How do you find the Maclaurin series for f(x) using the definition of a Maclaurin series, of 4 sinh(4x)?

Answer 1

#4sinh(4x) = sum_(n=0)^oo 2^(4n+4)(x^(2n+1))/((2n+1)!)#

Take the MacLaurin series of #sinh t#:
#sinht = sum_(n=0)^oo (t^(2n+1))/((2n+1)!)#
substitute #t=4x# and multiply by #4#:
#4sinh(4x) = 4sum_(n=0)^oo ((4x)^(2n+1))/((2n+1)!)=sum_(n=0)^oo 2^2*2^(2(2n+1))(x^(2n+1))/((2n+1)!)=sum_(n=0)^oo 2^(4n+4)(x^(2n+1))/((2n+1)!)#
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Answer 2

# 4sinh(4x) = 16x + 128/3 x^3 + 512/15 x^5 + ... + (4^(2n+2))/((2n+1))x^(2n+1) + ...#

We derive a Maclaurin series from the infinite series

# f(x) = f(0) + f'(0)x + (f''(0))/(2!)x^2 + (f^((3))(0))/(3!)x^3 + ... + (f^((n))(0))/(n!)x^n + ... #

We have:

# \ \ \ \ \ f(x) = 4sinh(4x) # # :. f(0) = 4sinh(0) = 0 #

Differentiate wrt to get the first derivative:

# \ \ \ \ \ f'(x) = 4cosh(4x) *4 # # :. f'(x) = 16cosh(4x) # # :. f'(0) = 16cosh(0) = 16 =4^2#

Differentiate again wrt to get the second derivative:

# \ \ \ \ \ f''(x) = 16sinh(4x)*4 # # :. f''(x) = 64sinh(4x) # # :. f''(0) = 0 #

Differentiate again wrt to get the third derivative:

# \ \ \ \ \ f^((3))(x) = 64cosh(4x)*4 # # :. f^((3))(x) = 256cosh(4x) # # :. f^((3))(0) = 256 =4^4#

Differentiate again wrt to get the fourth derivative:

# \ \ \ \ \ f^((4))(x) = 256sinh(4x)*4 # # :. f^((4))(x) = 1024sinh(4x) # # :. f^((4))(0) = 1024sinh(0) = 0 #

Differentiate again wrt to get the fifth derivative:

# \ \ \ \ \ f^((5))(x) = 1024cosh(4x)*4 # # :. f^((5))(x) = 4096cosh(4x) # # :. f^((5))(0) = 4096 = 4^6 #
# vdots #

And we can see a clear pattern forming, where

# f^((n))(0) = { (0, n " even"), (4*4^n, n " odd") :} #

So we can now form the Maclaurin series:

# f(x) = 0 + 16x + 0x^2 + 256/6 x^3 + 0x^4 +4096/120 x^5 + ... # # \ \ \ \ \ \ \ = 16x + 128/3 x^3 + 512/15 x^5 + ... + (4*4^(2n+1))/((2n+1))x^(2n+1) + ...# # \ \ \ \ \ \ \ = 16x + 128/3 x^3 + 512/15 x^5 + ... + (4^(2n+2))/((2n+1))x^(2n+1) + ...#
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Answer 3

To find the Maclaurin series for ( f(x) = 4\sinh(4x) ), we first need to determine the Maclaurin series expansion for ( \sinh(x) ). The Maclaurin series for ( \sinh(x) ) is:

[ \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdots ]

Since we have ( 4\sinh(4x) ), we substitute ( 4x ) for ( x ) in the series expansion of ( \sinh(x) ):

[ 4\sinh(4x) = 4(4x) + \frac{(4x)^3}{3!} + \frac{(4x)^5}{5!} + \frac{(4x)^7}{7!} + \cdots ]

Simplify this expression to obtain the Maclaurin series for ( f(x) = 4\sinh(4x) ):

[ f(x) = 16x + \frac{64x^3}{3!} + \frac{256x^5}{5!} + \frac{1024x^7}{7!} + \cdots ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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