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.
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…
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
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!
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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.
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.