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###Question

Question

Consider the following two lines of code, which differ only by replacing 1 by y:

SeriesCoefficient[
  (Exp[x + 1 + a] - 1)/(x + 1 + a), {a, 0, 0}, Assumptions -> x + 1 == 0]

SeriesCoefficient[
  (Exp[x + y + a] - 1)/(x + y + a), {a, 0, 0}, Assumptions -> x + y == 0]

The first version yields 1, which is the correct answer (just the limit of the function as "a" goes to zero). The second version yields:

(-1 + Exp[x + y])/(x + y)

which will cause problems for me later because x + y == 0.

Assuming x and y are real does not change the result.

###Background

Background

I want to take the limit of a large number of continuous functions of several variables:

Limit[
  Limit[Limit[Limit[f[y1, y2, y3, y4], y1 -> x1], y2 -> x2], y3 -> x3],
  y4 -> x4]

where each function f is similar to the one used above (ie it naively seems to have singularities but is actually continuous). I obtain much faster results using SeriesCoefficient:

SeriesCoefficient[f[y1 + a, y2 + a, y3 + a, y4 + a], {a, 0, 0}]

I want to be able to evaluate this with assumptions. For example:

SeriesCoefficient[
  f[x1 + a, x2 + a, x3 + a, x4 + a], {a, 0, 0}, 
  Assumptions -> x1 + x2 + x3 + x4 == 0]

but the assumptions are ignored.

###Question

Consider the following two lines of code, which differ only by replacing 1 by y:

SeriesCoefficient[
  (Exp[x + 1 + a] - 1)/(x + 1 + a), {a, 0, 0}, Assumptions -> x + 1 == 0]

SeriesCoefficient[
  (Exp[x + y + a] - 1)/(x + y + a), {a, 0, 0}, Assumptions -> x + y == 0]

The first version yields 1, which is the correct answer (just the limit of the function as "a" goes to zero). The second version yields:

(-1 + Exp[x + y])/(x + y)

which will cause problems for me later because x + y == 0.

Assuming x and y are real does not change the result.

###Background

I want to take the limit of a large number of continuous functions of several variables:

Limit[
  Limit[Limit[Limit[f[y1, y2, y3, y4], y1 -> x1], y2 -> x2], y3 -> x3],
  y4 -> x4]

where each function f is similar to the one used above (ie it naively seems to have singularities but is actually continuous). I obtain much faster results using SeriesCoefficient:

SeriesCoefficient[f[y1 + a, y2 + a, y3 + a, y4 + a], {a, 0, 0}]

I want to be able to evaluate this with assumptions. For example:

SeriesCoefficient[
  f[x1 + a, x2 + a, x3 + a, x4 + a], {a, 0, 0}, 
  Assumptions -> x1 + x2 + x3 + x4 == 0]

but the assumptions are ignored.

Question

Consider the following two lines of code, which differ only by replacing 1 by y:

SeriesCoefficient[
  (Exp[x + 1 + a] - 1)/(x + 1 + a), {a, 0, 0}, Assumptions -> x + 1 == 0]

SeriesCoefficient[
  (Exp[x + y + a] - 1)/(x + y + a), {a, 0, 0}, Assumptions -> x + y == 0]

The first version yields 1, which is the correct answer (just the limit of the function as "a" goes to zero). The second version yields:

(-1 + Exp[x + y])/(x + y)

which will cause problems for me later because x + y == 0.

Assuming x and y are real does not change the result.

Background

I want to take the limit of a large number of continuous functions of several variables:

Limit[
  Limit[Limit[Limit[f[y1, y2, y3, y4], y1 -> x1], y2 -> x2], y3 -> x3],
  y4 -> x4]

where each function f is similar to the one used above (ie it naively seems to have singularities but is actually continuous). I obtain much faster results using SeriesCoefficient:

SeriesCoefficient[f[y1 + a, y2 + a, y3 + a, y4 + a], {a, 0, 0}]

I want to be able to evaluate this with assumptions. For example:

SeriesCoefficient[
  f[x1 + a, x2 + a, x3 + a, x4 + a], {a, 0, 0}, 
  Assumptions -> x1 + x2 + x3 + x4 == 0]

but the assumptions are ignored.

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Series -- one assumption works but a slight variation does not?

