The number
e=2.7182818284... can be shown to be the limit shown above as x
goes to zero (from above). Write a program that shows how this
expression converges to e as x gets closer and closer to zero.
MATLAB
Homework Answers
the matlab for the above question is as follows:
//
Code to copy:
x=vpa(2,20); %% setting the precision
for i=1:50 %%loop for 50 times
st = sprintf('value of x');
disp(st);
x
st = sprintf('value of e for above value of x');
disp(st);
val = vpa(power(vpa((1+x),20),vpa((1/x),20)))
x=vpa(x/2,20); %% half the value of x
end
//Sample Output:
value of x
x =
2.0
value of e for above value of x
val =
1.7320508075688772935274463415059
value of x
x =
1.0
value of e for above value of x
val =
2.0
value of x
x =
0.5
value of e for above value of x
val =
2.25
value of x
x =
0.25
value of e for above value of x
val =
2.44140625
value of x
x =
0.125
value of e for above value of x
val =
2.565784513950347900390625
value of x
x =
0.0625
value of e for above value of x
val =
2.6379284973665998587631122129782
value of x
x =
0.03125
value of e for above value of x
val =
2.6769901293781826846423269494124
value of x
x =
0.015625
value of e for above value of x
val =
2.6973449525650988556024092745751
value of x
x =
0.0078125
value of e for above value of x
val =
2.7077390196880204920249853588651
value of x
x =
0.00390625
value of e for above value of x
val =
2.7129916242534343342857206707389
value of x
x =
0.001953125
value of e for above value of x
val =
2.7156320001689911665000557306844
value of x
x =
0.0009765625
value of e for above value of x
val =
2.7169557294664355449931552836268
value of x
x =
0.00048828125
value of e for above value of x
val =
2.7176184823368797609194672010495
value of x
x =
0.000244140625
value of e for above value of x
val =
2.7179500811896658725414143932781
value of x
x =
0.0001220703125
value of e for above value of x
val =
2.7181159362657970889147508982157
value of x
x =
0.00006103515625
value of e for above value of x
val =
2.7181988777219706710538002447564
value of x
x =
0.000030517578125
value of e for above value of x
val =
2.7182403519302939500582814250454
value of x
x =
0.0000152587890625
value of e for above value of x
val =
2.7182610899046034180605234581998
value of x
x =
0.00000762939453125
value of e for above value of x
val =
2.7182714591093061987333633780002
value of x
x =
0.000003814697265625
value of e for above value of x
val =
2.7182766437660459870080734138591
value of x
x =
0.0000019073486328125
value of e for above value of x
val =
2.7182792361080131539186100566889
value of x
x =
0.00000095367431640625
value of e for above value of x
val =
2.7182805322823960772285508101668
value of x
x =
0.000000476837158203125
value of e for above value of x
val =
2.7182811803704373765548802288986
value of x
x =
0.0000002384185791015625
value of e for above value of x
val =
2.7182815044146704859743472408775
value of x
x =
0.00000011920928955078125
value of e for above value of x
val =
2.7182816664368401556654642137453
value of x
x =
0.000000059604644775390625
value of e for above value of x
val =
2.7182817474479382692616570578932
value of x
x =
0.0000000298023223876953125
value of e for above value of x
val =
2.718281787953490645748073131057
value of x
x =
0.00000001490116119384765625
value of e for above value of x
val =
2.7182818082062676639134437131369
value of x
x =
0.000000007450580596923828125
value of e for above value of x
val =
2.7182818183326563804766799696431
value of x
x =
0.0000000037252902984619140625
value of e for above value of x
val =
2.7182818233958507906284371303994
value of x
x =
0.0000000018626451492309570312
value of e for above value of x
val =
2.7182818259274480086718506302954
value of x
x =
0.00000000093132257461547851562
value of e for above value of x
val =
2.7182818271932466209354411302969
value of x
x =
0.00000000046566128730773925781
value of e for above value of x
val =
2.7182818278261459278777073203327
value of x
x =
0.00000000023283064365386962891
value of e for above value of x
val =
2.7182818281425955815514581506747
value of x
x =
0.00000000011641532182693481445
value of e for above value of x
val =
2.718281828300820408438987999716
value of x
x =
0.000000000058207660913467407227
value of e for above value of x
val =
2.7182818283799328218954165327092
value of x
x =
0.000000000029103830456733703613
value of e for above value of x
val =
2.7182818284194890286267967013246
value of x
x =
0.000000000014551915228366851807
value of e for above value of x
val =
2.718281828439267131993278261162
value of x
x =
0.0000000000072759576141834259033
value of e for above value of x
val =
2.7182818284491561836767169099632
value of x
x =
0.0000000000036379788070917129517
value of e for above value of x
val =
2.7182818284541007095184857015844
value of x
x =
0.0000000000018189894035458564758
value of e for above value of x
val =
2.7182818284565729724393824642001
value of x
x =
0.00000000000090949470177292823792
value of e for above value of x
val =
2.7182818284578091038998339372093
value of x
x =
0.00000000000045474735088646411896
value of e for above value of x
val =
2.7182818284584271696300604466392
value of x
x =
0.00000000000022737367544323205948
value of e for above value of x
val =
2.7182818284587362024951738945855
value of x
x =
0.00000000000011368683772161602974
value of e for above value of x
val =
2.7182818284588907189277306668665
value of x
x =
0.00000000000005684341886080801487
value of e for above value of x
val =
2.7182818284589679771440090650839
value of x
x =
0.000000000000028421709430404007435
value of e for above value of x
val =
2.7182818284590066062521482672119
value of x
x =
0.000000000000014210854715202003717
value of e for above value of x
val =
2.7182818284590259208062178690307
value of x
x =
0.0000000000000071054273576010018587
value of e for above value of x
val =
2.7182818284590355780832526701288
value of x
x =
0.0000000000000035527136788005009294
value of e for above value of x
val =
2.718281828459040406721770070725
Add Answer to:
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The system above has two distinguishable particles, each can be
in either of two boxes. The system is in thermal equilibrium with a
heat bath at temperature, T. The energy of the particle is zero
when it's in the left box, and it is
when it is in the right box. There is a correlation energy term
that increases the system energy by
if the particles are in the same box.
If the particles are indistinguishable how many microstates will...
lim
x -> infinity
calculate the about with algebra and limit laws only. Please
show work and step. Please screen shot work so it's easier to read
than trying to type it out.
I can get to
but then i get stuck. please help
We were unable to transcribe this imageWe were unable to transcribe this image
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Show the setup and round the answer to threedecimal places.
(b) Find M4 for the integral
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decimal places.
Sketch the approximating rectangles on the graph.
(c) Compare the estimates with the actual value
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10.243 . Which estimate is the most accurate?
(d) Express the integral from...
A hockey puck is sliding on the smooth ice surface of a skating
rink as shown in the figure above.
The figure shows the puck’s position on the ice surface at a
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t2, t3 and t4. The x axis
shown is parallel to the rink’s length and the y axis to
its width. The marks along the axes are spaced 1 meter apart as are
the successive positions of the puck. The unseen...
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H e will be recognised as Euler's number. • Use to explicitly indicate multiplication between variables and other variables, between variables and brackets, or betwe (e.g. x'y rather than xy, 2*(x+y) rather than 2(x+y)) . Put arguments of functions in brackets (e.g. sin(x) rather than sin x). Let g(z,y) = –2-23-2-22 +4.72 +3. Show/hide sin(a) too a 2 Back Quit & Save Submit Assignment Show/hide sin (a)...
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