Mathematics Question Review (2025-10-03)

Refer to the previous review:

Below is the formal proof:

(using the inequalities from the hint 1)

a_(n+1) - a_n =

[1 + 1/2 + ... + 1/n + 1/(n+1) - ln(n+1)]

-[1 + 1/2 + ... + 1/n - ln(n)]

=1/(n+1) + ln (n/(n+1))

=1/(n+1) - ln ((n+1)/n)

= 1/(n+1) - ln (1 + 1/n) <0

So {a_n} is strictly decreasing

Also, a_n

= 1 + 1/2 + ... + 1/n - ln(n)

> ln(1 + 1) + ln (1 + 1/2) + ... ln (1 + 1/n) - ln(n)

= ln(n+1) - ln(n) > 0

So {a_n} has a lower bound 0.

Q.E.D.

Now another question:

Consider n as a positive integer of 1, 2, 3, ...),

and {x_n} exists for 0 < x_1< ⫪,

where x_(n+1) = sin(x_n).

(a) Compute x_(+∞)

(b) Denote f(n) = (x_(n+1)/x_n)^(1/(x^2)_n),

Compute f(+∞).

Solution for (a):

Note that the question wants us to compute the following limit:

sin(sin(sin(...(sin(x))...)))

which is equivalent to

x = sin(x)

The limit x_(+∞) is 0.

You may also try to use mathematical induction to prove it,

where x_(n+1)/x_n = sin(x_n)/x_n < 0 as x>0.

Solution for (b):

Using (a),

we are computing f(n) = (sin(x_n)/x_n)^(1/(x^2)_n)

Let t = x_n, as n --> ∞, t --> 0

So f(n) = g(t) = (sin(t)/t)^(1/t^2)

= [e^ln((sin(t)/t)^(1/t^2)

= e^(ln(sin(t)/t^3)

By L'Hospital's Rule,

g(0) = e^(ln(cos(t)/3t^2))

= e^(ln(-sin(t))/6t))

= e^(-1/6)

更正：

g(t) = (sin(t)/t)^(1/t^2)

= e^[ln((sin(t)/t)^(1/t^2)]

By small value approximation,

ln(t) ~ t-1

So g(t) becomes:

e^[[sin(t-t)]/t^3]

Note: Do no resolve [sin(t-t)]/t^3 into partial fractions, i.e. sin(t)/t^3 - 1/t^2

By L'Hospital's Rule,

g(0) = e^((cos(t)-1)/3t^2))

= e^((-sin(t))/6t))

= e^(-1/6)

So g(t) becomes:

e^[[sin(t)-t)]/t^3]

So g(t) becomes:

e^[[sin(t)-t]/t^3]

By L'Hospital's Rule,

g(0) = e^((cos(t)-1)/3t^2)

= e^((-sin(t))/6t)

= e^(-1/6)

g(t) = (sin(t)/t)^(1/t^2)

= e^[ln(sin(t)/t)^(1/t^2)]

recall sin(t)~t,

but the base sin(t)/t cannot be 1 when the exponent 1/t^2 become infinity. In this case, L'Hospital's Rule must be applied to find removable singularities.

Next question:

Give a formal proof of option A.

Solution:

(1+1/x)^x

= e^[x ln(1+1/x)]

= e^ [ln(1+1/x)/(1/x)]

Note that we cannot use small value approximation because 1/0 = ∞.

Since the index is ∞/∞ mode of indeterminate, we can apply L'Hospital's Rule. Now denote y = 1/x, the index becomes:

e^[ln(1+y)/y]

Taking limit on the index, we have:

e^[(1/(y+1))/1] = e^(1/(y+1))

= e^(1/∞) = e^0 = 1

Note that x>0

where x_(n+1)/x_n = sin(x_n)/x_n < 1 as x>0.

Summary: Properties of limits:

and for continuous functions:

Another question on small value approximation:

Show that x - sin(x) ~ x^3/3!

Solution:

sin(x) = x - x^3/3! + O(x^5)

So sin(x) - x = - x^3/3! = -x^3/6

Multiplying both side by -1, we get the desired result.

sin(x) - x = - x^3/3! = -x^3/6 + O(x^5) ~ -x^3/6

From #5,

by recalling ln(x+1) ~ x,

we have ln(x) ~ (x-1),

then we can transform an exponent of any integral base into base e, for example:

So no more logarithm to fuck with.

Another question: Show that e^x, ln(x), sin(x) are dimensionless.

Solution:

They are not polymonial, so their power series is infinite, i.e. no order.

An elementary function of no order has infinite derivatives, say:

(e^x)' = e^x

no roots

(ln(x))' = 1/x

(ln(x))" = (1/x)' = -1/x^2,

no roots

(sin(x))' = cos(x)

(sin(x))" =(cos(x))' = -sin(x).

infinite roots

Notice the order of polynomials from both numerator and denominator:

This time, if we take limit on x --> ∞,

notice that if p > 2, the limit = +∞,

because the order of the denominator is 2.

