Description

Devansh Jain
IIT Bombay
Questions to be Discussed
Sheet 2
8 (ii), (iii) Finding Functions with given Conditions
10 (i) Sketching curve with given properties
11 Sketching curve with given properties Sheet 3
1 (ii) Taylor Series for arctanx
2 Taylor Series for a polynomial
4 Convergence of Maclaurin Series of ex
5 Integration using Taylor Series
8. (ii). f 00(x) > 0 for all x ∈ R, f 0(0) = 1,f 0(1) = 2 f 00(x) > 0 for all x ∈ R
We know such a curve!!!
Can try quadratic =⇒ Let f (x) = ax2 + bx + c f 0(x) = 2ax + b f 00(x) = 2a
f 00(x) > 0 =⇒ a > 0 f 0(0) = 1 =⇒ b = 1 f 0(1) = 2 =⇒ 2a + b = 2
One such function is: f (x) = x22 + x

8. (iii). f 00(x) ≥ 0 for all x ∈ R, f 0(0) = 1,f (x) < 100 for all x > 0
Idea. Derivative at 0 is positive.
f 00 non negative, means derivative not decreasing, always positive after 0.
f must be strictly increasing with derivative > 1, which means, as x goes to x + 1, f (x) goes to something > f (x) + 1.
(That’s what derivative tells us, right?)
But f (x) is bounded above for positive x. Contradiction!!!
Such a function cannot exist!
Let’s write a formal argument.
8. (iii). Proof.
f 00(x) ≥ 0 ∀x ∈ R =⇒ f 0 not decreasing, so ∀x > 0,f 0(x) ≥ 1 To prove: f must exceed 100 at some point.
Can write, using MVT f (x) − f (0) ≥ f 0(c) ≥ 1 ∵ c ≥ 0 x − 0
Thus, f (x) ≥ x + f (0)
Just take x = 101 − f (0) =⇒ f (x) ≥ 101 .
10. (i). Graphing the polynomial f (x) = 2×3 + 2×2 − 2x − 1.
f 0(x) = 6×2 + 4x − 2 = 2(x + 1)(3x − 1) f 0(x) > 0 in (−∞,−1) ∪ (1/3,∞); so f (x) is strictly increasing here.
and f 0(x) < 0 in (-1, 1/3) so f (x) is strictly decreasing here.
Thus, x = −1 is a local maximum, and x = is a local minimum. f 00(x) = 12x + 4
Thus, f (x) is concave in (−∞,1/3) and convex in (−1/3,∞), with a point of inflection at x .
10. Graph.

11.

11. (contd.)
I have actually graphed a polynomial that satisfies the given properties.
Can you come up with it?
Is there a unique such polynomial? (We discussed this) What’s the minimum degree of such a polynomial?
Is there a unique polynomial with that degree? (This you should think about)
Suppose you have two distinct polynomials f and g that satisfy the given conditions. Can you come up with a distinct third polynomial such that it satisfies the conditions as well?

1. (ii). Taylor Series for arctan x at x=0.

1 X −1)kx2k
g(x) = = (
1 + x2 k=0
(This is only true for |x| < 1, so we restrict ourselves to that domain.) g(2k+1)(0) = 0 g(2k)(0) = (−1)k(2k)! ; k ≥ 0 (Do verify this!) but these are nth derivatives of (arctan(x))0, so are (n + 1)th derivatives of (arctan(x)).
∞ (n)
X f (0) n
Thus the series for arctan x , using the formula P(x) = (x) , is :
n!
n=0
∞ 2n+1 X n x
arctanx = (−1) (Do verify this!)
2n + 1
n=0
2. Taylor Series of f (x) = x3 − 3×2 + 3x − 1 = (x − 1)3 Can already guess!!! f (x) = x3 − 3×2 + 3x − 1 =⇒ f (1) = 0 f 0(x) = 3×2 − 6x + 3 =⇒ f 0(1) = 0 f 00(x) = 6x − 6 =⇒ f 00(1) = 0 f 000(x) = 6 =⇒ f 000(1) = 6 f (n)(x) = 0 for all n > 3.
f
X∞ (n)(x0) − x0)n
The Taylor series is P(x) = (x n!
n=0
Here, only the third derivative is non-zero! Only one term in the Taylor Series! f 000(0) 6
P
∞ k X x
4. Proving convergence of the series .
k!
k=0
We’ll follow steps given in the question, i.e. prove Cauchy.
Let us denote the partial sums of of the given series by sm(x).
We should show that for every > 0, there is a N ∈ N, such that for all m,n > N,
.
It can be shown that,
xn+1 1 xn xn+k 1 xn
for n > N0 = d2xe + 1 > 2x, (n + 1)! < 2 · n! (Iteratively, (n + k)! < 2k · n! )
Observe that (assuming W.L.O.G. m > n > N0),
m xkxn|
k!n!
k=n+1
4. (contd.)
|xn|
We have, |sm(x) − sn(x)| ≤ for all m > n
n!
|x|N0+k 1 |x|N0 Also, by induction, for all k > 0, we have, < · (N0 + k)! 2k N0!
1 |x|N0 For a given x and , for N0 = d2xe + 1 choose a k such that 2 k N0! .
Choose N = N0 + k.
Hence this sequence of partial sums of the series an = sn(x) is cauchy for every x ∈ R and thus the series is convergent.

Sheet 3
Z ex
5. To integrate: dx
x
∞ n
x = X x =⇒
e n!
n=0
Z ex! dx dx dx x
n=0 n=1 n=1
xn X∞ xn
= log(x) +
n! n! n n · n! n=1 n=1 n=1
References