- THE TAYLOR SERIES 
DEFINITION AND DISCUSSION
ABOUT INTERVAL OF CONVERGENCE.
SO, WE SAW THAT 
THE TAYLOR POLYNOMIAL
TO THE NTH DEGREE FOR F OF X 
AT X EQUALS "A."
IT WAS EQUAL TO THIS.
AND WHERE THOSE COEFFICIENTS 
WERE THE DERIVATIVES OF F
EVALUATE AT "A" DIVIDED BY 
THE POWER FACTORIAL,
THE K FACTORIAL.
SO, WRITTEN OUT, YOU GET THIS.
SO, THAT'S WHAT 
THE TAYLOR POLYNOMIAL WAS OR IS.
THE SERIES, ALL THAT DOES IS 
EXTENDS THESE TERMS TO INFINITY.
SO, YOU JUST PUT THE PLUS DOT, 
DOT, DOT,
YOU KEEP ON GOING IN THAT SAME 
PATTERN AND YOU WRITE IT
IN SERIES NOTATION F OF X EQUALS 
THIS SERIES RIGHT HERE.
YOUR DERIVATIVES OF F, 
EVALUATE AT THE "A"
OVER THE K FACTORIAL, 
WHICH MATCHES UP
TO THE POWER X MINUS "A".
BUT THAT EQUAL SIGN RIGHT HERE 
IS ONLY GOOD FOR X VALUES
JUST AS THE SERIES CONVERGES.
SO, IF THE SERIES DOESN'T 
CONVERGE FOR A GIVEN X VALUE,
THEN THAT EQUAL SIGN 
IS NOT EQUAL ANYMORE.
SOME OF THESE SERIES 
WE'LL SEE F OF X
DOES EQUAL THIS SERIES 
FOR ALL VALUES OF X.
OTHERS WILL SEE THEY DON'T, 
ONLY ON CERTAIN Xs.
AND ALSO, YOU MUST HAVE, 
ALL DERIVATIVES MUST EXIST
AT X EQUALS "A", 
OTHERWISE THESE COEFFICIENTS
DON'T EXIST.
SO, EXAMPLE ONE, 
FOR F OF X EQUALS SIGN X,
WE FOUND TAYLOR POLYNOMIALS 
FOR SIGN X AT X = 0
AND WE EXTENDED THAT 
AND THIS WAS THE PATTERN.
AND WE JUST EXTEND THAT 
TO INFINITY AND YOU GET
THAT SERIES RIGHT THERE.
BUT WE CAN'T JUST GO AROUND 
SAYING EQUALS.
WE HAVE TO KNOW 
WHERE IT IS EQUAL.
SO, FOR WHICH VALUES OF X 
DOES THIS SERIES CONVERGE?
WE DO THE RATIO TEST. THERE'S 
YOUR RATIO TEST IN ACTION THERE.
YOU JUST HAVE TO BE CAREFUL 
ABOUT THESE FACTORIALS.
ABSOLUTE VALUE 
OF THE FACTORIAL,
THESE ARE POSITIVE NUMBERS, 
THESE FACTORIALS
SO THAT'S WHY THERE'S 
NO ABSOLUTE VALUE THERE.
COMES OVER TO BEING THIS, 
WHICH EQUALS THIS
AND THEN WE TAKE THE LIMIT 
AS K GOES TO INFINITY,
X SQUARED FACTOR OUT LIKE THIS.
GOES TO X SQUARED x 0, 
WHICH EQUALS 0
NO MATTER WHAT X IS.
SO, THE LIMIT OF THE RATIO 
OF ABSOLUTE VALUE OF SUCCESSIVE
TERMS IN THE SERIES EQUALS 0 
AND 0 IS LESS THAN 1
IN THE RATIO TEST 
AND IS SUCH FOR ALL VALUES OF X
BECAUSE X SQUARED x 0 = 0 
NO MATTER WHAT THE VALUE OF X.
THEREFORE, SIGN X ACTUALLY DOES 
EQUAL THIS SERIES
FOR ALL VALUES OF X. 
SO, THAT EQUALS IS SOLID.
THE RADIUS OF CONVERGENCE 
IS INFINITY.
THE INTERVAL OF CONVERGENCE 
IS NEGATIVE INFINITY
TO INFINITY OR ALL REAL NUMBERS.
EXAMPLE TWO, WE SAW 
THAT THE TAYLOR POLYNOMIALS
FOR F OF X EQUALS LOG X 
AND X = 1 HAD THIS PATTERN,
WHICH I'LL REMIND YOU LATER 
IN THIS VIDEO WHY THAT WAS.
SO, EXTENDING THE TERMS 
TO INFINITY,
WE GET THIS SERIES HERE.
BUT YOU CAN'T JUST GO AROUND 
SAYING THIS IS EQUAL
AND WE HAVE TO FIND OUT 
WHAT X VALUES IT'S EQUAL FOR.
SO, WE SAW THAT THE INTERVAL OF 
CONVERGENCE WAS 0 LESS THAN X,
LESS THAN OR EQUAL TO 2.
THE RADIUS OF CONVERGENCE 
WAS 1.
HENCE, THE EQUAL SIGN 
IN THE LOG X EQUALS THE SERIES
IS ONLY GOOD FOR X SUCH THAT X 
IS IN THIS INTERVAL OF 0
LESS THAN X, 
LESS THAN OR EQUAL TO 2.
SO, WHEN WE PUT EQUALS, 
IT'S UNDERSTOOD
THAT ITS ONLY EQUALS 
WHEN THIS SERIES CONVERGES.
HERE'S A REMINDER OF WHERE THOSE 
TERMS CAME FROM FOR LOG X.
I'M JUST GONNA SHOW 
THE SCREEN HERE.
YOU CAN PAUSE IT HERE.
SO, THIS IS WHERE 
THEY CAME FROM,
THEN WHY THE INTERVAL 
OF CONVERGENCE IS THIS
WAS BECAUSE WE DID 
THE RATIO TEST
AND WE GOT THIS 
FOR THE PRELIMINARY.
AND THEN WE TESTED 
THE ENDPOINTS,
TESTING THE ENDPOINTS AT 0; 
WE DID THIS ALREADY
SO I'M JUST GOING TO SHOW IT 
HERE AGAIN.
WE GOT ALL THIS.
CAME OUT THAT IT WAS 
THE -1 FACTORED OUT.
-1 TIMES THE HARMONIC SERIES, 
WHICH WAS THIS,
SO THAT DIVERGED. 
SO X IS NOT 0.
THEN X EQUALS 2.
WE WENT THROUGH 
AND FOUND OUT THAT IT'S
THE ALTERNATING HARMONIC SERIES 
AND THAT DID CONVERGE.
SO X = 2 IS GOOD.
HENCE, THE INTERVAL OF 
CONVERGENCE WAS 0 LESS THAN X,
LESS THAN OR EQUAL TO 2.
SO, THAT EQUAL SIGN 
RIGHT UP HERE ONLY GOOD
ON THAT SMALL INTERVAL, 
0 LESS THAN X,
LESS THAN OR EQUAL TO 2.
