Foundations of Data Science

these equalities by derivatives.
∞∑
i=1
1
i = 1 + 1 + ( 1
+ ( 1
2 3 4)
+
1 + 1 + 1 + ) 1
5 6 7 8 + · · · ≥ 1 +
1
+ 1 + · · · and thus diverges.
2 2
The summation n ∑
i=1
1
grows as ln n since ∑ n
i
i=1
γ where γ ∼ = 0.5772 is Euler’s constant. Thus,
Truncated Taylor series
1
≈ ∫ n 1
i x=1 x
n∑
i=1
( n
)
∑
1
dx. In fact, lim − ln(n) =
i→∞ i
i=1
1 ∼ i
= ln(n) + γ for large n.
If all the derivatives of a function f(x) exist, then we can write
f(x) = f(0) + f ′ (0)x + f ′′ (0) x2
2 + · · · .
The series can be truncated. In fact, there exists some y between 0 and x such that
f(x) = f(0) + f ′ (y)x.
Also, there exists some z between 0 and x such that
f(x) = f(0) + f ′ (0)x + f ′′ (z) x2
2
and so on for higher derivatives. This can be used to derive inequalities. For example, if
f(x) = ln(1 + x), then its derivatives are
f ′ (x) = 1
1 + x ; f ′′ 1
(x) = −
(1 + x) ; f ′′′ (x) =
2
2
(1 + x) 3 .
For any z, f ′′ (z) < 0 and thus for any x, f(x) ≤ f(0) + f ′ (0)x hence, ln(1 + x) ≤ x, which
also follows from the inequality 1 + x ≤ e x . Also using
for z > −1, f ′′′ (z) > 0, and so for x > −1,
f(x) = f(0) + f ′ (0)x + f ′′ (0) x2
2 + f ′′′ (z) x3
3!
Exponentials and logs
ln(1 + x) > x − x2
2 .
a log b = b log a
e x = 1 + x + x2
2! + x3
3! + · · · e = 2.7182 1
e = 0.3679
Setting x = 1 in the equation e x = 1 + x + x2
2!
+ x3
3!
+ · · · yields e = ∞ ∑
i=0
1
. i!
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