Question

to my greatest discomfort I have to solve this exersise right here: http://prntscr.com/bcd827 (exercise 9) ... Can someone please lend a hand ?

Answer 1

cantfind the pattern..

Answer 2

You mean solve 5x=25?

Answer 3

no

Answer 4

wait

Answer 5

http://prntscr.com/bcd827

Answer 6

lol you have been posting calc questions and this is what shows up

Answer 7

because it was a random screenshot, my mistake :D

Answer 8

I uploaded the correct one

Answer 9

It's all good! :) But oh my gosh I did that last year and it is a pain. @agent0smith @Mehek14 @samanthagreer @sammixboo @sleepyjess @uri

some of these good people can help you better than I can.

Answer 10

http://prntscr.com/bcdn6x here is the taylor expansion , basically I want to know the nth partial sum of this thing , or the general term of the corresponding sequence ...

Answer 11

So there are ten questions. Which one can you not do?

Answer 12

I can't find the general term for exersise 9

Answer 13

for the mac laurin series that is

Answer 14

wolfram alpha prints a result, but if I plug in 0 I dont get 0

Answer 15

and if I plug in 1 i dont get 0

Answer 16

Are you applying the this formula for the MacLaurin series?

Answer 17

yes

Answer 18

http://mathworld.wolfram.com/MaclaurinSeries.html

Answer 19

I know the formulas , just something doesnt make sense

Answer 20

the taylor expansion for xsinx starts with 0 ,0 , x^2 .....

Answer 21

Can you show your work to see how far you have gone (before hitting trouble)?

Answer 22

so I have this http://prntscr.com/bcdn6x

Answer 23

I did it by hand

Answer 24

by hand is ok.

Answer 25

did you get the one up to O7 with wolfram?

Answer 26

yes thats correct

Answer 27

Did you use the formula at the Wolfram link? This is what you're supposed to use.

Answer 28

http://prntscr.com/bcdwip

Answer 29

my send result should be something like the above

Answer 30

what values did you get for
\[f'(0)=? \\ f''(0)=? \\ f^{(3)}(0)=? \\ f^{(4)}(0)=? \\ \\ f^{(5)}(0)=? \\ f^{(6)}(0)=? \\ f^{(7)}(0)=?\\ f^{(8)}(0)=?\]
you should see a pattern eventually

Answer 31

0 0 2 0 -4

Answer 32

can you make a guess for the next successive derivatives at 0

Answer 33

You would need to use the formula:
f(x)=f'(0)+f'(0)+f""(0)/2!+...
and use the derivatives as shown by @freckles

Answer 34

i can but it doesnt matter

Answer 35

since I cant see a pattern on the first 2 elements

Answer 36

ok then if you can't see the pattern yet then find the next few derivatives and evaluate it at x=0

Answer 37

The pattern comes from the list of derivatives!

Answer 38

how can you model a sequence that starts with 0 and has another consecutive 0

Answer 39

the pattern should be more obvious to you at the sixth derivative

Answer 40

I'll do the first four derivatives for you:
\(sin(x)+x*cos(x)\)
\(2*cos(x)-x*sin(x)\)
\(-3*sin(x)-x*cos(x)\)
\(x*sin(x)-4*cos(x)\)

Answer 41

0 , 0 , 1 , 2 , 3 whats the general term for this

Answer 42

mathmale I did all that

Answer 43

I am just stuck at finding the infinite series

Answer 44

You have to look at the odd and even derivatives separately.

Answer 45

hint: @Christos ignore the 0's
find a pattern with teh non-zeros

Answer 46

And substitute cos(0)=1, and sin(0)=0!

Answer 47

*sin(0)=0 !

Answer 48

freckles if I find a pattern without the zeroes , then my formula is gonna ignore the first 2 elements

Answer 49

Pretend I give you 4 terms of a sequence
\[a_1=2 \\ a_2=-4 \\ a_3=6 \\ a_4=-8\]

can you give me the nth term of the sequence based on the behavior of the first 4 terms ?

Answer 50

meaning that when I set n = 0 on my formula I will get the 3rd element

Answer 51

(-1)^(n-1) * 2 * (n-1)2

Answer 52

from n=1 to inf

Answer 53

here are even derivatives, 2,4,6,8,10

\(2\cos(x)-x\sin(x)\)
\(x\sin(x)-4\cos(x)\)
\(6\cos(x)-x\sin(x)\)
\(x\sin(x)-8\cos(x)\)
\(10\cos(x)-x\sin(x)\)

and bear in mind that sin(0)=0
You should find the pattern to apply to the equation.
You cannot easily see the pattern of the coefficients.

Answer 54

I'll let @freckles to continue before I confuse you.

Answer 55

"(-1)^(n-1) * 2 * (n-1)2
"
where does the end 2 come from?

