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# Taylor Expansion Watch

1. Hello,

I am awful at maths and I have to take a compulsory calculus course at uni. Unfortunately while talking about taylor expansion my teacher focused more on quantum physics and particles in a box than maths (I've never studied physics so you can imagine the dumb-founded expression on my face during that class!).

When would you need to use taylor expansion? And more so can someone explain to me how to do taylor series around a non-zero point? I really didn't understand this!

2. (Original post by Scotty Bear)
Hello,

I am awful at maths and I have to take a compulsory calculus course at uni. Unfortunately while talking about taylor expansion my teacher focused more on quantum physics and particles in a box than maths (I've never studied physics so you can imagine the dumb-founded expression on my face during that class!).

When would you need to use taylor expansion? And more so can someone explain to me how to do taylor series around a non-zero point? I really didn't understand this!

The taylor series of an infinitely differentiable function about a non-zero point, a, is given by

3. (Original post by Scotty Bear)
When would you need to use taylor expansion?
Some applications of Taylor Series
4. (Original post by Indeterminate)
The taylor series of an infinitely differentiable function about a non-zero point, a, is given by

So if I had something like cosx (where x was near 3) would I begin like this:

cosx=cos[3+(x-3)]= -3 + 3/2!(x-3)^2 - 3/4!(x-3)^4 ....
5. (Original post by Scotty Bear)
So if I had something like cosx (where x was near 3) would I begin like this:

cosx=cos[3+(x-3)]= -3 + 3/2!(x-3)^2 - 3/4!(x-3)^4 ....
No

Now use what I posted.

Some notation:

means "f differentiated n times", and this is can also be denoted by n dashes
6. (Original post by Indeterminate)
No

Now use what I posted.

Some notation:

means "f differentiated n times", and this is can also be denoted by n dashes
So I have to differentiate cosx first? so then

cosx=cos[3-(x-3)]= cos3cos(x-3)-sin3sin(x-3)= -cos(x-3)

And f '(a) = -cos(x-3) ?

Then the formula filled in would be:

3 - cos(x-3)/2! (x-3)^2 + cos (x-3)/4! (x-3)^4
7. (Original post by Scotty Bear)
So I have to differentiate cosx first? so then

cosx=cos[3-(x-3)]= cos3cos(x-3)-sin3sin(x-3)= -cos(x-3)

And f '(a) = -cos(x-3) ?

Then the formula filled in would be:

3 - cos(x-3)/2! (x-3)^2 + cos (x-3)/4! (x-3)^4
You don't write

Just keep differentiating

and evaluate derivatives at x=3 (keep the exact value).

Now can you do it?
8. (Original post by Scotty Bear)
So I have to differentiate cosx first? so then

cosx=cos[3-(x-3)]= cos3cos(x-3)-sin3sin(x-3)= -cos(x-3)

And f '(a) = -cos(x-3) ?

Then the formula filled in would be:

3 - cos(x-3)/2! (x-3)^2 + cos (x-3)/4! (x-3)^4
for you'd simply input 3 into the equation at the start. From then on you'd differentiate it and put 3 into the differentiate that. You'd then multiply that by and divide by

You differentiate it again put 3 into the second differential and multiply the value by and divide it by
So the is .

Cos x would be a fairly simple one because the constant in front of the term cycles between 4 values.
So the expansion for cos x about 3 would be:
9. Lets hope this is third time lucky lol...

f(3)= cos3
f ' (3)= -sin3
f ''(3)= -cos3

so then the series is

cos3 - sin3/1! (x-3) -sin3/2! (x-3)^2 -cos3/3! (x-3) ^3
10. (Original post by Scotty Bear)
Lets hope this is third time lucky lol...

f(3)= cos3
f ' (3)= -sin3
f ''(3)= -cos3

so then the series is

cos3 - sin3/1! (x-3) -sin3/2! (x-3)^2 -cos3/3! (x-3) ^3
yes, though that's just the first three terms of the expansion. Because the expansion carries on infinitely, unless it only asks for 3 terms, always add at the end.
See if you can give the first 5 terms in the expansion of about 5.
11. (Original post by Rainingshame)
yes, though that's just the first three terms of the expansion. Because the expansion carries on infinitely, unless it only asks for 3 terms, always add at the end.
See if you can give the first 5 terms in the expansion of about 5.
Great thanks

so sinx at 5 would be:

Sin5 + cos5/1! (x-5) - sin5/2! (x-5) ^2 - cos5/3! (x-5)^3 + sin5/4! (x-5)^4 + cos5 (x-5)^5
12. (Original post by Scotty Bear)
Great thanks

so sinx at 5 would be:

Sin5 + cos5/1! (x-5) - sin5/2! (x-5) ^2 - cos5/3! (x-5)^3 + sin5/4! (x-5)^4 + cos5 (x-5)^5
You missed the divided by on the SIXTH term lol. Sorry if that seemed a bit harsh pointing it out but the last thing you want is a mistake like that costing you mark on an exam that matters. But other than that it's correct.
13. Thanks for all the help from all of you! Much appreciated! and yepp definately don't want to lose a mark on an exam for such a silly mistake!

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