Appreciate Some Guidance, Help, Steps on this question

**jono1966** · Aug 20th 2012, 12:12 PM

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

**Soroban** · Aug 20th 2012, 05:15 PM

Hello, jono1966!

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

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

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

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

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

**dean54321** · Mar 25th 2015, 01:15 PM

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

**skeeter** · Mar 25th 2015, 04:57 PM

Originally Posted by dean54321

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

post #2 is a reply ...

**dean54321** · Apr 20th 2015, 11:10 AM

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

**skeeter** · Apr 20th 2015, 02:28 PM

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.

$\sinh(x) = \frac{1}{2}(e^x - e^{-x})$

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

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

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

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

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

see a pattern?

the general Maclaurin series is ...
$f(x) = f(0) + f'(0)x + \frac{f''(0) x^2}{2!} + \frac{f'''(0) x^3}{3!} + ...$

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

now find the composite function ...

$\sinh(2x) = ?$

**dean54321** · Apr 21st 2015, 01:19 AM

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! .........

**skeeter** · Apr 21st 2015, 04:21 AM

No, it is not ... please read the last instruction I gave you regarding finding the composite function.

**dean54321** · Apr 21st 2015, 10:33 AM

Sorry I don't get what the composite function is.

**skeeter** · Apr 21st 2015, 11:17 AM

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

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

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

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

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

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

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

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

$g(x) = 3$ ...

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

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

Thread: Appreciate Some Guidance, Help, Steps on this question

LinkBack

Thread Tools

Display

Appreciate Some Guidance, Help, Steps on this question

Re: Appreciate Some Guidance, Help, Steps on this question

Re: Appreciate Some Guidance, Help, Steps on this question

Re: Appreciate Some Guidance, Help, Steps on this question

Re: Appreciate Some Guidance, Help, Steps on this question

Re: Appreciate Some Guidance, Help, Steps on this question

Re: Appreciate Some Guidance, Help, Steps on this question

Re: Appreciate Some Guidance, Help, Steps on this question

Re: Appreciate Some Guidance, Help, Steps on this question

Re: Appreciate Some Guidance, Help, Steps on this question

Similar Math Help Forum Discussions

question on one of the steps in a weak solution proof

Missing steps in probability question.

Research Sampling Techniques - Question : Need some guidance

Please help me with this multi-product break even question. please show me the steps

[SOLVED] [SOLVED] Need guidance in getting to solve a PERT question

Search Tags