why do we expand any function by taylor's theorem or by maclaurin's theorem?

Answer 1

The main reason why we expand a function using Taylor's series & more frequently the MacLaurin expansion is because there are so many mathematical functions, integrals etc. which exist, but we are not sure as to how to calculate the values of those functions readily. The MacLaurin's expansion helps "decompose" the whole complicated function into a sum of a simple algebraic or polynomial series, which can be easily calculated.

Take for example, I asked you to find e^2. What you would do is reach out for your scientific calculator and press the "ex" key or the "Shift + ln" key and then "2". It gives you the result immediately. But now, suppose you did NOT have a Scientific calculator with you, then what would you do? You will have to work out everything by hand - this is where you apply your knowledge of the MacLaurin series:

f(x) = f(0) + {f'(0)*x} + {f''(0)*(x^2)/2!} + {f'''(0)*(x^3)/3!} + ....

Here your function is e^x, where x=2. We have to find out the derivatives of the function as well, so:

f(x) = e^x
f'(x) = e^x
f''(x) = e^x
.....

Now, apply the Mclaurin series (I suggest upto atleast 9 terms, to get good accuracy):

e^x = e^0 + {(e^0)*x} + {(e^0)*(x^2)/2!} + {(e^0)*(x^3)/3!} + ....

e^x = 1 + x + {(x^2)/(1*2)} + {(x^3)/(1*2*3)} + ....

e^2 = 1 + 2 + (4/2) + (8/6) + (16/24) + (32/120) + (64/720) + (128/5040) + (256/40320) + ...negligible terms

Now, all you have to do is add up the terms on the RHS:

e^2 = 1 + 2 + 2 + 1.333 + 0.667 + 0.267 + 0.089 + 0.025 + 0.006

e^2 = 7.387

This is the answer for e^2, you will get by working out manually using McLaurin Series. If you used a Scientific calculator, it would give you a value of 7.389, which is almost the same.

Now, you understand why the McLaurin series is so important. You can use it to find the values of almost any function you want, provided you know their derivatives as well. In fact, this McLaurin Series is exactly what the Scientific Calculator itself does when you press those keys, it just expands it upto 50 terms or more!

Answer 2

First of all, Taylor's and Maclaurin's series can only expand those functions which are differentiable and hence not any function.

These series are very useful in finding patterns among seemingly diverse and totally unrelated functions and thus are a great tool in research and analysis.

They are not the only ones, more powerful is the Fourier series which can decompose functions into series of sines and cosines and is used in the encoding of video and audio files.

Answer 3

These are the way a computer or calculator can
compute the functions. Recall that a computer
can only handle polynomials, so to compute
sin x, for instance, requires a series expansion.
Taylor's or Maclaurin's series are an excellent
(but not the only) way to do this.

Answer 4

f(x) = e^x so f(0) = a million f'(x) = e^x so f'(0) = a million f"(x) = e^x so f"(0) = a million .................................... f^n(x) = e^x so f^n(0) =a million for this reason the Maclaurin improve is: e^x = a million+x+x^2/2! +x^3/3! +....+ x^n/n! +... by utilising the ratio try, this sequence converges for all values of x f(x) = sin x so f(0) = 0 f'(x) = cosx so f'(0) = a million f"(x) = -sinx so f"(0) = 0 f"'(x) = -cosx so f"'(0) = -a million f""(x) = sinx so f""(0)= 0 ......................................... f^n(x) = sin(x + a million/2npi) so F^n(0) = sin a million/2npi So Maclaurin improve is: sinx = x-x^3/3! +x^5/5! - ... +(-a million)^m+a million x^(2m-a million)/(2m-a million)! +... to locate the Taylor sequence related to the element pi/3, replace pi /3 for each and all of the derivatives rather of 0. cos x equivalent to sin x

Answer 5

Mainly for the purpose of finding limits.

Answer 6

to find the values of functions
for example if we have to find the value of e^x
let f(x=e^x F(0)=1
f'(x)=e^x f'(0)=1
and so on
maclaurin's series is
f(x)=f(0)+xf'(0)+X^2/2!f''(0)+.............
so e^x=1+x+x^2/2!+x^3/3!+....................