Taylor series finding sin(x^2)+cos(x) from sin(x^2) and cos(x) alone

In summary, the conversation discusses finding the Taylor series at x = 0 for sin(x^2) and cos(x), and how to combine these two series to get the series for sin(x^2) + cos(x). The summary also mentions the importance of adding the fractions with a common denominator and the process of adding polynomials when combining the series.
  • #1
IntegrateMe
217
1
If I want to find the taylor series at x = 0 for sin(x^2)+cos(x)...

sin(x^2) = x^2 - x^6/3! + x^10/5! - x^14/7! ...
cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! ...

So why does sin(x^2) + cos(x) = 1 + x^2/2! + x^4/4! + 121x^6/6! ...?

Thanks!
 
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  • #2
I get -121x^6/6!, other then that just give the fractions a common denominator and add.
 
  • #3
yes, sorry, it is a minus sign...can you explain that more elaborately? I think I'm starting to understand, but I'm not quite there yet.
 
  • #4
Well the 1 + x^2/2! + x^4/4! is hopefully obvious. As for the -121x^6/6!, you need to add those terms in your two taylor series which have an x^6 in them, so that is - x^6/3! and - x^6/6!. But before you can add these fractions you need to give them a common denominator. Remember 6! = 6*5*4*3*2*1 and 3! = 3*2*1.
 
  • #5
Well the 1 + x^2/2! + x^4/4! is hopefully obvious.

Nope :\
 
  • #6
Ok forgetting these series are infinite for a minute just consider the numbers you wrote without the ... . Then you are basically adding polynomials correct? So what do you do when you add polynomials well you add the x^2 terms together and you add the x terms together etc. Maybe you could be more specific about what you aren't understanding since I'm not sure how weak/strong your foundation is.
 

FAQ: Taylor series finding sin(x^2)+cos(x) from sin(x^2) and cos(x) alone

How do I find the Taylor series for sin(x^2)+cos(x) using only sin(x^2) and cos(x)?

To find the Taylor series for a function, we use the formula: f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ... where a is the center of the series. In this case, we can use the identities sin(x^2) = sin(a^2) + 2xcos(a^2) and cos(x) = cos(a) - xsin(a) to substitute for sin(x^2) and cos(x), respectively. Then, we can expand each term using the above formula to get the Taylor series for sin(x^2)+cos(x).

What is the center of the Taylor series for sin(x^2)+cos(x)?

The center of the series is the value of x that we are expanding around. In this case, since we are using the identities sin(x^2) and cos(x), the center would be x = 0, since these identities are centered at 0. Therefore, the Taylor series for sin(x^2)+cos(x) using only sin(x^2) and cos(x) would be centered at x = 0.

Can I use other trigonometric identities to find the Taylor series for sin(x^2)+cos(x)?

Yes, there are many other trigonometric identities that can be used to manipulate sin(x^2) and cos(x) to find the Taylor series for sin(x^2)+cos(x). Some common ones include the double angle identities, sum and difference identities, and half angle identities.

How many terms do I need to include in the Taylor series for accurate approximation?

The number of terms needed to accurately approximate a function using its Taylor series depends on the function and the desired level of accuracy. Generally, the more terms included, the more accurate the approximation will be. However, for simpler functions like sin(x^2)+cos(x), a few terms may be enough to get a good approximation.

Can I use the Taylor series for sin(x^2)+cos(x) to evaluate the function at any value of x?

Yes, that is the purpose of the Taylor series - to approximate a function at any given x value. However, the accuracy of the approximation may vary depending on the number of terms included in the series.

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