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# Tricky Taylor Series watch

1. The question is: Find the first four non-vanishing terms of:

iii.

So I've tried two methods, one method I substituted but that got complicated quickly and I wasn't sure what to do with it once I subbed back in for next I tried the old fashion differentiate it each time method and that didn't work because I kept ending up with 3 products to differentiate (and we haven't learnt a product rule for 3 things multiplied together).

Attached is the worked solution which I don't even understand, come someone please help me with this question as I am well and truly stuck!

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2. (Original post by physicsboy1)
The question is: Find the first four non-vanishing terms of:

iii.

So I've tried two methods, one method I substituted but that got complicated quickly and I wasn't sure what to do with it once I subbed back in for next I tried the old fashion differentiate it each time method and that didn't work because I kept ending up with 3 products to differentiate (and we haven't learnt a product rule for 3 things multiplied together).

Attached is the worked solution which I don't even understand, come someone please help me with this question as I am well and truly stuck!

write the exponential series at the bottom
cancel the 1
then cancel an x all they through
then ...
3. (Original post by TeeEm)
write the exponential series at the bottom
cancel the 1
then cancel an x all they through
then ...
It's still , so we need some way to convert the to the numerator.

OP, the product rule for 3 is , you can prove this by doing .

Now for the answer you posted, the writer just substituted in , because the taylor series of , so at the boundary, it is bound to stop (you won't need to consider any higher powers).

I tried integrating , and then get the taylor series of the integral and then differentiate, but the integral is not nice at all, so that fails. I think the writer of the solution just wanted to force a substitution. So you can attempt various substitution via trial and error, for which in terms of the substitution has a nice taylor series, and you can sub your values back in and do that hog-mess of algebra.
4. (Original post by 0x2a)
It's still , so we need some way to convert the to the numerator.

OP, the product rule for 3 is , you can prove this by doing .

Now for the answer you posted, the writer just substituted in , because the taylor series of , so at the boundary, it is bound to stop (you won't need to consider any higher powers).

I tried integrating , and then get the taylor series of the integral and then differentiate, but the integral is not nice at all, so that fails. I think the writer of the solution just wanted to force a substitution. So you can attempt various substitution via trial and error, for which in terms of the substitution has a nice taylor series, and you can sub your values back in and do that hog-mess of algebra.
do you know the binomial expansion?
5. (Original post by TeeEm)
do you know the binomial expansion?
Ohhhh of course
6. (Original post by 0x2a)
Ohhhh of course
7. in fact I am saying silly things too

factorize minus out
-x/(1- ex) then binomial

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