Using Taylor series colnt Find the Taylor series for f (x) cs centered at a T You can simply write out the first 3 4 terms of the series Use your Tavlor series from...

Use the definition of Taylor series to find the Taylor series, centered at $c$ for the function. $$f(x)=\sec x, \quad c=0 \text { (first three nonzero terms) }$$

Alright for this problem we are asked to use known taylor series to find the first four non zero terms of the taylor series. About zero for the function. Eat the power of tee times. Coast of T. So to begin we should know the taylor series for both. Eat the power of tea and co safety. The taylor series for each party is going to be one plus T plus t squared over two plus t cubed over six plus T. To the power of 4/4 factorial plus dot dot dot. We're multiplying that by the series of cossacks which is going to be one or sorry coast T which is going to be one minus t squared over two factorial plus T to the power of 4/4 factorial minus teeth power of 6/6 factorial was dot dot dot. So so again we would have one plus T plus t squared over two plus t cubed over six plus dot dot dot from doing the one or sorry the won the first set of brackets times the one there. Then we would have minus 1/2 factorial times. Um Or sorry it would be minus t squared over two factorial times one plus t plus t squared over two plus t cubed over six plus dot dot dot. And then in addition we have t to the power 4/4 factorial times one plus t plus t squared over two plus t cubed over three and so on. Sorry t cubed over six and so on. Yeah so the reason why we're doing this we should be able to see what the highest order that will be non zero here. Or sorry the not highest order that will be non zero. But what the what the first four orders will be. So if we look well have we should definitely have a one plus a t plus t squared. It's a T cubed. Except for the fact that we do have this minus T squared over two. So the t squared over two will end up cancelling out. Then we'll have to go up to t to the power of 4/4 factorial. We're all here so we we look we will only have one term at the order of a constant or T. To 50 which is going to be one. We have one term at the order of T. To the power of one which is just plus T. Then we have no terms at the order of T squared. Then if we look at the order of T cubed we should have well we'd have won over six from this guy and then minus T cubed over two. So 1/6 minus one half. Mhm. And then at the order of teach the power of four we look and have plus to the power for over four factorial. So we can write teaches power four times 1/4 factorial. Then we'd have minus. So there is this t squared over two times t squared over two, so we'd have minus 1/4. Then we'd have teach the power of 4/4 factorial times one, so plus 1/4 factorial and I'm going to pause and simplify this off screen. This should come out to one plus t minus one third T cubed minus 1/6 T to the power of four plus dot dot dot.

Alright for this problem we are asked to find the first four non zero terms of the taylor series. About zero for the function T sign of three T. Using known taylor series. So to begin we know that the taylor series of sine of X is equal to x minus x cubed over three factorial plus X. The power of five. 5/5 factorial minus X. The power of seven over seven factorial. So that means that sign of three T would be three t minus three T cubed over three factorial plus three T. Power of 5/5 factorial minus minus three T. To the power of seven over seven factorial. Which means then that t sign of three T. It's going to be three t squared minus t times three T cube plus t times three teeth. Our five plus or a minus t times three T. To the power of seven should include the plus dot dot dot here. So you can simplify that down a little bit. I'll do that off screen. All right. It's not necessarily simpler, but expanding everything out that should come to three T squared minus nine t. To the power of 4/2 plus 81 t. To the power of 6/40 minus 243 t. To the power of eight over 560. It's not about that might be worthwhile. Keeping it in the form of above here, depends on your professor teachers expectations.

For this problem we are asked to find the first non first four non zero terms of the taylor series. About zero for the function E to the power of negative X. Using known taylor series. So to begin, we should know that the taylor series for the function E to the power of X. Or the first four terms at least B one plus X plus X squared over two factorial. Which is just to plus X cubed over three factorial. Now all we need to do to find the taylor series for E to the negative acts is wherever we had the X. In that original taylor series, we replace it with a negative act. So it becomes one minus X plus and now it's going to be plus, it would be negative X squared over two factorial but negative X squared is just positive X squared. So we have one minus X plus X squared over two factorial and it would be minus X cubed over three factorial. Because negative x cubed gives us minus x cubed and that is the first four terms. And then it would be plus with a dot dot dot

Alright for this problem we are asked to find the first four non zero terms of the taylor series. About zero for the functions E over E to the power of zed should be eat powers that squared. Using known taylor series Over off to the side here, I have some of the known taylor series. The obviously relevant one to us here is E to the power of X. Excuse me. Where we can note that we can rewrite zet overeat carbs each square. Excuse me, I'm going to pause. All right. We can write zet overeat powers that squared as Zed times E to the power of negative Z squared. Which we can then right as Ed times. Whatever the taylor series expansion of eat the power of negatives Ed squared is going to be so we can see all we need to do is put in negative Z squared. Where we wherever we have an X in R E. To the power of X expansion there. So we have one minus zero squared plus negative zed squared squared over two factorial plus negative zed squared cubed over three factorial. So that will become z times one minus zero squared plus Z. To the power of 4/2 minus z to the power of six over. That would be six. I should include a plus dot dot dot because we are leaving out the extra terms there. So we should get zen minus z cubed plus zed to the power of 5/2 minus Z to the power of 7/6 plus dot dot dot.