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
Hello, jono1966!
$\displaystyle \text{8.1 Use Maclaurin’s series to find the first 4 terms of the expansion of }\sinh(2x)$
$\displaystyle \text}8.2. Use this expansion to evaluate }\sinh(2x)\text{ when }x = \frac{1}{2}\text{, to 4 decimal places.}$
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: .$\displaystyle \sinh(1) \:\approx\:1.1752$
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
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
$\displaystyle \sinh(x) = \frac{1}{2}(e^x - e^{-x})$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 ...
$\displaystyle \sinh(2x) = ?$
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! .........
$\displaystyle f(x) = \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + ...$
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.