Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

Maclaurin series question a level further maths?

The first 3 terms of the Maclaurin series for (1+sinx)e^(2x) are identical to the first 3 terms of the binomial series for (1+ax)^n. Find the constants a and n.

I've expanded e^2x to 1+2x+2x^3+...

and (1+an)^n to 1+nax-0.5n^2a^2na^2x^2 but I'm not sure how to use the (1+sinx) bracket.

Any help appreciated

Relevance
• 2 months ago

You don't need any terms beyond x^2 to get the first three terms of that expression.

e^(2x) ≈ 1 + 2x + 4x²/2 = 1 + 2x + 2x²

1 + sin x ≈ 1 + x      [the cubic term isn't needed]

(1 + sin x) e^(2x) ≈ (1 + 2x + 2x²)(1 + x)

≈ (1 + 2x + 2x²) + (x + 2x² + [don't need the cubic])

(1 + sin x) e^(2x) ≈  1 + 3x + 4x²

There's the first expression.  Then second is

(1 + ax)^n ≈  1 + n*ax + (n² - n)/2 * a²x²

Equate coefficients of like powers (obviously 1=1), to get:

na = 3

(n² - n)a² / 2 = 4

Simplify the second one to:

n²a² - na² = 8

(na)² - (na)a = 8

9 - 3a = 8

a = 1/3

n = 3/a = 9

You really ought to plug and check that, since I haven't been in a calculus classroom for nearly half a century.

• Ian H
Lv 7
2 months ago

e^x ~ 1 + x + x^2/2 + ....

e^2x ~ 1 + 2x + 2x^2 + ............................ (1)

1 + sin(x) ~ 1 + x - x^3/6 + ......................(2)

The first 3 terms of the Maclaurin series for [1 + sin(x)]e^(2x) are

1 + 3x + 4x^2 + .... ...............................(3)

Compare the first 3 terms with those in

(1 + ax)^n ~ 1 + nax + [n(n – 1)/2]a^2x^2

n = 3/a, and n – 1 = (3 – a)/a

[n(n – 1)/2]*a^2 = 3(3 – a)/2 = 9/2 – 3a/2 = 4

a = 1/3, n = 9 and to check that,

[1 + (x/3)]^9 = 1 + 3x + (9*8/2)(1/9) x^2 = 1 + 3x + 4x^2

• 2 months ago

sin(x) = x - x^3 / 3! + x^5 / 5! - x^7 / 7! + ...

e^(2x) = 1 + 2x + (2x)^2 / 2! + (2x)^3 / 3! + ....

(1 + sin(x)) * e^(2x) =>

(1 + x - x^3 / 6 + x^5 / 120 - ....) * (1 + 2x + 2x^2 + (4/3) * x^3 + ....)

1 + 2x + 2x^2 + .... + x + 2x^2 + 4x^3 + .... - x^3 / 6 - x^4 / 3 - x^5 / 3 - ....

So we just need up to the quadratic

1 + 2x + x + 2x^2 + 2x^2 =>

1 + 3x + 4x^2

(1 + ax)^n =>

1^n + n * 1^(n - 1) * ax + n * (n - 1) * (1/2) * 1^(n - 2) * (ax)^2 =>

1 + anx + (1/2) * (n^2 - n) * a^2 * x^2

1 = 1

anx = 3x

an = 3

a = 3/n

(1/2) * (n^2 - n) * a^2 * x^2 = 4x^2

(1/2) * (n^2 - n) * a^2 = 4

(n^2 - n) * a^2 = 8

(n^2 - n) * (3/n)^2 = 8

(n^2 - n) * 9 / n^2 = 8

((n - 1) / n) * 9 = 8

9 * (n - 1) = 8n

9n - 9 = 8n

9n - 8n = 9

n = 9

a = 3/n

a = 3/9

a = 1/3