogym27101990]

the first half doesnt hold until n=5
somethings wrong

 

[u-borat]

uh what?

do you read?

its not a sum to n terms.

its 1 + 1/2^2 + all the way to 1/99^2

 

[tacogym27101990]

haha yeah sorry
heads in english

 

[3.14159potato26]

1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2 <1.99

Well, to prove that 1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2,
consider the graph of 1/x^2.
From the graph,
I{1->100} 1/x^2 < 1 + 1/2^2 + 1/3^2 + ... + 1/99^2,
that is taking the higher/left value of the function values as the retangle height, i.e. if function values are 1/n^2 and 1/(n+1)^2, retangle length = 1/n^2.
Therefore, 1 + 1/2^2 + 1/3^2 + ... + 1/99^2 represents the summation of the retangular areas, which is more than the actual value of the integral.
However, since the graph is a decreasing function, therefore approximation using trapezium gives an area value that is less than the area value from approximation using retangles, that is:
Summation of trapezium areas from 1->100 < 1 + 1/2^2 + 1/3^2 + ... + 1/99^2.
At the x-values n and n+1, the function values are 1/n^2 and 1/(n+1)^2.
Area of trapezium
= 1/2(Sum of length of parallel sides)(Width)
= 1/2(1/n^2 + 1/(n+1)^2)(1)
= 1/2(1/n^2 + 1/(n+1)^2)
Therefore, summation of trapezium areas from 1->100
= 1/2(1/1^2 + 1/2^2) + 1/2(1/2^2+1/3^2) + 1/2(1/3^2+1/4^2) + ... + 1/2(1/98^2+1/99^2)
= 1/2(1/1^2 + 2/2^2 + 2/3^2 + 2/4^2 + ... + 2/98^2 + 1/99^2)
= 1/2(1/1^2 + 1/99^2) + (1/2^2 + 1/3^2 + 1/4^2 + ... + 1/98^2)
= 1/2(1/1^2 + 1/99^2) + (Sum of retangles - 1/1^2 - 1/99^2)
= (Sum of retangles) - 1/2(1/1^2 + 1/99^2)
< 1.99 - 1/2(1/1^2 + 1/99^2)
= 1.489948985
Therefore, summation of trapezium areas from 1->100 < 1.489948985
However, consider summation up to the 6th term = 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 = 1.511797092 > 1.489948985
Since 1 + 1/2^2 + 1/3^2 + ...+1/99^2 > 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36,
Therefore, 1 + 1/2^2 + 1/3^2 + ...+1/99^2 > summation of trapezium areas.
1.489948985 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2, and since 1.45 < 1.489948985,
1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2.

Note: It 's a very messy method, but since you can't immediately generalize that summation of trapezium areas from 1->100 < 1.489948985 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2, its the only way i can get the answer for now.

 

[duy.le]

Well, to prove that 1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2,
consider the graph of 1/x^2.
From the graph,
I{1->100} 1/x^2 < 1 + 1/2^2 + 1/3^2 + ... + 1/99^2,
that is taking the higher/left value of the function values as the retangle height, i.e. if function values are 1/n^2 and 1/(n+1)^2, retangle length = 1/n^2.
Therefore, 1 + 1/2^2 + 1/3^2 + ... + 1/99^2 represents the summation of the retangular areas, which is more than the actual value of the integral.
However, since the graph is a decreasing function, therefore approximation using trapezium gives an area value that is less than the area value from approximation using retangles, that is:
Summation of trapezium areas from 1->100 < 1 + 1/2^2 + 1/3^2 + ... + 1/99^2.
At the x-values n and n+1, the function values are 1/n^2 and 1/(n+1)^2.
Area of trapezium
= 1/2(Sum of length of parallel sides)(Width)
= 1/2(1/n^2 + 1/(n+1)^2)(1)
= 1/2(1/n^2 + 1/(n+1)^2)
Therefore, summation of trapezium areas from 1->100
= 1/2(1/1^2 + 1/2^2) + 1/2(1/2^2+1/3^2) + 1/2(1/3^2+1/4^2) + ... + 1/2(1/98^2+1/99^2)
= 1/2(1/1^2 + 2/2^2 + 2/3^2 + 2/4^2 + ... + 2/98^2 + 1/99^2)
= 1/2(1/1^2 + 1/99^2) + (1/2^2 + 1/3^2 + 1/4^2 + ... + 1/98^2)
= 1/2(1/1^2 + 1/99^2) + (Sum of retangles - 1/1^2 - 1/99^2)
= (Sum of retangles) - 1/2(1/1^2 + 1/99^2)
< 1.99 - 1/2(1/1^2 + 1/99^2)
= 1.489948985
Therefore, summation of trapezium areas from 1->100 < 1.489948985
However, consider summation up to the 6th term = 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 = 1.511797092 > 1.489948985
Since 1 + 1/2^2 + 1/3^2 + ...+1/99^2 > 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36,
Therefore, 1 + 1/2^2 + 1/3^2 + ...+1/99^2 > summation of trapezium areas.
1.489948985 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2, and since 1.45 < 1.489948985,
1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2.

Note: It 's a very messy method, but since you can't immediately generalize that summation of trapezium areas from 1->100 < 1.489948985 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2, its the only way i can get the answer for now.

there is a more quick algebraic way

 

[jkwii]

or even better, "by inspection this statement is true for n = k +1 if it is true for n = k."

AW YEAH

 

[u-borat]

duyle can you please post the solution?

anyone with the 1987 paper solutions paste the solution?