how to i solve this t1, stuck at this.while i pressing the calculator couldnt..

[longcrystal]

https://lh3.ggpht.com/-CwcTPsCelkY/T2TXSccJ28I/AAAAAAAAEtE/OywHPH8r92E/s1600-h/image%255B35%255D.png



know the step how to do ~~~but cannot solve ...weird !!!~~~

 

[etmabreu]

That was solved with numeric aproximation, such as Newton-Raphson or other.

 

[longcrystal]

That was solved with numeric aproximation, such as Newton-Raphson or other.

but do u know how to describe in simple step...to get approximation ....this answer....such as u mean in newton or other ~~~~

 

[albbg]

You can approximate sin(x)≈x if x is close to 0 (it is the first term of McLaurin series), then:

1250•t2 - 100•Π•t2 -1.25 = 0 ==> t2 = 1.25/(1250 - 100•Π) = 1.336 ms

however if x (in your case 100•Π•t2) is not close to zero the accuracy is bad.

You may want to use the iterative Newton method. We need the function f(x) = 0 and its first derivative f'(x)
Starting from a first guess xo (arbitrary choosen) we can find a new solution approximation as:

x1 = xo - f(xo)/f'(xo)

then

x2 = x1 - f(x1)/f'(x1)

continuing, after n steps:

xn+1 = xn - f(xn)/f'(xn)

Your function is f(t2) = 1250•t2 - sin(100•Π•t2) -1.25 = 0
the first derivative is f'(t2) = 1250 - 100•Π•cos(100•Π•t2)

starting from t2(0) = 1 arbitrary choosen we will have

t2(1) = -0.334362
t2(2) = -0.014242
t2(3) = 0.000881
t2(4) = 0.001326
t2(5) = 0.001323
t2(6) = 0.001323

 

[longcrystal]

You can approximate sin(x)≈x if x is close to 0 (it is the first term of McLaurin series), then:

1250•t2 - 100•Π•t2 -1.25 = 0 ==> t2 = 1.25/(1250 - 100•Π) = 1.336 ms

however if x (in your case 100•Π•t2) is not close to zero the accuracy is bad.

You may want to use the iterative Newton method. We need the function f(x) = 0 and its first derivative f'(x)
Starting from a first guess xo (arbitrary choosen) we can find a new solution approximation as:

x1 = xo - f(xo)/f'(xo)

then

x2 = x1 - f(x1)/f'(x1)

continuing, after n steps:

xn+1 = xn - f(xn)/f'(xn)

Your function is f(t2) = 1250•t2 - sin(100•Π•t2) -1.25 = 0
the first derivative is f'(t2) = 1250 - 100•Π•cos(100•Π•t2)

starting from t2(0) = 1 arbitrary choosen we will have

t2(1) = -0.334362
t2(2) = -0.014242
t2(3) = 0.000881
t2(4) = 0.001326
t2(5) = 0.001323
t2(6) = 0.001323

albbg ...u are very intelligent to explain easy step for me..let me easy to understanding about that right now ...without u explanation maybe i will stuck at this....and u provide me a two method to solve....very thankq for you.........

really thank for u .......dontknow how to describe this expression feeling....because i was not science student in my secondary school....now study undergraduate in engineering..so i always stuck some trickly equation ............but now i fully i understand~~your step ....and ~what is the step ...simplify one by one !!!!!!!!!!pretty understanding right now!!!!!!!!!!!!!!!thankq u ***!!!!!.............................can make a friend ?? in facebook or twitter also can haha !!!! ..!!!!!!!!~~~next time may i ask ...some question if i stuck some math ??????????



may i ask one more thing,., starting from t2(0) = 1 arbitrary ....mean is beginning point...so this beginning point just guess to choose a value to starting in 1......so this t2(0) = 1 can be calculated or cannot be calculated?? but other is fine ...very good....just dontknow how to do get t2(0) = 1........................................if i use t2=0 mean x=0 mean as u explain the McLaurin series.......so the newton method ..is depend what value do u want try to put at starting point?????................. anyway really thankq u ~~~~~~~~!!!!!

 

[albbg]

You can choose any point you want. Usually is only matter of how many iterations are needed to converge, but the Netwon method is always quite fast. If it doesn't converge you can choose another initial guess.