This power point highlights the way of solving log(1+x) using Taylor's expansion. Also there are brief discussion about the Formula and certain examples of other such series. The slides are good enough for an engineering term paper of mathematics.

1. Represented By : Arijit Dhali

5. • Continuity:
Continuity is the formulation of the intuitive concept of a function that varies with no abrupt
breaks or jumps.
• Derivative:
Derivative of a function of a real variable measures the sensitivity to change of the function
value with respect to a change in its argument.
• Taylor Series:
Taylor series of a function is an infinite sum of terms that are expressed in terms of the
function's derivatives at a single point.
• Lagrange Remainder:
Lagrange remainder is: where M is the maximum of the absolute value of the (n + 1)th
derivative of function on the interval from x to c. The error is bounded by this remainder.

9. • Infinite Series for the function Log(1+x) around x=5
Here the function f(x) = log(1+x), -1<x<∞ is continuous and derivable at each point of (-1, ∞)
Now,
f(x) = log(1-x)
f’(x) =
𝟏
𝟏+𝐱
f’’(x) =
−𝟏
(𝟏+𝐱)𝟐
f’’’(x) =
𝟐
(𝟏+𝐱)𝟑 and so on……..
Now using Taylor’s formula for the function f(x) about x=5 is,
f(x) = f(5) + (x-5) f’(5) +
(𝒙−𝟓)𝟐
𝟐!
f’’(5) +
(𝒙−𝟓)𝟑
𝟑!
f’’’(5) + …+
(𝒙−𝟓)𝒏
𝒏!
𝒇𝒏 𝟓 + θ 𝐧 − 𝟓 , 0 < θ < 1
Since, f(5) = log(6), f’(5) =
𝟏
𝟔
, f’’(5) = −
𝟏
𝟑𝟔
, f’’’(5) =
𝟐
𝟐𝟏𝟔
…………..and so on.
Therefore, f(x) = log 6 +
𝟏
𝟔
(x-5) +
(𝒙−𝟓)𝟐
𝟐!
(
−𝟏
𝟑𝟔
) +
(𝒙−𝟓)𝟑
𝟑!
(
𝟐
𝟐𝟏𝟔
) +……+
(𝒙−𝟓)𝒏
𝒏!
(−𝟏)𝒏−𝟏 𝒏−𝟏 !
𝟔𝒏
Log(11) = 1 +
𝟏
𝟔
(x-10) -
(𝒙−𝟏𝟎)𝟐
𝟕𝟐
+
(𝒙−𝟏𝟎)𝟑
𝟔𝟒𝟖
-……+
(𝒙−𝟏𝟎)𝒏
𝒏
(−𝟏)𝒏−𝟏
𝟔𝒏

10. • Infinite Series for the function Log(1+x) around x=10
Here the function f(x) = log(1+x), -1<x<∞ is continuous and derivable at each point of (-1, ∞)
Now,
f(x) = log(1-x)
f’(x) =
𝟏
𝟏+𝐱
f’’(x) =
−𝟏
(𝟏+𝐱)𝟐
f’’’(x) =
𝟐
(𝟏+𝐱)𝟑 and so on……..
Now using Taylor’s formula for the function f(x) about x=5 is,
f(x) = f(10) + (x-10) f’(10) +
(𝒙−𝟓)𝟐
𝟐!
f’’(10) +
(𝒙−𝟏𝟎)𝟑
𝟑!
f’’’(10) + …+
(𝒙−𝟏𝟎)𝒏
𝒏!
𝒇𝒏 𝟏𝟎 + θ 𝐧 − 𝟏𝟎 , 0 < θ < 1
Since, f(10) = log(10), f’(10) =
𝟏
𝟏𝟏
, f’’(10) = −
𝟏
𝟏𝟐𝟏
, f’’’(10) =
𝟐
𝟏𝟑𝟑𝟏
…………..and so on.
Therefore, f(x) = 1 +
𝟏
𝟏𝟏
(x-10) +
(𝒙−𝟏𝟎)𝟐
𝟐!
(
−𝟏
𝟏𝟐𝟏
) +
(𝒙−𝟏𝟎)𝟑
𝟑!
(
𝟐
𝟏𝟑𝟑𝟏
) +……+
(𝒙−𝟏𝟎)𝒏
𝒏!
(−𝟏)𝒏−𝟏 𝒏−𝟏 !
𝟏𝟏𝒏
Log(11) = 1 +
𝟏
𝟏𝟏
(x-10) -
(𝒙−𝟏𝟎)𝟐
𝟐𝟒𝟐
+
(𝒙−𝟏𝟎)𝟑
𝟑𝟗𝟗𝟑
-……+
(𝒙−𝟏𝟎)𝒏
𝒏
(−𝟏)𝒏−𝟏
𝟏𝟏𝒏

11. • This project has been beneficial for us , as it enabled us to gain a lot of knowledge about the use
of Taylor Series about a appoint , as well as its applications. It also helped us to develop a better
coordination among us as we shared different perspectives and ideas regarding the sub-topics
we organized in this project .

12. • https://en.wikipedia.org/wiki/Taylor_series
• https://math.stackexchange.com
• https://www.quora.com/What-is-the-expansion-of-log-1-x
• Engineering Mathematics I – B.K.Pal & K.Das