Answer 56

end2 = step

Answer 57

of the numeric sequence

Answer 58

middle2 = starting point

Answer 59

typo

Answer 60

(-1)^(n-1) * [2 + (n-1) * 2]

Answer 61

I think the nth term of the sequence 2,-4,6,-8,...
notice you have even numbers so think 2k from k=1 to k=n
the terms alternate (-1)^(k-1)
so put together you could write \[2k \cdot (-1)^{k-1}\]



and you already know the powers of the x that correspond to these ... they are 2,4,6,8,...

and now you have to look at that denominator (the factorial part) that corresponds to each

Answer 62

and 2+(n-1)*2
works to
that simplifies to 2n

Answer 63

so what is the end series

Answer 64

just put it all together

\[\sum_{n=1}^{\infty} \frac{2n (-1)^{n-1} x^{2n}}{(2n)!}\]

Answer 65

now lets evaluate this

Answer 66

from n = 1 to k = 3

Answer 67

I should get 0 + 0 + x^2

Answer 68

no?

Answer 69

no you don't care about the 0's

Answer 70

you only care about the terms that actually mean something
0 doesn't contribute anything

Answer 71

it contributes the zero value

Answer 72

if you do from n=1 to n=3 you should get x^2-x^4/6+x^6/120
hopefully

Answer 73

so if I evaluate n=1 to k=2

Answer 74

shouldn't that implies that I take the 2 first elements and add them together, whatever those may be

Answer 75

i'm not sure what you mean

Answer 76

i think I am a little stupid

Answer 77

I dont get this

Answer 78

:D

Answer 79

throw the 0's out
they contribute nothing

Answer 80

find the pattern between the things that aren't 0

Answer 81

@mathmate I actually have to go
if you are still here maybe you can explain more

I think also @blurbendy is pretty smart at math I think

Answer 82

so whenever I have 0's (even one) at the beginning I always toss it away

Answer 83

yes you can

Answer 84

ok thanks for your help

Answer 85

good luck

Answer 86

if you are asking how to write the "pattern"
\[ x^2 - \frac{x^4}{6 }+\frac{x^6}{120} - ...\]

First, we don't try to put in terms such as x^1 or x^3 (with coeffs of 0). Rather, we use various "techniques"

I would change the denominators into factorials:
\[ x^2 - \frac{x^4}{3! }+\frac{x^6}{5!

} - ...\]

Answer 87

next, we notice that the exponents are the even numbers starting with 2
2,4,6,...

the "technique" or trick is to use 2n as an exponent
if we start with n=1, this will generate 2,4,6, which is what we want.

Answer 88

if we write the pattern as
\[ \frac{x^2}{1! } - \frac{x^4}{3! }+\frac{x^6}{5!} - \frac{x^8}{7! }+...\]
we see the factorial is one less than the exponent, so 2n-1

the last thing is to get the correct sign.
we see the first term is plus, and they alternate:
n 1 +
n 2 -
n 3 +

if we use (-1)^(n-1) (or (n+1) for the exponent), we will get the correct sign.

putting it together, each term has the form
\[ (-1)^{(n-1)} \frac{x^{2n}}{(2n-1)!} \]
starting with n=1
thus
\[ \sum_{n=1}^\infty (-1)^{(n-1)} \frac{x^{2n}}{(2n-1)!} \]
as a check, the fourth term (i.e. n=4)
we get
\[ - \frac{x^8}{7!}\]
which is correct.

Answer 89

if you evaluate that series at x=0
you will get
0 - 0 + 0-0 +... = 0
at x= pi/6, we expect to get pi/6 * sin(pi/6) = pi/12 =0.261799387799149
using the first four terms, we get
2.7416e-01 - 1.2527e-02 + 1.7172e-04 - 1.1209e-06= 0.261799383541780

Answer 90

I'm tempted to say that this could be a lot easier depending on how you're expected to find these.

For instance, if you know the Maclauren series for \(e^x\), \(\sin x\), and a few others you are really plugging in a few things to get the answers.

Like if we look at 6, we know the geometric series is

\[\frac{1}{1-y} = \sum_{n=0}^\infty y^n\]
so if you plug in y=-x you get:
\[\frac{1}{1+x} = \sum_{n=0}^\infty (-1)^n x^n\]
And you can do similar stuff for all the rest.

Answer 91

The "easiest" method will always depend on the background of the person working on the problem, AND the tools available.

If Wolfram is not available (as in a classroom test), then we don't get to see the pattern of the coefficient, so we need the basics, which is the pattern of the derivatives, by hand, as in old school.

If Wolfram is available, we get to see the pattern of the coefficients, as the OP has done, or even easier, the pattern of the derivatives (no worries about sign, factorials, etc.).
Noting that sin(0)=0, cos(0)=1, the first 10 derivatives (using Wolfram) can be easily evaluated to:
\(\sin(x)+x \cos(x)\) = 0
\(2 \cos(x)-x \sin(x)\) =2
\(-3 \sin(x)-x \cos(x)\)=0
\(x \sin(x)-4 \cos(x)\)=-4
\(5 \sin(x)+x \cos(x)\)=0
\(6 \cos(x)-x \sin(x)\)=6
\(-7 \sin(x)-x \cos(x)\)=0
\(x \sin(x)-8 \cos(x)\)=-8
\(9 \sin(x)+x \cos(x)\)=0
\(10 \cos(x)-x \sin(x)\)=10
from which we conclude that odd derivatives are zero, and even derivatives are \((-1)^{n/2+1}n\)
Now all we have to do is to divide by the corresponding factorials to get the complete series.

I would recommend the basic method starting from the definition of MacLaurin series. Once the foundations are made, it would be easy to find/devise other shortcuts according to one's experience.