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junior college 1 | H2 Maths
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Hi, I need help with these questions please
Date Posted: 3 years ago
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i) From MF26, we know that the Maclaurin expansion of the following are :
e^x = 1 + x + x²/2! + x³/3! + ...
ln (1 + x) = x - x²/2 + x³/3 - ...
We only want up to x³.
e^(2x) ln (1 + ax)
= (1 + 2x + (2x)²/2! + ...)(ax - (ax)²/2 + (ax)³/3 - ...)
= (1 + 2x + 2x² + ...)(ax - a²x²/2 + a³x³/3 - ...)
= ax - a²x²/2 + a³x³/3 + 2ax² - a²x³ + 2ax³ + ...
= ax + (2a - a²/2)x² + (a³/3 - a² + 2a)x³ + ...
= ax + a(2 - a/2)x² + a(a²/3 - a + 2)x³ + ...
For there to be no term in x², the coefficient of the x² term must be 0.
So, a(2 - a/2) = 0
a = 0 (N.A since a is non-zero) or 2 - a/2 = 0
2 = a/2
a = 4
part ii) in the next post.
e^x = 1 + x + x²/2! + x³/3! + ...
ln (1 + x) = x - x²/2 + x³/3 - ...
We only want up to x³.
e^(2x) ln (1 + ax)
= (1 + 2x + (2x)²/2! + ...)(ax - (ax)²/2 + (ax)³/3 - ...)
= (1 + 2x + 2x² + ...)(ax - a²x²/2 + a³x³/3 - ...)
= ax - a²x²/2 + a³x³/3 + 2ax² - a²x³ + 2ax³ + ...
= ax + (2a - a²/2)x² + (a³/3 - a² + 2a)x³ + ...
= ax + a(2 - a/2)x² + a(a²/3 - a + 2)x³ + ...
For there to be no term in x², the coefficient of the x² term must be 0.
So, a(2 - a/2) = 0
a = 0 (N.A since a is non-zero) or 2 - a/2 = 0
2 = a/2
a = 4
part ii) in the next post.
ii)
From MF26, we can obtain the expansion of cosx to be :
1 - x²/2! + x⁴/4! - x⁶/6! + ...
So ln (cos2x) = ln (1 - (2x)²/2! + (2x)⁴/4! - (2x)⁶/6! +...)
= ln (1 + (-2x² + ⅔x⁴ - 4/45 x⁶ + ...) )
Combine this with the expansion of ln(1 + x) by replacing x with (-2x² + ⅔x⁴ - 4/45 x⁶ + ...)
Then expand up to term in x⁶
So ln (1 + (-2x² + ⅔x⁴ - 4/45 x⁶ + ...) )
= (-2x² + ⅔x⁴ - 4/45 x⁶ + ...) - (-2x² + ⅔x⁴ - 4/45 x⁶ + ...)²/2 + (-2x² + ⅔x⁴ - 4/45 x⁶ + ...)³/3
= -2x² + ⅔x⁴ - 4/45 x⁶ + ... - ½(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)(-2x² + ⅔x⁴ - 4/45 x⁶ + ...) + ⅓(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)
= -2x² + ⅔x⁴ - 4/45 x⁶ + ... - ½( -2x²(-2x²) - 2x²(⅔x⁴) + ⅔x⁴(-2x²) + ...) + ⅓((-2x²)(-2x²)(-2x²) + ...)
= -2x² + ⅔x⁴ - 4/45 x⁶ - 2x⁴ + 4/3 x⁶ - 8/3 x⁶ + ...
= -2x² + ⅔x⁴ - 2x⁴ - 4/45 x⁶ + 4/3 x⁶ - 8/3 x⁶ + ...
= -2x² - 4/3 x⁴ - 64/45 x⁶ + ...
Now, ln x is undefined for x ≤ 0. Likewise, ln(cos2x) is undefined when cos2x ≤ 0
0 ≤ x ≤ ¼π → 0 ≤ 2x ≤ ½π
For this domain, cos2x is always positive, EXCEPT :
When x = ¼π, 2x = ½π
cos2x = cos(½π) = 0
So ln(cos2x) is undefined when x = ¼π, which in turn means that the expansion is not valid.
From MF26, we can obtain the expansion of cosx to be :
1 - x²/2! + x⁴/4! - x⁶/6! + ...
So ln (cos2x) = ln (1 - (2x)²/2! + (2x)⁴/4! - (2x)⁶/6! +...)
= ln (1 + (-2x² + ⅔x⁴ - 4/45 x⁶ + ...) )
Combine this with the expansion of ln(1 + x) by replacing x with (-2x² + ⅔x⁴ - 4/45 x⁶ + ...)
Then expand up to term in x⁶
So ln (1 + (-2x² + ⅔x⁴ - 4/45 x⁶ + ...) )
= (-2x² + ⅔x⁴ - 4/45 x⁶ + ...) - (-2x² + ⅔x⁴ - 4/45 x⁶ + ...)²/2 + (-2x² + ⅔x⁴ - 4/45 x⁶ + ...)³/3
= -2x² + ⅔x⁴ - 4/45 x⁶ + ... - ½(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)(-2x² + ⅔x⁴ - 4/45 x⁶ + ...) + ⅓(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)
= -2x² + ⅔x⁴ - 4/45 x⁶ + ... - ½( -2x²(-2x²) - 2x²(⅔x⁴) + ⅔x⁴(-2x²) + ...) + ⅓((-2x²)(-2x²)(-2x²) + ...)
= -2x² + ⅔x⁴ - 4/45 x⁶ - 2x⁴ + 4/3 x⁶ - 8/3 x⁶ + ...
= -2x² + ⅔x⁴ - 2x⁴ - 4/45 x⁶ + 4/3 x⁶ - 8/3 x⁶ + ...
= -2x² - 4/3 x⁴ - 64/45 x⁶ + ...
Now, ln x is undefined for x ≤ 0. Likewise, ln(cos2x) is undefined when cos2x ≤ 0
0 ≤ x ≤ ¼π → 0 ≤ 2x ≤ ½π
For this domain, cos2x is always positive, EXCEPT :
When x = ¼π, 2x = ½π
cos2x = cos(½π) = 0
So ln(cos2x) is undefined when x = ¼π, which in turn means that the expansion is not valid.
thank you!
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