Show that the MacLaurin series for e^x is..
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Show that the MacLaurin series for e^x is..

[From: ] [author: ] [Date: 11-08-05] [Hit: ]
com/watch?f(x)= f(0)+f (0) x/1!+ f (0)x^2/2! + f (0) x^3/3! + f (0) x^4 /4! + .......
Show that the MacLaurin series for e^x is;

1 + x + (x^2)/2! + (x^3)/3! ....

Hence evaluate;

Integral from 0 to 0.2 of e^ (-x^2)

Correct to 5 decimal places.

Thank you for your help in advace.

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If you cannot do this, the answer alone is not going to help much. Look at this great video and you'll be able to do it yourself in 10 minutes.
http://www.youtube.com/watch?v=cjPoEZ0I5…

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The infinite series expansion for f(x) about x = 0 becomes:MacLaurin series ie

f(x) = f(0)+f ' (0) x/1! + f ''(0) x^2/2! + f '''(0) x^3/3! + f ''''(0) x^4 /4! + ......

Recall that the derivative of f(x) = e^x is f '(x) = ex. In fact, all the derivatives are e^x.

f '(0) = e0 = 1

f ''(0) = e0 = 1

f '''(0) = e0 = 1

We see that all the derivatives, when evaluated at x = 0, give us the value 1.

Also, f(0) = 1, so we have:

The Maclaurin Series expansion will be simply:

e^x = 1+ x+ x^2/2! +x^3/3! +x^4/4! + ...... so on

put x = -t^2 and integrate with respect to t

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I hate MacLaurin series lol. Here's how to do the question:

The MacLaurin series is just the Taylor series but when a = 0. As you know, the Taylor Series is:

f(x) = f(a) + f' '(a)(x - a) + [f ''(a)(x-a)²]/2! + [f '''(a)(x - a)³]/3! ....


In this question, as it's MacLaurin series, let a = 0 and f(x) = e^x

f(x) = e^x ....... at a, f(a) = 1
f '(x) = e^x ..... at a, f '(a) = (1)(x - 0) = x
f ''(x) = e^x ..... at a, f ''(a) = [(1)(x - 0)²]/2! = x²/2!
f '''(x) = e^x ..... at a, f '''(a) = [(1)(x - 0)³]/3! = x³/3!
.
.
.

This carries on and that's the MacLaurin series for e^x. If they'd asked for the Taylor Series at a = y, you simply put in your value 'y' into the series formula.


For the second part, ,as I can't be bothered to work everything out, what you do is this:

Let -x² = y. So now you e^y. Use the MacLaurin series and you get the MacLaurin series for e^y. I would say use it upto y³. You therefore have:

e^y = 1 + y + y²/2! + y³/3! ....... No working out necessary as you have already done it for e^x and just swap the letters.

Now integrate 1 + y + y²/2! + y³/3! between 2 limits. My mind has gone blank and I can't think of what you need to change the limits to, so when you have integrated this function, I would say to swap the 'y' back to '-x²' and then put in your original limits.
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