深夜數學題
Without using calculator, compare the following 3 real values:
a=31/32
b=cos(1/4)
c=4 sin(1/4)
draw circles
draw circles
No. Try to use CE level limits on the 3 basic trigonometric functions.
draw circles
No. Try to use CE level limits on the 3 basic trigonometric functions.
In radian measures
Note that 1 - [(1/4)^2] /2 = 31/32
Now plug it with a suitable half angle formula.
cos(x) = 1 - 2 sin(x/2)
Find the limits and then straightforward.
另一題：
Given y=|x|.
Find the tangents passing through the origin.
另一題：
Given y=|x|.
Find the tangents passing through the origin.
Given y=ln|x|.
第三題：
Solve |Z| < 0
Solutions:
第一題：
c/b = tan(1/4)/(1/4) >0
So c > b
Denote cos(x) = 1 - 2 sin(x/2)^2
When x tends to zero, taking the limit, we have:
cos(x) > 1 -2(x/2)^2
Pit x= 1/4, now:
cos(1/4) > 1 - 2 (1/8)^2
= 1 - 2(1/64)
= 1- 1/32
= 31/32
Clearly b>a
QED
另一題：
Given y=|x|.
Find the tangents passing through the origin.
Given y=ln|x|.
T: (y - y_0) = dy/dx (x - x_0)
d(ln|x|)/d|x|= 1/|x|
= 1/x for x>0 and 1/(-x) for x <0
Put x_0 =0 and y_0 = 0, x = 1/e or -1/e
QED
第三題：
Solve |Z| < 0
(x - iy) (x+ iy) = x^2 + y^2 < 0
Denote x and y as real values, we have:
x^2 < (iy)^2
This implies that y is purely imaginary, which contradicts with given conditions above.
So, no solution.
Solutions:
第一題：
c/b = tan(1/4)/(1/4) >0
So c > b
Denote cos(x) = 1 - 2 sin(x/2)^2
When x tends to zero, taking the limit, we have:
cos(x) > 1 -2(x/2)^2
Pit x= 1/4, now:
cos(1/4) > 1 - 2 (1/8)^2
= 1 - 2(1/64)
= 1- 1/32
= 31/32
Clearly b>a
QED
c/b = tan(1/4)/(1/4) > 1
另一題：
Given y=|x|.
Find the tangents passing through the origin.
Given y=ln|x|.
T: (y - y_0) = dy/dx (x - x_0)
d(ln|x|)/d|x|= 1/|x|
= 1/x for x>0 and 1/(-x) for x <0
Put x_0 =0 and y_0 = 0, x = 1/e or -1/e
QED
Note that dy/dx = 1/x for all real x=/=0
I.e. d ln(x) / dx = 1/x for x > 0
And d ln(-x) / dx
= [d ln(-x) / d(-x) ] • [d(-x)]/dx
= 1/(-x) • -1
= 1/x for x < 0
y is always positive for both cases
第四題：
Explain why for degree for a polynomial >4,
the product of roots will tend to be simple.
第四題：
Explain why for degree for a polynomial >4,
the product of roots will tend to be simple.
Solution:
Denote product of roots = Π z_n
Since i^5 = i, i^6 = -1 and so on,
the principal value of each z_n will be uncertain.
QED
第四題：
Explain why for degree for a polynomial >4,
the product of roots will tend to be simple.
Hints:
A complex logarithm is multi-valued.
No need to use Galois group. Just a little trick.
第四題：
Explain why for degree for a polynomial >4,
the product of roots will tend to be simple.
Solution:
Denote product of roots = Π z_n
Since i^5 = i, i^6 = -1 and so on,
the principal value of each z_n will be uncertain.
QED
Π z_n = e^(∑ ln(z_n))
Since complex logarithms are involved, the polynomial will then tend to be unsolvable.
第四題：
Explain why for degree for a polynomial >4,
the product of roots will tend to be simple.
Solution:
Denote product of roots = Π z_n
Since i^5 = i, i^6 = -1 and so on,
the principal value of each z_n will be uncertain.
QED
Π z_n = e^(∑ ln(z_n))
Since complex logarithms are involved, the polynomial will then tend to be unsolvable.
當每個complex logarithm 都限於acute angle
只要有5個或以上嘅complex logarithm,
個total angle就有機會過哂360度，
即係只要degree達5或以上，
個polynomial 就肯定有機會無general solution.
Example of unsolvable polynomial:
x^5 × x^4 + x^3 + x^2 + x + i = 0
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