Use Maclaurin’s series:

8.1. to find the first 4 terms of the expansion of Sinh(2x), then

8.2. use this expansion to evaluate Sinh(2x) when x = 0.5, to 4 decimal places

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- Aug 20th 2012, 12:12 PMjono1966Appreciate Some Guidance, Help, Steps on this question
Use Maclaurin’s series:

8.1. to find the first 4 terms of the expansion of Sinh(2x), then

8.2. use this expansion to evaluate Sinh(2x) when x = 0.5, to 4 decimal places - Aug 20th 2012, 05:15 PMSorobanRe: Appreciate Some Guidance, Help, Steps on this question
Hello, jono1966!

Quote:

$\displaystyle \text{8.1 Use Maclaurin’s series to find the first 4 terms of the expansion of }\sinh(2x)$

$\displaystyle \text}8.2. Use this expansion to evaluate }\sinh(2x)\text{ when }x = \frac{1}{2}\text{, to 4 decimal places.}$

The Maclaurin series begins: .$\displaystyle \sinh(2x) \;=\;(2x) + \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} + \frac{(2x)^7}{7!} + \cdots$

If $\displaystyle x = \tfrac{1}{2}$, we have: .$\displaystyle \sinh(1) \;=\;1 + \frac{1^3}{3!} + \frac{1^5}{5!} + \frac{1^7}{7!} \;=\; 1.175198413 $

Therefore: .$\displaystyle \sinh(1) \:\approx\:1.1752$

- Mar 25th 2015, 01:15 PMdean54321Re: Appreciate Some Guidance, Help, Steps on this question
Hi did you get a reply for this question please ?

Use Maclaurin’s series:

8.1. to find the first 4 terms of the expansion of Sinh(2x), then

8.2. use this expansion to evaluate Sinh(2x) when x = 0.5, to 4 decimal places

I have the same question and am struggling thanks for any advice - Mar 25th 2015, 04:57 PMskeeterRe: Appreciate Some Guidance, Help, Steps on this question
- Apr 20th 2015, 11:10 AMdean54321Re: Appreciate Some Guidance, Help, Steps on this question
I need to use the workings below for 8.1 could you give me any advice please. This is so the assessment criteria is covered.

Question 8 – There are standard differential coefficients for both Cosh and Sinh, or you may wish to use their exponential definitions to successively differentiate. It may help to use the exponential definitions when evaluating f(0), f’(0), f’’(0) etc.

I have tried with this so far but am struggling from here thanks.

f"(x)=2^2sinh(2x) so f"(0)=0

f"'(x)=2^3cosh(2x), f"'(0)=2^3

fv(x)=2^4sinh(2x), fv(0)=0

Thanks very much - Apr 20th 2015, 02:28 PMskeeterRe: Appreciate Some Guidance, Help, Steps on this questionQuote:

There are standard differential coefficients for both Cosh and Sinh, or**you may wish to use their exponential definitions to successively differentiate**. It may help to use the exponential definitions when evaluating f(0), f’(0), f’’(0) etc.

$\displaystyle f(x) = \frac{1}{2}(e^x - e^{-x}) \text{ ... } f(0) = 0$

$\displaystyle f'(x) = \frac{1}{2}(e^x + e^{-x}) \text{ ... } f'(0) = 1$

$\displaystyle f''(x) = \frac{1}{2}(e^x - e^{-x}) \text{ ... } f''(0) = 0$

$\displaystyle f'''(x) = \frac{1}{2}(e^x + e^{-x}) \text{ ... } f'''(0) = 1$

$\displaystyle f^{iv}(x) = \frac{1}{2}(e^x - e^{-x}) \text{ ... } f^{iv}(0) = 0$

see a pattern?

the general Maclaurin series is ...

$\displaystyle f(x) = f(0) + f'(0)x + \frac{f''(0) x^2}{2!} + \frac{f'''(0) x^3}{3!} + ...$

$\displaystyle \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + ...$

now find the composite function ...

$\displaystyle \sinh(2x) = ?$ - Apr 21st 2015, 01:19 AMdean54321Re: Appreciate Some Guidance, Help, Steps on this question
Hi

yes i can see a pattern is the following correct

Sinh (2x) = 2 x 1/2 (ex - e-x)

f (x) = 2 x 1/2 (ex - e-x)..... f (0) = 0

fi (x) = 2 x 1/2 (ex + e-x)..... f (0) = 2

fii (x) = 2 x 1/2 (ex - e-x)..... f (0) = 0

fiii (x) = 2 x 1/2 (ex + e-x)..... f (0) = 2

fiv (x) = 2 x 1/2 (ex - e-x)....... f (0) = 0

putting these values into maclaurins series

Sinh (2x) = 2x + 2x^3/3! + 2 x^5/5! + 2x^7/7! + 2x^9/9! ......... - Apr 21st 2015, 04:21 AMskeeterRe: Appreciate Some Guidance, Help, Steps on this question
No, it is not ... please read the last instruction I gave you regarding finding the

**composite**function. - Apr 21st 2015, 10:33 AMdean54321Re: Appreciate Some Guidance, Help, Steps on this question
Sorry I don't get what the composite function is.

- Apr 21st 2015, 11:17 AMskeeterRe: Appreciate Some Guidance, Help, Steps on this question
$\displaystyle f(x) = \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + ...$

the following are various examples of composite functions of $\displaystyle f[g(x)] = \sinh[g(x)]$ ...

$\displaystyle g(x) = 2x$ ...

$\displaystyle f(2x) = \sinh(2x) = 2x + \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} + \frac{(2x)^7}{7!} + ...$

$\displaystyle g(x) = x+1$ ...

$\displaystyle f(x+1) = \sinh(x+1) = (x+1) + \frac{(x+1)^3}{3!} + \frac{(x+1)^5}{5!} + \frac{(x+1)^7}{7!} + ...$

$\displaystyle g(x) = x^2$ ...

$\displaystyle f(x^2) = \sinh(x^2) = x^2 + \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!} + \frac{(x^2)^7}{7!} + ...$

$\displaystyle g(x) = 3$ ...

$\displaystyle f(3) = \sinh(3) = 3 + \frac{(3)^3}{3!} + \frac{(3)^5}{5!} + \frac{(3)^7}{7!} + ...$

... if you still don't understand, then I recommend you research composite functions.