## 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 โ R, f 0(0) = 1,f 0(1) = 2 f 00(x) > 0 for all x โ R

We know such a curve!!!

Can try quadratic =โ Let f (x) = ax2 + bx + c f 0(x) = 2ax + b f 00(x) = 2a

f 00(x) > 0 =โ a > 0 f 0(0) = 1 =โ b = 1 f 0(1) = 2 =โ 2a + b = 2

One such function is: f (x) = x22 + x

8. (iii). f 00(x) โฅ 0 for all x โ 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.

(Thatโs what derivative tells us, right?)

But f (x) is bounded above for positive x. Contradiction!!!

Such a function cannot exist!

Letโs write a formal argument.

8. (iii). Proof.

f 00(x) โฅ 0 โx โ R =โ f 0 not decreasing, so โx > 0,f 0(x) โฅ 1 To prove: f must exceed 100 at some point.

Can write, using MVT f (x) โ f (0) โฅ f 0(c) โฅ 1 โต c โฅ 0 x โ 0

Thus, f (x) โฅ x + f (0)

Just take x = 101 โ f (0) =โ f (x) โฅ 101 .

10. (i). Graphing the polynomial f (x) = 2×3 + 2×2 โ 2x โ 1.

f 0(x) = 6×2 + 4x โ 2 = 2(x + 1)(3x โ 1) f 0(x) > 0 in (โโ,โ1) โช (1/3,โ); 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 = โ1 is a local maximum, and x = is a local minimum. f 00(x) = 12x + 4

Thus, f (x) is concave in (โโ,1/3) and convex in (โ1/3,โ), 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) Whatโs 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 โ1)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) = (โ1)k(2k)! ; k โฅ 0 (Do verify this!) but these are nth derivatives of (arctan(x))0, so are (n + 1)th derivatives of (arctan(x)).

โ (n)

X f (0) n

Thus the series for arctan x , using the formula P(x) = (x) , is :

n!

n=0

โ 2n+1 X n x

arctanx = (โ1) (Do verify this!)

2n + 1

n=0

2. Taylor Series of f (x) = x3 โ 3×2 + 3x โ 1 = (x โ 1)3 Can already guess!!! f (x) = x3 โ 3×2 + 3x โ 1 =โ f (1) = 0 f 0(x) = 3×2 โ 6x + 3 =โ f 0(1) = 0 f 00(x) = 6x โ 6 =โ f 00(1) = 0 f 000(x) = 6 =โ f 000(1) = 6 f (n)(x) = 0 for all n > 3.

f

Xโ (n)(x0) โ 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

โ k X x

4. Proving convergence of the series .

k!

k=0

Weโll 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 โ 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 ยท n! (Iteratively, (n + k)! < 2k ยท 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) โ sn(x)| โค for all m > n

n!

|x|N0+k 1 |x|N0 Also, by induction, for all k > 0, we have, < ยท (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 โ R and thus the series is convergent.

Sheet 3

Z ex

5. To integrate: dx

x

โ n

x = X x =โ

e n!

n=0

Z ex! dx dx dx x

n=0 n=1 n=1

xn Xโ xn

= log(x) +

n! n! n n ยท n! n=1 n=1 n=1

References

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