Then if p=2, the limit = 1

if p<2, the limit = 0.

Recall that 1/∞ = 0.

Note that for binomial expansion for any real power n:

Then when we operate small value approximation on a binomial below:

Put x = 1,

then 1^n = 1 (The principal value of roots of unity)

Then put Δx = 1/x, we soon notce that:

(1 + 1/x)^(1/6) ~ 1 + (1/6)(1/x)

for |1/x|< 1

Another question: Try to compute the limit below:

Solution: Rewrite the polynomial inside the bracket:

Define the series a_1 > a_2 > a_3 > ... > a_n

Then a_2/_1 > 1, a_3/a_1 > 1 and so on.

By the discussion in #21,

we soon notice that the limit become a_1 ‧ (1+0+0+0)^(0) = a_1‧1 = a_1

好多 MK, 尤其係 Gooder,

覺得烏大龜開親 post 都係認叻,

這只反映佢哋小器同極端無知

以下是烏大龜一貫嘅學習法:

Richard Feynman 學習法是由諾貝爾物理學獎得主開發的一種強大學習方法，以其簡化複雜概念的能力而聞名。這位被稱為「偉大解釋者」的物理學家相信，如果你無法用簡單的話語解釋某個概念，那麼你就還沒有真正理解它。

這個方法的核心原則是：透過教學來學習。Feynman 認為複雜性和術語往往掩蓋了理解的不足，真正的掌握在於能夠將複雜的想法分解成任何人都能理解的基本組成部分。

費曼學習法的核心理念

學習過程中最大的挑戰之一就是確保自己真正理解所學內容。許多人都聽過這句經常被歸屬於愛因斯坦的名言：「如果你無法簡單地解釋一件事，那就代表你還沒有完全理解它。」雖然這句話的真實來源可能並非愛因斯坦，但它所蘊含的學習智慧卻是不可否認的。

費曼學習法正是基於這個理念發展而來的強大學習技巧。這個方法的核心思想是：想要真正理解某個概念，最好的方式就是嘗試解釋它。透過教授他人的過程，我們能夠深入檢視自己的知識結構，發現理解上的盲點。

Another Question:

We know that the small value approximation below:

ln(x+1) ~ x

ln(x+1) ~ x - x^2/2

ln(x+1) ~ x - x^2/2 + x^3/3!

is equivalent to the Taylor series with Big O notation:

ln(x+1) = x + O(x)

ln(x+1) = x - x^2/2 + O(x^2)

ln(x+1) = x - x^2/2 + x^3/3! + O(x^3)

So which one of them should be used when evaluating a limit?

Erratum:

ln(x+1) = x + O(x)

ln(x+1) = x - x^2/2 + O(x^2)

ln(x+1) = x - x^2/2 + x^3/3 + O(x^3)

Solution:

Using the discussion in #21 again, we have:

Below is an example:

Note that n[n(-1/2)(1/n^2)] = -1/2, which is the desired limit.

Note that there are many gimmicks when expanding a polynomial using Big O notation.

1- (1-x) (1-x^2)

= 1 - 1 + x + x^2 - x^3

= x + x^2 +O(x^2)

Then we observe that

So the expansions yields the same result with O(x^2).

Sorry, should be little-o notation instead:

x + x^2 + o(x^2)

Now we apply the gimmick discussed in #29 when taking small value approximation:

by cos(x) ~ 1- x^2/2, we have:

by eliminating the common factor x^2 from the numerator and denominator, we have the following limit when x approaches 0:

Alternative solution: Transform the triginometric expression into exponents

Now the required becomes:

Then by (e^lnr)' = d(e^lnr)/dx

= (e^lnr) (d(lnr)/dr) (dr/dx) = (e^lnr) (lnr)' r',

we have:

The remaining steps are the same as #31.

Another gimmick to be introduced:

∑ f(n+i) = f(n+1) + f(n+2) + ... + f(n+n)

Then when squeezing a function g(n), we often write:

nf(n+1) < g(n) < nf(n+n)

嗱, 你依家就可以喺預科純數卷瘋狂執分:

Last gimmick today:

Just read the section If we're asked to write a Riemann sum from a definite integral...

In higher math, we do the reverse so that the limit of a series becomes a definite integral.

Below is an example:

Notice the gimmick of integration by part:

Erratum:

(2^2 sin2x / 2x) (1/ cos2x)

Erratum:

(2^2 sin2x / 2x) (1/ cos2x)

note that if n is not a positive integer,

the expansion will be infinite series.

For fractional indics:

So long as the index is rational,

we always use binomial expansion to solve it.

More erratum:

(n^2 sinnx/nx) (1/cosnx)

This gimmick is very useful when you observe the object to be integrated seems hard to substitute:

Another application of the gimmick is deriving reduction formula to simplify a definite integral:

See, HKDSE provides nothing on this damn important skillset.