Question ###Question

Consider the following two lines of code, which differ only by replacing 1 by yy:

SeriesCoefficient[
  (Exp[x+1+a]Exp[x + 1 + a] - 1)/(x+1+ax + 1 + a), {a, 0, 0}, Assumptions ->x+1> x + 1 == 0]

SeriesCoefficient[
  (Exp[x+y+a]Exp[x + y + a] - 1)/(x+y+ax + y + a), {a, 0, 0}, Assumptions ->x+y> x + y == 0]

The first version yields 1, which is the correct answer (just the limit of the function as "a" goes to zero). The second version yields:

(-1 + Exp[x+y]Exp[x + y])/(x+yx + y)

which will cause problems for me later because x + y = 0 x + y == 0.

Assuming xx and yy are real does not change the result.

Background ###Background

I want to take the limit of a large number of continuous functions of several variables:

Limit[Limit[Limit[Limit[f[y1Limit[
  Limit[Limit[Limit[f[y1, y2, y3, y4], y1 ->x1]> x1], y2 ->x2]> x2], y3 ->x3]> x3],
  y4 ->x4]> x4]

where each function ff is similar to the one used above (ie it naively seems to have singularities but is actually continuous). I obtain much faster results using SeriesCoefficientSeriesCoefficient:

SeriesCoefficient[f[y1+aSeriesCoefficient[f[y1 + a,y2+a y2 + a,y3+a y3 + a,y4+a] y4 + a], {a, 0, 0}]

I want to be able to evaluate this with assumptions. For example:

SeriesCoefficient[f[x1+aSeriesCoefficient[
  f[x1 + a,x2+a x2 + a,x3+a x3 + a,x4+a] x4 + a], {a, 0, 0}, 
  Assumptions ->x1+x2+x3+x4> x1 + x2 + x3 + x4 == 0]

but the assumption isassumptions are ignored.

Series -- one assumption works but slight variation does not?

Question

Consider the following two lines of code, which differ only by replacing 1 by y:

SeriesCoefficient[(Exp[x+1+a]-1)/(x+1+a), {a,0,0},Assumptions->x+1 == 0]

SeriesCoefficient[(Exp[x+y+a]-1)/(x+y+a), {a,0,0},Assumptions->x+y == 0]

The first version yields 1, which is the correct answer (just the limit of the function as "a" goes to zero). The second version yields:

(-1 + Exp[x+y])/(x+y)

which will cause problems for me later because x + y = 0.

Assuming x and y are real does not change the result.

Background

I want to take the limit of a large number of continuous functions of several variables:

Limit[Limit[Limit[Limit[f[y1,y2,y3,y4], y1->x1],y2->x2],y3->x3],y4->x4]

where each function f is similar to the one used above (ie it naively seems to have singularities but is actually continuous). I obtain much faster results using SeriesCoefficient:

SeriesCoefficient[f[y1+a,y2+a,y3+a,y4+a],{a,0,0}]

I want to be able to evaluate this with assumptions. For example:

SeriesCoefficient[f[x1+a,x2+a,x3+a,x4+a],{a,0,0},Assumptions->x1+x2+x3+x4 == 0]

but the assumption is ignored.

Series -- one assumption works but a slight variation does not

###Question

Consider the following two lines of code, which differ only by replacing 1 by y:

SeriesCoefficient[
  (Exp[x + 1 + a] - 1)/(x + 1 + a), {a, 0, 0}, Assumptions -> x + 1 == 0]

SeriesCoefficient[
  (Exp[x + y + a] - 1)/(x + y + a), {a, 0, 0}, Assumptions -> x + y == 0]

The first version yields 1, which is the correct answer (just the limit of the function as "a" goes to zero). The second version yields:

(-1 + Exp[x + y])/(x + y)

which will cause problems for me later because x + y == 0.

Assuming x and y are real does not change the result.

###Background

I want to take the limit of a large number of continuous functions of several variables:

Limit[
  Limit[Limit[Limit[f[y1, y2, y3, y4], y1 -> x1], y2 -> x2], y3 -> x3],
  y4 -> x4]

where each function f is similar to the one used above (ie it naively seems to have singularities but is actually continuous). I obtain much faster results using SeriesCoefficient:

SeriesCoefficient[f[y1 + a, y2 + a, y3 + a, y4 + a], {a, 0, 0}]

I want to be able to evaluate this with assumptions. For example:

SeriesCoefficient[
  f[x1 + a, x2 + a, x3 + a, x4 + a], {a, 0, 0}, 
  Assumptions -> x1 + x2 + x3 + x4 == 0]

but the assumptions are ignored.

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